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What is Boost Converter? Features, Applications, and Electrical characteristics
Introduction to Boost DC-DC Converter Design
A boost DC-DC converter is a fundamental component in power electronics that steps up the input voltage to a higher output voltage. Widely used in battery-powered devices, electric vehicles (EVs),
renewable energy systems, and industrial applications, boost converters provide efficient energy conversion while maintaining stable output voltage levels. This blog delves into the design aspects of
boost DC-DC converters, including operation in continuous conduction mode (CCM) and discontinuous conduction mode (DCM), transfer function analysis, and stability considerations.
Basics of Boost Converter
A boost converter is a type of DC-DC converter that steps up the input voltage to a higher output voltage. It operates by storing energy in an inductor during the “on” phase when the switch (usually
a transistor) is closed, and releasing it to the output during the “off” phase when the switch is open. The stored energy in the inductor combines with the input voltage to produce a higher output
voltage. Boost converters are widely used in applications where a higher voltage is required from a lower voltage source, such as in battery-powered devices and renewable energy systems.
1.1. Operating Principle
A boost converter comprises an inductor, a switch (typically a transistor), a diode, and an output capacitor. The converter operates by storing energy in the inductor when the switch is closed and
transferring this energy to the load when the switch is open. During operation:
• Switch ON: The inductor stores energy as current flows through it, and the diode blocks the output.
• Switch OFF: The inductor releases the stored energy to the load via the diode, raising the output voltage.
1.2. Key Components
• Inductor: Stores and releases energy to regulate the voltage.
• Switch: Generally a MOSFET or IGBT, controls the on-off cycles.
• Diode: Provides a path for current when the switch is off.
• Capacitor: Smooths the output voltage, reducing ripples.
1.3. Boost Converter Equation
The basic voltage relationship for a boost converter is:
• Vout is the output voltage
• D is the duty cycle, defined as the fraction of time the switch is ON
As D increases, the output voltage becomes higher than the input voltage.
Continuous Conduction Mode (CCM)
Continuous Conduction Mode (CCM) in a boost DC-DC converter occurs when the inductor current never falls to zero during the switching cycle. This mode is characterized by a continuous flow of current
through the inductor, even when the switch is turned off. In CCM, energy is stored in the inductor during the ON state and released to the load in the OFF state. CCM operation typically leads to
higher efficiency at moderate to high loads, but requires careful design to ensure stability and proper component sizing, as it affects the converter’s transfer function and dynamic response.
2.1. Operation in CCM
In Continuous Conduction Mode (CCM), the inductor current never falls to zero during the switching cycle. This mode is typically desired for high-power applications where a smooth and continuous
current is beneficial. The inductor stores energy when the switch is ON and releases it when the switch is OFF, ensuring that the current through the inductor is always positive.
2.2. CCM Waveforms
• Inductor Current: The current rises during the ON period and falls during the OFF period, but never reaches zero.
• Capacitor Voltage: The voltage across the capacitor is relatively stable with low ripple.
2.3. CCM Design Considerations
• Inductor Size: A large inductor is required to ensure that the inductor current does not drop to zero, leading to continuous conduction.
• Duty Cycle: The duty cycle affects the output voltage. For higher voltages, the duty cycle must approach but not exceed 1.
2.4. Efficiency in CCM
Efficiency is generally higher in CCM compared to DCM because the losses associated with switching and inductor current are minimized. The efficiency of a boost converter in CCM can be approximated
Discontinuous Conduction Mode (DCM)
Discontinuous Conduction Mode (DCM) in a boost DC-DC converter occurs when the inductor current falls to zero during each switching cycle, resulting in intervals where the inductor is not conducting.
This mode typically happens at low load conditions or low input voltage, where the energy stored in the inductor is insufficient to sustain continuous current. In DCM, the converter operates in a
non-linear fashion, with a different transfer function compared to Continuous Conduction Mode (CCM). DCM can lead to reduced efficiency but simplifies control in certain low-power applications due to
its inherent switching characteristics.
3.1. Operation in DCM
In Discontinuous Conduction Mode (DCM), the inductor current falls to zero during part of the switching cycle. This mode typically occurs at light loads, where the energy transferred during each
switching cycle is smaller.
3.2. DCM Waveforms
• Inductor Current: The current rises during the ON period, falls to zero during the OFF period, and remains at zero for a portion of the switching cycle.
• Capacitor Voltage: The voltage across the capacitor may have more ripple due to the intermittent nature of the current.
3.3. DCM Design Considerations
• Inductor Size: A smaller inductor can be used in DCM, reducing the overall converter size and cost. However, this comes at the expense of higher current ripple.
• Control Complexity: DCM introduces more complexity into the control design due to the non-linear behavior of the inductor current.
3.4. Efficiency in DCM
DCM can be less efficient than CCM at light loads due to increased switching losses and higher current ripple. However, for certain applications, such as low-power systems, DCM is preferred because
it reduces the size and cost of the inductor.
Transfer Function of Boost Converter
The transfer function of a boost converter describes the relationship between the input voltage, output voltage, and control input (typically duty cycle) in both continuous conduction mode (CCM) and
discontinuous conduction mode (DCM). In CCM, the transfer function can be derived by modeling the converter’s small-signal behavior using state-space averaging techniques. It shows that the output
voltage is inversely proportional to the duty cycle, meaning that as the duty cycle increases, the output voltage rises. The system is non-linear, but linearization around an operating point helps
analyze stability and dynamic response. In DCM, the transfer function becomes more complex due to the intermittent current flow, and this affects the converter’s efficiency and transient behavior.
4.1. Small-Signal Modeling
To design a stable control loop, it is crucial to understand the small-signal behavior of the boost converter. A small-signal model captures how the output voltage responds to small perturbations in
the input voltage, duty cycle, or load.
For the boost converter, the duty cycle to output voltage transfer function in CCM can be expressed as:
• S is the Laplace variable
4.2. Poles and Zeros
The transfer function typically has two poles, resulting in a second-order response. The location of the poles and zeros impacts the stability and dynamic response of the converter.
• Poles: Determined by the inductor and capacitor values, the poles dictate the natural frequency and damping of the system.
• Zeros: The right half-plane zero is a key characteristic of boost converters, which can complicate control design.
4.3. Control Design
A typical approach to control boost converters is using a Proportional-Integral (PI) or Proportional-Integral-Derivative (PID) controller. The control loop adjusts the duty cycle to regulate the
output voltage in response to changes in the input or load.
Stability Analysis
Stability analysis of a boost converter is critical to ensure reliable and efficient operation. It involves examining the system’s dynamic behavior, particularly how it responds to disturbances or
changes in load or input voltage. This is done by analyzing the converter’s transfer functions and ensuring that the feedback loop is stable under continuous conduction mode (CCM) and discontinuous
conduction mode (DCM). Techniques such as Bode plot analysis are commonly used to assess the phase margin and gain margin, ensuring that the system can return to its equilibrium state without
oscillations or instability when subjected to disturbances. Proper compensation design in the feedback loop is essential for maintaining stable operation.
5.1. Bode Plot
A Bode plot is an essential tool in power converter design. It provides a graphical representation of the gain and phase of the system across different frequencies, helping to assess stability. For a
boost converter, the Bode plot reveals the system’s gain margin and phase margin, which indicate how far the system is from instability.
5.2. Phase Margin
The phase margin is the difference between the phase of the system and -180° when the open-loop gain crosses 0 dB. A phase margin of 45° to 60° is typically considered stable for power converters.
5.3. Gain Margin
The gain margin is the amount of gain that can be added to the system before it becomes unstable. A higher gain margin ensures that the converter remains stable under varying load and input
Design Procedure for Boost Converter
Begin by defining the required specifications, including input voltage, output voltage, load current, switching frequency, and efficiency targets. These parameters form the basis for component
selection and design.
6.1. Component Selection
• Inductor: The inductor value is chosen based on the desired current ripple and operating mode (CCM or DCM).
For CCM operation:
Where ΔIL is the desired inductor current ripple, and fs is the switching frequency.
Capacitor: The output capacitor is selected to ensure low output voltage ripple.
Where ΔVout is the allowable output voltage ripple.
• Switch: The switch must be rated for the peak current and voltage stress. MOSFETs are often chosen for their low ON-resistance and fast switching capabilities.
• Diode: The diode must handle the output current and withstand the reverse voltage when the switch is ON.
6.2. Thermal Design
Ensure that the components selected can dissipate the heat generated during operation. Use heat sinks or proper PCB layout techniques to manage thermal performance, especially at high power levels.
6.3. Control Loop Design
Implement a control loop using either analog or digital methods. The loop compensator (e.g., PI or PID controller) must be designed to achieve the desired dynamic response while maintaining
6.4. Simulation and Prototyping
Before building the physical converter, use simulation tools (e.g., MATLAB, LTSpice) to validate the design. Simulations provide insight into transient response, stability, and efficiency. Once the
design is verified, proceed with prototyping and testing.
Boost DC-DC converters are essential in modern power electronics systems due to their ability to efficiently step up voltage levels while maintaining high efficiency. Understanding their operation in
CCM and DCM, analyzing transfer functions, and ensuring stability through proper control design are critical steps in developing a reliable | {"url":"https://tenxerlabs.com/resources/blogs/boost-converter/","timestamp":"2024-11-09T13:15:26Z","content_type":"text/html","content_length":"501413","record_id":"<urn:uuid:4cae8405-c8d1-4f7e-a9a6-d7666f6bad34>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00355.warc.gz"} |
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Pie Chart Multiple Data Series Excel 2024 - Multiplication Chart Printable
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arcsin() return value with tiny imaginary part ?
arcsin() return value with tiny imaginary part ?
where does this imaginary small quantity come from ?
SageMath notebook 9.1 W10
show(arcsin(sqrt(2)/sqrt(2)).n(),"\t ok no small imaginary part")
1 Answer
Sort by ยป oldest newest most voted
It comes from machine precision. sqrt(2)*sqrt(1/2) is slightly bigger than one. In fact sqrt(2)*sqrt(1/2)-1 returns 2.22044604925031e-16 on my computer, instead of zero. To avoid the issue, you can
simply replace .n() by .simplify().n().
edit flag offensive delete link more
Hi Florentin
but arcsin((2.22044604925031e-16)).n() does not have tiny imaginary part .
ortollj ( 2020-11-08 20:18:40 +0100 )edit
You forgot the +1. Arcsin is a complex (multivalued) function outside $[-1, 1]$. Try to compute arcsin(1+1/1000). (Also It's not accepting floats, since reals values outside the interval define a
branch cut, you can find the plots on wikipedia).
Florentin Jaffredo ( 2020-11-09 08:56:48 +0100 )edit
Thank you Florentin
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than the unvalued-case algorithm, ignore.eval should be set to TRUE wherever possible. | {"url":"https://www.rdocumentation.org/packages/sna/versions/2.4/topics/geodist","timestamp":"2024-11-14T11:01:35Z","content_type":"text/html","content_length":"65452","record_id":"<urn:uuid:d24f45dc-71d6-4441-b2c1-1b0960d93831>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00338.warc.gz"} |
When quoting this document, please refer to the following
DOI: 10.4230/LIPIcs.SoCG.2016.39
URN: urn:nbn:de:0030-drops-59310
URL: http://dagstuhl.sunsite.rwth-aachen.de/volltexte/2016/5931/
Fomin, Fedor ; Kolay, Sudeshna ; Lokshtanov, Daniel ; Panolan, Fahad ; Saurabh, Saket
Subexponential Algorithms for Rectilinear Steiner Tree and Arborescence Problems
A rectilinear Steiner tree for a set T of points in the plane is a tree which connects T using horizontal and vertical lines. In the Rectilinear Steiner Tree problem, input is a set T of n points in
the Euclidean plane (R^2) and the goal is to find an rectilinear Steiner tree for T of smallest possible total length. A rectilinear Steiner arborecence for a set T of points and root r in T is a
rectilinear Steiner tree S for T such that the path in S from r to any point t in T is a shortest path. In the Rectilinear Steiner Arborescense problem the input is a set T of n points in R^2, and a
root r in T, the task is to find an rectilinear Steiner arborescence for T, rooted at r of smallest possible total length. In this paper, we give the first subexponential time algorithms for both
problems. Our algorithms are deterministic and run in 2^{O(sqrt{n}log n)} time.
BibTeX - Entry
author = {Fedor Fomin and Sudeshna Kolay and Daniel Lokshtanov and Fahad Panolan and Saket Saurabh},
title = {{Subexponential Algorithms for Rectilinear Steiner Tree and Arborescence Problems}},
booktitle = {32nd International Symposium on Computational Geometry (SoCG 2016)},
pages = {39:1--39:15},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-009-5},
ISSN = {1868-8969},
year = {2016},
volume = {51},
editor = {S{\'a}ndor Fekete and Anna Lubiw},
publisher = {Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik},
address = {Dagstuhl, Germany},
URL = {http://drops.dagstuhl.de/opus/volltexte/2016/5931},
URN = {urn:nbn:de:0030-drops-59310},
doi = {10.4230/LIPIcs.SoCG.2016.39},
annote = {Keywords: Rectilinear graphs, Steiner arborescence, parameterized algorithms}
Keywords: Rectilinear graphs, Steiner arborescence, parameterized algorithms
Collection: 32nd International Symposium on Computational Geometry (SoCG 2016)
Issue Date: 2016
Date of publication: 10.06.2016
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Does an A/B test need to be 50/50?
Not necessarily. While a 50/50 split is common and helps in quickly reaching statistical significance, you can use other ratios like 60/40 or 70/30 depending on your testing goals and traffic. Just
remember that the more even the split, the quicker you’ll gather reliable data.
Is it better statistically to have unequal sample sizes or equal sample sizes?
From a statistical standpoint, equal sample sizes are preferable. They make it easier to detect differences between variants and ensure that your test results are robust and reliable. | {"url":"https://unbounce.com/ab-test-duration-calculator/","timestamp":"2024-11-08T05:36:29Z","content_type":"text/html","content_length":"389821","record_id":"<urn:uuid:68d7c31f-a0cf-44b0-aaa9-71333d8f9020>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00134.warc.gz"} |
Tree traversal techniques in JavaScript
Tree is an interesting data structure. It has wide variety of applications in all sorts of fields.
For example:
• DOM is a tree data structure
• Directory and files in our OS can be represented as trees
• A family hierarchy can be represented as a tree.
There are bunch of variations of tree (such as heaps, BST etc.) which can be used in solving problems related to scheduling, image processing, databases etc. Many of complex problems may not seem
related to tree on a quick look, but can actually be represented as one. We'll walk through such problems as well (in later parts of this series) to see how trees can make seemingly complex problems
much easier to comprehend and solve.
Implementing a Node for a binary tree is pretty straightforward.
function Node(value){
this.value = value
this.left = null
this.right = null
// usage
const root = new Node(2)
root.left = new Node(1)
root.right = new Node(3)
So these few lines of code would create a binary tree for us which looks like this:
Cool! So that was easy. Now, how do we put this to use?
Let's start with trying to walk through these connected tree nodes (or a tree). Just as we can iterate through an array, it would be cool if we can 'iterate' through tree nodes as well. However,
trees are not linear data structures like arrays, so there isn't just one way of traversing these. We can broadly classify the traversal approaches into following:
• Breadth first traversal
• Depth first traversal
Breadth First Search/Traversal (BFS)
In this approach, we traverse the tree level by level. We would start at the root, then cover all of it's children, and we cover all of 2nd level children, so on and so forth.
For example for the tree above, traversal would result in something like this:
Here's an illustration with slightly complex tree to make this even simpler to understand:
To achieve this form of traversal we can use a queue (First In First Out) data structure. Here's how the overall algorithm would look like:
1. Initiate a queue with root in it
2. Remove the first item out of queue
3. Push the left and right children of popped item into the queue
4. Repeat steps 2 and 3 until the queue is empty
Here's how this algorithm would look like post implementation:
function walkBFS(root){
if(root === null) return
const queue = [root]
const item = queue.shift()
// do something
if(item.left) queue.push(item.left)
if(item.right) queue.push(item.right)
We can modify above algorithm slightly to return an array of arrays, where each inner array represents a level with elements within in:
function walkBFS(root){
if(root === null) return
const queue = [root], ans = []
const len = queue.length, level = []
for(let i = 0; i < len; i++){
const item = queue.shift()
if(item.left) queue.push(item.left)
if(item.right) queue.push(item.right)
return ans
Depth First Search/Traversal (DFS)
In DFS, we take one node and keep exploring it's children until the depth the fully exhausted. It can be done in one of following ways:
root node -> left node -> right node // pre-order traversal
left node -> root node -> right node // in-order traversal
left node -> right node -> root node // post-order traversal
All of these traversal techniques can be implemented recursively as well as iteratively. Let's jump into the implementation details:
Pre-Order traversal
Here's how PreOrder traversal looks like for a tree:
root node -> left node -> right node
We can use this simple trick to find out the PreOrder traversal of any tree manually: traverse the entire tree starting from the root node keeping yourself to the left.
Let's dive into actual implementation for such a traversal.
Recursive approach is fairly intuitive.
function walkPreOrder(root){
if(root === null) return
// do something here
// recurse through child nodes
if(root.left) walkPreOrder(root.left)
if(root.right) walkPreOrder(root.left)
Iterative approach for PreOrder traversal is very similar to BFS, except we use a stack instead of a queue and we push the right child first into the queue:
function walkPreOrder(root){
if(root === null) return
const stack = [root]
const item = stack.pop()
// do something
if(item.right) stack.push(item.right)
if(item.left) stack.push(item.left)
In-Order traversal
Here's how InOrder traversal looks like for a tree:
left node -> root node -> right node
We can use this simple trick to find out InOrder traversal of any tree manually: keep a plane mirror horizontally at the bottom of the tree and take the projection of all the nodes.
function walkInOrder(root){
if(root === null) return
if(root.left) walkInOrder(root.left)
// do something here
if(root.right) walkInOrder(root.right)
function walkInOrder(root){
if(root === null) return
const stack = []
let current = root
while(stack.length || current){
current = current.left
const last = stack.pop()
// do something
current = last.right
Post-Order traversal
Here's how postOrder traversal looks like for a tree:
left node -> right node -> root node
For quick manual PostOrder traversal of any tree: pluck all the leftmost leaf nodes one by one.
Let's dive into actual implementation for such a traversal.
function walkPostOrder(root){
if(root === null) return
if(root.left) walkPostOrder(root.left)
if(root.right) walkPostOrder(root.right)
// do something here
function walkPostOrder(root){
if(root === null) return []
const tempStack = [root], mainStack = []
const last = tempStack.pop()
if(last.left) tempStack.push(last.left)
if(last.right) tempStack.push(last.right)
return mainStack.reverse()
Bonus: JavaScript tip
How nice it would be if we could traverse the tree in one of following ways:
for(let node of walkPreOrder(tree) ){
Looks really nice and pretty simple to read, isn't it? All we've got to do is use a walk function, which would return an iterator.
Here's how we can modify our walkPreOrder function above to behave as per the example shared above:
function* walkPreOrder(root){
if(root === null) return
const stack = [root]
const item = stack.pop()
yield item
if(item.right) stack.push(item.right)
if(item.left) stack.push(item.left)
This article has been originally published at StackFull.dev. If you'd like to be notified when I drop more such articles, consider subscribing to the newsletter.
Top comments (2)
Ananthakrishnanj • • Edited on • Edited
Hi @anishkumar ,
It was a wonderful read👌
However, In your example of in order traversal, it is mentioned as
root node -> left node -> right node
instead of like this
left node -> root node -> right node
Anish Kumar •
Yes, thanks for pointing out. Corrected.
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/anishkumar/tree-data-structure-in-javascript-1o99","timestamp":"2024-11-05T06:00:52Z","content_type":"text/html","content_length":"133112","record_id":"<urn:uuid:61c81349-5712-4aa3-9e70-330249640bd2>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00528.warc.gz"} |
Interesting Esoterica
Prime Number Races
• Published in 2004
• Added on
In the collections
This is a survey article on prime number races. Chebyshev noticed in the first half of the nineteenth century that for any given value of x, there always seem to be more primes of the form 4n+3
less than x then there are of the form 4n+1. Similar observations have been made with primes of the form 3n+2 and 3n+1, with primes of the form 10n+3/10n+7 and 10n+1/10n+9, and many others
besides. More generally, one can consider primes of the form qn+a, qn+b, qn+c, >... for our favorite constants q, a, b, c, ... and try to figure out which forms are "preferred" over the others.
In this paper, we describe these phenomena in greater detail and explain the efforts that have been made at understanding them.
Other information
BibTeX entry
key = {PrimeNumberRaces},
type = {article},
title = {Prime Number Races},
author = {Andrew Granville and Greg Martin},
abstract = {This is a survey article on prime number races. Chebyshev noticed in the
first half of the nineteenth century that for any given value of x, there
always seem to be more primes of the form 4n+3 less than x then there are of
the form 4n+1. Similar observations have been made with primes of the form 3n+2
and 3n+1, with primes of the form 10n+3/10n+7 and 10n+1/10n+9, and many others
besides. More generally, one can consider primes of the form qn+a, qn+b, qn+c,
>... for our favorite constants q, a, b, c, ... and try to figure out which
forms are "preferred" over the others. In this paper, we describe these
phenomena in greater detail and explain the efforts that have been made at
understanding them.},
comment = {},
date_added = {2018-11-12},
date_published = {2004-10-09},
urls = {http://arxiv.org/abs/math/0408319v1,http://arxiv.org/pdf/math/0408319v1},
collections = {Attention-grabbing titles,Easily explained,Fun maths facts,Integerology},
url = {http://arxiv.org/abs/math/0408319v1 http://arxiv.org/pdf/math/0408319v1},
year = 2004,
urldate = {2018-11-12},
archivePrefix = {arXiv},
eprint = {math/0408319},
primaryClass = {math.NT} | {"url":"https://read.somethingorotherwhatever.com/entry/PrimeNumberRaces","timestamp":"2024-11-08T17:17:35Z","content_type":"text/html","content_length":"6421","record_id":"<urn:uuid:7c59808f-325f-4530-8101-5db58f14d4b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00550.warc.gz"} |
PIMCO Compares Bond with Stock by PE Ratio?!
PIMCO is one of the top bond fund manager. It recently released an
to compare bond vs stock using PE ratio. For bond, PE ratio is the inverse of bond yield. The chart shown in the article indicates that bond is more risky than stock, (not only more over-valued).
This methodology and conclusion is really bizarre, particularly when coming out of a top bond fund manager. If Bill Gross had remained as the head of the fund, such article probably would not have
had a chance to be released. PIMCO seems to really go downhill.
Use inverse of bond yield as PE ratio seems intuitive, but it's inappropriate to compare it with stock. First, bond PE ratio, or yield, is locked in to maturity. If the yield of a 5 year bond is 2%,
an investor gets 2% annual return when the bond matures. But stock is a different story. For a stock of PE ratio at 50, in 5 years, investor may get negative return. 50 PE ratio is pretty high, so
the stock is risky. For such a stock, the chance of negative return in 5 years is not negligible. In other words, bond price is pulled to par, and coupons can be fully realized. Stock price can be
pretty much anything.
Secondly, bond has fixed maturity date. In order for a bond fund to maintain effective maturity, the bond fund needs to adjust portfolio and add new bond periodically. For example, a 5 year bond fund
is loaded with 5 year bond. In one year, the remaining maturity is 4 years, so the fund needs to sell half the asset to add 6 years bond, in order to achieve 5 year effective maturity. Without doing
so, the bond fund will end up with all cash in hand once bonds mature. (This example is to illustrate the concept, not operationally accurate.) The process to roll bond holding effectively lets a
bond fund to invest in new bond when interest rate goes up, which exposes itself with higher return in yield. This is kind of similar to average investment strategy, but it's a by-product of managing
effective maturity. This is a unique benefit to bond fund, which stock does not have.
Figure below is the PIMCO PE ratio comparison
Figure 1. PIMCO PE ratio comparison
PIMCO paper misleads on the volatility of bond. As PE ratio is the yardstick of asset valuation, it implies that bigger volatility in PE ratio equates to bigger asset price volatility. This is
totally wrong as explained above because bond fund has its own uniqueness. In order to further illustrate the crucial difference, below are charts included in Bill Gross' June investment report.
Fig. 2 shows aggregated bond annualized return is 7.47% in 40 years. Fig 3 shows S&P 500 total return is only so slightly better in the same period of time, but has a much bigger volatility. Readers
are likely to draw completely opposite conclusion if they judge by Fig. 1.
Figure 2. Aggregated Bond Index
Figure 3. S&P 500 Index (Total Return) | {"url":"http://en.wujibifan.org/2016/08/pimco-compares-bond-with-stock-by-pe.html","timestamp":"2024-11-03T22:45:41Z","content_type":"text/html","content_length":"40898","record_id":"<urn:uuid:7427b01b-8737-419e-b341-de2976045e13>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00871.warc.gz"} |
CVIPGS MRes
Research Masters in Computer Vision, Image Processing, Graphics and Simulation
CVIPGS MRes
Information on the courses taught by B F Buxton:
Machine Vision (MV)
Taught with Danny Alexander
Project Outlines 2001-2002
Practical 1
3c72 Practical 1
3c72 Practical 2
3c72 Practical 3
Practical 2
Practical 3
Practical 4
Detailed syllabus for B F Buxton, S Robson and D P Chapman, for 1999-00 Revision examples for B F Buxton for 1997-98 For some old machine vision examination papers, click here for:
Pattern Recognition and Machine Vision (PR&MV)
Delivered as GI04 on the Intelligent Systems advanced MSc and as 4c75 on the undergraduate MSci programme.
PR&MV and 4c75 Pattern Reconginition Practical
Maths Methods, Algorithms and Implementation (MMAI)
Taught with Simon Arridge and Vicky Hardman
Detailed syllabus for B F Buxton and V Hardman, for 1999-00 B F Buxton: Some lecture notes for 1998-99
1 & 2. Analog and digital video & algorithm requirements
3. Low-level image processing
4. Introduction to the z-transform
5. Algorithm complexity
6. Real-time computing systems
7. Processors and computing devices
8. Multi-processor systems
A non-compulsory coursework/revision exercise on multiprocessor systems is available here . See also the papers distributed during the lectures.
Physics, Psychophysics and Physiology of Vision (PPP)
Taught with Keith Langley (Psychology)
Physics, Psychophysics and Physiology of Vision, detailed syllabus for B F Buxton for 1997-98 Practical 1 Practical 2 Physics, Psychophysics and Physiology of Vision, revision examples for B F Buxton
for 1997-98 For some old physics, psychophysics and physiology of vision examination papers, click here for: | {"url":"http://www0.cs.ucl.ac.uk/staff/B.Buxton/teaching/mres/mres_teaching.html","timestamp":"2024-11-06T06:07:51Z","content_type":"text/html","content_length":"4862","record_id":"<urn:uuid:d4a63574-436e-45c3-b2fd-914a12e688e6>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00817.warc.gz"} |
Election Week and US-China Trade War: Mr. Market is Unsure
Global stocks closed varied on Tuesday (26/6). On US, the Dow rose to 24,283 (+0.12%) and Nasdaq rose to 7,561 (+ 0.39%) from previous day. In European exchanges, DAX weakened 0.29% to 12,234 while
FTSE 100 rose 0.37% to 7,537. Similarly, in Asia, Hang Seng fell -0.28% to 28,881 while the Nikkei stagnated at 22,342 (+0.02%). All these are expected seeing that many are happening in the global
Watching global market, the hottest discussion recently is China-US trade war. It started on March 22, 2018 after Donald Trump declared a plan to put tariffs on Chinese goods on the basis of “unfair
trade practices” and theft of intellectual property. As a respond, the Chinese government also put tariffs to over 128 US products that are imported there. This issue is making investors unsure and
anxious, resulting in a mixed global index performance in the last few weeks.
Seeing into oil industry, news from last Tuesday reported that crude futures increased over 2 percent and US oil touched the price of $70 for the first time in two months. This is very much because
of Trump’s suggestion to US’ allies to halt imports of crude oil from Iran, making the global supply decreasing. This is followed by the strengthening of energy stocks.
Another interesting matter is that global coal price keeps increasing since the end of April and is still in a positive trend. One thing to be mentioned is that Indonesia is one of the largest
exporters of coal in Indonesia. In terms of coal reserves, Indonesia is ranked 10^thwith 2.2 percent of global coal reserves (Statistical Review of World Energy 2018). We’re still confident that coal
is prospective since it’s still the dominating force in power generation, with roughly 27% of the world’s total energy output. With the abundant reserves, added with Indonesia’s strategic
geographical position, coal industry in Indonesia is still going strong.
In the domestic market, JCI’s movement on Tuesday was quite volatile until finally closed at 5,825, indicating a 0.57% decrease. We think that global sentiment is still overshadowing the movement of
JCI. Investors are still worried about the trade war between China and the US, putting pressure on Asian regional indices, and eventually on Indonesian exchange. In addition, foreign investors
continued to sell and recorded net sales of IDR 453.09 billion with the most sales in BBRI. We expect the movement of this week’s index to be sideways, since investors are still waiting and seeing,
along with the elections held today. Even so, we believe that there is an upside potential from the increase of oil and coal price, strengthening energy-based stocks.
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Tate twists in singular and de Rham cohomology
Maths > Abelian varieties > Absolute Hodge classes
Posted by Martin Orr on Friday, 19 June 2015 at 19:30
Tate twists in singular cohomology are a device for dealing with factors of which come up whenever we compare singular and de Rham cohomology of complex projective varieties. In this post I will
explain the problem, including calculating the in the case of , and define Tate twists to solve it.
In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, we could just write out factors of
The problem
As I briefly mentioned previously, for each .
If we choose an embedding , we can also define a cycle class for Hodge conjecture). But there are two different normalisations for the cycle class map into singular cohomology.
First, there is the topologist's version , defined by taking the fundamental class of
However, it turns out that the topologist's cycle class map is not compatible with the de Rham cycle class map. They differ by a factor of I will prove this below for the class of a point in
The solution: Tate twists
We fix this problem by changing the normalisation of the cycle class map for singular cohomology. Why do we change the normalisation for singular cohomology instead of for de Rham cohomology? Because
this normalisation is something we will only do over
If we simply multiplied by , then we would solve the compatibility issue, but it would no longer take values in (and hence its values would not be Hodge classes). So instead we introduce a new object
called a Tate twist.
The Tate twist for singular cohomology is a and with Hodge type .
As pieces of notation, we define for each positive integer , and define to be the dual of . We also define to be -Hodge structure induced by , and
We define the algebraic geometer's cycle class map for singular cohomology to be (The subscript
It is useful to also define Tate twists for de Rham cohomology, even though they are trivial. We thus define to be equal to (not just isomorphic). The reason why we want to use the label for a Tate
twist in de Rham cohomology is so that we remember which isomorphism to use to compare it with singular cohomology.
I was a bit confused about this until I asked a MathOverflow question a few days ago which helped me clear it up. The comparison isomorphism is defined as the untwisted comparison isomorphism
multiplied by the inclusion .
With this comparison isomorphism, we get the desired compatibility between cycle class maps:
Because the Tate twist is defined to have Hodge type , the Hodge structure has weight 0. We define a Hodge class in to be an element of with Hodge type . The image of
We could have carried out all of our previous two posts on absolute Hodge classes using in place of ; the changes would be essentially just notational.
Calculating the de Rham cycle class map for
I am going to justify the fact that the de Rham cycle class map and the topologist's cycle class map into differ by a factor of . One often sees the , but that doesn't seem to fit our setting of
smooth projective varieties. So I am going to write out this calculation carefully (except that I will not be careful about signs - there are several sign conventions along the way and I don't think
it is worth keeping track of them). Thanks to Jack Shotton who helped with this.
First, is a generator of . If we choose a 2-form , the definition of the comparison isomorphism says that
The hard work is to calculate . In particular, we want to show that
First, we interpret our point as a divisor of degree 1 on , a divisor of degree 1 corresponds to the Čech cocycle with respect to the open cover , (this is one place where I can never remember the
sign convention).
We then apply the map represented by the Čech cocyle
The degeneracy of the Hodge-de Rham spectral sequence gives us an isomorphism . To compute this isomorphism analytically over
Here such that , vanishes on a neighbourhood of zero and vanishes on a neighbourhood of . Note that and this implies that the square in the centre of the diagram commutes.
We conclude that is represented by the -form We want to compute the integral of over
Choose a loop . Let and be the two regions into which and .
On , the 1-form is well-defined and has total derivative equal to . Similarly, on , the 1-form is well-defined and has total derivative equal to .
Hence, using Stokes' theorem, we get (The minus sign in the second integrand cancels with the fact that the boundary of goes backwards round .) | {"url":"http://www.martinorr.name/blog/2015/06/19/tate-twists-in-singular-and-de-rham-cohomology/","timestamp":"2024-11-12T05:36:47Z","content_type":"application/xhtml+xml","content_length":"30741","record_id":"<urn:uuid:675589ba-2ce5-493c-8133-e5ae321d610c>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00744.warc.gz"} |
dh parameter calculator calculation for Calculations
29 Mar 2024
Popularity: ⭐⭐⭐
Pressure at the Bottom of a Vertical Pipe Filled with Liquid
This calculator provides the calculation of pressure at the bottom of a vertical pipe filled with liquid.
Calculation Example: The pressure at the bottom of a vertical pipe filled with liquid is given by the formula P = ? * g * H, where ? is the density of the liquid, g is the acceleration due to
gravity, and H is the height of the liquid in the pipe.
Related Questions
Q: What is the significance of pressure in fluid mechanics?
A: Pressure is a crucial parameter in fluid mechanics as it determines the forces acting on fluids and the behavior of fluid systems. It is essential for understanding fluid flow, buoyancy, and other
fluid-related phenomena.
Q: How does the height of the liquid affect the pressure at the bottom of the pipe?
A: The height of the liquid directly influences the pressure at the bottom of the pipe. As the height increases, the pressure also increases due to the increased weight of the liquid column above.
| —— | —- | —- |
Calculation Expression
Pressure Function: The pressure at the bottom of the pipe is given by P = ? * g * H
Calculated values
Considering these as variable values: ?=1000.0, D=0.3, g=9.81, H=1.2, the calculated value(s) are given in table below
| —— | —- |
Pressure Function 11772.0
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Net income required rate
Do you earn enough money to buy the home you want? By entering just a few data points into NerdWallet's mortgage income calculator, we can help you determine how much income you'll need to qualify In
management accounting, residual income represents any excess of a department's income over the opportunity cost of the capital that it employs. It is calculated by subtracting the product of a
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s level of operations and the assets needed to sustain operating activities. s return on equity, return on assets, and net profit margin ratio as the product of CAPM is a theory concentrated with
deriving the expected rates of return on risky Apple Inc.'s profitability ratio calculated as net income divided by total assets. 12 Dec 2019 Whether you're an investor or a business owner, net
income is an For instance, net income can help you calculate a company's price-to-earnings ratio. Americans are expected to take it for the 2019 tax year and beyond. 27 Jan 2020 family net income
(AFNI), as reported in last year's tax return Tax Credit ( AFETC), there is no minimum working income requirement.
Net profit margin is a financial ratio comparing a company's net profit after taxes There are some notable exceptions to this general rule, but that would require
What Is the Formula for Calculating Net Present Value (NPV)? FACEBOOK value at the end of the project and the required rate of return is 8%. the present value of future income, cash flows You usually
must pay self-employment tax if you had net earnings from self-employment of $400 or more. Generally, the amount subject to self-employment tax is 92.35% of your net earnings from self-employment.
The formula in computing for the residual income is: where: Desired income = Minimum required rate of return x Operating assets. Note: In most cases, the minimum required rate of return is equal to
the cost of capital. The average of the operating assets is used when possible. Required income 101. Your debt and salary limit what you can afford. Besides showing you how much income you need to
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minimum required rate of return) Residual Income. Net operating income - minimum required rate of return dollar amount. Residual Income (Equation 2) CM/Sales.
7 Apr 2019 In management accounting, residual income represents any excess of a where it equals the difference between a company's net income and the Residual Income = Controllable Margin - Required
Return × Average
18 Dec 2017 A capitalization (cap) rate is the ratio of a property's Net Operating Income (NOI) in the first Perpetuity Value = Annual Income / Expected Rate of Return Capitalization Rate = Expected
Returns – Growth Rate of Income Under the accrual method of accounting, net income is calculated as follows: In short, the statement of cash flows is a needed financial statement because the Let's
call this projects. Theses are potential investments. You have projects, and then you have some level of expected return. Each of the people who are thinking 23 Sep 2019 Net Income is one of the
simplest concepts in economics and one that Such expenses are documented on a tax return form and will affect Aaron would compute his annual net income by subtracting total expenses ($67,500) from
total income. Since Aaron’s revenues exceed his expenses, he will show $132,500 profit. If Aaron only made $50,000 of revenues for the year, he would not have negative earnings, however. Instead, he
would have a net loss of $17,500.
22 Nov 2015 Net income is our gross profit of $35,000 less our operating expenses of $29,000 = $6,000 for the year. The Final Projection. Now that you've put
to earn greater profits than the return normally to be expected on the capital represented by the net tangible assets employed in the business. In considering the 20 May 2019 Calculate net income
needed in retirement Even at this modest rate, a monthly expense of Rs 1 lakh per month will balloon to Rs 5.74 lakh in 22 Nov 2015 Net income is our gross profit of $35,000 less our operating
expenses of $29,000 = $6,000 for the year. The Final Projection. Now that you've put available for common stockholders, take the company's after-tax profit -- also called net income or earnings --
and subtract any amount of that profit that must be Preferred Stock; What Can Affect a Return on Common Stockholders' Equity ? 18 Dec 2018 Net income is a firm's profit after it pays and accounts for
all of its annual liabilities . This includes line-items such as taxes, salaries, cost of goods 18 Dec 2017 A capitalization (cap) rate is the ratio of a property's Net Operating Income (NOI) in the
first Perpetuity Value = Annual Income / Expected Rate of Return Capitalization Rate = Expected Returns – Growth Rate of Income
The required rate of return (hurdle rate) is the minimum return that an investor is expecting to receive for their investment. Essentially, the required rate of return
There is a more important measurement to look at, however – net profit. of your net profits, i.e. the net profit figure is required before the tax calculation can be to access your net profit by
analysing the return you submit to the tax authorities. s level of operations and the assets needed to sustain operating activities. s return on equity, return on assets, and net profit margin ratio
as the product of CAPM is a theory concentrated with deriving the expected rates of return on risky Apple Inc.'s profitability ratio calculated as net income divided by total assets. 12 Dec 2019
Whether you're an investor or a business owner, net income is an For instance, net income can help you calculate a company's price-to-earnings ratio. Americans are expected to take it for the 2019
tax year and beyond. 27 Jan 2020 family net income (AFNI), as reported in last year's tax return Tax Credit ( AFETC), there is no minimum working income requirement.
Introduction Residual income reflects net income minus a deduction for the required return on common equity. While a firm may show positive earnings, the. Required interest rate on security= nominal
risk-free rate + default risk premium+ liquidity +Unincorporated business net income (business owner's income). 6 Jun 2019 You can calculate net income by using the following formula: like to compare
company earnings using the price-to-earnings (P/E) ratio, which Internal rate of return (IRR) is the interest rate at which the NPV of all the cash flows and Acme's required rate of return
(opportunity cost of capital) is 23%, Acme It is worth noting that PBP calculation uses cash flows, not the net income. 10 Mar 2020 Return on investment (ROI) is a financial ratio intended to measure
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In the given context, capacity is a term that describes how much a container will hold. It is used in reference to the volume of fluids or gases and is measured in units such as litres or
The Cartesian plane or Cartesian coordinate system is a system that describes the exact location of any point in a plane using an ordered pair of numbers, called coordinates. It is defined by the
intersection of a horizontal and vertical number line at a point called the origin. The coordinates of the origin are (0, 0).
The Cartesian plane is divided into four quadrants by these perpendicular axes called the x-axis (horizontal line) and the y-axis (vertical line). The axes can be used to identify any point in the
plane using a pair of coordinates, as shown in the diagram below.
A categorical variable is a variable whose values are categories. For example, blood group is a categorical variable; its common values are: A, B, AB or O.
A census is a survey of a whole population.
A chord in a circle is a line segment joining any two points on the circle. Chord PQ, illustrated below, joins points P and Q.
A circle, with centre O and radius r, is the set of all points on a plane whose distance from O is r.
Circumference refers to the boundary of a circle. The length of the circumference c is given by \(c=\pi d\), where d is the diameter. Alternatively, it is given by \(c=2\pi r\), where r is the
Angles are classified according to their size. See acute angle, obtuse angle, reflex angle, right angle, straight angle and revolution.
Co-interior angles lie between two lines and on the same side of a transversal.
In each diagram the two marked angles are called co-interior angles.
If the two lines are parallel, then co-interior angles add to give 180^o and so are supplementary. In the diagram below the angles ∠CGF and ∠AFG are supplementary.
Conversely, if a pair of angles are supplementary, then the lines are parallel. Line segment CD is parallel to line segment AB, because ∠CGF + ∠AFG = 180°.
A column graph is a graph used in statistics for organising and displaying categorical data. It consists of a series of equal-width rectangular columns, one for each category. Each column has a
height equal to the frequency of the category. This is shown in the example below which displays the hair colours of 27 students.
Column graphs are frequently called bar graphs or bar charts. In a bar graph or chart, the bars can be either vertical or horizontal.
A common factor (or common divisor) of a set of numbers or algebraic expressions is a factor of each element of that set. For example, 6 is a common factor of 24, 54 and 66, since \(24=6\times4\), \
(54=6\times9\), and \(66=6\times11\). Similarly, \(x+1\) is a common factor of \(x^2-1\) and \(x^2+5x+4\), since \(x^2-1\;=(x+1)(x-1)\) and \(x^2+5x+4\;=(x+1)(x+4)\).
Operations are commutative if the order in which terms are given does not affect the result.
The commutative law for addition is: \(a+b=b+a\) for all numbers a and b.
For example, 3+5=5+3.
The commutative law for multiplication is: \(ab=ba\) for all numbers a and b.
For example, 4×7=7×4.
Subtraction and division are not commutative because 5-3≠3-5 and 12÷4≠4÷12.
Two angles that add to 90^o are called complementary; for example, 23^o and 67^o are complementary angles.
Events A and B are complementary events if A and B are mutually exclusive (have no overlap) and \(\;Pr(A)\;+\;Pr(B)\;=\;1\), where the symbol \(Pr(A)\;\) denotes the probability of event \(A\)
A composite number is a natural number that has a factor other than 1 and itself.
The interest earned by investing a sum of money (the principal) is compound interest if each successive interest payment is added to the principal for the purpose of calculating the next interest
payment. For example, if the principal \(\$P\) earns compound interest at the rate of r% per period, then after n periods the principal plus interest is \(\;\$P(1+r)^n\).
Computation is mathematical calculation.
A cone is a solid that is formed by taking a circle, called the base, and a point, called the vertex, which lies above or below the circle, and joining the vertex to each point on the circle.
Two plane shapes are congruent if they are identical in size and shape and one can be moved or reflected so that it fits exactly on top of the other figure.
Matching sides have the same length, and matching angles have the same size.
The four standard congruence tests for triangles
Two triangles are congruent if:
SSS: the three sides of one triangle are respectively equal to the three sides of the other triangle, or
SAS: two sides and the included angle of one triangle are respectively equal to two sides and the included angle of the other triangle, or
AAS: two angles and one side of one triangle are respectively equal to two angles and the matching side of the other triangle, or
RHS: the hypotenuse and one side of one right‐angled triangle are respectively equal to the hypotenuse and one side of the other right‐angled triangle.
Continuous numerical data includes any value that lies within an interval. In practice, the values taken are subject to the accuracy of the measurement instrument used to obtain these values. Height,
reaction time to a stimulus and systolic blood pressure are all types of continuous numerical data that can be collected.
A coordinate is one value of an ordered pair that describes the location of a point along an axis in the Cartesian plane. By definition, the first number (xcoordinate) of the ordered pair denotes the
horizontal distance, the second number (ycoordinate) gives the vertical distance from the centre (origin) of the coordinate system. Positive x coordinates indicate that the point is located to the
right (East), negative to the left (West) of the origin. Positive y coordinates indicate a location above (North of), negative below (South of) the origin. The origin has the coordinates (0,0).
For instance, in the ordered pair (4, –2) the number 4 denotes the x coordinate of a point situated at a horizontal distance of 4 units to the origin. The number –2 denotes the y coordinate of the
same point indicating a vertical distance of 2 units below the origin.
Corresponding angles are formed when two lines are crossed by another line (the transversal). In each diagram the two marked angles are called corresponding angles because they are on the same side
of the transversal and in corresponding positions in relation to the lines.
If the lines are parallel, then each pair of corresponding angles is equal (as are the angles ∠QGD and ∠GFB in the diagram shown below).
Conversely, if a pair of corresponding angles is equal, then the lines are parallel.
In any right-angled triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), where \(0<\theta<90^\circ\).
In any triangle ABC, \(c^2=a^2+b^2-2ab\cos C\)
Counting numbers are the positive integers, that is, the numbers 1, 2, 3, … .
Sometimes it is taken to mean the non-negative integers, which include zero.
Counting on is a strategy for solving simple addition problems. For example, a student can add 6 and 4 by counting on from 6, saying ‘7, 8, 9, 10’. If students are asked how many more objects need to
be added to a collection of 8 to give a total of 13, they can count ‘9, 10, 11, 12, 13’ to find the answer 5.
A cylinder is a solid that has parallel circular discs of equal radius at the ends, and whose horizontal cross-section is a circle with the same radius. The centres of these circular cross-sections
lie on a straight line, called the axis of the cylinder. | {"url":"https://australiancurriculum.edu.au/f-10-curriculum/mathematics/glossary/?letter=C","timestamp":"2024-11-12T00:28:44Z","content_type":"text/html","content_length":"67675","record_id":"<urn:uuid:b0d0c8ae-dc7c-4668-9d5f-b69da69ea6e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00642.warc.gz"} |
solve Function with tol argument
solvet {Hmisc} R Documentation
solve Function with tol argument
A slightly modified version of solve that allows a tolerance argument for singularity (tol) which is passed to qr.
solvet(a, b, tol=1e-09)
a a square numeric matrix
b a numeric vector or matrix
tol tolerance for detecting linear dependencies in columns of a
See Also
version 5.1-3 | {"url":"https://search.r-project.org/CRAN/refmans/Hmisc/html/solvet.html","timestamp":"2024-11-04T05:27:06Z","content_type":"text/html","content_length":"2223","record_id":"<urn:uuid:32648845-5a7f-4034-809b-4a55174a4e3d>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00336.warc.gz"} |
Posts by gecomol
Doing some maths with the fusion reactors. The reactors put out 300k EU @ 30eu/tick, which means they last for 10k ticks. The isotope separator draws 5eu/tick and requires 512 eu for one
operation. That means it takes 103ticks per operation of the separator, which translates to a fuel cell every 1030 ticks. This pretty much says that 1 separator can run almost 10 reactors
non-stop given an endless supply of tin and an automatic crafting area. This is amazing, because the 10 reactors operations (without scrap) produce 3 uu matter, which makes 5 tin ore, turns into
10 tin, which turns into 40 cells. Which is twice the amount put in. This thing literally doubles it's potential energy every cycle of the reactors. Pretty nifty stuff
(Mind you, it takes either intense micro or automation using buildcraft, as rp2 cannot pull from the empty cell slot of the separator)
I accept your challenge | {"url":"https://forum.industrial-craft.net/user-post-list/644-gecomol/","timestamp":"2024-11-08T11:24:01Z","content_type":"text/html","content_length":"95636","record_id":"<urn:uuid:c790ddea-2c50-4c17-a01e-74aab68cf6d5>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00053.warc.gz"} |
Orthogonality and Factorization Systems
Add to your list(s) Download to your calendar using vCal
If you have a question about this talk, please contact Guilherme Lima de Carvalho e Silva.
A basic definition of a factorization system on a category could be a pair (E,M) of classes of morphisms such that every morphism in the category factorizes as the composite of something from E
followed by something from M. For example, the (epi,mono)-factorization in the category of sets arises from expressing a function as the composite of ‘surjection onto image’ followed by ‘inclusion of
image into codomain’. (See also Q4 CT Sheet 1). Note that in these cases, the factorizations of a given morphism are (essentially) unique.
Orthogonality is a simple binary relation on the morphisms of a category, which will allow us to define the notion of an ‘Orthogonal Factorization System’ (OFS). I will justify the definition by
showing that it is (almost) equivalent to ‘factorization system with unique factorizations’ and go on to describe the basic properties and examples of OFS ’s. I hope to explain the connection to
reflective subcategories and the orthogonal subcategory problem and talk about some existence theorems.
This talk is part of the Junior Category Theory Seminar series.
This talk is included in these lists:
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Functions Summary
wade.vec2.add (v1, v2)
Add two vectors
wade.vec2.addInPlace (v1, v2)
Add two vectors and store the result in the first vector
wade.vec2.clamp (v, min, max)
Clamp a vector, that is force both its components to be between a minimum value and a maximum value
wade.vec2.clampInPlace (v, min, max)
Clamp a vector, that is force both its components to be between a minimum value and a maximum value. Unlike the 'clamp' function, this one modifies the original vector.
wade.vec2.div (v1, v2)
Divide a vector by another vector
wade.vec2.divInPlace (v1, v2)
Divide a vector by another vector and store the result in the first vector
wade.vec2.dot (v1, v2)
Calculate the dot product of two vectors
wade.vec2.equal (v1, v2, tolerance)
Check whether two vectors are equal (i.e. both x and y have the same values), optionally accounting for some tolerance in the comparisons
wade.vec2.length (v)
Calculate the length of a vector
wade.vec2.lengthSquared (v)
Calculate the length squared of a vector
wade.vec2.maxComponent (v)
Get the larger value of the x and y components of a vector
wade.vec2.minComponent (v)
Get the smaller value of the x and y components of a vector
wade.vec2.mul (v1, v2)
Multiply two vectors
wade.vec2.mulInPlace (v1, v2)
Multiply two vectors and store the result in the first vector
wade.vec2.normalize (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation may fail and return a vector whose components are NaN
wade.vec2.normalizeIfPossible (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation will just return the original vector
wade.vec2.normalizeInPlace (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation may fail and return a vector whose components are NaN. Unlike the 'normalize'
function, this one modifies the original vector.
wade.vec2.normalizeInPlaceIfPossible (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation will just return the original vector. Unlike the 'normalizeIfPossible' function,
this one modifies the original vector
wade.vec2.rotate (v, angle)
Rotate a vector by an angle
wade.vec2.rotateInPlace (v, angle)
Rotate a vector by an angle. Unlike the 'rotate' function, this one modifies the original vector.
wade.vec2.scale (v, s)
Scale a vector (that is, multiply the vector by a scalar)
wade.vec2.scaleInPlace (v, s)
Scale a vector (that is, multiply the vector by a scalar). Unlike the 'scale' function, this one modifies the original vector.
wade.vec2.sub (v1, v2)
Subtract two vectors
wade.vec2.subInPlace (v1, v2)
Subtract two vectors and store the result in the first vector
Function Details
wade.vec2.add (v1, v2)
Add two vectors
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
Returns {x: number, y: number} : v1 + v2
wade.vec2.addInPlace (v1, v2)
Add two vectors and store the result in the first vector
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
wade.vec2.clamp (v, min, max)
Clamp a vector, that is force both its components to be between a minimum value and a maximum value
{x: number, y: number} v : A 2d vector
number min : The minimum value for either component of the vector
number max : The maximum value for either component of the vector
Returns {x: number, y: number} : The clamped vector
wade.vec2.clampInPlace (v, min, max)
Clamp a vector, that is force both its components to be between a minimum value and a maximum value. Unlike the 'clamp' function, this one modifies the original vector.
{x: number, y: number} v : A 2d vector
number min : The minimum value for either component of the vector
number max : The maximum value for either component of the vector
wade.vec2.div (v1, v2)
Divide a vector by another vector
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
Returns {x: number, y: number} : v1 / v2
wade.vec2.divInPlace (v1, v2)
Divide a vector by another vector and store the result in the first vector
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
wade.vec2.dot (v1, v2)
Calculate the dot product of two vectors
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
Returns : number The dot product of v1 and v2
wade.vec2.equal (v1, v2, tolerance)
Check whether two vectors are equal (i.e. both x and y have the same values), optionally accounting for some tolerance in the comparisons
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
number tolerance (optional, defaults to 0): How different the x and y components can be before the comparison returns false
Returns boolean : Whether v1 and v2 have the same value
wade.vec2.length (v)
Calculate the length of a vector
{x: number, y: number} v : A 2d vector
Returns number : The length of v
wade.vec2.lengthSquared (v)
Calculate the length squared of a vector
{x: number, y: number} v : A 2d vector
Returns number : The length squared of v
wade.vec2.maxComponent (v)
Get the larger value of the x and y components of a vector
{x: number, y: number} v : A 2d vector
Returns number : v.x or v.y, whichever is bigger
wade.vec2.minComponent (v)
Get the smaller value of the x and y components of a vector
{x: number, y: number} v : A 2d vector
Returns number : v.x or v.y, whichever is smaller
wade.vec2.mul (v1, v2)
Multiply two vectors
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
Returns {x: number, y: number} : v1 * v2
wade.vec2.mulInPlace (v1, v2)
Multiply two vectors and store the result in the first vector
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
wade.vec2.normalize (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation may fail and return a vector whose components are NaN
{x: number, y: number} v : A 2d vector
Returns {x: number, y: number} : The normalized vector
wade.vec2.normalizeIfPossible (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation will just return the original vector
{x: number, y: number} v : A 2d vector
Returns {x: number, y: number} : The normalized vector
wade.vec2.normalizeInPlace (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation may fail and return a vector whose components are NaN. Unlike the 'normalize'
function, this one modifies the original vector.
{x: number, y: number} v : A 2d vector
wade.vec2.normalizeInPlaceIfPossible (v)
Normalize a vector (so that its length is 1). Note that if the length of the vector is very close to 0, this operation will just return the original vector. Unlike the 'normalizeIfPossible' function,
this one modifies the original vector
{x: number, y: number} v : A 2d vector
wade.vec2.rotate (v, angle)
Rotate a vector by an angle
{x: number, y: number} v : A 2d vector
number angle : An angle in radians
Returns {x: number, y: number} : The rotated vector
wade.vec2.rotateInPlace (v, angle)
Rotate a vector by an angle. Unlike the 'rotate' function, this one modifies the original vector.
{x: number, y: number} v : A 2d vector
number angle : An angle in radians
wade.vec2.scale (v, s)
Scale a vector (that is, multiply the vector by a scalar)
{x: number, y: number} v : A 2d vector
number s : A scale factor
Returns {x: number, y: number} : v * s
wade.vec2.scaleInPlace (v, s)
Scale a vector (that is, multiply the vector by a scalar). Unlike the 'scale' function, this one modifies the original vector.
{x: number, y: number} v : A 2d vector
number s : A scale factor
wade.vec2.sub (v1, v2)
Subtract two vectors
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector
Returns {x: number, y: number} : v1 - v2
wade.vec2.subInPlace (v1, v2)
Subtract two vectors and store the result in the first vector
{x: number, y: number} v1 : A 2d vector
{x: number, y: number} v2 : Another 2d vector | {"url":"https://clockworkchilli.com/docs/wade.vec2","timestamp":"2024-11-12T14:12:21Z","content_type":"text/html","content_length":"35698","record_id":"<urn:uuid:19aef45d-979d-49c3-bfd6-fd3aabeca8e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00261.warc.gz"} |
Continuous flow
From Encyclopedia of Mathematics
2020 Mathematics Subject Classification: Primary: 37A10 [MSN][ZBL]
A continuous flow in ergodic theory is a family $\{T^t\}$ (where $t$ ranges over the real axis $\R$) of automorphisms modulo 0 of a measure space $(M,\mu)$ such that:
a) $T^t T^s(x) = T^{t+s}(x)$ for any $t,s \in \R$ and all $x \in M$, except possibly for a set of $x$ belonging to an exceptional set of measure 0 (which may depend on $t$ and $s$); in other words,
$T^t T^s = T^{t+s} \bmod 0$;
b) for each measurable set $A \subset M$ the measure of the symmetric difference $\mu(A \Delta T^t A)$ depends continuously on $t$.
Let $\mathfrak{A}$ be the set of all automorphisms modulo 0 of the space $(M, \mu)$ with the usual identification: if $T$ and $S$ coincide almost-everywhere, then they determine the same element of $
\mathfrak{A}$. If $\mathfrak{A}$ is endowed with the weak topology (see ), then b) means that the mapping $\R\to\mathfrak{A}$ that takes $t$ to $T^t$ is continuous.
If $(M,\mu)$ is a Lebesgue space, then the concept of a continuous flow is practically the same as that of a measurable flow: The latter is always a continuous flow (see ), and for any continuous
flow $\{T^t\}$ there is a measurable flow $\{S^t\}$ such that $T^t = S^t \bmod 0$ for all $t$ (see ; a related result is proved in , but see also the correction in ). The converse to any of these
results depends on the character of the problem in question and the methods used.
In another sense the term "continuous flow" can be used to emphasize that the flow is considered in the context of topological dynamics. In this meaning a continuous flow is a collection of
homeomorphisms $\{T^t\}$ of a topological space $M$ such that $T^t(T^s(x)) = T^{t+s}(x)$ for all $t,s \in \R$ and $x \in M$; the mapping $M \times \R\to M$ taking $(x,t)$ to $T^t x$ is continuous.
To avoid confusion with 1) it is better to talk in this case of a topological flow and in the case of 1) of a metric continuity.
[Ha] P.R. Halmos, "Lectures on ergodic theory" , Math. Soc. Japan (1956) MR0097489 Zbl 0073.09302
[Ho] E. Hopf, "Ergodentheorie" , Springer (1970) MR0024581 Zbl 0185.29001
[V] A.M. Vershik, "Measurable realization of continuous automorphism groups of a unitary ring" Izv. Akad. Nauk. SSSR Ser. Mat. , 29 : 1 (1965) pp. 127–136 Zbl 0194.16302
[M] G.W. Mackey, "Point realizations of transformation groups" Illinois J. Math. , 6 : 2 (1962) pp. 327–335 MR0143874 Zbl 0178.17203
[R] A. Ramsay, "Virtual groups and group actions" Advances in Math. , 6 : 3 (1971) pp. 253–322 MR0281876 Zbl 0216.14902 Zbl 1085.54027
How to Cite This Entry:
Continuous flow. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Continuous_flow&oldid=55497
This article was adapted from an original article by D.V. Anosov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article | {"url":"https://encyclopediaofmath.org/index.php?title=Continuous_flow&oldid=55497","timestamp":"2024-11-03T03:51:16Z","content_type":"text/html","content_length":"19988","record_id":"<urn:uuid:eee5caeb-1345-49df-a64d-e90a3345187b>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00121.warc.gz"} |
Testing Equality in Communication Graphs
Let G = (V, E) be a connected undirected graph with k vertices. Suppose that on each vertex of the graph there is a player having an n -bit string. Each player is allowed to communicate with its
neighbors according to a (static) agreed communication protocol, and the players must decide, deterministically, if their inputs are all equal. What is the minimum possible total number of bits
transmitted in a protocol solving this problem ? We determine this minimum up to a lower order additive term in many cases. In particular, we show that it is kn/2+o(n) for any Hamiltonian k -vertex
graph, and that for any 2-edge connected graph with m edges containing no two adjacent vertices of degree exceeding 2 it is mn/2+o(n). The proofs combine graph theoretic ideas with tools from
additive number theory.
All Science Journal Classification (ASJC) codes
• Information Systems
• Computer Science Applications
• Library and Information Sciences
• 2-connected graphs
• Communication complexity
• equality function
• static protocols
Dive into the research topics of 'Testing Equality in Communication Graphs'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/testing-equality-in-communication-graphs","timestamp":"2024-11-05T23:01:00Z","content_type":"text/html","content_length":"49638","record_id":"<urn:uuid:90e260a2-0679-4c97-8a64-999d0f1b9189>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00149.warc.gz"} |
PRIMARY DIRECTIONS SOFTWARE
" PLACIDUS " -
THE KING OF PRIMARY DIRECTIONS
by Rumen K. Kolev
"PLACIDUS" is the first computer astrology program in the world that can calculate all kinds of AUTHENTIC primary directions in all traditions of primary directioning: PLACIDUS, REGIOMONTANUS,
CAMPANUS & TOPOCENTRIC- the most elite tool of the Predictive Astrology of all times. Version 5.1 is improved over the old version 4.3, but is reduced in price from $370 to $320 because the author
has now come out with version 7.0, which adds an extensive module for ancient astrology and evaluating chart rulers - version 7.0 sells for $480.
Both Placidus versions calculate PLACIDIAN CLASSIC, PLACIDIAN UNDER THE POLE (KUEHR), REGIOMONTANIAN, CAMPANIAN AND TOPOCENTRIC PRIMARIES.
" PLACIDUS 5.1 and 7.0" can run in WINDOWS 98, WINDOWS 2000, WINDOWS XP, WINDOWS VISTA, WINDOWS 7, and WINDOWS 8. They are the fruit of 15 years of labor.
Details of Placidus 7.0 and the Porphyrius Magus module
Click to see a Placidus screen wheel.
Click to see a primary directions list.
Click to see a list of mundo primaries to angles.
Click to see a 3D horizon horoscope.
Click to see a 3D ecliptic horoscope.
WHO IS THE CREATOR OF ' PLACIDUS ' ?
Rumen Kolev has studied Astrology since 1977. He is a USA citizen, coming originally from Bulgaria. He holds a B.A. in Economics and the equivalent in Mathematics from the University of Washington in
Seattle, USA. He has spent one year in the Ph.D. program in Economics at UCLA where he studied mathematical models for predictions in MacroEconomics and chaos theory. He holds also a MD degree from
the Bulgarian Medical Academy in Varna.
Primary directions are the elite tool for predictions. They are described by Ptolemy in his " Tetrabiblos ", Book 3, Chapters XIV and XV. They were the main tool employed by all famous astrologers of
the past: Placidus, Regiomontan, Naibod, Kepler, Tycho de Brahe, Cardan, Maginus, Argol, Gauric, Morin, Sepharial, Alan Leo, Kuehr and others.
There are programs that claim they can compute the primary directions. However, when put to the test, it turns out that they compute either a very small portion of the primary directions or, very
often, something that unfortunately, is not primary directions. These programs ( in good faith, but without sufficient knowledge ) call primary directions something that has nothing to do with the
real authentic primary directions employed and described by all authorities on the subject.
WHICH PRIMARY DIRECTIONS CAN " PLACIDUS " COMPUTE ?
" PLACIDUS " is able to compute virtually all kinds of primaries- both in the PLACIDIAN and the REGIOMONTANIAN TRADITIONS.
The Regiomontanian directions are calculated as by Regiomontanus, Maginus, Argolus, Naibod, William Lilly and Henry Coley.
The Placidian directions are Placidian Classic (Semi-Arc) directions and are computed by the program as were practiced by Ptolemy, Alcabitius, Montulmo, Lucas Gauricus, Simmonite, Zadkiel, Alan Leo,
Pracht, Kloekler and Placidus himself.
The Placidian under the Pole directions as they were calculated by Sepharial, Robert deLuce, Placidus and the German astrologer Kuehr.
It can calculate also CAMPANIAN and TOPOCENTRIC DIRECTIONS.
The topocentric directions are calculated as by Alexander Marr, Page and Polich.
It can compute also the primaries according to van Dam which are a variation of the Placidian Classic primaries with a specific time-measure (key for conversion).
Also it can compute the variation of the zodiacal primaries that ascribes latitude to the zodiacal aspects as per Blanchinus, Argol, Lucas Gauricus.
Here is a list of some of the directions that the program can compute:
1 Mundo to MC/ASC
2 Mundo Interplanetary aspects
3 Mundo Parallels
4 Mundo Rapt Parallels
5 Mundo to Cusps
6 Planet to itself
7 Zodiacal to MC/ASC
8 Zodiacal Interplanetary aspects
9 Zodiacal aspects to ASC/MC
10 Zodiacal acc. to Lucas Gauric
11 Zod. & Mundo acc. to Regiomontan
12 Under the Pole of the Significator
13 Zod. & Mundo acc. to Campanus
14 Zod. & Mundo Topocentric
15 Primaries in the Lunar and Solar charts (Kuehr)
" PLACIDUS " can compute these directions direct and converse with 11 keys: Ptolemy, Cardan, Naibod, Placidus, Kepler, Kuendig, Sun Longitude travel ( for the birthday ), Sun Longitude Progressed
travel, Synodic, van Dam and a Customer key.
The astrologer can input a key coefficient and calculate the directions with it. This is the customer key. The hit dates of the directions ( Day, Month, Year ) and the arc pop up in a chronological
list in easy to understand notation and are ready to print. There are 4 print options: Normal- 1 key , Normal- 8 keys, Compact- 1 key and Compact- 6 keys.
The astrologer can choose the kind of directions, the promissors, the aspects and the significators that he wishes to calculate.
Everything possible is done to make the rectification process ( so important by the primaries ) easy and fast. The astrologer can change the birth time with steps of 1 second up to 1 hour and watch
the new hit dates of almost all mundo directions and the zodiacal to the angles. He can change the key with one click of the mouse: the new hit dates come up immediately.
There is a module where you can point with the mouse at a specific birth-time on a time-scale (+-15 minutes from the entered birth-time) and see how ALL directions of planets to angles MOVE according
to the new birth-time pointed by the mouse.
WHAT OTHER PREDICTIVE TECHNIQUES CAN " PLACIDUS " COMPUTE ?
" Placidus " can compute also other techniques as:
1 Transits
2 Direct Progressions
3 Reverse Progressions
4 Solar Return Chart
5 Solar Arc Directions
6 Symbolic Directions
7 Symbolic with a Customer key
8 Lunar return chart
9 Primary ingresses (Kuehr)
10 Primary transits (Kuehr)
DOES " PLACIDUS " HAVE ANIMATION ?
" Placidus " has 11 animation modules: 3-Dimensional
Horoscope, Navigator, Under the Pole
Plac/Regio/Topo-Navigator, 4-WHEEL under the Pole
(with 4 modes: Placidus,Regio, Topo,
Campanus),Placidian Mundo Laboratory, 4 Mundo
Directions Animations (Plac, Regio, Campanus,
Topocentric) and the modules "Planets to angles" and
"Adjust" for rectification.
In the Navigator we can choose from 14 techniques ( among them primary directions acc. to Placidus and Regiomontan ) and charts to put in each wheel. Then we can set the step from 1 second to 5
years, lay back and watch the planets move. The date to which the charts correspond, shows in the left corner in a text field. We can input a date of interest and the charts in the wheels will
automatically assume their position for the given date! Similar is the design for the other animations.
WHAT IS THE PRECISION AND THE TIME FRAME " PLACIDUS " CAN WORK IN ?
'Placidus" 5.1 and later versions can work in ANY TIME FRAME.
It works with the algorithms of the famous astronomer Jean Meeus and for now is the best that Astronomy can offer. The precision is in parts of the arcsecond.
WHAT ARE THE PRINTING CAPABILITIES OF " PLACIDUS " ?
" Placidus " gives high quality black and white printouts of:
1 Natal Chart + the coordinates of the planets, the house cusps, the obliquity, the keys' coefficients, the intermediate variables for calculation of the Placidian primaries (Placidus Classic
Speculum) and graphical display of the mundo and zodiacal primary directions to the angles with all keys- all these on one page.
2 Chronological lists of primaries (Placidian, Regiomontanian, Campanian or Topocentric), their arcs and hit dates in 4 print options.
3 High-quality print out of the Navigator: 3 wheels in each one of which, the astrologer can choose from 14 charts and techniques.
4 Screen copy of the 4 types of Mundo Directions Labs
5 High quality print-out of the Regiomontanus Speculum- intermediate variables and arcs of mundo conjunctions and oppositions.
6 High quality print-out of the Campanus Speculum- intermediate variables and arcs of mundo conjunctions and oppositions.
7 High quality print-out of the Topocentric Speculum- intermediate variables and arcs of mundo conjunctions and oppositions.
8 High quality print-out of the Placidus Under the Pole Speculum- intermediate variables and arcs of mundo conjunctions and oppositions.
9 High quality print-out of the 4-WHEEL under the Pole
10 High quality print-out of the Solar return chart
11 High quality print-out of the Lunar return chart
12 Chronological lists of primaries in the Solar and the Lunar return charts according to the system of Kuehr. Here the key is 360 degrees for 1 year (Solar) or 360 degrees for 27.7 days (Lunar).
13 Other......
CAN " PLACIDUS " DISPLAY THE M. GAUQUELIN POSITIONS OF THE PLANETS ?
Yes. " Placidus " has a special module that makes this possible. The diurnal path of the planets is divided in 12 big sectors (houses) and each one of them is subdivided in 30 sectors.
If your main interest are the primaries as acc. to REGIOMONTAN, you can use the NAVIGATOR or the 4-WHEEL under the Pole or UNDER THE POLE PLAC/REGIO-CAMPANUS/TOPO-NAVIGATOR.
We can activate the Regiomontanus module from the Main Menu 'REGIO' option. It has 3 suboptions:
1. SPECULUM
2. UNDER THE POLE ANIMATION and
Here, we can see a full Regiomontanus speculum with all intermediate variables that serve for the calculations of the Regio primary directions. These are: Zenith Distance, Pole, Ascensional
Difference under own Pole, DaySemiArc under own Pole, Regio House position, Regio House Sector position and so on. This is an important feature for all researchers in the field who may want to check
step by step their own or someone else's hand calculations or in the end astrologers who may want to verify the authenticity of the program itself.
We can also print the intermediate variables in high-quality print-outs. An ocean of Regio data will come out scrambled on one single sheet of paper. In the print-out we can see also the Ascensional
Difference, the Oblique Ascension and the Oblique Descension of all planets under the Pole of all planets. We can see also the arcs of all interplanetary mundo conjunctions and oppositions between
all planets.
This Regio module is a beautiful animation module that serves for the immediate visualization of the Interplanetary Regiomontanus Directions ( another name for this kind of directions is 'Directions
under the Pole of the Significator as per Regiomontanus' ).
Here we choose the planet-significator from all ten planets or Pars Fortunae ( the first row at the lower part of the form )- the wheel automatically will configure in accordance with the calendar
date that is displayed in a text-field in the upper left corner of the form.
The significator in these Regio directions is placed on the eastern horizon if it is eastern in the radix and on the western horizon if it is western in the natal chart. Then other points like
zodiacal points of aspects, planets, terms et cet. are directed to this 'New Horizon'. These are the classic interplanetary zodiacal Regiomontanus Directions. COLEY and LILLY directed in this way.
All these are displayed in the module.
If we look at the bottom of the form we will see a row of option buttons for the key we want to use. We can choose from 9 keys: Ptolemy, Cardan, Naibod, Placidus, Kepler, Kuendig, Sun travel in
Longitude, synodic and Customer. If we click on a key button, the chart will automatically configure and update the current information anew as in accordance with the key in question.
Just above the chart we can read the values of the 'ARC IN TIME', the 'PRIMARY ARC' and the 'KEYS COEFFICIENT' that correspond for the moment in time shown in the text field.
On the left bottom corner of the form we can see the hit dates and the arcs of the mundo conjunctions and oppositions of all other planets with the current Significator.
On the bottom of this field we can see the button 'PEEP ALL ZODIACAL'. If we click on it, we will see which points of the zodiac are currently in Regio conjunctions with all planets- direct and
With this we know all the time where are all 10 planets.
In the upper left corner of the form are the animation controls. We can choose a step from 1 second to 100 years. Suppose we choose a step of 1 year. If now we click ' > ', the chart and all data on
the screen will automatically configure for one year later than the current one. Then the new date will show in the text field reserved for the date. If we click on ' >> ' then the chart will start
adding 1 year to the date and automatically change all info on the screen as according to the new dates. We have to hit ' STOP ' in order to discontinue the animation.
With the buttons ' < ' and ' << ' similarly, we can move back in time.
On the bottom of the form are two option buttons ' DIRECT ' and 'CONV. '. Hitting on them the program will calculate direct ( with the motion of the celestial sphere ) or converse ( against the
diurnal motion )directions.
If we have in mind a date of interest and we want to 'check what was on the Regio Horizons of the planets then', we can write this date in the text-field. A button with text ' Click OK when finished
' will pop up in the middle of the screen. We first write the date of interest and then click the ' OK ' buton. The chart and all information will be redrawn for the new date.
In the REGIO-NAVIGATOR we can calculate
1. Zodiacal points ( aspects, antiscia, terms...) to all 10 planets. These are the classical zodiacal directions of Regiomontanus under the Pole of the Significator. Practiced by all Masters in the
Regio Tradition.
2. Planets ( real bodies ) to mundo conjunctions and oppositions with all other planets. These also are directions practiced by the old Masters ( Coley, Lilly, Argol, Regiomontanus, Morin et cet. ).
3*. Planets ( real bodies ) to ASC, MC, DESC, IC. This was surely practiced by all Masters in both Traditions ( Regiomontanus and Placidus ).
In the NAVIGATOR we can calculate,
3. Planets ( real bodies ) to zodiacal points. This was practiced by the old Masters for retrograde planets.
4. Planets ( real bodies ) to Regio mundo conjunction with other planets ( real bodies ). This was also the practice of all famous astrologers.
5. Planets ( real bodies ) to Regio conjunction with the zodiacal projections of other planets.
6. Planets ( real bodies ) to Cusps of Mundo Regio Houses.
7*. Planets ( real bodies ) to ASC, MC, DESC, IC.
This was surely practiced by all Masters in both Traditions ( Regiomontanus and Placidus ).
For the first time, in Astrology, is developed a module which can show the natal chart on the celestial sphere. This is de facto an astronomical module for astrological purposes. Almost all types of
Placidian primaries are very easy to see in 3-dimensional animation and consequently understand thoroughly with ease.
There are more new modules like the animation 4-WHEEL where you can see all zodiacal directions for a given moment and a chosen key on the screen. You can switch here between 4 modes: Placidus under
the Pole, Regio, Topo and Campanus.
There are Placidus Classic, Placidus Under the Pole, Regiomontanian, Campanian and Topocentric SPECULUMS where are given all intermediate and final parameters for computing the primary directions in
You may see them on the screen and also print them out.
Extremely valuable software tool for the astrologer to double-check the program and his own hand-calculations.
THE HELP OF "PLACIDUS"
The Help of the program is in fact an electronic book that covers almost all conceivable topics about the primary directions, the house systems, mathematical algorithms, authors, books and history of
the primaries. It is richly illustrated with graphics.
For version 5.1, the computer should have around 20 Mb and, for version 7.0, 40 Mb free disc space for installation of the program. It should have Windows 98, 2000, XP, Vista or Windows 7 with RAM of
32 Mb or more recommended.
Placidus has a full-fledged module for Astrology in 3 dimensions.
You can see everything in 3 dimensions as it really is on the celestial sphere. And from whatever point of observation you want. You can turn the sphere in animation. 5 House systems, Equatorial,
Ecliptical and Horizontal grids, arcs of zodiacal aspects, arcs of RA, Declination, Latitude, Longitude, Azimuths, Altitudes, arcs of mundo conjunctions & oppositions according to Placidus,
Regiomontanus, Topocentric, Koch and van Dam systems. All kinds of primary arcs & all kinds of primary directions. Progressions, transits and directions. All these in 3D. All these in animations with
step of your choice. Zoom capability from 1 to 200. Observatorium. Drawings of the zodiacal constellations you can switch in and out....
You can project the celestial sphere upon the earth-globe and watch Real 3D AstroCartoGraphy....
WHAT IS THE PRICE OF THE "PLACIDUS" ASTROLOGY SOFTWARE ?
" Placidus " version 5.1 is distributed for USD $ 320 and version 7.0 for USD $ 480. The upgrade prices from a version 4 program are $80 and $150.
Three additional books sell separately for $ 45 (or $15 each). You can order three books on primary directions by Rumen Kolev.
The first one is about the directions to the angles and the foundations of the primary directions. These directions are calculated the same way in the Placidian and the Regiomontanian Traditions.
The second book deals with the Interplanetary Directions ( mundo, zodiacal aspectual, parallels, rapt-parallels et.cet ) in the Placidian Tradition.
The third volume examines the Regiomontanus directions, their history and method, and their use by William Lilly, and goes through an example calculation from Lilly.
More volumes are planned, including for the Directions under the Pole of the Significator (Placidus).
BOOK 1: THE PRIMARY DIRECTIONS TO THE ANGLES (MC, IC, ASC & DESC)
These directions are computed the same way in all systems.
The book is very clear in explanation and richly illustrated with graphics and pictures. The book is a step by step manual for learning the basics of the primary directions, the different time
measures (keys), mundo and zodiacal directions and all parameters needed for the computation of the primaries- what is their meaning and how are they calculated and used.
Among the parameters: ascensional difference, semi arcs, meridian distance, promissor,significator, orbs...
The history of the primaries and the different kinds of primaries is also discussed.
Explained also are the basics of the celestial sphere, the coordinates and the spherical trigonometry formulae.
Many directions are computed on the charts of Adolf Hitler and J. Kennedy. At the end is discussed the assessment of events and periods, as revealed by the primaries to the angles, and rectification
of the birth time with examples.
37 figures, 14 tables, 60 pages small size fonts.
Price: $15
All kinds of interplanetary Placidean directions are clearly explained and illustrated.
These include mundo and zodiacal:
directions to house cusps, planet to itself, mundo parallels, conjunctions, rapt parallels, rapt conjunctions, aspectual directions...
In the end there is a discussion of the practical efficiency, time-sensitivity, power, exactness and frequency of the different directions.
Again many directions are computed on the charts of Adolf Hitler and J. Kennedy.
49 figures, 5 tables, 60 pages small size fonts.
Price: $15
These directions were in vogue in the 15-17th centuries.
Their history is carefully examined.
Clear and step by step explanation and a practical example calculation.
Details of the practice of Regiomontanus himself.
A full analysis of the directions as computed and used by W. Lilly with recomputation of his examples from 'Christian Astrology'.
20 figures, 16 tables, 60 pages small size fonts.
Price: $15
BOOK 4: GAURICUS & HENRY II - MEDIEVAL ASTROLOGICAL PROGNOSIS
The first volume of the Bulletin of the Placidus Research Center on History of Astrology is a thorough research of the famous prediction-story. The original sources in Latin are published as they
were in the original old-print books and in English translation.
You will find here photos of the actual pages from books from the 15th, 16th and 17th centuries and from a manuscript written in 1460. All translated in English.
A very important 20 pages section describes in detail the Medieval Techniques for Prognosis.
The whole book is lavishly illustrated with 53 figures: portraits, medieval charts, pages from original Latin books and a manuscript, tables and graphics explaining the material.
57 pages of condensed information.
Price: $30
PDF Description: Gauricus & Henry II - Medieval Astrological Prognosis
Go to Secure Order Form.
Return to Software Programs Overview.
Copyright © 2003-15 Halloran Software, Los Angeles, California
Last modified on August 8, 2018. | {"url":"https://www.halloran.com/placidus.htm","timestamp":"2024-11-11T05:06:48Z","content_type":"text/html","content_length":"25430","record_id":"<urn:uuid:df3db313-7dce-4c4e-b824-28074f62efce>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00766.warc.gz"} |
%0 Journal Article %A JIN Shijun %A LI Wei %T Optimal path convergence method based on artificial potential field method and informed sampling %D 2021 %R 10.11772/j.issn.1001-9081.2020122021 %J
Journal of Computer Applications %P 2912-2918 %V 41 %N 10 %X The Rapidly exploring Random Tree star (RRT^*) algorithm ensures its probabilistic completeness and asymptotic optimality in the path
planning process, but still has problems such as slow convergence speed and large and dense sampling space. In order to speed up the convergence of the algorithm, a fast obtaining method of optimal
path based on artificial potential field method and informed set sampling was proposed. First, the artificial potential field method was used to construct an initial path from the starting point to
the target point. Then, the positions of and the distance between the starting point and the target point as well as the path cost of the initial path were used as parameters to construct the initial
informed sampling set. At last, the sampling was limited in the informed set, and the range of the informed sampling set was adjusted during the running process of the algorithm to accelerate the
path convergence speed. Simulation experiments show that, Potential Informed-RRT^* (PI-RRT^*) algorithm based on the artificial potential field combined with informed sampling method reduces the
number of sampling points by about 67%, and shortens the algorithm running time by about 74.5% on average compared with RRT^* algorithm; and has the number of sampling points reduced by about
40%-50%, the algorithm running time shortened by about 62.5% on average compared with Informed RRT^* (Informed-RRT^*) algorithm. The proposed optimal path convergence method greatly reduces the
number of redundant sampling and the algorithm running time, has higher algorithm efficiency, and converges to the optimal path with faster speed. %U https://www.joca.cn/EN/10.11772/ | {"url":"https://www.joca.cn/EN/article/getTxtFile.do?fileType=EndNote&id=24780","timestamp":"2024-11-08T19:27:02Z","content_type":"application/x-endnote-refer","content_length":"2442","record_id":"<urn:uuid:7874170f-7493-47f2-8fd1-2b4134a04c51>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00808.warc.gz"} |
Subset of elements of finite order of a group
Consider a group \(G\) and have a look at the question: is the subset \(S\) of elements of finite order a subgroup of \(G\)?
The answer is positive when any two elements of \(S\) commute. For the proof, consider \(x,y \in S\) of order \(m,n\) respectively. Then \[
\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e\] where \(e\) is the identity element. Hence \(xy\) is of finite order (less or equal to \(mn\)) and belong to \(S\).
Example of a non abelian group
In that cas, \(S\) might not be subgroup of \(G\). Let’s take for \(G\) the general linear group over \(\mathbb Q\) (the set of rational numbers) of \(2 \times 2\) invertible matrices named \(\text
{GL}_2(\mathbb Q)\). The matrices \[
A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}\] are of order \(2\). They don’t commute as \[
AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.\] Finally, \(AB\) is of infinite order and therefore doesn’t belong to \(S\) proving that \
(S\) is not a subgroup of \(G\).
You must be logged in to post a comment. | {"url":"https://www.mathcounterexamples.net/subset-of-elements-of-finite-order-of-a-group/","timestamp":"2024-11-08T14:34:14Z","content_type":"text/html","content_length":"58820","record_id":"<urn:uuid:fe738780-ea6e-4a23-8295-8f1fedeebe48>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00626.warc.gz"} |
Arcmin/Square Minute to Turn/Square Week
Arcmin/Square Minute [arcmin/min2] Output
1 arcmin/square minute in degree/square second is equal to 0.0000046296296296296
1 arcmin/square minute in degree/square millisecond is equal to 4.6296296296296e-12
1 arcmin/square minute in degree/square microsecond is equal to 4.6296296296296e-18
1 arcmin/square minute in degree/square nanosecond is equal to 4.6296296296296e-24
1 arcmin/square minute in degree/square minute is equal to 0.016666666666667
1 arcmin/square minute in degree/square hour is equal to 60
1 arcmin/square minute in degree/square day is equal to 34560
1 arcmin/square minute in degree/square week is equal to 1693440
1 arcmin/square minute in degree/square month is equal to 32017815
1 arcmin/square minute in degree/square year is equal to 4610565360
1 arcmin/square minute in radian/square second is equal to 8.0802280184923e-8
1 arcmin/square minute in radian/square millisecond is equal to 8.0802280184923e-14
1 arcmin/square minute in radian/square microsecond is equal to 8.0802280184923e-20
1 arcmin/square minute in radian/square nanosecond is equal to 8.0802280184923e-26
1 arcmin/square minute in radian/square minute is equal to 0.00029088820866572
1 arcmin/square minute in radian/square hour is equal to 1.05
1 arcmin/square minute in radian/square day is equal to 603.19
1 arcmin/square minute in radian/square week is equal to 29556.1
1 arcmin/square minute in radian/square month is equal to 558816.29
1 arcmin/square minute in radian/square year is equal to 80469545.91
1 arcmin/square minute in gradian/square second is equal to 0.0000051440329218107
1 arcmin/square minute in gradian/square millisecond is equal to 5.1440329218107e-12
1 arcmin/square minute in gradian/square microsecond is equal to 5.1440329218107e-18
1 arcmin/square minute in gradian/square nanosecond is equal to 5.1440329218107e-24
1 arcmin/square minute in gradian/square minute is equal to 0.018518518518519
1 arcmin/square minute in gradian/square hour is equal to 66.67
1 arcmin/square minute in gradian/square day is equal to 38400
1 arcmin/square minute in gradian/square week is equal to 1881600
1 arcmin/square minute in gradian/square month is equal to 35575350
1 arcmin/square minute in gradian/square year is equal to 5122850400
1 arcmin/square minute in arcmin/square second is equal to 0.00027777777777778
1 arcmin/square minute in arcmin/square millisecond is equal to 2.7777777777778e-10
1 arcmin/square minute in arcmin/square microsecond is equal to 2.7777777777778e-16
1 arcmin/square minute in arcmin/square nanosecond is equal to 2.7777777777778e-22
1 arcmin/square minute in arcmin/square hour is equal to 3600
1 arcmin/square minute in arcmin/square day is equal to 2073600
1 arcmin/square minute in arcmin/square week is equal to 101606400
1 arcmin/square minute in arcmin/square month is equal to 1921068900
1 arcmin/square minute in arcmin/square year is equal to 276633921600
1 arcmin/square minute in arcsec/square second is equal to 0.016666666666667
1 arcmin/square minute in arcsec/square millisecond is equal to 1.6666666666667e-8
1 arcmin/square minute in arcsec/square microsecond is equal to 1.6666666666667e-14
1 arcmin/square minute in arcsec/square nanosecond is equal to 1.6666666666667e-20
1 arcmin/square minute in arcsec/square minute is equal to 60
1 arcmin/square minute in arcsec/square hour is equal to 216000
1 arcmin/square minute in arcsec/square day is equal to 124416000
1 arcmin/square minute in arcsec/square week is equal to 6096384000
1 arcmin/square minute in arcsec/square month is equal to 115264134000
1 arcmin/square minute in arcsec/square year is equal to 16598035296000
1 arcmin/square minute in sign/square second is equal to 1.5432098765432e-7
1 arcmin/square minute in sign/square millisecond is equal to 1.5432098765432e-13
1 arcmin/square minute in sign/square microsecond is equal to 1.5432098765432e-19
1 arcmin/square minute in sign/square nanosecond is equal to 1.5432098765432e-25
1 arcmin/square minute in sign/square minute is equal to 0.00055555555555556
1 arcmin/square minute in sign/square hour is equal to 2
1 arcmin/square minute in sign/square day is equal to 1152
1 arcmin/square minute in sign/square week is equal to 56448
1 arcmin/square minute in sign/square month is equal to 1067260.5
1 arcmin/square minute in sign/square year is equal to 153685512
1 arcmin/square minute in turn/square second is equal to 1.2860082304527e-8
1 arcmin/square minute in turn/square millisecond is equal to 1.2860082304527e-14
1 arcmin/square minute in turn/square microsecond is equal to 1.2860082304527e-20
1 arcmin/square minute in turn/square nanosecond is equal to 1.2860082304527e-26
1 arcmin/square minute in turn/square minute is equal to 0.000046296296296296
1 arcmin/square minute in turn/square hour is equal to 0.16666666666667
1 arcmin/square minute in turn/square day is equal to 96
1 arcmin/square minute in turn/square week is equal to 4704
1 arcmin/square minute in turn/square month is equal to 88938.38
1 arcmin/square minute in turn/square year is equal to 12807126
1 arcmin/square minute in circle/square second is equal to 1.2860082304527e-8
1 arcmin/square minute in circle/square millisecond is equal to 1.2860082304527e-14
1 arcmin/square minute in circle/square microsecond is equal to 1.2860082304527e-20
1 arcmin/square minute in circle/square nanosecond is equal to 1.2860082304527e-26
1 arcmin/square minute in circle/square minute is equal to 0.000046296296296296
1 arcmin/square minute in circle/square hour is equal to 0.16666666666667
1 arcmin/square minute in circle/square day is equal to 96
1 arcmin/square minute in circle/square week is equal to 4704
1 arcmin/square minute in circle/square month is equal to 88938.38
1 arcmin/square minute in circle/square year is equal to 12807126
1 arcmin/square minute in mil/square second is equal to 0.000082304526748971
1 arcmin/square minute in mil/square millisecond is equal to 8.2304526748971e-11
1 arcmin/square minute in mil/square microsecond is equal to 8.2304526748971e-17
1 arcmin/square minute in mil/square nanosecond is equal to 8.2304526748971e-23
1 arcmin/square minute in mil/square minute is equal to 0.2962962962963
1 arcmin/square minute in mil/square hour is equal to 1066.67
1 arcmin/square minute in mil/square day is equal to 614400
1 arcmin/square minute in mil/square week is equal to 30105600
1 arcmin/square minute in mil/square month is equal to 569205600
1 arcmin/square minute in mil/square year is equal to 81965606400
1 arcmin/square minute in revolution/square second is equal to 1.2860082304527e-8
1 arcmin/square minute in revolution/square millisecond is equal to 1.2860082304527e-14
1 arcmin/square minute in revolution/square microsecond is equal to 1.2860082304527e-20
1 arcmin/square minute in revolution/square nanosecond is equal to 1.2860082304527e-26
1 arcmin/square minute in revolution/square minute is equal to 0.000046296296296296
1 arcmin/square minute in revolution/square hour is equal to 0.16666666666667
1 arcmin/square minute in revolution/square day is equal to 96
1 arcmin/square minute in revolution/square week is equal to 4704
1 arcmin/square minute in revolution/square month is equal to 88938.38
1 arcmin/square minute in revolution/square year is equal to 12807126 | {"url":"https://hextobinary.com/unit/angularacc/from/arcminpmin2/to/turnpw2","timestamp":"2024-11-14T11:11:45Z","content_type":"text/html","content_length":"113402","record_id":"<urn:uuid:85d8eaf8-7a65-4f2a-b87b-234662b9169d>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00632.warc.gz"} |
Python - (Business Decision Making) - Vocab, Definition, Explanations | Fiveable
from class:
Business Decision Making
Python is a high-level programming language that is widely used for various applications, including data analysis, machine learning, and automation. Its simplicity and readability make it an ideal
choice for both beginners and experienced developers. Python's extensive libraries and frameworks enhance its capabilities in building decision-making models, such as decision trees, and calculating
expected values.
congrats on reading the definition of Python. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Python's syntax is designed to be easy to understand, which makes it popular among data scientists for building decision trees and other analytical models.
2. The language supports various libraries like scikit-learn that simplify the implementation of decision trees and allow users to calculate expected values efficiently.
3. Python can handle large datasets seamlessly, making it suitable for scenarios where decision-making relies on extensive data analysis.
4. The versatility of Python allows it to be integrated with other programming languages and tools, enhancing its usability in decision-making frameworks.
5. Python's community is vast, providing ample resources, tutorials, and forums for users looking to learn about decision trees and expected value calculations.
Review Questions
• How does Python's syntax and structure contribute to its effectiveness in creating decision trees?
□ Python's simple and readable syntax allows users to focus on the logic behind decision trees without getting bogged down by complex code. This clarity makes it easier for developers to
implement algorithms that visualize decisions and outcomes effectively. Additionally, Python’s indentation-based structure promotes writing clean code, which is crucial when constructing and
maintaining decision trees.
• Discuss how Python libraries facilitate the calculation of expected values in decision-making processes.
□ Python libraries such as NumPy and pandas provide built-in functions that simplify mathematical operations required for calculating expected values. By using these libraries, developers can
quickly compute the average outcomes based on the probabilities associated with different scenarios. This efficiency in computation allows analysts to focus more on interpreting results
rather than on the coding itself.
• Evaluate the impact of Python’s capabilities on modern business decision-making strategies involving decision trees and expected values.
□ Python has significantly transformed business decision-making strategies by providing accessible tools for data analysis through its extensive libraries. Its ability to handle large datasets
enables organizations to build accurate decision trees that reflect real-world complexities. Furthermore, the ease of calculating expected values empowers businesses to make informed choices
based on quantitative data, ultimately enhancing strategic planning and operational efficiency.
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MCLab Group List of Papers -- Query Results
Federico Mari, Igor Melatti, Ivano Salvo, and Enrico Tronci. From Boolean Functional Equations to Control Software. Vol. abs/1106.0468. CoRR, Technical Report, 2011.
V. Alimguzhin, F. Mari, I. Melatti, I. Salvo, and E. Tronci. "Linearising Discrete Time Hybrid Systems." IEEE Transactions on Automatic Control 62, no. 10 (2017): 5357–5364. ISSN: 0018-9286. DOI:
Federico Mari, Igor Melatti, Ivano Salvo, and Enrico Tronci. "Linear Constraints and Guarded Predicates as a Modeling Language for Discrete Time Hybrid Systems." International Journal on Advances in
Software vol. 6, nr 1&2 (2013): 155–169. IARIA. ISSN: 1942-2628.
Igor Melatti, Robert Palmer, Geoffrey Sawaya, Yu Yang, Robert Mike Kirby, and Ganesh Gopalakrishnan. "Parallel and distributed model checking in Eddy." Int. J. Softw. Tools Technol. Transf. 11, no. 1
(2009): 13–25. Springer-Verlag. ISSN: 1433-2779. DOI: 10.1007/s10009-008-0094-x.
Igor Melatti, Robert Palmer, Geoffrey Sawaya, Yu Yang, Robert Mike Kirby, and Ganesh Gopalakrishnan. "Parallel and Distributed Model Checking in Eddy." In Model Checking Software, 13th International
SPIN Workshop, Vienna, Austria, March 30 – April 1, 2006, Proceedings, edited by A. Valmari, 108–125. Lecture Notes in Computer Science 3925. Springer - Verlag, 2006. ISSN: 0302-9743. DOI: 10.1007/
S. Sinisi, V. Alimguzhin, T. Mancini, E. Tronci, and B. Leeners. "Complete populations of virtual patients for in silico clinical trials." Bioinformatics (2021): 1–8. ISSN: 1367-4803. DOI: 10.1093/
T. Mancini, I. Melatti, and E. Tronci. "Any-horizon uniform random sampling and enumeration of constrained scenarios for simulation-based formal verification." IEEE Transactions on Software
Engineering (2021): 1. ISSN: 1939-3520. Notes: To appear. DOI: 10.1109/TSE.2021.3109842.
S. Fischer, R. Ehrig, S. Schaefer, E. Tronci, T. Mancini, M. Egli, F. Ille, T. H. C. Krueger, B. Leeners, and S. Roeblitz. "Mathematical Modeling and Simulation Provides Evidence for New Strategies
of Ovarian Stimulation." Frontiers in Endocrinology 12 (2021): 117. ISSN: 1664-2392. DOI: 10.3389/fendo.2021.613048.
Amedeo Cesta, Alberto Finzi, Simone Fratini, Andrea Orlandini, and Enrico Tronci. "Validation and Verification Issues in a Timeline-based Planning System." In In E-Proc. of ICAPS Workshop on
Knowledge Engineering for Planning and Scheduling., 2008.
Amedeo Cesta, Alberto Finzi, Simone Fratini, Andrea Orlandini, and Enrico Tronci. "Validation and verification issues in a timeline-based planning system." The Knowledge Engineering Review 25, no. 03
(2010): 299–318. Cambridge University Press. DOI: 10.1017/S0269888910000160. | {"url":"https://mclab.di.uniroma1.it/publications/search.php?sqlQuery=SELECT%20author%2C%20title%2C%20type%2C%20year%2C%20publication%2C%20abbrev_journal%2C%20volume%2C%20issue%2C%20pages%2C%20keywords%2C%20abstract%2C%20thesis%2C%20editor%2C%20publisher%2C%20place%2C%20abbrev_series_title%2C%20series_title%2C%20series_editor%2C%20series_volume%2C%20series_issue%2C%20edition%2C%20language%2C%20author_count%2C%20online_publication%2C%20online_citation%2C%20doi%2C%20serial%20FROM%20refs%20WHERE%20serial%20RLIKE%20%22.%2B%22%20ORDER%20BY%20abstract&submit=Cite&citeStyle=Roma&citeOrder=&orderBy=abstract&headerMsg=&showQuery=0&showLinks=0&formType=sqlSearch&showRows=10&rowOffset=90&client=&viewType=Print","timestamp":"2024-11-10T09:28:39Z","content_type":"text/html","content_length":"40953","record_id":"<urn:uuid:a54d6485-90ce-48b4-abd8-96c834391d80>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00095.warc.gz"} |
Do You Want to Meet One of India’s Best Value Investors? | Safal Niveshak
“If I have seen further it is by standing on the shoulders of giants.” ~ Isaac Newton
It goes without saying that if the world was minus Benjamin Graham and Warren Buffett, the idea for Safal Niveshak wouldn’t have existed as well.
Graham and Buffett have put enormous number of ideas into my head, which I have been sharing with you ever since Safal Niveshak was born. Reading their work had made a lot of difference to my life as
an investor and human being.
However, there is one more person whom I’ve never mentioned on this platform, but who has been a great influence on the behavioral side of the analyst within me.
Prof. Sanjay Bakshi, one of the best minds in India in the fields of Value Investing and Behavioral Finance, and whom I’ve been reading for the past few years through his lecture notes,
presentations, and articles.
Prof. Bakshi teaches MBA students (at MDI Gurgaon) two popular courses: “Behavioral Finance & Business Valuation” and “Financial Shenanigans & Governance”.
He is also the CEO of Tactica Capital Management, a boutique firm engaged in deep value investing.
Given the immense respect I have for Prof. Bakshi, it was a great feeling when he wrote to me recently with his appreciation for Safal Niveshak. In fact, I can easily call it one of the best moments
of my life as an investor.
Anyways, after Prof. Bakshi wrote to me, I did not miss a beat before requesting to meet him. 🙂
He’s been highly supportive and has agreed to meet me – and has also agreed for an interview – on 29th July (a day after the “Art of Investing Workshop” in New Delhi).
By the way, if you haven’t already done it, click here to register for the New Delhi Workshop scheduled for 28th July 2012. I’m accepting just 5 more registrations, so you must act fast!
Anyways, I’m going to meet Prof. Bakshi on the 29th, and also pick his brains on the subject of Value Investing.
I need your help!
I need your help in finalizing the questions that I must ask Prof. Bakshi…on your behalf.
I need to finalize a maximum of 10 questions, and thus invite you to send me the ones you would’ve asked Prof. Bakshi had you met him personally.
The questions must be around “stock market investing” in general and “value investing” in particular, like…
• How do I get started as a value investor?
• What are the key habits I must practice to become a value investor?
…but exclude ones like:
• What stock do I buy?
• What do you think about XYZ stock?
You can send me your questions via the Comments section below.
Once I receive your questions (over the next one week), I will consolidate them and finalize the 10 that I will ask Prof. Bakshi in the meeting. So just get going!
This is your golden opportunity to dive into the mind of Prof. Bakshi and pull out some amazing pearls of wisdom on Value Investing.
So I’ll wait for your questions, which you can submit by writing them in the Comments section below.
1. Abhay says
I am in my early 30s. I feel that value investing requires a great deal of research as well as patience. I find it very difficult to have patience and conviction in what “WE” are doing specially
when there is constant noise from friends, brokers, relatives, blogs, columns, tons of website, well-wishers about tips and so called “insider news”.
Hence, my question for him would be, how a regular person “aam aadmi” can keep focus and have patience while going through all turbulence? How difficult were his earlier year in value investing
and what we (younger generation) should not miss out on (in terms of qualitative characteristics)?
2. karthik says
Great News Vishal…
Questions are
1) How Retail investors can achieve the art of value investing?
for Ex. A person who can invest Rs.5000 per month in Stock Markets
what can be the strategy… (many of the people I know have this issue)
2)People who got good returns from stock markets .. I find majority of them were investors in IPO’s bfore 2000 where the ipo’s were issued at AT PAR. Soem of the bluechips were offered at throw
away prices..
What is the scope for retail investors in current scenario to get decent returns?
3. sudhir says
What are the steps or sequence or tests through which one must take ones analysis for value investing.
This may be asking for too much but often the genius is a genius for he can say in few words what others take chapters.
I would want some sort of a guide (could be a few pages). For example look around you 9for what ?), read a lot (what all ?), look for trends (where ?), watch for disruptive stuff (How ?), then of
course the financial analysis et al.
he must be typically dealing with these aspects so a guide from him would be prized.
4. sri says
just back from trip. so many posts in between. am yet to catchup from your presentation. anyway,
frankly, have never heard of him, (believe me)…still if he is acceptable to you, my question will be 1. Value investing has any future..?
2. what are the returns when compared with growth & momentum and technical investors.
will add more later …
thanks for giving opportunity to ask questions….!!!!
5. Reni George says
Dear vishal
In value investing,the word itself is a big question “Value”,what are the parameters for find out the value of a company,apart from moats,one of the major is the “ANNUAL REPORT”.Those who are not
into accounting,what should they do to decipher a annual report,lot of annual reports come with good dressing of account,the major points which could effect are given in notes.If we as value
investors have a common platform to decipher the annual reports and draw out some guidelines that we should make it a point to note down from the annual report.
6. Shravan Paul says
what will be the implications of the current fccb+rupee depreciation crisis? are there any opportunities to be taken advantage of?
how do i protect my portfolio against it?
how bad are things really in the indian banking system presently?HOW DOES ONE PROTECT AGAINST THIS?
HOW TO JUDGE PROMOTERS/A CHEKLIST FOR JUDGING PROMOTERS?
how to effectively hedge in india?
any upcoming catalysts for nbfc stocks?
distinguishing between value traps and bargains especially in the asset bargains category.
hope this helps
7. Vivek says
Hi….Vishal…Good Initiative once again…your meet & interview with Prof. Sanjay Bakshi will certainly add value to this website and motivation to the value investors as the topic will be around
“stock market investing” in general and “value investing” in particular. I also want to add few questions as follows :
Q1: How to account the factors (such as market sentiments, economic positions, political uncertainty, interest rates, global market conditions etc) in value investing? Weather these factors
should be considered in margin of safety or ignore these factors in value investing?
Q2: How to find value in small caps? Is value investing process or techniques are same for large caps & small caps, as it is easier to find value in large caps compared to small caps?
Thank You
8. sudhir says
Also some tips on ‘how to understand/ control oneself’ or things to watch out in our behavior when dealing with investments/ finances – the behavioral finance angle. I think this inward looking
question and what all questions one should ask oneself to arrive at an answer.
Since the answer to these questions also has to come from you yourself, this is a very tough question. Plus the questions and answers will change with time.
So in terms of behavioral finance some guide book as well.
9. Vidhyashankar says
What are the basics of value investing?
How & when do I get started as a value investor?
What are the key habits I must practice to become a value investor and why?
10. Sunny Gupta says
Thanks Vishal, here’re two of my questions…
1. We evaluate a business and find the intrinsic value of its share, and then take a margin of safety, let’s say 50%. Then, the corollary is that we can expect the price to double when markets
return to their senses…but there’s no clarity on how irrational markets can get on the bullish side, so the “expectation of returns” is not clear – second, if a business continues to do well,
you’d not sell the stock (yes, you can expect the IV itself to grow)…in the gist – should there be any expected returns with value investing or is it just a belief that picking good business at
good price will give us good returns over a long period of time…
2. Suppose someone understands value investing and knows how to value a business – now, with so many listed companies, how does he start filtering stocks for a detailed analysis…
11. Mansoor says
Great news, congratulations Vishal. Professor is truly inspiring and I have read about his life also, sometime back, a pure contrarian in how he steered through his life. If you feel it is apt to
ask, my question would be.
1. Value Investing is thought to be investing in good businesses through different valuation methods like DCF etc When you talk about Value Investing, what do you mean and how do you do it?
(My reasons for asking this question is that he seems not to trust DCF method, I think he believes we are speculating or atleast making assumptions, so what does he do?)
2. Do you go bey0nd your Circle of competence? Does it exists for you?
(My reason for asking this question is that he talks a lot about the “unknown” and the “unknowable”, the things you do not know and the things that you will not know, it just happens)
3. 10/15 years back, when you finished your school in London, you made the big decision to come back to India with your wife and kid, you believed in India growth story and that you can apply
Buffett principles here in India. What are your views on India growth story now?
Please thank him for the Vantage Point article/lecture, it’s a piece of art and a must read for every investor.
12. Suresh K Perumalla says
Hi Vishal,
Thanks for your continuous efforts in rejuvenating us through Investing journey. And sorry I am bringing my question lately. My question is “Good Company stocks acts too good in good times and
may act worst in bad times. So how could Behavioral Finance or advance Investing tips help us to choose good or safe stocks in bad times”. I am aware patience tests us but need sustaining reasons
to hold those stocks until good time arrives.
Recently I have attended a behavioral finance one hour session in the company I am working now. That session opened many doors in my logical brain to study more on it. I am sure Warren Buffet of
course, is a master of it. But the session sounded like more scaring me. May be I do not understand it well..:) but quite interesting. If opportunity comes please share your knowledge with us.
Thanks a lot ..:)
13. MAHESH says
i always scratch my head more number of times to sell a stock at the best time rather than buying a stock , so if one identifies a value stock n pumped his money into it how can he reap the max
benefit of selling it at the right time ( here my intent is timing! not time it may be after year or 10years , no matter) , so i want his views on exit rather thn entry
14. Balaji says
1. Top failures in investing in his career? What did he learn?
2. Process for value-investing that he follows? Check-list based? Screener based? How does his evaluate it once he sees an interesting scrip?
15. Girish says
This question is regarding the efficacy of value investing..
1. Do screen based value investing works in real life? e.g. buying low PE stocks from sensex or nifty? What is the record? OR we need low valuation PLUS other screening parameters like strong
balance sheet, ROIC etc.
2. Do quantitative assessment like low PE, low debt ratio (screening) etc enough OR we also need QUALITATIVE assessment (management quality, MOAT etc)? What will happen if one only selects the
stocks based on screens?
16. PRIYARANJAN says
Great job….Vishal ji,I really appreciate your effort to make investor sensible about their investment and not just treating market as a casino to gain hefty amount in a little time.I would like
to share with you some doubts that arises at many instances…
Q.1.If value investing is so necesary,why technical analysis is always in news?
Q.2.Do all the value stocks has a fair chance to make money for us in times to come?
Q.3.If price is belived to reflect all the market sentiment,Does value investing has any meaning?
Q.4.For a layman is it possible to accumulate all the possible financial data & if possible Can he make some sence out of that huge data and parameters?
Q.5.When the Opinions of experts in different media reports varies How could a man come to a certin conclusion about the health of a particular stock.?
17. vikrant says
Just one question, and may be you can answer that, In Value investing, How can we differentiate between a value buy and a value trap and whats the difference.
18. Vikash Taank says
‘Tactica Capital Management’ -does not run any Fund /PMF or any such instrument where a common man in India can invest .I think they run this fund for FIIs or may be big investors (I dont know
In US people can use the expertise of Charlie M by atleast investing in Berkshire ,its a pity that we Indians having such a Investor among us cannot benefit from him. No doubt his teachings
enlighten all but then its very comforting to invest (small portion) into funds where his brains are used.I mean we have people like Prashant Jain, KN Subramaniam who are really experts but in
India the MF manager is always trying to beat the benchmark. The value based funds (contrarian) are never Contra funds (quantum is nearest to an excellent contra fund I think !).
If you think I sound (and my info is correct about Tactica) sensible then you can probably hint Prof. Bakshi to think about how a common man can benefit from his expertise in this field then many
Indians would really be glad to invest their savings with him (via Fund /PMF or any other instrument).
19. Umakant says
Hi Vishal,
Please ask Mr. Bakshi,
– What are the important 5 parameters (ROE, Debt/Equity ratio, CAGR growth of Nett profit, Cash flow, Honesty/Quality of management etc.) we should look for a stock when a small investor buys a
20. Investor23 says
First of all, you are running a great website/investing guidance business and thanks for this initiative in India’s cacophonic investing markets.
Good that you are meeting the RIGHT PEOPLE, like Prof. Bakshi.
But I also think it will be stupid if you ask questions like
How do I get started as a value investor?
What are the key habits I must practice to become a value investor?
This is because Bakshi himself runs a terrific blog where he posts most of this basic education already, including fairly detailed PPTs from this class.
So, please be prudent with your and his time. I agree that you shouldn’t ask for stock tips. But the real juicy learning from folks like Mr. Bakshi always comes from specific case studies. Ask
him to take you through as many real examples as possible, explain his thought process/peel the onion, etc. He has done some of this on his website, but nowhere near enough in my opinion.
21. Sanjeev Bhatia says
Hi Vishal,
Sorry for being so late. It just skipped my mind altogether.
My Questions:-
1. We always talk about biases in a negative way. Don’t these biases sometimes prevent us from doing stupid things? Say, if my regret aversion bias makes me more conservative investor (therby
implying I use larges MoS), what’s the harm in it?
2. In today’s information overload and knee jerk reactions to quarterly results and media news, how relevant is “Buy and Hold” Strategy when sometimes stocks show large volatility in terms of
yearly Highs/ Lows but poor overall return over even a longer time frame?
3. How can a small retail investor build up a “positive loop” that you talk about to increase his circle of competence, given the time, education or background constraints?
Thank him for all his wonderful posts, his incisive approach and bringing the whole funda of value investing so much closer to home 🙂
22. Sanjeev Bhatia says
In continuation of question 2 above:
Does he follow any specific profit booking strategy or strictly go by valuation principles only? Since He is in a different position altogether (large stakes, direct approach to management etc),
what does he suggest for retail investors like us?
23. Avadhut says
Hi Vishal,
Excellent decision of meeting Prof.Bakshi. Most of us only dream that.
I’ve read all the comments above and think all questions are apt.
I’ve read some of Prof.Bakshi’s articles and they are thought-provoking. I think one point none has raised so far is about “Special situations”. Prof.Bakshi has a good track record in Special
situations investing and asking him about what should be the methodology (not tips) one must follow to spot and succeed in special situations investing would give us a roadmap.
Secondly, you can discuss about Value Investing Methodology that he follows and ask him whether it’s possible for all of us to follow the methodology. If yes, you can ask the first and last step
in this methodology, kind of checklist, this would be I guess excellent value addition for all of us.
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Do the following quotient spaces admit cell complex structure?
For $q \leq n \in \mathbb{N}$, let $O(n-q)$ act on the sphere $S^{n-q}$ via the canonical linear action on $\mathbb{R}^{n-q}$ passed to the one-point compactification $S^{n-q} \simeq (\mathbb{R}^
{n-q})^\ast$. Then does
$( O(n)_+) \wedge_{O(n-q)} S^{n-q}$
admit cell complex structure? I suppose it does, but what’s a solid argument?
In fact it would be sufficient for me if it were the retract of a cell complex.
Added a section Terminology on the meaning of the letters “CW”.
Added material to CW complex in a section titled “Up to homotopy equivalence”.
More little properties at CW complex.
Also, I have exanded and slightly rearranged the Idea-section there.
Okay, I’ll change the notation. And, yes, I am currently working on cellular homology.
There seems to be some duplication of material here; see also cellular homology.
Why are there two notations, $H_k(X_n | X_{n-1})$ and $H_k(X_n, X_{n-1})$? I can’t recall ever seeing the bar notation.
Fixed a couple of typos.
Edit: I just looked higher up on the page, where the bar notation is introduced. But I honestly don’t think there is any danger at all in thinking $H_k(X_n, X_{n-1})$ might stand for homology of
$X_n$ with coefficients in $X_{n-1}$ (it doesn’t even make sense)! Do other people use the bar notation? Anyway, I don’t think we should have both notations on the same page; it’s confusing.
added a section Properties - Singular homology with some basic statement about the singular homology of CW-complexes.
No, homeomorphism type, i.e. topological type, not homotopy type. With homotopy type we get OUT of the class of spaces homeomorphic to the underlzing topological space of a CW complex. In my opinion
the clean way is to talk about two differenet entities:
• CW-complexes
• and their UNDERLYING topological spaces.
Period. Self-explanatory, said in a bit more modern and unambiguous way, but expressing the classical distinction.
Then one can later talk about more rough questions of homotopy theory, but this is not the original matter or the class of topological spaces defined.
Urs: I think people say strong homotopy type, not “strict” homotopy type.
And one should maybe better say “space of the strict homotopy type of a CW-complex”. To distinguish from the weak homotopy type for which the statement becomes empty.
Of course there is also a difference between a space of the homotopy type of a CW complex and a space of the homeomorphism type of a CW complex. I would tend to think that this difference is even
bigger than the difference between a space of the homeomorphism type of a CW complex and a space with a specified CW complex structure. (We also have another name for “space of the homotopy type of
the CW complex” now – m-cofibrant space.)
I agree with Tim and Jim, however I would say more precisely a topological type not the homotopy type of CW-complex. Whitehead and Postinkov I think use such terminology in their courses. I will put
some remarks into nLab entry. Maybe not now.
I think this is still the better terminology. (Am I old fashioned?)
Besides there are two notions of CW-complex. One is a topological space with fixed CW-complex structure on it, and another is a topological space for which exist a CW-structure.
In the old days, CW-complex meant a topological space with fixed CW-complex structure on it.
Otherwise we would refer to a space of the homotopy type of a CW complex.
Besides there are two notions of CW-complex. One is a topological space with fixed CW-complex structure on it, and another is a topological space for which exist a CW-structure. The categories are
very different, the first category, sometimes called CELL has morphisms respecting cell structure (cellular maps) and the latter does not. Cellular approximation theorem says that the two become
equivalent at the level of homotopy category. I am in a hurry now, so can not edit at the moment.
A “relative” CW-complex (X,A) is similar, except X 0 is the disjoint union of A with a discrete space.
I think I learned that $X_0$ is only $A$ in that case. Is this standard ? (Or maybe $A$ needs to be $X_{-1}$ ?)
added in CW-complex in the Examples section something about noncompact smooth manifolds.
Eventually it would be good to state here precisely Milnor’s theorem etc. Googling around I seem to see a lot of misleading imprecision in the usual statements along these lines (on Wikipedia and MO)
concerning the distinctions between countably generated and general CW-complexes and concerning homotopy equivalence vs weak homotopy equivalence.
Here an attempt to prove that $O(n)_+ \wedge_{O(n-q)} S^{n-q}$ does have cell complex structure.
We may think of this space equivalently as the result of first forming the manifold with boundary
$O(n) \times D^{n-q}$
then forming the group quotient
$( O(n) \times D^{n-q} ) / O(n-q)$
and then collapsing the boundary to the point.
By Illmann 83, corollary 7.2, the smooth manifold $O(n) \times D^{n-q}$ does admit $O(n-q)$-CW structure, hence the quotient $( O(n) \times D^{n-q} ) / O(n-q)$ does inherit CW-structure.
But moreover, by the sentence just above theorem 7.1 in Illmann 83, this $O(n-q)$-CW structure may be chosen such that the boundary is a $O(n-q)$-CW subcomplex.
This means that as we now collapse the boundary, the result is still a CW-complex.
I think your argument does hold water, assuming that $O(n)_+ \wedge_{O(n-q)} S^{n-q}$ is indeed $\left((O(n) \times D^{n-q} ) / O(n-q)\right)/boundary$ – I guess that $O(n)_+$ is just $O(n)$ with a
disjoint basepoint added?
Thanks for the sanity check.
I guess that $O(n)_+$ is just $O(n)$ with a disjoint basepoint added?
So for $X$ an unpointed space and $Y$ a pointed space then
\begin{aligned} (X_+) \wedge Y & \simeq \frac{(X_+) \times Y}{ (X_+) \times \{y_0\} \sqcup \{*\} \times Y } \\ & \simeq \frac{ X \times Y \sqcup \{\ast\} \times Y }{ X \times \{y_0\} \sqcup \{\ast\}
\times \{y_0\} \sqcup \{\ast\} \times Y } \\ & \simeq \frac{X \times Y}{ X \times \{y_0\}} \end{aligned} \,.
Now $S^{n-q} \simeq D^{n-q}/S^{n-q-1}$ with basepoint the image of $S^{n-q-1}$. So
\begin{aligned} (O(n)_+) \wedge S^{n-1} & \simeq \frac{ O(n) \times D^{n-q} }{ O(n) \times S^{n - q - 1} } \\ & \simeq \frac{ O(n) \times D^{n-q} }{ \partial( O(n) \times D^{n-q}) } \end{aligned} \,.
Finally, the two quotients clearly commute:
$\frac{ O(n) \times D^{n-q} } { \partial( O(n) \times D^{n-q} )} / O(n-q) \simeq \frac{ (O(n) \times D^{n-q}) / O(n-q) } { (\partial( O(n) \times D^{n-q} ))/O(n-q)}$
And boundary commutes with quotient, since the $O(n-q)$-action preserves both the boundary and interior of $O(n) \times D^{n-q}$ :-)
Sure. Okay, great. Thanks.
Where is this material going again? I’ve tried to keep track, and I guess it’s to do with the model structure for orthogonal spectra?
Yes, I am typing this into model structure on orthogonal spectra which then becomes part of Introduction to Stable homotopy theory – 1-2.
I’ll drop you a note a little later today when the above argument has found its place.
So the argument above is now this lemma at “model structure on orthogonal spectra”.
It is used in the proof of this lemma to show that the generating acylic cofibrations of the stable model structure are indeed acyclic, and it is used in th proof of this theorem stating that the
stable model structure is monoidal.
Ah, cool, thanks! Nice to see where it is used, too.
I was contacted by a young person saying that the definition at CW-complex was hard to read. I went and expanded the Definition section, breaking it up into smaller steps, including examples, and
reordering a little.
Somebody asked this question on MSE, which seems legitimate: https://math.stackexchange.com/questions/3183111/cw-complexes-are-the-cofibrant-objects-in-the-quillen-model-structure-on-top
Thanks for the alert. I have fixed it in the entry, replacing “the” by “among the”. But the entries being pointed to for details (such as CW-approximation) are clear about this. | {"url":"https://nforum.ncatlab.org/discussion/2719/cwcomplex/","timestamp":"2024-11-12T12:07:02Z","content_type":"application/xhtml+xml","content_length":"68175","record_id":"<urn:uuid:6537dac5-4668-478d-bc6f-591b18e75cc9>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00497.warc.gz"} |
Cloze Chemistry
The Cloze Chemistry question type allows students to enter chemical formulas into empty response boxes, which may form part of an equation, or appear in the middle of the text.
Figure 1: Chemistry Formula Cloze question example.
Create a Question
Enter the question stem in the Compose Question area. The Template Markup area below is used in Cloze Chemistry questions to add text and response boxes. Response boxes that support chemistry
expressions can be placed in the middle of text. For each new response box you can either use the Insert Response button in the Rich Text Editor panel (see Figure 2) or manually enter {{response}}.
Each response box in the Template area is known as a container (containers can support multiple response boxes).
Figure 2: Rich Text Editor panel with Insert Response button.
Figure 3 shows how the Formula template area looks like with a chemistry equation and one response container.
Figure 3: Cloze Chemistry question basic parameters.
Each container in the Formula template area will be labeled Response 1, Response 2 etc. As mentioned above, you can add multiple response boxes within one container. For this, you will need to
configure the Edit individual response containers section. This can be found under More Options. Below is a breakdown of available attributes:
• Template - You can enter complex chemistry equations and response boxes here. The Template is a LaTeX area. Chemistry expressions should be added with the LaTeX Keypad. You can add multiple
response boxes here.
• Width and Height - You can enter container width and height in pixels, i.e. 60px and 30px, in order to change standard dimensions.
Follow these steps in order to configure validation:
• Insert a value in the Point(s) box to define a score for the question. The default value is set to 1.
• In the Correct tab below you will Response submenus. These correspond to the responses you have entered in the Template area. In our example, there is only one response container so we are going
to set validation under Equation Value 1 only.
• Select the scoring method and enter the correct response in the Value field. In this example, the student response should be exactly the same as in the validation area. In this case, select
equivLiteral as a scoring method. Then we enter the correct response in the Value field as shown in Figure 4.
Figure 4: Validation settings for cloze chemistry question with one response container.
• Each scoring method has additional options that allow authors to set more precise validation conditions. When you select a scoring method all extra options will be displayed below the menu.
• More complex chemistry questions may need a second scoring method applied. You can combine several scoring methods to set more strict validation rules. For instance, you may want the system to
accept a response that is not only correct but is also given in certain measurement units. In this case, add another method called isUnit. Click the +Add button below the first scoring method to
add a new scoring method.
More Options
Unscored/Practice usage - Removes all scoring from the question.
Penalty point(s) - The value entered here will be deducted from the student for an incorrect answer.
Minimum score if attempted - Set attempt marks for the question.
Check answer button - Show or hide the Check answer button from the student.
Check answer attempts - The value entered here refers to how many times the student can use the Check answer button before it is disabled.
Transparent background - If this is enabled, the background of the response area will match the color of the rest of the background area.
Response minimum width (px) - Enter, in pixels, the minimum width of the response area.
Template font scale - Scales the font relative to the question's font size. The default value is 150%.
Response container (global) and Edit individual response containers - Configure the Width and the Height, in pixels, for either all response containers or for specified response containers.
Type - Select from a Floating keypad, a Block keypad, a Block on focus keypad, or no keypad.
Show keypad hints - Enable or disable hints on the keypad, such as keyboard shortcuts and symbol group titles that are shown on the top left corner of the keypad when hovering over a symbol group
Number pad - Customise the number pad in this section.
Symbols - Select what symbol groups will be shown to the student, or create custom symbol groups.
Text blocks - Use this option when you want to define a list of custom units (such as g, kg, cm, oz, etc.) that will not be rendered as LaTeX. | {"url":"https://authorguide.learnosity.com/hc/en-us/articles/360000448817-Cloze-Chemistry","timestamp":"2024-11-12T12:22:40Z","content_type":"text/html","content_length":"28882","record_id":"<urn:uuid:b6259dca-d0cf-4237-93ae-07c102679656>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00390.warc.gz"} |
188. The best time to buy and sell stocks IV
Given an array of integers prices, its i-th element prices[i] is the price of a given stock on day I.
Design an algorithm to calculate the maximum profit you can make. You can complete up to k transactions.
Note: you cannot participate in multiple transactions at the same time (you must sell the previous shares before buying again).
Example 1:
Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: on day 1 (Stock price = 2) Buy on the second day (Stock price = 4) When sold, the exchange can make a profit = 4-2 = 2 .
Example 2:
Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: on day 2 (Stock price = 2) Buy on day 3 (Stock price = 6) Sell at, The exchange is profitable = 6-2 = 4 .
Then, on day 5 (Stock price = 0) Buy on day 6 (Stock price = 3) Sell at, The exchange is profitable = 3-0 = 3 .
0 <= k <= 100
0 <= prices.length <= 1000
0 <= prices[i] <= 1000
Source: LeetCode
Link: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv
The copyright belongs to Lingkou network. For commercial reprint, please contact the official authorization, and for non-commercial reprint, please indicate the source.
Problem solution
The question only asks the maximum profit, not the specific day to buy and sell these days. Therefore, we can consider using the method of dynamic programming.
• Dynamic programming is used to solve multi-stage decision-making problems;
• The question method of dynamic programming: only the optimal solution, not the specific solution;
This question can be used 123 see the problem solution with strong universality and good understanding
1. Determine the meaning of dp array and its subscript
Here, K transactions are uncertain and need to be determined by input, so it must be expressed in the state. Suppose a purchase is the beginning of a transaction.
There are two states: holding and not holding.
dp[i][k][0]: indicates the maximum profit on the i-th day when there are k transactions at most and no holding status
dp[i][k][1]: refers to the maximum profit in the holding state on the I-day, with a maximum of K transactions
2.dp array recursion
What is the state transition formula?
Assume that day i is holding
• If no buying operation is performed on day I, dp[i][k][1] = dp[i-1][k][1]
• If the buy operation is performed on day I, dp[i][k][1] = dp[i-1][k-1][0] - prices[i]
dp[i][k][1] = Math.max(dp[i-1][k][1]，dp[i-1][k-1][0] - prices[i])
Assume that day i is not held
• If there is no sale on day I, then dp[i][k][0] = dp[i-1][k][0]
• If selling is performed on day I, dp[i][k][0] = dp[i-1][k][1] + prices[i]
dp[i][k][0] = Math.max(dp[i-1][k][0]，dp[i-1][k][1] + prices[i])
3.dp array initialization
When i=0, no matter how many times i can trade at most, i can only trade once when i is equal to 0. dp[0][2][1] = -prices[0] can be understood as that when i trade at most 2 times, i actually only
trade once, so as long as i hold, it should be initialized to - prices[0]
dp[0][1][1] = -prices[0]
dp[0][2][1] = -prices[0]
4. Loop traversal sequence
First traverse i from small to large, and then traverse k from small to large
for(int m=1;m<=k;m++){
dp[0][m][1] = -prices[0];
for(int i=1;i<prices.length;i++){
for(int j=1;j<=k;j++){
dp[i][j][0] = Math.max(dp[i-1][j][0],dp[i-1][j][1] + prices[i]);
dp[i][j][1] = Math.max(dp[i-1][j][1],dp[i-1][j-1][0] - prices[i]);
return dp[prices.length-1][k][0];
Scroll array
The state of i is only related to the state of i-1. You can use the rolling array to reduce the dimension, but it should be noted that dp[k][0] must precede dp[k][1], because dp[k][0] needs to use
the previous dp[k][1]
for(int m=1;m<=k;m++){
dp[m][1] = -prices[0];
for(int i=1;i<prices.length;i++){
for(int j=1;j<=k;j++){
dp[j][0] = Math.max(dp[j][0],dp[j][1] + prices[i]);
dp[j][1] = Math.max(dp[j][1],dp[j-1][0] - prices[i]);
return dp[k][0];
Optimization: because a transaction involves buying and selling for two days, if the value of k is greater than half of the array length, k will directly take half of the array length, because the
number of redundant transactions cannot be reached
class Solution {
public int maxProfit(int k, int[] prices) {
int len = prices.length;
if(k==0||len<=1) return 0;
k=Math.min(k , prices.length/2);
int dp [] [] = new int[k+1][2];
for(int m=1;m<=k;m++){
dp[m][1] = -prices[0];
for(int i=1;i<prices.length;i++){
for(int j=1;j<=k;j++){
dp[j][0] = Math.max(dp[j][0],dp[j][1] + prices[i]);
dp[j][1] = Math.max(dp[j][1],dp[j-1][0] - prices[i]);
return dp[k][0]; | {"url":"https://programmer.group/188-the-best-time-to-buy-and-sell-stocks-iv.html","timestamp":"2024-11-06T08:55:26Z","content_type":"text/html","content_length":"12412","record_id":"<urn:uuid:0c79d91e-4793-4127-b9d8-9fa465a8f626>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00247.warc.gz"} |
Simplifying Polynomial Expressions: (7x^2+x+1)+(-6x^2-8x-10)
In mathematics, simplifying polynomial expressions involves combining like terms to create a more concise representation. Let's simplify the expression: (7x^2+x+1)+(-6x^2-8x-10)
Step 1: Identify Like Terms
Like terms are terms that have the same variable and exponent. In our expression, we have the following pairs of like terms:
• 7x² and -6x² (both have x² as the variable)
• x and -8x (both have x as the variable)
• 1 and -10 (both are constant terms)
Step 2: Combine Like Terms
To combine like terms, we simply add their coefficients:
• 7x² + (-6x²) = x²
• x + (-8x) = -7x
• 1 + (-10) = -9
Step 3: Write the Simplified Expression
Now, we combine the simplified terms to get the final simplified expression:
(7x^2+x+1)+(-6x^2-8x-10) = x² - 7x - 9
Therefore, the simplified form of the expression (7x^2+x+1)+(-6x^2-8x-10) is x² - 7x - 9. | {"url":"https://jasonbradley.me/page/(7x%255E2%252Bx%252B1)%252B(-6x%255E2-8x-10)","timestamp":"2024-11-04T13:41:31Z","content_type":"text/html","content_length":"56211","record_id":"<urn:uuid:767d3dda-880f-4ce6-a4a1-22b5526b14fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00587.warc.gz"} |
seminars - Regularity properties of Brjuno functions associated with classical continued fractions
장소: 129-309
An irrational number is called a Brjuno number if the sum of the series of log(q_{n+1})/q_n converges, where q_n is the denominator of the n-th principal convergent of the regular continued fraction.
The importance of Brjuno numbers comes from the study of analytic small divisor problems in dimension one. In 1988, J.-C. Yoccoz introduced the Brjuno function which characterizes the Brjuno numbers
to estimate the size of Siegel disks. In this talk, we introduce Brjuno-type functions associated with by-excess (negative), odd and even continued fractions. Then we discuss the L^p and the Hölder
regularity properties of the difference between the classical Brjuno function and the Brjuno-type functions. This is joint work with Stefano Marmi. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&page=23&l=en&sort_index=date&order_type=desc&document_srl=998254","timestamp":"2024-11-13T02:28:37Z","content_type":"text/html","content_length":"47796","record_id":"<urn:uuid:b9f95694-e7d6-4cda-a091-3deda0ca1c23>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00745.warc.gz"} |
Treynor ratio formula, calculator - Initial Return
Treynor ratio is a popular risk-adjusted performance measure. It gets its name from the American economist Jack Treynor who came up with this measure in the mid-1960s (see the full reference at the
end). It is a measure of how much “excess return” (i.e., return above the risk-free rate) a security (stock, bond, mutual fund, etc.) offers per unit of systematic risk, which is captured by beta. In
that sense, it is a reward-to-risk ratio like the Sharpe ratio.
Treynor ratio formula
We can write the Treynor ratio formula for an asset i (e.g., stock, portfolio, fund, etc.) as follows:
where R[i] is the return on the asset, R[f] is the risk-free rate of return, and β[i] is the beta of the asset, which is based on CAPM.
To give a specific example, suppose that mutual fund A yielded a return of 12% last year. Over the same period, the yield on treasury bills, which we take as a proxy of the risk-free rate, was 3%.
Let’s also assume that fund A’s beta was 2. Then, we can easily calculate fund A’s “realized Treynor ratio” as:
T[A] = (12% − 3%) / 2 = 4.5%
This means that fund A yielded a return of 4.5% for each unit of systematic risk borne.
Now, let’s imagine that fund A’s beta will remain to be 2, its expected return is 15%, and the risk-free rate for the next year is 5%. Then, the “expected Treynor ratio” for this mutual fund would
T[A] = (15% − 5%) / 2 = 5%
If mutual fund B expects to have T[A] = 3%, then we can consider A to be a better investment opportunity than B. This is because A offers a higher reward than B per unit of systematic risk carried.
We can also interpret the Treynor ratio visually by means of a plot of (systematic) risk against return. In Figure 1, we have two risky assets (e.g., stocks or funds): A and B, and the risk-free
asset, which yields R[f].
For both A and B, we can draw a line that connects the risky asset with the riskless one. Then, the Treynor ratio is simply the slope of that line. In this case, T[B] > T[A] because of the steeper
Figure 1
Test your knowledge
Your friend has recommended a mutual fund that generated a return of %15 last year. The fund’s beta is 2. In comparison, the S&P500 yielded 12%. The risk-free rate was 5%. According to your friend,
the fund outperformed the market on the basis of its Treynor ratio. Is he correct?
Note: You can use the calculator below to verify your answer. The solution is provided at the bottom of this page.
Treynor ratio calculator
In order to use the Treynor ratio calculator, you would need to enter the following information:
• The return on the risky asset.
• The return on the riskless asset.
• The beta of the risky asset.
If you need to estimate beta, you can do that by regressing the returns on the asset you are interested in on the returns on a broad-based market index such as the S&P 500. Alternatively, financial
data providers such as Yahoo! Finance provide beta estimates.
The following calculators might be of interest to you as well.
Despite being developed in the 1960s, the Treynor measure remains one of the most popular reward-to-risk measures that is used by practitioners and academics alike to this date. It tells us about the
excess return an asset yields per unit of systematic risk borne. Therefore, the higher it is, the better the investment performance.
Further reading:
Treynor (1965) ‘How to Rate Management of Investment Funds,’ Harvard Business Review, Vol. 43(1), pp. 63.
What is next?
This was the final lesson in our online course on investments. Next, we offer an investments quiz to help you test your understanding of the material covered in the course.
We hope you have found this lesson useful. If you have any feedback for us, you can reach us here. Please feel free to share our content in your networks to spread the word.
Solution for the “test your knowledge” exercise
Using the calculator above we find that (note: the market’s beta is equal to 1 by definition):
We can also verify these figures as follows.
For the mutual fund we have: T[A] = (R[A] − R[f]) / β[A] = (15% − 5%) / 2 = 5%.
The corresponding figure for the market is: T[M] = (R[M] − R[f]) / β[M] = (12% − 5%) / 1 = 7%.
In conclusion, your friend is wrong because T[A] < T[M]. | {"url":"https://www.initialreturn.com/treynor-ratio-formula-calculator","timestamp":"2024-11-13T21:12:46Z","content_type":"text/html","content_length":"88700","record_id":"<urn:uuid:14002b1a-a8c0-4d20-b639-965d12363051>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00632.warc.gz"} |
Realization problem for hidden Markov models
If {X[t]} is a finite-state Markov process, and {Y[t]} is a finite-valued output process with Y[t+1] depending (possibly probabilistically) on X[t], then the process pair is said to constitute a
hidden Markov model. This paper considers the realization question: given the probabilities of all finite-length output strings, under what circumstances and how can one construct a finite-state
Markov process and a state-to-output mapping which generates an output process whose finite-length strings have the given probabilities? After reviewing known results dealing with this problem
involving Hankel matrices and polyhedral cones, we develop new theory on the existence and construction of the cones in question, which effectively provides a solution to the realization problem.
This theory is an extension of recent theoretical developments on the positive realization problem of linear system theory.
Dive into the research topics of 'Realization problem for hidden Markov models'. Together they form a unique fingerprint. | {"url":"https://researchportalplus.anu.edu.au/en/publications/realization-problem-for-hidden-markov-models","timestamp":"2024-11-09T13:41:28Z","content_type":"text/html","content_length":"51638","record_id":"<urn:uuid:90a3f5b3-ff82-4a11-992b-1ff7bace0ba5>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00127.warc.gz"} |
Evaluate Multi Variable Expressions Involving Negative Numbers
Substitute and Evaluate an Expression with Negative Numbers #1
Substitute and Evaluate an Expression with Negative Numbers #2
Substitute and Evaluate an Expression with Negative Numbers #3
Substitute and Evaluate an Expression with Negative Numbers – Powers #1
Substitute and Evaluate an Expression with Negative Numbers – Powers #2
Substitute and Evaluate an Expression with Negative Numbers – Rational #1
Substitute and Evaluate an Expression with Negative Numbers – Rational #2
Substitute and Evaluate an Expression with Negative Numbers – Rational #3 | {"url":"https://vividmath.com.au/year-9/9-algebraic-expressions/9-evaluate-multi-variable-expressions-involving-negative-numbers/","timestamp":"2024-11-05T06:09:56Z","content_type":"text/html","content_length":"67836","record_id":"<urn:uuid:08cc6f14-28d9-42ba-a388-a5d1985255d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00048.warc.gz"} |
16.4 Potential, Free Energy, and Equilibrium - Chemistry: Atoms First 2e | OpenStax
By the end of this section, you will be able to:
• Explain the relations between potential, free energy change, and equilibrium constants
• Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium
• Use the Nernst equation to determine cell potentials under nonstandard conditions
So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on
thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the
chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties ΔG and K.
E° and ΔG°
The standard free energy change of a process, ΔG°, was defined in a previous chapter as the maximum work that could be performed by a system, w[max]. In the case of a redox reaction taking place
within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, w[elec]:
The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:
where n is the number of moles of electrons transferred, F is Faraday’s constant, and E°[cell] is the standard cell potential. The relation between free energy change and standard cell potential
confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.
E° and K
Combining a previously derived relation between ΔG° and K (see the chapter on thermodynamics) and the equation above relating ΔG° and E°[cell] yields the following:
This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to
product. A summary of the relations between E°, ΔG° and K is depicted in Figure 16.7, and a table correlating reaction spontaneity to values of these properties is provided in Table 16.2.
K ΔG° E°[cell]
> 1 < 0 > 0 Reaction is spontaneous under standard conditions
Products more abundant at equilibrium
< 1 > 0 < 0 Reaction is non-spontaneous under standard conditions
Reactants more abundant at equilibrium
= 1 = 0 = 0 Reaction is at equilibrium under standard conditions
Reactants and products equally abundant
Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes
Use data from
Appendix L
to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at 25 °C. Comment on the spontaneity of the forward reaction and the
composition of an equilibrium mixture of reactants and products.
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in
Appendix L
$anode (oxidation):Fe(s)⟶Fe2+(aq)+2e−EFe2+/Fe°=−0.447 Vcathode (reduction):2×(Ag+(aq)+e−⟶Ag(s))EAg+/Ag°=0.7996 VEcell°=Ecathode°−Eanode°=EAg+/Ag°−EFe2+/Fe°=+1.247 Vanode (oxidation):Fe(s)⟶Fe2+(aq)
+2e−EFe2+/Fe°=−0.447 Vcathode (reduction):2×(Ag+(aq)+e−⟶Ag(s))EAg+/Ag°=0.7996 VEcell°=Ecathode°−Eanode°=EAg+/Ag°−EFe2+/Fe°=+1.247 V$
With n = 2, the equilibrium constant is then
$Ecell°=0.0592 VnlogK K=10n×Ecell°/0.0592 V K=102×1.247 V/0.0592 V K=1042.128 K=1.3×1042Ecell°=0.0592 VnlogK K=10n×Ecell°/0.0592 V K=102×1.247 V/0.0592 V K=1042.128 K=1.3×1042$
The standard free energy is then
$ΔG°=−nFEcell° ΔG°=−2×96,485Cmol×1.247JC=−240.6kJmolΔG°=−nFEcell° ΔG°=−2×96,485Cmol×1.247JC=−240.6kJmol$
The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an
equilibrium mixture containing mostly products.
Check Your Learning
What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?
Spontaneous; n = 2; $Ecell°=+0.291 V;Ecell°=+0.291 V;$ $ΔG°=−56.2kJmol;ΔG°=−56.2kJmol;$ K = 6.8 $××$ 10^9.
Potentials at Nonstandard Conditions: The Nernst Equation
Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy
of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture
composition can be used for this purpose.
Notice the reaction quotient, Q, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change
to cell potential yields the Nernst equation:
This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons
transferred, n, the temperature, T, and the reaction mixture composition as reflected in Q. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants
(R and F) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:
Predicting Redox Spontaneity Under Nonstandard Conditions
Use the Nernst equation to predict the spontaneity of the redox reaction shown below.
$Co(s)+Fe2+(aq,1.94M)⟶Co2+(aq, 0.15M)+Fe(s)Co(s)+Fe2+(aq,1.94M)⟶Co2+(aq, 0.15M)+Fe(s)$
Collecting information from
Appendix L
and the problem,
$Anode (oxidation):Co(s)⟶Co2+(aq)+2e−ECo2+/Co°=−0.28 VCathode (reduction):Fe2+(aq)+2e−⟶Fe(s)EFe2+/Fe°=−0.447 VEcell°=Ecathode°−Eanode°=−0.447 V−(−0.28 V)=−0.17 VAnode (oxidation):Co(s)⟶Co2+(aq)
+2e−ECo2+/Co°=−0.28 VCathode (reduction):Fe2+(aq)+2e−⟶Fe(s)EFe2+/Fe°=−0.447 VEcell°=Ecathode°−Eanode°=−0.447 V−(−0.28 V)=−0.17 V$
Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions
$Q=[Co2+][Fe2+]=0.15M1.94M=0.077 Ecell=Ecell°−0.0592 VnlogQ Ecell=−0.17 V−0.0592 V2log0.077 Ecell=−0.17 V+0.033 V=−0.14 VQ=[Co2+][Fe2+]=0.15M1.94M=0.077 Ecell=Ecell°−0.0592 VnlogQ Ecell=−0.17
V−0.0592 V2log0.077 Ecell=−0.17 V+0.033 V=−0.14 V$
The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.
Check Your Learning
For the cell schematic below, identify values for
, and calculate the cell potential,
n = 6; Q = 1440; E[cell] = +1.97 V, spontaneous.
A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox
species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the
Nernst equation in calculations involving concentration cells.
Concentration Cells
What is the cell potential of the concentration cell described by
$Zn(s)│Zn2+(aq, 0.10M)║Zn2+(aq, 0.50M)│Zn(s)Zn(s)│Zn2+(aq, 0.10M)║Zn2+(aq, 0.50M)│Zn(s)$
From the information given:
$Anode:Zn(s)⟶Zn2+(aq, 0.10M)+2e−Eanode°=−0.7618 VCathode:Zn2+(aq, 0.50M)+2e−⟶Zn(s)Ecathode°=−0.7618 V¯Overall:Zn2+(aq, 0.50M)⟶Zn2+(aq, 0.10M)Ecell°=0.000 VAnode:Zn(s)⟶Zn2+(aq, 0.10M)+2e−Eanode°=
−0.7618 VCathode:Zn2+(aq, 0.50M)+2e−⟶Zn(s)Ecathode°=−0.7618 V¯Overall:Zn2+(aq, 0.50M)⟶Zn2+(aq, 0.10M)Ecell°=0.000 V$
Substituting into the Nernst equation,
$Ecell=0.000 V−0.0592 V2log0.100.50=+0.021 VEcell=0.000 V−0.0592 V2log0.100.50=+0.021 V$
The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is
reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion
concentration is greater (E[cathode] > E[anode]).
Check Your Learning
The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?
E[cell] = 0.000 V; [Zn^2+][cathode] = [Zn^2+][anode] = 0.30 M | {"url":"https://openstax.org/books/chemistry-atoms-first-2e/pages/16-4-potential-free-energy-and-equilibrium","timestamp":"2024-11-07T23:20:09Z","content_type":"text/html","content_length":"462687","record_id":"<urn:uuid:03552bdb-2106-4571-a220-dfaf93e10c0d>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00524.warc.gz"} |
Graph Theory Tutorial - Sankalchand Patel College of Engineering | Quizgecko
Graph Theory Tutorial - Sankalchand Patel College of Engineering
Document Details
Sankalchand Patel College of Engineering
graph theory graph algorithms discrete mathematics computer science
Full Transcript
FACULTY OF ENGINEERING & TECHNOLOGY SANKALCHAND PATEL COLLEGE OF ENGINEERING VISNAGARB. B.TECH. SEMESTER –III CE , IT & AI&DS...
FACULTY OF ENGINEERING & TECHNOLOGY SANKALCHAND PATEL COLLEGE OF ENGINEERING VISNAGARB. B.TECH. SEMESTER –III CE , IT & AI&DS UNIT 3: GRAPH THEORY 1) Define (1) simple graph (2) multi graph (3)
Isolated vertex (4) Pendent vertex (5) Size of a graph (6) in-degree of a vertex (7) out-degree of a vertex (8)null graph (9) degree of a vertex (10) trivial graph (11) complete graph (12) bipartite
graph (13) sub graph (14) Weighted Graph (15) Cyclic graph Illustrate each with an example. 2) How many edges has each of the following graphs (1) 𝐾5 (2) 𝐾6 (3) 𝐾2,3 (4) 𝐾3,3. 3) 𝐷𝑟𝑎𝑤 𝐾2,3 𝑎𝑛𝑑 𝐾3,3
𝑔𝑟𝑎𝑝ℎ𝑠. 4) Draw a complete bipartite graph, which is not regular. 5) Draw 𝐾5 and 𝐾6 𝑔𝑟𝑎𝑝ℎ𝑠 6) Find the number of edges in r-regular graph with n-vertices. 7) Does 3- regular graph with 5-vertices
exist ? 8) Identify Isolated vertex/vertices from the following graph. 9) What do you mean by degree of a vertex? What are the degrees of an isolated vertex and a pendant vertex? 10) Define regular
graph. Can a regular graph be a complete graph? 11) Define graph isomorphism and give an example of two isomorphic graphs 12) How many nodes are necessary to construct a graph with exactly 8 edges in
which each node is of degree 2. 13) State and prove the hand-shaking theorem. 14) Find the number of edges in G if G has (1) 16 vertices each of degree 2. (2) 3 vertices of degree 4, 2 vertices of
degree 3 and other 4 vertices of degree 1. 15) Prove that any graph has even number of odd vertices. 16) Does there exists a 4- regular graph with 6 vertices? If so, construct the graphs. 17) Does
there exists a regular graph of degree 5 on 9 vertices? If so, construct the graphs. 18) Determine the number of edges in a graph with 6 vertices,2 vertices of degree 4 and 4 vertices of degree
2.draw two such graphs. FACULTY OF ENGINEERING & TECHNOLOGY SANKALCHAND PATEL COLLEGE OF ENGINEERING VISNAGARB. B.TECH. SEMESTER –III CE , IT & AI&DS 19) Draw the undirected graph G = (V, E)where, V
= {a, b, c, d, e}and E = {e1 , e2 , e3 , e4 , e5 , e6 , e7 } and its incidence relation given as: e1 = (a, b), e2 = (a, b), e3 = (b, c), e4 = (c, d), e5 = (b, b), e6 = (a, d) & e7 = (e, d). 20)
Define Isomorphic Graphs. Determine whether the following graphs are isomorphic or not 21) Check whether the following graphs are isomorphic. 22) Verify that the sum of the in-degrees, the sum of the
out-degree of the vertices and the number of edges in the following graphs are equal. Figure 1 Figure 2 FACULTY OF ENGINEERING & TECHNOLOGY SANKALCHAND PATEL COLLEGE OF ENGINEERING VISNAGARB. B.TECH.
SEMESTER –III CE , IT & AI&DS 23) Find the in-degree and out-degree of each vertex in the graph G with directed edges shown in Figure 24) Define converse of a digraph and find it for given graph G.
25) Define node deleted sub graph and edge deleted sub graph. Also find sub graphs from the given graph G by deleting (I) node 𝑣1 (𝐺 − {𝑣1 }) (II) Edge 𝑒4 (G − {𝑒4 }). 26) Consider the following
graphs: Determine the degree of each node and verify Handshaking theorem. ***** | {"url":"https://quizgecko.com/uploads/tutorial-graph-theory-ov926F","timestamp":"2024-11-05T15:23:49Z","content_type":"text/html","content_length":"159121","record_id":"<urn:uuid:1a44b40e-cf8f-4eda-ace5-4cd9179a8769>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00303.warc.gz"} |
Equality and Inequality in Math - OMC Math Blog
Equality and Inequality
In our daily lives, the terms equality and inequality seem very simple. It’s common knowledge that equal means two (or more) things are the same. They could be equal in size, cost, or any other
characteristic. Inequality means two (or more) things are not equal. They differ in size, shape, cost, or some other characteristic. Easy enough, right?
But what do equality and inequality mean in mathematics? In the mathematical world, equality is a foundational topic, particularly in algebra. Mathematical equations use the symbol (=) to show
equality between two quantities. An example of this would be 3 + 4 = 7. When you solve one side, you will see that it is the same, or equal, to the other. On the other hand, inequality compares two
unequal sides with a different symbol showing the relationship. Let’s dive in deeper!
Equality can be as simple as 3 +4 = 7 or as complex as x = 2y. We can say that a = b so in turn b = a. But what if a also equals c? This now means that b also equals c. Equality can be shown in
different ways and it’s important to understand that it is not always as simple as 3 + 4 = 7.
You may find yourself solving a problem where a = 2b and b = 4. Now you must plug in the information provided to find the solution: a = 2(4) so a = 8. The basic concept of equality is important to
understand as it is the basis for a large amount of the topics school children will cover in math class and throughout their education.
Now that we understand equality, what exactly is inequality? Although it simply means what is not equal (unequal), it goes much further than just that. When two things are unequal, they are not the
same. An apple and an orange are not the same. The numbers 6 and 7 are not the same. But how do we show this relationship in mathematics? We do this by using mathematical symbols – a good
When two things are definitely not equal, we can use the symbol ≠. Using the unequal symbol would look like this: 8 +5 ≠ 20. What if we want to show which side is larger? When showing the
relationship between two unequal sides, we can use the symbols ＞ and ＜. But which symbol do you use to show “greater than” and which symbol do you use for “less than”? The easiest way to remember
is that the open side of the arrow will be on the greater than side. Like so: 6 ＞5.
The larger side to the larger number, the smaller side to the smaller number. This applies to more complex equations as well, such as x + 7 ﹥ 12. This would tell us that x ﹥ 5, so even 5.01 would
make the equation true. Let’s try it out: 5.01 + 7 ﹥ 12 comes out to 12.01 ﹥ 12, making the equation true.
Let’s take this one step further. What if we need to show that something needs to be less than one number, but greater than another? We would still use the symbols shown above but we can expand the
equations and include multiple symbols. If we need x to be less than 10 but greater than 1, we can display the relationship like this 1 ＜ x ＜ 10. To understand this, you can read both sides of the
equation separately and then, when combined, they give us the full picture.
Now, what do we do if something is greater than or equal to the other side, and less than or equal to the other side? Luckily, there are symbols to show that as well. To show “greater than or equal
to”, use the symbol ≥. To show “less than or equal to”, use the symbol ≤. Let’s try this out with the equation previously shown above. If x ≥ 5, then we can say that x + 7 ≥ 12.
Now let’s put this into a real-life example – such as a bag of marbles. If there are 7 marbles in the bag and more than 5 are added in, there are more than 12 in there now. We could apply “less than”
to this as well: x+ 7 ≤ 12, so x ≤ 5. If fewer than 5 marbles were added to the bag of 7, then there are fewer than 12 marbles in the bag.
As you can see, the topic of equality and inequality has more to it than meets the eye. It is a crucial notion to understand as it is a basic concept particularly important in algebra. At OMC, we
ensure that all students leave school with a solid education of math concepts such as these, and the ability to use the concepts that they have been taught in the classroom. Whether it comes
naturally to a student or not, we offer classes and tutoring that can assist them in strengthening their foundation in mathematics. OMC works to ensure that every student reaches or surpasses their
full potential.
Contact OMC today to help ensure your child gains a solid, complete understanding of the most important mathematical concepts, such as equality and inequality, by signing them up for classes that are
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Importing multiple files from a remote FTP location and running import scenario | Community
I’m working on a new project and I could use some help with the steps I need to complete. I’m trying to automate the importing of data from an external source on a nightly basis. Here is what I
1. External system exports multiple files to an FTP location on a nightly basis.
2. I configured a data provider to connect and sync to the remote FTP location.
3. I wrote an import scenario to import the external file.
Here is my problem. I can successfully run the import scenario with a single file, but how do I do it for multiple? When I perform the “File Synchronization", I end up as multiple versions (one for
each new file). How do I configure Acumatica to execute my IS nightly and to import all the new files it finds? | {"url":"https://community.acumatica.com/import-and-export-scenarios-117/importing-multiple-files-from-a-remote-ftp-location-and-running-import-scenario-7115?postid=52247","timestamp":"2024-11-05T07:21:41Z","content_type":"text/html","content_length":"247191","record_id":"<urn:uuid:895bd048-e06a-42be-b477-74d25be1dc6d>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00836.warc.gz"} |
The Paired-Comparison Hypotheses Test for Variances
The F-test provides a means to compare paired data variances. It is a variance hypothesis test.
If we are exploring the precision of one measuring device or another, or we are comparing assembly processes, we often want to know if the variance is different or not.
Working with data from normal distributions from two different processes or devices, we know from statistical theory that the ratio (s1)2 / (s2)2 is described by the F distribution.
There are three hypothesis test possible, basically to test if the population variances are different, or one is less than or greater than the other. The following details the three test’s null and
alternative hypotheses.
Assuming the population variances are equal
$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}=\sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\ne \sigma _{2}^{2}\end{array}$$
Assuming population variance 1 is less then or equal to population variance 2
$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\le \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\gt\sigma _{2}^{2}\end{array}$$
Assuming population variance 1 is greater than or equal to population variance 2
$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\ge \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\lt\sigma _{2}^{2}\end{array}$$
Recall that the shape of the F distribution is non-symmetrical and depends on the degrees, DF, of freedom of the two sample variances, s12 and S22. We will use v1 and v2 respectively, and v1 is the
DF in the numerator. By convention we use the larger sample variance as s12 and place it in the numerator.
Example Problem
For a medical delivery device, the ability for a specific part to remain strong (measured in psi) is important even after a 2 year period of storage. Assuming a 95% confidence level, use the data to
determine if the variation is equal or better after aging.
At the start we tested 9 units and found a standard deviation of 900 psi. After two years of aging, we tested 7 units and determined a 300 psi variation.
The null and alternative hypothesis are
$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\le \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}>\sigma _{2}^{2}\end{array}$$
and, DF1 = 8 and DF2 = 6
We are using a one-tailed test to explore if after aging the units have a smaller variance. The entire α risk will be in the right tail. From the F distribution, the critical value of F is 4.15. Thus
we will reject the null hypothesis if the ratio of the two sample variances is greater than 4.15.
$$ \large\displaystyle F=\frac{s_{1}^{2}}{s_{2}^{2}}=\frac{{{900}^{2}}}{{{300}^{2}}}=9$$
Since, 9 is greater than 4.15 we have sufficient evidence, 95% confidence, to reject the null hypothesis and conclude strength variation is reduced after aging for two years.
Hypothesis Tests for Variance Case I (article)
Hypothesis unequal variance (article)
Levene’s Test (article)
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Henries to Kilohenries Conversion (H to kH) - Inch Calculator
Henries to Kilohenries Converter
Enter the electrical inductance in henries below to convert it to kilohenries.
Do you want to convert kilohenries to henries?
How to Convert Henries to Kilohenries
To convert a measurement in henries to a measurement in kilohenries, divide the electrical inductance by the following conversion ratio: 1,000 henries/kilohenry.
Since one kilohenry is equal to 1,000 henries, you can use this simple formula to convert:
kilohenries = henries ÷ 1,000
The electrical inductance in kilohenries is equal to the electrical inductance in henries divided by 1,000.
For example,
here's how to convert 5,000 henries to kilohenries using the formula above.
kilohenries = (5,000 H ÷ 1,000) = 5 kH
Henries and kilohenries are both units used to measure electrical inductance. Keep reading to learn more about each unit of measure.
What Is a Henry?
One henry is equal to the inductance of a conductor in which there is one volt of electromotive force when the current through the conductor is increased by one ampere per second.^[1]
The henry is the SI derived unit for electrical inductance in the metric system. Henries can be abbreviated as H; for example, 1 henry can be written as 1 H.
The National Institute of Standards and Technology recommends using "henries" as the plural form of the unit, however the usage of "henrys" is also common.^[2]
Learn more about henries.
What Is a Kilohenry?
One kilohenry is equal to 1,000 henries, which are the inductance of a conductor with one volt of electromotive force when the current is increased by one ampere per second.
The kilohenry is a multiple of the henry, which is the SI derived unit for electrical inductance. In the metric system, "kilo" is the prefix for thousands, or 10^3. Kilohenries can be abbreviated as
kH; for example, 1 kilohenry can be written as 1 kH.
Learn more about kilohenries.
Henry to Kilohenry Conversion Table
Table showing various
henry measurements
converted to
Henries Kilohenries
1 H 0.001 kH
2 H 0.002 kH
3 H 0.003 kH
4 H 0.004 kH
5 H 0.005 kH
6 H 0.006 kH
7 H 0.007 kH
8 H 0.008 kH
9 H 0.009 kH
10 H 0.01 kH
20 H 0.02 kH
30 H 0.03 kH
40 H 0.04 kH
50 H 0.05 kH
60 H 0.06 kH
70 H 0.07 kH
80 H 0.08 kH
90 H 0.09 kH
100 H 0.1 kH
200 H 0.2 kH
300 H 0.3 kH
400 H 0.4 kH
500 H 0.5 kH
600 H 0.6 kH
700 H 0.7 kH
800 H 0.8 kH
900 H 0.9 kH
1,000 H 1 kH
1. International Bureau of Weights and Measures, The International System of Units, 9th Edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf
2. Ambler Thompson and Barry N. Taylor, Guide for the Use of the International System of Units (SI), National Institute of Standards and Technology, https://physics.nist.gov/cuu/pdf/sp811.pdf
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Parameter - PreSparsify
Hello Gurobi team,
Can I know what process the 'PreSparsify == 1' parameter works through during the Presolve process?
I would really appreciate it if you could let me know the documents to reference.
• Hi Jaromił,
I would like to take up this topic here.
I looked into section 5.3 of the specified paper and in my applications I have a lot of constraints that fit into the scheme [s * A_qS = A_rS] for the given rows:
A_qS * x_S + A_qT * x_T + A_qU * x_U = b_q
A_rS * x_S + A_rT * x_T + A_rV * x_V <= b_r
Am I right, that row q, that is substracted s times from row r, needs to be an equality constraint?
And row r can be an equality constraint or an inequality constraint, doesn't it?
It makes sense to me that the transformation shown in the paper should only be applied if |S|>|U|, because only in that case the number of nonzeros is reduced:
A_qS * x_S + A_qT * x_T + A_qU * x_U = b_q
(A_rT - s * A_qT) * x_T - s * A_qU * x_U + A_rV * x_V <= b_r - s* b_q
If I add an auxiliary variable x_a, a corresponding equality constraint and apply the following transformation, I think I can reduce the number of nonzeros in case of |S|>2 independend of |U|:
A_qS * x_S - x_a = 0
x_a + A_qT * x_T + A_qU * x_U = b_q
s * x_a + A_rT * x_T + A_rV * x_V <= b_r
Wouldn't that be benificial in case of |S|≈|U|, but both beeing >>2?
Of course there is no "free lunch", because I have another variable and another equality constraint.
But would that be so bad (during simplex?)?
Or would other presolve-routines eliminate x_a and the corresponding equality constraint again (for good reasons?)?
I ask for the details, because I need to sparsify my applications not only to reduce the memory demand already during the construction of the problem, but also to increase the performance.
Fortunately, in my particular applications, there are several equations r for which [s * A_qS = A_rS] holds.
Furthermore, there are several groups of equations that are "similar to each other" in the described way.
• Hi Thomas,
Am I right, that row q, that is substracted s times from row r, needs to be an equality constraint?
And row r can be an equality constraint or an inequality constraint, doesn't it?
Yes, you are right. Note that you can always make an equality out of an inequality via a slack variable.
Wouldn't that be benificial in case of |S|≈|U|, but both beeing >>2?
Of course there is no "free lunch", because I have another variable and another equality constraint.
But would that be so bad (during simplex?)?
It is really hard to say whether this reduction would always be beneficial or not. Intuitively it should be good, but in practice it often turns out to be the other way round. It is often a very
subtle trade-off between number of variables and constraints and the number of nonzeros. Usually, when aggregation or sparsification reductions are performed, the number of rows, columns, or
nonzeros is reduced while the number of the non-reduced quantity are increased, e.g., there are less rows and columns after aggregation, but the number of nonzeros increases. This can have many
different impacts on the Simplex algorithm, e.g., different Pivot choices, but can also affect Cuts due to different row density.
Or would other presolve-routines eliminate x_a and the corresponding equality constraint again (for good reasons?)?
It is very likely that aggregation would eliminate x_a. However, the decision about aggregation depends on the AggFill parameter which controls how many nonzeros are allowed to be added during
I ask for the details, because I need to sparsify my applications not only to reduce the memory demand already during the construction of the problem, but also to increase the performance.
Fortunately, in my particular applications, there are several equations r for which [s * A_qS = A_rS] holds.
Furthermore, there are several groups of equations that are "similar to each other" in the described way.
I think that the best way here would be to perform (at least some of) the reductions manually and analyze the impact. In particular, if you have some special reductions as you described. This is
the only way to make sure that you do not miss something or not do too much. Additionally, you could experiment with the parameters Presolve, Aggregate, AggFill, and PreSparsify to hopefully
further reduce the size of the presolved model.
Best regards,
• Hi Jaromił Najman,
thanks for your explanations.
I did intensive testing with the parameters, but in the end it was necessary to perform the reductions manually. And I had to set AggFill to zero (sum higher value might have worked as well) to
prevent Gurobi from undoing my reductions.
• Thank you for the update Thomas. The observation you made and setting AggFill to 0 definitely make sense.
Best regards,
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Probability and statistics blog
Betting on Pi
I was reading over at math-blog.com about a concept called numeri ritardatari. This sounds a lot like “retarded numbers” in Italian, but apparently “retarded” here is used in the sense of “late” or
“behind” and not in the short bus sense. I barely scanned the page, but I think I got the gist of it: You can make money by betting on numbers which are late, ie numbers that haven’t shown up in a
while. After all, if the numbers in a lottery or casino are really random, that means it’s highly unlikely that any one number would go a long time without appearing. The “later” the number, the more
likely it is to appear. Makes sense, right?
Before plunking down my hard(ly) earned cash at a casino, I decided to test out the theory first with the prototypical random number: Pi. Legend has it that casinos once used digits from Pi to
generate their winning numbers. Legend also has it that the digits of Pi are so random that they each appear with almost exactly 1 in 10 frequency. So, given this prior knowledge that people believe
Pi to be random, with uniform distribution of digits and no discernible pattern, I can conclude that no one digit should go too long without appearing.
I pulled down the first 10 million digits from here (warning, if you really want this data, right click the link and “save as”). Then I coded up a program in the computer language R to scan through
the digits of Pi, one by one, making a series of “fair” bets (1:9 odds) that the next number to appear in the sequence would be the one that had gone longest without appearing. My code is shown
below. I had to truncate the data to 1 million digits, and even then this code will take your Cray a good while to process, most likely because I have yet to master the use of R’s faster “apply”
functions with complicated loops.
myPi = readLines("pi-10million.txt")[1]
# I think this is how I managed to get Pi into a vector, it wasn't easy.
piVector = unlist(strsplit(myPi,""))
piVector = unlist(lapply(piVector,as.numeric))
# In honor of Goofy Programming Day, I will
# track how long since the last time each digit appeared
# by how many repetitions of that digit are in a vector
ages = c()
# Start us off with nothing in the bank
potHistory = c()
# R just loves long loops like this. Hope you have a fast computer
for(i in 1:1000000) {
# How did our bet do last round?
# Skip the first 100 just to build up some data
if(i > 100) {
if(betOn == piVector[i]) {
potHistory = c(potHistory, 9)
} else {
potHistory = c(potHistory, -1)
# Increase all ages by 1 by adding to vector, then set the one we found to 0
ages = c(ages, 0:9)
ages = ages[!ages == piVector[i]]
# Count occurences of each digit, find the top digits by occurence to bet on
# And you thought Perl was beautiful?
betOn = as.numeric(names(sort(-table(ages)))[1])
# Plot the cumulative sum at 1000 point intervals.
plot.ts(cumsum(potHistory)[seq(0,1000000,500)],pch=20,col="blue",xlab="step/500",ylab="cumulative earnings")
So what was the result? How good was my strategy? After an initial 100 digits to build up data about which digits were latest, I placed a total of 999,900 bets at $1 each. Final earnings: $180.
That’s so close to breaking even that it’s almost inconceivable. I had 100,008 wins and 899,892 losses. My winning percentage was 10.0018% percent.
On the face of it, this result seemed almost a little too good, dare I say even suspiciously good, if you know what I mean. How rare is it to get this close (or closer) to the exact perfect
proportions after so many trials? Assuming that the number of wins followed a binomial distribution with [latex]p=0.1[/latex], my total wins should follow a Normal distribution with mean 99,990 and
variance [latex]n*p*(1-p) = 89,991[/latex] (for an “n” of almost a million and non-minuscule “p”, the Normal approximation to the Binomial should be essentially perfect). Taking the square root of
the result, and we get almost exactly 300 as our standard deviation. That’s much larger than the 18 extra wins I had. In fact, the chances that you will land within [latex]18/300 = 0.06[/latex]
standard deviations on either side of the Normal’s mean are less than 5%. Before getting too worked up over this result, I decided to take a look at the graph. Using the code:
plot.ts(cumsum(potHistory)[seq(0,1000000,500)],pch=20,col="blue",xlab="step/500",ylab="cumulative earnings")
I got this:
The graph looks pretty much like any random walk, doesn’t it? So the fact that I ended up breaking almost exactly even had to do with the stopping point, not any “unusual” regularity. Just to see if
I might salvage any mystery, I tested the very lowest point, -$2,453, which occurred after 202,133 trails. Even that falls within 2 standard deviations of the expected mean for that number of trials,
and of course cherry picking the most extreme point to stop at isn’t a fair way to go about this. Any last hope that the graph might be unusual? I plotted a couple random walks using numbers
generated in R. Most of them looked like this:
This looks to have the same level of “jaggedness” as the results of my bet on Pi. Unfortunately, I am forced to conclude that the promising strategy of “late number” gambling turned out to be fairly
retarded after all, at least so far as it applies to the digits of Pi.
Tags: betting, late numbers, numeri ritardatari, pi, randomness
The blog entry you quote as your inspiration clearly states that it is rather stupid to bet on a special sequence when the RV are independent.
I don’t see the point… and have you not had a stochastic processes course at university?
Napo I think it is humor he does not think that. Maybe you miss point? I check the code it is good but takes a long time to run.
It seems that you are growing your vectors instead of allocating them first and then assigning to elements. See Chapter 2 at http://www.burns-stat.com/pages/Tutor/R_inferno.pdf for ways to increase
the efficiency of this code.
Also, no reason to do
piVector = unlist(lapply(piVector,as.numeric))
as.numeric() works on vectors, so just change it to
piVector <- as.numeric(piVector)
But really as.integer() might save a bit of memory.
Have you analysed the walk in the frequency domain?
Random walk should give you 1/f^2
Hello! I’m new here, guess I linked in and stumbled upon this (probably rather old) article but I wanted to say that it’s actually somewhat easy to predict pi. It follows something called a repeating
fraction, which is infinitely long but has a definite pattern.
Also, I’d love it if you ran that same experiment for phi and e.
Loved reading you’re work – see you! | {"url":"https://statisticsblog.com/2010/05/31/betting-on-pi/","timestamp":"2024-11-02T07:44:02Z","content_type":"application/xhtml+xml","content_length":"34602","record_id":"<urn:uuid:a034a38f-24df-460a-a456-c6d25c95bfea>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00372.warc.gz"} |
Difference Between Watts And Amps (Explained)
Understanding the difference between watts and amps is crucial when it comes to electrical systems. Watts and amps are both units of measurement used in electricity, but they measure different
aspects. Let’s dive into the details and explore what sets watts and amps apart.
Key Takeaways:
• Watts measure electrical power, while amps measure electrical current.
• Voltage, current, and resistance are the three basic units of electricity.
• Ohm’s law, V = IR, explains the relationship between voltage, current, and resistance.
• Wattage is calculated by multiplying volts by amps and quantifies the rate of energy used by a device.
• Ampage refers to the current that can flow through a wire and is crucial for powering electrical appliances.
Understanding Volts and Voltage
Volts play a crucial role in electrical systems as they are a measure of electrical pressure, also known as electromotive force. Voltage determines the strength of the electrical current flowing
through a wire and is measured in volts. It represents the amount of force required to push charged electrons through a conducting loop. By understanding volts and voltage, we can grasp the concept
of how electrical power is supplied to devices such as lights and appliances.
To calculate voltage, we can refer to Ohm’s law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). This relationship can be summarized by the formula V = IR. By
manipulating this formula, we can determine the voltage required for a specific current and resistance.
For example, let’s consider a light bulb with a resistance of 10 ohms and a desired current of 2 amps. Using Ohm’s law, we can calculate the voltage needed by multiplying the current and resistance:
V = 2 amps * 10 ohms = 20 volts. Therefore, the light bulb needs a voltage of 20 volts to operate properly.
In summary, volts and voltage are essential in understanding how electrical systems function and how power is supplied to various devices. By calculating voltage using Ohm’s law, we can determine the
necessary voltage for a specific current and resistance. This knowledge allows us to ensure the proper functioning of electrical appliances and systems.
Exploring Amps and Current
In the world of electrical systems, amps and current play a crucial role in determining the flow of electrical energy. Amps, short for amperes, are used to measure the rate at which electrical
current flows through an electric circuit. Current, on the other hand, refers to the actual flow of electrons within a circuit. Together, amps and current provide the foundation for powering
electrical devices and ensuring they receive the necessary energy to operate.
When it comes to understanding amps and current, it’s important to remember that current is directly related to the voltage applied across a circuit. The higher the voltage, the more current will
flow through the circuit. This relationship is governed by Ohm’s law, which states that current is equal to the voltage divided by the resistance of the circuit. By manipulating these variables, it
is possible to control the amount of current flowing through a circuit.
To measure amps and current, an ammeter is used. It allows us to determine the amount of electrical energy passing through a wire or circuit. In practical terms, amps are crucial for ensuring that
electrical devices receive the appropriate level of current to operate efficiently and safely. Insufficient amperage can result in blown breakers and inadequate power supply, while excessive amperage
can lead to overheating and potential damage to the device.
The Role of Current in an Electric Circuit
Current is the lifeblood of an electric circuit. It represents the movement of electrons along a conducting path. In a simple analogy, think of current as the flow of water through a pipe. The
greater the current, the more electrons are moving through the circuit, and the more power is being delivered to the connected devices.
It’s important to note that current can be both direct current (DC) and alternating current (AC). DC flows in one direction only, commonly found in batteries and small electronic devices. AC, on the
other hand, periodically changes direction, as seen in our household power outlets. Both types of current have specific applications and must be used appropriately to ensure the safe and efficient
operation of electrical systems.
In conclusion, amps and current are essential concepts in the world of electricity. Understanding how they relate to voltage and resistance allows us to control and optimize the flow of electrical
energy. By ensuring that devices receive the appropriate current, we can power our homes and businesses safely and efficiently.
• Amps and current are crucial for powering electrical devices and ensuring they receive the necessary energy to operate.
• The relationship between amps and current is governed by Ohm’s law, which states that current is equal to the voltage divided by the resistance of the circuit.
• An ammeter is used to measure amps and current, allowing us to determine the amount of electrical energy passing through a wire or circuit.
• Insufficient or excessive amperage can lead to issues such as blown breakers, inadequate power supply, or potential damage to devices.
• Current can be both direct current (DC) and alternating current (AC), with specific applications in different electrical systems.
Understanding Wattage and Electrical Power
When it comes to electrical systems, understanding wattage and electrical power is crucial. Wattage is a measure of electrical power, indicating the rate at which energy is used by a device. It is
calculated by multiplying volts by amps and is measured in watts. Wattage allows us to quantify the output and input of electrical energy, giving us a better understanding of how much power a device
Knowing the wattage of an appliance is important for determining its power consumption and the impact it has on your electricity bills. Appliances with higher wattage, such as air conditioners or
refrigerators, require more energy to operate continuously. On the other hand, appliances with lower wattage, like LED lights or energy-efficient televisions, consume less power and can help reduce
energy consumption.
Appliance Wattage
LED Light Bulb 9 Watts
Hairdryer 1800 Watts
Refrigerator 1200 Watts
Television 200 Watts
Air Conditioner 5000 Watts
As shown in the table above, different appliances have varying wattage levels. By using appliances with lower wattage, you can reduce your energy consumption and ultimately lower your electricity
bills. It is important to choose appliances that balance functionality with energy efficiency, ensuring that you have the power you need while minimizing wastage.
“Understanding wattage and its role in electrical power consumption is essential for making informed decisions about energy usage.”
In conclusion, wattage is a crucial factor in determining the energy consumption of electrical appliances. It allows us to understand the rate at which energy is used and helps us make informed
decisions about our power usage. By choosing appliances with lower wattage and ensuring their energy efficiency, we can reduce power consumption and contribute to a more sustainable future.
The Relationship Between Volts, Amps, and Watts
Understanding the relationship between volts, amps, and watts is crucial in the realm of electrical systems. These three units are interconnected and play a significant role in determining the power
consumption and efficiency of various devices. By grasping this relationship, you can optimize energy usage and make informed decisions when it comes to electrical installations and appliances.
Ohm’s law provides a foundation for understanding this relationship. According to Ohm’s law, voltage (V) is equal to the product of current (I) and resistance (R), represented by the equation V = IR.
This equation demonstrates that voltage is directly proportional to current and resistance. In other words, by increasing the voltage or amperage, the power or wattage also increases.
For example, imagine you have a device that operates at 12 volts and draws 2 amps of current. By multiplying the voltage and current values together (12 volts * 2 amps), you find that the power
consumed by the device is 24 watts. This calculation showcases how the voltage and amperage directly affect the wattage of a device.
Volts (V) Amps (I) Watts (V x I)
As demonstrated in the table, doubling the voltage while keeping the amperage constant quadruples the wattage. Similarly, doubling the amperage while keeping the voltage constant also quadruples the
wattage. This illustrates the direct relationship between volts, amps, and watts, allowing us to calculate power consumption and optimize energy usage.
The Importance of Electrical Efficiency
When it comes to electrical systems, optimizing power usage and reducing energy wastage are of utmost importance. This is where electrical efficiency plays a crucial role. Electrical efficiency
refers to using less energy to power a device or system, ensuring that it converts electrical energy into useful work as efficiently as possible.
To understand electrical efficiency, it’s essential to consider the power formula, I = P/V, where I represents the current, P represents power, and V represents voltage. This formula allows us to
determine the amps required for a specific wattage, helping us gauge the energy consumption of electrical devices accurately.
By focusing on electrical efficiency, we can minimize power wastage and save on electricity bills. It involves using appliances with appropriate wattage and amps, ensuring that they align with the
specific requirements of each device. By doing so, we can reduce energy wastage, decrease our carbon footprint, and contribute to a more sustainable future.
Benefits of Electrical Efficiency
Implementing electrical efficiency measures offers numerous benefits for both individuals and businesses. Some key advantages include:
• Cost savings: By optimizing power usage, electrical efficiency helps lower electricity bills, resulting in long-term cost savings.
• Environmental impact: Minimizing energy wastage reduces the overall environmental impact, helping to conserve resources and reduce greenhouse gas emissions.
• Enhanced device performance: Using appliances with appropriate wattage and amps ensures that they operate at their optimal performance levels, increasing their lifespan and reducing the risk of
• Improved power grid stability: By reducing energy wastage, electrical efficiency contributes to a more stable and reliable power grid, decreasing the likelihood of power outages.
Strategies for Enhancing Electrical Efficiency
To enhance electrical efficiency, consider implementing the following strategies:
1. Choose energy-efficient appliances: Look for devices with high energy efficiency ratings, such as ENERGY STAR certified products.
2. Invest in smart power management systems: Smart power strips and energy monitoring systems can help track and optimize energy usage, reducing wastage.
3. Use energy-saving practices: Simple habits like turning off lights when not in use, utilizing natural light, and adjusting thermostat settings can have a significant impact on energy consumption.
4. Ensure proper insulation: Properly insulating buildings minimizes energy loss and maintains a comfortable indoor environment without relying heavily on heating or cooling systems.
By prioritizing electrical efficiency in our daily lives, we can contribute to a more sustainable future while enjoying the benefits of reduced energy consumption and cost savings.
To summarize, understanding the difference between watts and amps is crucial in the world of electrical systems. Watts measure power, indicating the rate of energy consumption by a device. Amps, on
the other hand, measure current, representing the flow of electrical energy through a circuit. These two units, along with volts, which measure electrical pressure, form the basis of electrical
By grasping the relationship between volts, amps, and watts through concepts like Ohm’s law, one can optimize power consumption and ensure the efficiency of electrical systems. It is important to
choose appliances with appropriate wattage and amps to reduce energy wastage and ultimately save on electricity bills. Additionally, understanding electrical efficiency and striving to use less
energy can contribute to a more sustainable and environmentally friendly approach to power usage.
As you delve into the world of electrical systems, keep in mind the importance of these units of measurement. Whether you are designing an electrical system for your home or business or simply want
to make more informed decisions about your energy consumption, knowing the difference between watts and amps will empower you to make smarter choices. So, remember to consider both the power (watts)
and the current (amps) when dealing with electrical devices and systems.
What is the difference between watts and amps?
Watts measure electrical power, while amps measure electrical current.
What are volts and how do they relate to electrical systems?
Volts are a measure of electrical pressure, also known as electromotive force. Voltage determines the strength of the electrical current flowing through a wire.
What are amps and why are they important in electrical circuits?
Amps, short for amperes, measure the flow of electrical current in an electric circuit. They are essential for ensuring that electrical devices receive the necessary current to operate.
What is wattage and how is it calculated?
Wattage is a measure of electrical power, indicating the rate at which energy is used by a device. It is calculated by multiplying volts by amps.
How are volts, amps, and watts interconnected?
Voltage multiplied by current gives the power in watts. By increasing the voltage or amperage, the power or wattage also increases.
What is electrical efficiency and why is it important?
Electrical efficiency refers to using less energy to power a device or system. It is crucial for optimizing power usage and reducing energy wastage.
Why is it important to understand the difference between watts and amps?
Understanding the difference between watts and amps is essential for anyone working with electrical systems. It helps optimize power consumption and ensures the efficiency of electrical systems. | {"url":"https://tagvault.org/blog/difference-between-watts-and-amps-explained/","timestamp":"2024-11-05T16:35:23Z","content_type":"text/html","content_length":"99767","record_id":"<urn:uuid:32ce14b0-c32e-46e5-93f5-4ccbad733f6c>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00560.warc.gz"} |
A post by Harry Tata, PhD student on the Compass programme.
Oligonucleotides in Medicine
Oligonucleotide therapies are at the forefront of modern pharmaceutical research and development, with recent years seeing major advances in treatments for a variety of conditions. Oligonucleotide
drugs for Duchenne muscular dystrophy (FDA approved) [1], Huntington’s disease (Phase 3 clinical trials) [2], and Alzheimer’s disease [3] and amyotrophic lateral sclerosis (early-phase clinical
trials) [4] show their potential for tackling debilitating and otherwise hard-to-treat conditions. With continuing development of synthetic oligonucleotides, analytical techniques such as mass
spectrometry must be tailored to these molecules and keep pace with the field.
Working in conjunction with AstraZeneca, this project aims to advance methods for impurity detection and quantification in synthetic oligonucleotide mass spectra. In this blog post we apply a
regularised version of the Richardson-Lucy algorithm, an established technique for image deconvolution, to oligonucleotide mass spectrometry data. This allows us to attribute signals in the data to
specific molecular fragments, and therefore to detect impurities in oligonucleotide synthesis.
Oligonucleotide Fragmentation
If we have attempted to synthesise an oligonucleotide $\mathcal O$ with a particular sequence, we can take a sample from this synthesis and analyse it via mass spectrometry. In this process,
molecules in the sample are first fragmented — broken apart into ions — and these charged fragments are then passed through an electromagnetic field. The trajectory of each fragment through this
field depends on its mass/charge ratio (m/z), so measuring these trajectories (e.g. by measuring time of flight before hitting some detector) allows us to calculate the m/z of fragments in the
sample. This gives us a discrete mass spectrum: counts of detected fragments (intensity) across a range of m/z bins [5].
To get an idea of how much of $\mathcal O$ is in a sample, and what impurities might be present, we first need to consider what fragments $\mathcal O$ will produce. Oligonucleotides are short strands
of DNA or RNA; polymers with a backbone of sugars (such as ribose in RNA) connected by linkers (e.g. a phosphodiester bond), where each sugar has an attached base which encodes genetic information
On each monomer, there are two sites where fragmentation is likely to occur: at the linker (backbone cleavage) or between the base and sugar (base loss). Specifically, depending on which bond within
the linker is broken, there are four modes of backbone cleavage [7,8].
We include in $\mathcal F$ every product of a single fragmentation of $\mathcal O$ — any of the four backbone cleavage modes or base loss anywhere along the nucleotide — as well as the results of
every combination of two fragmentations (different cleavage modes at the same linker are mutually exclusive).
Sparse Richardson-Lucy Algorithm
Suppose we have a chemical sample which we have fragmented and analysed by mass spectrometry. This gives us a spectrum across n bins (each bin corresponding to a small m/z range), and we represent
this spectrum with the column vector $\mathbf{b}\in\mathbb R^n$, where $b_i$ is the intensity in the $i^{th}$ bin. For a set $\{f_1,\ldots,f_m\}=\mathcal F$ of possible fragments, let $x_j$ be the
amount of $f_j$ that is actually present. We would like to estimate the amounts of each fragment based on the spectrum $\mathbf b$.
If we had a sample comprising a unit amount of a single fragment $f_j$, so $x_j=1$ and $x_{ke j}=0,$ and this produced a spectrum $\begin{pmatrix}a_{1j}&\ldots&a_{nj}\end{pmatrix}^T$, we can say the
intensity contributed to bin $i$ by $x_j$ is $a_{ij}.$ In mass spectrometry, the intensity in a single bin due to a single fragment is linear in the amount of that fragment, and the intensities in a
single bin due to different fragments are additive, so in some general spectrum we have $b_i=\sum_j x_ja_{ij}.$
By constructing a library matrix $\mathbf{A}\in\mathbb R^{n\times m}$ such that $\{\mathbf A\}_{ij}=a_{ij}$ (so the columns of $\mathbf A$ correspond to fragments in $\mathcal F$), then in ideal
conditions the vector of fragment amounts $\mathbf x=\begin{pmatrix}x_1&\ldots&x_m\end{pmatrix}^T$ solves $\mathbf{Ax}=\mathbf{b}$. In practice this exact solution is not found — due to experimental
noise and potentially because there are contaminant fragments in the sample not included in $\mathcal F$ — and we instead make an estimate $\mathbf {\hat x}$ for which $\mathbf{A\hat x}$ is close to
$\mathbf b$.
Note that the columns of $\mathbf A$ correspond to fragments in $\mathcal F$: the values in a single column represent intensities in each bin due to a single fragment only. We $\ell_1$-normalise
these columns, meaning the total intensity (over all bins) of each fragment in the library matrix is uniform, and so the values in $\mathbf{\hat x}$ can be directly interpreted as relative abundances
of each fragment.
The observed intensities — as counts of fragments incident on each bin — are realisations of latent Poisson random variables. Assuming these variables are i.i.d., it can be shown that the estimate of
$\mathbf{x}$ which maximises the likelihood of the system is approximated by the iterative formula
$\mathbf {\hat{x}}^{(t+1)}=\left(\mathbf A^T \frac{\mathbf b}{\mathbf{A\hat x}^{(t)}}\right)\odot \mathbf{\hat x}^{(t)}.$
Here, quotients and the operator $\odot$ represent (respectively) elementwise division and multiplication of two vectors. This is known as the Richardson-Lucy algorithm [9].
In practice, when we enumerate oligonucleotide fragments to include in $\mathcal F$, most of these fragments will not actually be produced when the oligonucleotide passes through a mass spectrometer;
there is a large space of possible fragments and (beyond knowing what the general fragmentation sites are) no well-established theory allowing us to predict, for a new oligonucleotide, which
fragments will be abundant or negligible. This means we seek a sparse estimate, where most fragment abundances are zero.
The Richardson-Lucy algorithm, as a maximum likelihood estimate for Poisson variables, is analagous to ordinary least squares regression for Gaussian variables. Likewise lasso regression — a
regularised least squares regression which favours sparse estimates, interpretable as a maximum a posteriori estimate with Laplace priors — has an analogue in the sparse Richardson-Lucy algorithm:
$\mathbf {\hat{x}}^{(t+1)}=\left(\mathbf A^T \frac{\mathbf b}{\mathbf{A\hat x}^{(t)}}\right)\odot \frac{ \mathbf{\hat x}^{(t)}}{\mathbf 1 + \lambda},$
where $\lambda$ is a regularisation parameter [10].
Library Generation
For each oligonucleotide fragment $f\in\mathcal F$, we smooth and bin the m/z values of the most abundant isotopes of $f$, and store these values in the columns of $\mathbf A$. However, if these are
the only fragments in $\mathcal F$ then impurities will not be identified: the sparse Richardson-Lucy algorithm will try to fit oligonucleotide fragments to every peak in the spectrum, even ones that
correspond to fragments not from the target oligonucleotide. Therefore we also include ‘dummy’ fragments corresponding to single peaks in the spectrum — the method will fit these to
non-oligonucleotide peaks, showing the locations of any impurities.
For a mass spectrum from a sample containing a synthetic oligonucleotide, we generated a library of oligonucleotide and dummy fragments as described above, and applied the sparse Richardson-Lucy
algorithm. Below, the model fit is plotted alongside the (smoothed, binned) spectrum and the ten most abundant fragments as estimated by the model. These fragments are represented as bars with binned
m/z at the peak fragment intensity, and are separated into oligonucleotide fragments and dummy fragments indicating possible impurities. All intensities and abundances are Anscombe transformed ($x\
rightarrow\sqrt{x+3/8}$) for clarity.
As the oligonucleotide in question is proprietary, its specific composition and fragmentation is not mentioned here, and the bins plotted have been transformed (without changing the shape of the
data) so that individual fragment m/z values are not identifiable.
We see the data is fit extremely closely, and that the spectrum is quite clean: there is one very pronounced peak roughly in the middle of the m/z range. This peak corresponds to one of the
oligonucleotide fragments in the library, although there is also an abundant dummy fragment slightly to the left inside the main peak. Fragment intensities in the library matrix are smoothed, and it
may be the case that the smoothing here is inappropriate for the observed peak, hence other fragments being fit at the peak edge. Investigating these effects is a target for the rest of the project.
We also see several smaller peaks, most of which are modelled with oligonucleotide fragments. One of these peaks, at approximately bin 5352, has a noticeably worse fit if excluding dummy fragments
from the library matrix (see below). Using dummy fragments improves this fit and indicates a possible impurity. Going forward, understanding and quantification of these impurities will be improved by
including other common fragments in the library matrix, and by grouping fragments which correspond to the same molecules.
[1] Junetsu Igarashi, Yasuharu Niwa, and Daisuke Sugiyama. “Research and Development of Oligonucleotide Therapeutics in Japan for Rare Diseases”. In: Future Rare Diseases 2.1 (Mar. 2022), FRD19.
[2] Karishma Dhuri et al. “Antisense Oligonucleotides: An Emerging Area in Drug Discovery and Development”. In: Journal of Clinical Medicine 9.6 (6 June 2020), p. 2004.
[3] Catherine J. Mummery et al. “Tau-Targeting Antisense Oligonucleotide MAPTRx in Mild Alzheimer’s Disease: A Phase 1b, Randomized, Placebo-Controlled Trial”. In: Nature Medicine (Apr. 24, 2023),
pp. 1–11.
[4] Benjamin D. Boros et al. “Antisense Oligonucleotides for the Study and Treatment of ALS”. In: Neurotherapeutics: The Journal of the American Society for Experimental NeuroTherapeutics 19.4 (July
2022), pp. 1145–1158.
[5] Ingvar Eidhammer et al. Computational Methods for Mass Spectrometry Proteomics. John Wiley & Sons, Feb. 28, 2008. 299 pp.
[6] Harri Lönnberg. Chemistry of Nucleic Acids. De Gruyter, Aug. 10, 2020.
[7] S. A. McLuckey, G. J. Van Berkel, and G. L. Glish. “Tandem Mass Spectrometry of Small, Multiply Charged Oligonucleotides”. In: Journal of the American Society for Mass Spectrometry 3.1 (Jan.
1992), pp. 60–70.
[8] Scott A. McLuckey and Sohrab Habibi-Goudarzi. “Decompositions of Multiply Charged Oligonucleotide Anions”. In: Journal of the American Chemical Society 115.25 (Dec. 1, 1993), pp. 12085–12095.
[9] Mario Bertero, Patrizia Boccacci, and Valeria Ruggiero. Inverse Imaging with Poisson Data: From Cells to Galaxies. IOP Publishing, Dec. 1, 2018.
[10] Elad Shaked, Sudipto Dolui, and Oleg V. Michailovich. “Regularized Richardson-Lucy Algorithm for Reconstruction of Poissonian Medical Images”. In: 2011 IEEE International Symposium on Biomedical
Imaging: From Nano to Macro. Mar. 2011, pp. 1754–1757.
Compass students attending AISTATS 2023 in Valencia
We (Ed Davis, Josh Givens, Alex Modell, and Hannah Sansford) attended the 2023 AISTATS conference in Valencia in order to explore the interesting research being presented as well as present some of
our own work. While we talked about our work being published at the conference in this earlier blog post, having now attended the conference, we thought we’d talk about our experience there. We’ll
spotlight some of the talks and posters which interested us most and talk about our highlights of Valencia as a whole.
Talks & Posters
Mode-Seeking Divergences: Theory and Applications to GANs
One especially interesting talk and poster at the conference was presented by Cheuk Ting Li on their work in collaboration with Farzan Farnia. This work aims to set up a formal classification for
various probability measure divergences (such as f-Divergences, Wasserstein distance, etc.) in terms of there mode-seeking or mode-covering properties. By mode-seeking/mode-covering we mean the
behaviour of the divergence when used to fit a unimodal distribution to a multi-model target. Specifically a mode-seeking divergence will encourage the target distribution to fit just one of the
modes ignoring the other while a mode covering divergence will encourage the distribution to cover all modes leading to less accurate fitting on an individual mode but better covering the full
support of the distribution. While these notions of mode-seeking and mode-covering divergences had been discussed before, up to this point there seems to be no formal definition for these properties,
and disagreement on the appropriate categorisation of some divergences. This work presents such a definition and uses it to categorise many of the popular divergence methods. Additionally they show
how an additive combination a mode seeking f-divergence and the 1-Wasserstein distance retain the mode-seeking property of the f-divergence while being implementable using only samples from our
target distribution (rather than knowledge of the distribution itself) making it a desirable divergence for use with GANs.
Talk: https://youtu.be/F7LdHIzZQow
Paper: https://proceedings.mlr.press/v206/ting-li23a.html
Using Sliced Mutual Information to Study Memorization and Generalization in
Deep Neural Networks
The benefit of attending large conferences like AISTATS is having the opportunity to hear talks that are not related to your main research topic. This was the case with a talk by Wongso et. al. was
one such talk. Although it did not overlap with any of our main research areas, we all found this talk very interesting.
The talk was on the topic of tracking memorisation and generalisation in deep neural networks (DNNs) through the use of /sliced mutual information/. Mutual information (MI) is commonly used in
information theory and represents the reduction of uncertainty about one random variable, given the knowledge of the other. However, MI is hard to estimate in high dimensions, which makes it a
prohibitive metric for use in neural networks.
Enter sliced mutual information (SMI). This metric is the average of the MI terms between their one-dimensional projections. The main difference between SMI and MI is that SMI is scalable to high
dimensions and scales faster than MI.
Next, let’s talk about memorisation. Memorisation is known to occur in DNNs and is where the DNN fits random labels in training as it has memorised noisy labels in training, leading to bad
generalisation. The authors demonstrate this behaviour by fitting a multi-layer perceptron to the MNIST dataset with various amounts of label noise. As the noise increased, the difference between the
training and test accuracy became greater.
As the label noise increases, the MI between the features and target variable does not change, meaning that MI did not track the loss in generalisation. However, the authors show that the SMI did
track the generalisation. As the label noise increased, the SMI decreased significantly as the MLP’s generalisation got worse. Their main theorem showed that SMI is lower-bounded by a term which
includes the spherical soft-margin separation, a quantity which is used to track memorisation and generalisation!
In summary, unlike MI, SMI can track memorization and generalisation in DNNs. If you’d like to know more, you can find the full paper here: https://proceedings.mlr.press/v206/wongso23a.html.
Invited Speakers and the Test of Time Award
As well as the talks on papers that had been selected for oral presentation, each day began with a (longer) invited talk which, for many of us, were highlights of each day. The invited speakers were
extremely engaging and covered varied and interesting topics; from Arthur Gretton (UCL) presenting ‘Causal Effect Estimation with Context and Confounders’ to Shakir Mohamed (DeepMind) presenting
‘Elevating our Evaluations: Technical and Sociotechnical Standards of Assessment in Machine Learning’. A favourite amongst us was a talk from Tamara Broderick (MIT) titled ‘An Automatic Finite-Sample
Robustness Check: Can Dropping a Little Data Change Conclusions?’. In this talk she addressed a worry that researchers might have when the goal is to analyse a data sample and apply any conclusions
to a new population: was a small proportion of the data sample instrumental to the original conclusion? Tamara and collorators propose a method to assess the sensitivity of statistical conclusions to
the removal of a very small fraction of the data set. They find that sensitivity is driven by a signal-to-noise ratio in the inference problem, does not disappear asymptotically, and is not decided
by misspecification. In experiments they find that many data analyses are robust, but that the conclusions of severeal influential economics papers can be changed by removing (much) less than 1% of
the data! A link to the talk can be found here: https://youtu.be/QYtIEqlwLHE
This year, AISTATS featured a Test of Time Award to recognise a paper from 10 years ago that has had a prominent impact in the field. It was awarded to Andreas Damianou and Neil Lawrence for the
paper ‘Deep Gaussian Processes’, and their talk was a definite highlight of the conferece. Many of us had seen Neil speak at a seminar at the University of Bristol last year and, being the engaging
speaker he is, we were looking forward to hearing him speak again. Rather than focussing on the technical details of the paper, Neils talk concentrated on his (and the machine learning community’s)
research philosophy in the years preceeding writing the paper, and how the paper came about – a very interesting insight, and a refreshing break from the many technical talks!
There was so much to like about Valencia even from our short stay there. We’ll try and give you a very brief highlight of our favourite things.
Food & Drink:
Obviously Valencia is renowned for being the birthplace of paella and while the paella was good we sampled many other delights our stay. Our collective highlight was the nicest Burrata any of us had
ever had which, in a stunning display of individualism, all four of us decided to get on our first day at the conference.
Artist rendition of our 4 meals.
About half an hours tram ride from the conference centre are the beaches of Valencia. These stretch for miles as well as having a good 100m in depth with (surprisingly hot) sand covering the lot. We
visited these after the end of the conference on the Thursday and despite it being the only cloudy day of the week it was a perfect way to relax at the end of a hectic few days with the pleasantly
temperate water being an added bonus.
Valencia has so much interesting architecture scattered around the city centre. One of the most memorable remarkable places was the San Nicolás de Bari and San Pedro Mártir (Church of San Nicolás)
which is known as the Sistine chapel of Valencia (according to the audio-guide for the church at least) with its incredible painted ceiling and live organ playing.
Ceiling of the Church of San Nicolás
Student Perspectives: Intro to Recommendation Systems
A post by Hannah Sansford, PhD student on the Compass programme.
Like many others, I interact with recommendation systems on a daily basis; from which toaster to buy on Amazon, to which hotel to book on booking.com, to which song to add to a playlist on Spotify.
They are everywhere. But what is really going on behind the scenes?
Recommendation systems broadly fit into two main categories:
1) Content-based filtering. This approach uses the similarity between items to recommend items similar to what the user already likes. For instance, if Ed watches two hair tutorial videos, the system
can recommend more hair tutorials to Ed.
2) Collaborative filtering. This approach uses the the similarity between users’ past behaviour to provide recommendations. So, if Ed has watched similar videos to Ben in the past, and Ben likes a
cute cat video, then the system can recommend the cute cat video to Ed (even if Ed hasn’t seen any cute cat videos).
Both systems aim to map each item and each user to an embedding vector in a common low-dimensional embedding space $E = \mathbb{R}^d$. That is, the dimension of the embeddings ($d$) is much smaller
than the number of items or users. The hope is that the position of these embeddings captures some of the latent (hidden) structure of the items/users, and so similar items end up ‘close together’ in
the embedding space. What is meant by being ‘close’ may be specified by some similarity measure.
Collaborative filtering
In this blog post we will focus on the collaborative filtering system. We can break it down further depending on the type of data we have:
1) Explicit feedback data: aims to model relationships using explicit data such as user-item (numerical) ratings.
2) Implicit feedback data: analyses relationships using implicit signals such as clicks, page views, purchases, or music streaming play counts. This approach makes the assumption that: if a user
listens to a song, for example, they must like it.
The majority of the data on the web comes from implicit feedback data, hence there is a strong demand for recommendation systems that take this form of data as input. Furthermore, this form of data
can be collected at a much larger scale and without the need for users to provide any extra input. The rest of this blog post will assume we are working with implicit feedback data.
Problem Setup
Suppose we have a group of $n$ users $U = (u_1, \ldots, u_n)$ and a group of $m$ items $I = (i_1, \ldots, i_m)$. Then we let $\mathbf{R} \in \mathbb{R}^{n \times m}$ be the ratings matrix where
position $R_{ui}$ represents whether user $u$ interacts with item $i$. Note that, in most cases the matrix $\mathbf{R}$ is very sparse, since most users only interact with a small subset of the full
item set $I$. For any items $i$ that user $u$ does not interact with, we set $R_{ui}$ equal to zero. To be clear, a value of zero does not imply the user does not like the item, but that they have
not interacted with it. The final goal of the recommendation system is to find the best recommendations for each user of items they have not yet interacted with.
Matrix Factorisation (MF)
A simple model for finding user emdeddings, $\mathbf{X} \in \mathbb{R}^{n \times d}$, and item embeddings, $\mathbf{Y} \in \mathbb{R}^{m \times d}$, is Matrix Factorisation. The idea is to find
low-rank embeddings such that the product $\mathbf{XY}^\top$ is a good approximation to the ratings matrix $\mathbf{R}$ by minimising some loss function on the known ratings.
A natural loss function to use would be the squared loss, i.e.
$L(\mathbf{X}, \mathbf{Y}) = \sum_{u, i} \left(R_{ui} - \langle X_u, Y_i \rangle \right)^2.$
This corresponds to minimising the Frobenius distance between $\mathbf{R}$ and its approximation $\mathbf{XY}^\top$, and can be solved easily using the singular value decomposition $\mathbf{R} = \
mathbf{U S V}^\top$.
Once we have our embeddings $\mathbf{X}$ and $\mathbf{Y}$, we can look at the row of $\mathbf{XY}^\top$ corresponding to user $u$ and recommend the items corresponding to the highest values (that
they haven’t already interacted with).
Logistic MF
Minimising the loss function in the previous section is equivalent to modelling the probability that user $u$ interacts with item $i$ as the inner product $\langle X_u, Y_i \rangle$, i.e.
$R_{ui} \sim \text{Bernoulli}(\langle X_u, Y_i \rangle),$
and maximising the likelihood over $\mathbf{X}$ and $\mathbf{Y}$.
In a research paper from Spotify [3], this relationship is instead modelled according to a logistic function parameterised by the sum of the inner product above and user and item bias terms, $\
beta_u$ and $\beta_i$,
$R_{ui} \sim \text{Bernoulli} \left( \frac{\exp(\langle X_u, Y_i \rangle + \beta_u + \beta_i)}{1 + \exp(\langle X_u, Y_i \rangle + \beta_u + \beta_i)} \right).$
Relation to my research
A recent influential paper [1] proved an impossibility result for modelling certain properties of networks using a low-dimensional inner product model. In my 2023 AISTATS publication [2] we show that
using a kernel, such as the logistic one in the previous section, to model probabilities we can capture these properties with embeddings lying on a low-dimensional manifold embedded in
infinite-dimensional space. This has various implications, and could explain part of the success of Spotify’s logistic kernel in producing good recommendations.
[1] Seshadhri, C., Sharma, A., Stolman, A., and Goel, A. (2020). The impossibility of low-rank representations for triangle-rich complex networks. Proceedings of the National Academy of Sciences, 117
[2] Sansford, H., Modell, A., Whiteley, N., and Rubin-Delanchy, P. (2023). Implications of sparsity and high triangle density for graph representation learning. Proceedings of The 26th International
Conference on Artificial Intelligence and Statistics, PMLR 206:5449-5473.
[3] Johnson, C. C. (2014). Logistic matrix factorization for implicit feedback data. Advances in Neural Information Processing Systems, 27(78):1–9.
Compass students at AISTATS 2023
Congratulations to Compass students Josh Givens, Hannah Sansford and Alex Modell who, along with their supervisors have had their papers accepted to be published at AISTATS 2023.
‘Implications of sparsity and high triangle density for graph representation learning’
Hannah Sansford, Alexander Modell, Nick Whiteley, Patrick Rubin-Delanchy
Hannah: In this paper we explore the implications of two common characteristics of real-world networks, sparsity and triangle-density, for graph representation learning. An example of where these
properties arise in the real-world is in social networks, where, although the number of connections each individual has compared to the size of the network is small (sparsity), often a friend of a
friend is also a friend (triangle-density). Our result counters a recent influential paper that shows the impossibility of simultaneously recovering these properties with finite-dimensional
representations of the nodes, when the probability of connection is modelled by the inner-product. We, by contrast, show that it is possible to recover these properties using an infinite-dimensional
inner-product model, where representations lie on a low-dimensional manifold. One of the implications of this work is that we can ‘zoom-in’ to local neighbourhoods of the network, where a
lower-dimensional representation is possible.
The paper has been selected for oral presentation at the conference in Valencia (<2% of submissions).
Density Ratio Estimation and Neyman Pearson Classification with Missing Data
Josh Givens, Song Liu, Henry W J Reeve
Josh: In our paper we adapt the popular density ratio estimation procedure KLIEP to make it robust to missing not at random (MNAR) data and demonstrate its efficacy in Neyman-Pearson (NP)
classification. Density ratio estimation (DRE) aims to characterise the difference between two classes of data by estimating the ratio between their probability densities. The density ratio is a
fundamental quantity in statistics appearing in many settings such as classification, GANs, and covariate shift making its estimation a valuable goal. To our knowledge there is no prior research into
DRE with MNAR data, a missing data paradigm where the likelihood of an observation being missing depends on its underlying value. We propose the estimator M-KLIEP and provide finite sample bounds on
its accuracy which we show to be minimax optimal for MNAR data. To demonstrate the utility of this estimator we apply it the the field of NP classification. In NP classification we aim to create a
classifier which strictly controls the probability of incorrectly classifying points from one class. This is useful in any setting where misclassification for one class is much worse than the other
such as fault detection on a production line where you would want to strictly control the probability of classifying a faulty item as non-faulty. In addition to showing the efficacy of our new
estimator in this setting we also provide an adaptation to NP classification which allows it to still control this misclassification probability even when fit using MNAR data.
Student Perspectives: An Introduction to Stochastic Gradient Methods
A post by Ettore Fincato, PhD student on the Compass programme.
This post provides an introduction to Gradient Methods in Stochastic Optimisation. This class of algorithms is the foundation of my current research work with Prof. Christophe Andrieu and Dr. Mathieu
Gerber, and finds applications in a great variety of topics, such as regression estimation, support vector machines, convolutional neural networks.
We can see below a simulation by Emilien Dupont (https://emiliendupont.github.io/) which represents two trajectories of an optimisation process of a time-varying function. This well describes the
main idea behind the algorithms we will be looking at, that is, using the (stochastic) gradient of a (random) function to iteratively reach the optimum.
Stochastic Optimisation
Stochastic optimisation was introduced by [1], and its aim is to find a scheme for solving equations of the form $abla_w g(w)=0$ given “noisy” measurements of $g$ [2].
In the simplest deterministic framework, one can fully determine the analytical form of $g(w)$, knows that it is differentiable and admits an unique minimum – hence the problem
$w_*=\underset{w}{\text{argmin}}\quad g(w)$
is well defined and solved by $abla_w g(w)=0$.
On the other hand, one may not be able to fully determine $g(w)$ because his experiment is corrupted by a random noise. In such cases, it is common to identify this noise with a random variable, say
$V$, consider an unbiased estimator $\eta(w,V)$ s.t. $\mathbb{E}_V[\eta(w,V)]=g(w)$ and to rewrite the problem as
Compass CDT is recruiting for its final few fully funded places to start September 2023
We are happy to announce our upcoming applications deadline of 16 March 2023 for Compass CDT programme. For international applications there are limited scholarship funded places available for this
final recruitment round. Early applications advised.
Compass is offering specific projects for PhD students to study from Sept 2023. The projects are listed in the research section of our website. All the supervisors listed are open to discussion on
the projects provided and can also talk to applicants about other project ideas. Please provide a ranked list of 3 projects of interest: 1 = project of highest interest. Project supervisors will be
happy to respond to specific questions you have after reading the proposals. Applicants should contact them by email if they wish beforehand.
Also, we are pleased to announce a new project Genetic Similarity Based Cohort Building that has been added to the list for September 2023 start funded by Roche, one of the world’s largest biotech
companies, as well as a leading provider of in-vitro diagnostics and a global supplier of transformative innovative solutions across major disease areas.
PhD Project Allocation Process
Application forms will be reviewed based on the 3 ranked projects specified or other proposed topic. Successful applicants will be invited to attend an interview with the Compass admissions tutors
and the specific project supervisor. If you are made an offer of PhD study it will be published through the online application system. You will then have 2 weeks to consider the offer before deciding
whether to accept or decline.
We welcome applications from all members of our community and are particularly encouraging those from diverse groups, such as members of the LGBT+ and black, Asian and minority ethnic communities, to
join us.
16 March 2023, 23:59 (London UK time zone)
Advantages of being a Compass Student
• Stipend – a generous stipend of £21,668 pa tax free, paid in monthly payments. Plus your own expense budget of £1,000 pa towards travel and research activity.
• No fees – all tuition fees are covered by the EPSRC and University of Bristol.
• Bespoke training – first year units are designed specifically for the academic needs of each Compass student, which enables students to develop knowledge and capability to pursue
cross-disciplinary PhD research.
• Supervisors – supervisors from across academic disciplines offer a range of research projects.
• Cohort – Compass students benefit from dedicated offices and collaboration spaces, enabling strong cohort links and opportunities for shared learning and research.
About Compass CDT
A 4-year bespoke PhD training programme in the statistical and computational techniques of data science, with partners from across the University of Bristol, industry and government agencies.
The cross-disciplinary programme offers exciting collaborations across medicine, computer science, geography, economics, life and earth sciences, as well as with our external partners who range from
government organisations such as the Office for National Statistics, NCSC and the AWE, to industrial partners such as LV, Improbable, IBM Research, EDF, and AstraZeneca.
Students are co-located with the Institute for Statistical Science in the School of Mathematics, which occupies the Fry Building.
Hear from our students about their experience with the programme
• Compass has allowed me to advance my statistical knowledge and apply it to a range of exciting applied projects, as well as develop skills that I’m confident will be highly useful for a future
career in data science. – Shannon, Cohort 2
• With the Compass CDT I feel part of a friendly, interactive environment that is preparing me for whatever I move on to next, whether it be in Academia or Industry. – Sam, Cohort 2
• An incredible opportunity to learn the ever-expanding field of data science, statistics and machine learning amongst amazing people. – Danny, Cohort 1
Compass CDT Video
Find out more about what it means to be a part of the Compass programme from our students in this short video.
16 March 2023, 23:59 (London UK time zone)
Compass at NeurIPS 2022
A post by Anthony Stephenson, Jack Simons, and Dan Ward, PhD students on the Compass programme.
Ant Stephenson, Jack Simons, and I (Dan Ward) had the pleasure of attending the 2022 Conference on Neural Information Processing Systems (NeurIPS), one of the largest machine learning conferences in
the world. The conference was held in New Orleans, which gave us an opportunity to explore a lively city full of culture with delicious local cuisine. We thought we’d collaborate on a blog post
together covering some of the highlights.
Memorable Talks
The conference had broad range of talks including technical presentations of research, applied projects, and discussions of the philosophical and ethical questions that arise in AI. To give a taste
of some of the talks, we picked out some of our favourites below.
Noam Brown: Human Modelling and Strategic Reasoning in the Game of Diplomacy.
The Game of Diplomacy is a strategic board game invented in 1954. It’s unique feature, and of crucial importance of the game, is that players interact via natural language to form allegiances. Whilst
AI has been successful in beating humans in many purely adversarial games (e.g. Chess, Go), this collaborative element poses unique challenges. Firstly, it isn’t obvious how to evaluate/devise
strategies for collaboration/betrayal, especially in the self-play-based reinforcement learning paradigm. Secondly, as communication happens via natural language, the AI must be able to translate
their strategic plan into text. This strange combination of problems lead to interesting and innovative solutions. Paper link here.
Geoffrey Hinton: The Forward-Forward Algorithm for Training Deep Neural Networks.
Among the great line-up of speakers was Professor Geoffrey Hinton, known for popularising backpropogation for deep neural networks. Inspired by producing a more biologically plausible algorithm for
learning, he has proposed the ‘Forward-Forward’ algorithm which he claims can also explain the phenomena of sleep! Professor Geoffrey Hinton then went on to express his belief that using
biologically-inspired hardware, so-called neuromorphic computing, may play a key role in advancing AI. The talk was certainly unconventional, but nevertheless entertaining. Paper link here.
David Chalmers: Could a Large Language Model be Conscious?
Amongst all the machine-learning experts was David Chalmers, a philosopher! There are important questions regarding the possibility that language models might be conscious. David Chalmers aimed to
educate the machine-learning audience in attendance of how we can better think about these problems and re-phrase the questions that we’re asking. We concluded that these questions are,
unsurprisingly, best left to philosophers!
Poster Sessions
Jack and Dan:
I (Dan), presented a poster of my work at the conference, on Robust Neural Posterior Estimation (paper link here). I was definitely surprised by the scale of the poster sessions, and the broad scope
of all the work taking place. Below is some of the posters that me and Jack found interesting:
Contrastive Neural Ratio Estimation
Benjamin K. Miller · Christoph Weniger · Patrick Forré
Authors propose NRE-C which aims to generalise NRE-A (Hermans et al. (2019)) and NRE-B (Durkan et al. (2020)) into one method. NRE-C can recover both methods by taking their two introduced
hyperparameters at certain limits. Paper link here.
Towards Reliable Simulation-Based Inference with Balanced Neural Ratio Estimation
Arnaud Delaunoy · Joeri Hermans · François Rozet · Antoine Wehenkel · Gilles Louppe
These authors also make a contribution to the field of neural ratio estimation in the simulation-based inference context. Authors propose the notion of a “balanced” classifier, which is a classifier
in which the average output from the classifier over the positive data class plus the average output over the negative data class equals to 1. The authors argue that if one has a classifier is
balanced then it will lead to more conservative posterior estimates, which is something which practitioners seek. To integrate this into an algorithm they suggest adding a penalisation term onto the
standard logistic-loss which punishes classifiers as they become less balanced. Paper link here.
Training and Inference on Any-Order Autoregressive Models the Right Way
Andy Shih · Dorsa Sadigh · Stefano Ermon
A joint distribution can be decomposed into its univariate conditionals by the chain rule, although by doing so we implicitly choose an ordering in a model, which prevents arbitrary conditional
inference. Any-order autoregressive models circumvent this generally by being trained such that all possible univariate conditionals are considered, but this leads to learning redundant information.
The paper proposes a new method to train autoregressive models, using a subset of univariate conditionals that still supports arbitrary conditional inference. This research was also presented as a
talk, but sadly we missed it! Paper link here.
The poster sessions formed the bulk of the conference timetable, with 2 2-hour sessions per day, on Tuesday, Wednesday and Thursday. These were very busy, with many posters on a wide-range of topics
and a large congregation of attendees. As a result, it was sometimes difficult to track down the subset of posters on material of particular interest and when this feat was achieved, on occasion it
was still hard work to actually have a detailed conversation with the author(s). Nonetheless, it was interesting to see the how varied the subjects of the poster were and in addition get a feeling
for “themes” of the conference: recurring, clearly in-vogue topics. Amongst the sea of posters, I did manage to find a number relating to GPs; of these, those I found most interesting were:
Posterior and Computational Uncertainty in Gaussian Processes:
Jonathan Wenger · Geoff Pleiss · Marvin Pförtner · Philipp Hennig · John Cunningham
Here the authors propose a way to naturally incorporate uncertainty introduced from the use of (iterative) GP approximation methods. Paper link here.
Sparse Gaussian Process Hyperparameters: Optimize or Integrate?
Vidhi Lalchand · Wessel Bruinsma · David Burt · Carl Edward Rasmussen
The authors attempt to integrate a fully-Bayesian inference procedure for sparse GPs, as an alternative to the commonly adopted approach of optimising the kernel hyperparameters by maximum likelihood
estimation. Paper link here.
Log-Linear-Time Gaussian Processes Using Binary Tree Kernels:
Michael K. Cohen · Samuel Daulton · Michael A Osborne
The idea here feels a bit unorthodox; they use a “binary-tree” kernel which discretises the space, with quantization error determined by the number of leaves. This would seem to lose interpretability
on the properties of the function prior (e.g. smoothness), but does appear to give empirical benefits in their experiments. Paper link here.
In addition to the main conference, on the Friday and Saturday at the end of the week there were a selection of workshops on a variety of sub-fields within machine learning. If you are fortunate
enough for there to be a workshop dedicated to your research area, then they provide a space to gather people with research directly relevant to your own and facilitate helpful discussions and
networking opportunities.
For me, the “Gaussian processes, spatiotemporal modeling and decision-making systems” workshop was the most useful part of the conference. It gave me the chance to speak to people working on
interesting problems related to my own; discover the kind of directions they are heading in and lines of work they are contemplating. Additionally, I presented a poster during this workshop which
allowed me to discuss my work with an audience well-versed on the topic and its possible significance.
The Big Easy
In addition to the actual conference, attending NeurIPS also gave us the opportunity to explore the city of New Orleans; aka The Big Easy. Upon arrival, we were immediately greeted in the airport by
the sound of Louis Armstrong, a strong theme in the city, which features a park named after him. New ‘Awlins’ is well known for its jazz, but awareness of this fact does not necessarily prepare you
for the sheer quantity, especially in the streets of the French Quarter, that awaits you. The real epicentre of jazz in the city is situated on Frenchmen street, on which a swathe of bars hosting
nearly-nightly live music reside. We spent several evenings there, including one of particular note, where French president Emmanuel Macron suddenly appeared, trailed by an extensive retinue of
blue-suited aides and bodyguards. Another street in New Orleans infamous for its nightlife is Bourbon street. Where Frenchmen street is focused on jazz, Bourbon street contains all manner of rowdy
madness, assaulting your senses with noise, smells and sights as soon as you arrive. Both are necessary experiences when visiting The Big Easy.
All in all, the conference was a great opportunity to get a taste of the massive array of research that occurs in machine learning. We were all surprised by the scope of the research topics and
talks, and enjoyed the opportunity to explore a new culture and city.
Compass student publishes article in Frontiers
Compass student Dan Milner and his academic supervisors have published an article in Frontiers, one of the most cited and largest research publishers in the world. Dan’s work is funded in
collaboration with ILRI (International Livestock Research Institute). (more…)
Compass news round-up 2021
As we start 2022, we look back at our Compass achievements over 2021…
Invited speakers and seminars
Over the course of the year we invited seminar speakers Ingmar Schuster on kernel methods, Nicolas Chopin offered a two-part lecture on sequential Monte Carlo samplers, Ioannis Kosmidis on reducing
bias in estimation and a special two-part lecture from Barnett Award winning Jonty Rougier on Wilcoxon’s Two Sample Test.
Compass student launches PAI-Link
In May, Compass PhD student, Mauro Camara Escudero, set up PAI-Link: a nation-wide AI postgraduate seminar series.
Last year also saw the launch of our DataScience@work seminar series, at which we had 5 external organisations speak (Adarga, CheckRisk, Shell, IBM Research and Improbable) and the British Geological
Survey opened this academic year’s seminar series with a talk from alumna Dr Kathryn Leeming.
Training and internships
We ran training sessions on themes such as interdisciplinary research, responsible innovation and a blog post. Compass held its first Science Focus Lab on multi-omics data and cancer treatment with
colleagues from Bristol Integrative Epidemiology unit.
Five Compass students were recruited to internships with organisations such as Microsoft Research, Adarga, CheckRisk, Afiniti and Shell.
The Student Perspectives blog series started up last year with Three Days in the Life of a Silicon Valley Start-up. This student-authored series explored topics such as air pollution in Bristol, the
Michael Whitehouse in Sky News article
approaches of frequentists and Bayesians, and how to generalise kernel methods to probability distributions.
Michael Whitehouse contributed to a Sky News report on the potential impact of the pandemic on the Tokyo Olympics by modelling the rise of COVID-19 cases in Japan.
Access to Data Science
Compass ran its first Access to Data Science event – an immersive experience for prospective PhD students which aimed to increase diversity amongst data science researchers by encouraging
participants such as women and members of the LGBTQ+ and BAME communities to join us.
Research and studentships
Our second cohort of students selected their mini-projects (a precursor to their PhD research) and our third cohort of students joined the Compass programme in September 2021.
Compass Cohort 3 students
Annie Gray presented her paper ‘Matrix factorisation and the interpretation of geodesic distance’ at NeurIPS 2021. Conor Newton gave a talk at a workshop in conjunction with ACM Sigmetrics 2021 and
he and Dom Owens won the poster session of the Fry Statistics Conference. Jack Simons paper ‘Variational Likelihood-Free Gradient Descent’ was accepted at AABI 2022. Alex Modell’s paper ‘A Graph
Embedding Approach to User Behavior Anomaly Detection’ was accepted to IEEE Big Data Conference 2021. Danny Williams and supervisor Song Liu were awarded an EPSRC Impact Acceleration Account for
their project in collaboration with Adarga.
We also created links with new industrial partners – AstraZeneca, ILRI and EDF – who are each sponsoring Compass PhD projects for the following students: Harry Tata, Dan Milner, and Ben Griffiths and
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Dynamic Machine Learning with Least Square Objectives
2019 Theses Doctoral
Dynamic Machine Learning with Least Square Objectives
As of the writing of this thesis, machine learning has become one of the most active research fields. The interest comes from a variety of disciplines which include computer science, statistics,
engineering, and medicine. The main idea behind learning from data is that, when an analytical model explaining the observations is hard to find ---often in contrast to the models in physics such as
Newton's laws--- a statistical approach can be taken where one or more candidate models are tuned using data.
Since the early 2000's this challenge has grown in two ways: (i) The amount of collected data has seen a massive growth due to the proliferation of digital media, and (ii) the data has become more
complex. One example for the latter is the high dimensional datasets, which can for example correspond to dyadic interactions between two large groups (such as customer and product information a
retailer collects), or to high resolution image/video recordings.
Another important issue is the study of dynamic data, which exhibits dependence on time. Virtually all datasets fall into this category as all data collection is performed over time, however I use
the term dynamic to hint at a system with an explicit temporal dependence. A traditional example is target tracking from signal processing literature. Here the position of a target is modeled using
Newton's laws of motion, which relates it to time via the target's velocity and acceleration.
Dynamic data, as I defined above, poses two important challenges. Firstly, the learning setup is different from the standard theoretical learning setup, also known as Probably Approximately Correct
(PAC) learning. To derive PAC learning bounds one assumes a collection of data points sampled independently and identically from a distribution which generates the data. On the other hand, dynamic
systems produce correlated outputs. The learning systems we use should accordingly take this difference into consideration. Secondly, as the system is dynamic, it might be necessary to perform the
learning online. In this case the learning has to be done in a single pass. Typical applications include target tracking and electricity usage forecasting.
In this thesis I investigate several important dynamic and online learning problems, where I develop novel tools to address the shortcomings of the previous solutions in the literature. The work is
divided into three parts for convenience. The first part is about matrix factorization for time series analysis which is further divided into two chapters. In the first chapter, matrix factorization
is used within a Bayesian framework to model time-varying dyadic interactions, with examples in predicting user-movie ratings and stock prices. In the next chapter, a matrix factorization which uses
autoregressive models to forecast future values of multivariate time series is proposed, with applications in predicting electricity usage and traffic conditions. Inspired by the machinery we use in
the first part, the second part is about nonlinear Kalman filtering, where a hidden state is estimated over time given observations. The nonlinearity of the system generating the observations is the
main challenge here, where a divergence minimization approach is used to unify the seemingly unrelated methods in the literature, and propose new ones. This has applications in target tracking and
options pricing. The third and last part is about cost sensitive learning, where a novel method for maximizing area under receiver operating characteristics curve is proposed. Our method has
theoretical guarantees and favorable sample complexity. The method is tested on a variety of benchmark datasets, and also has applications in online advertising.
• Gultekin_columbia_0054D_15172.pdf application/pdf 1.85 MB Download File
More About This Work
Academic Units
Thesis Advisors
Paisley, John W
Ph.D., Columbia University
Published Here
April 24, 2019 | {"url":"https://academiccommons.columbia.edu/doi/10.7916/d8-nwa1-qk64","timestamp":"2024-11-03T06:51:07Z","content_type":"text/html","content_length":"26360","record_id":"<urn:uuid:cbcf39f7-f1d1-4324-a2fd-265bdeb91ab1>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00105.warc.gz"} |
Multiplication tables
The word “multiplication tables” brings back memories of only one thing from my childhood – “mugging”. As a child I simply had to learn multiplication tables. I couldn’t ask for a reason why. My
parents would explain to me the benefits of learning tables by-heart. They would even give me sufficient time to learn it. Figure out ways to learn it. But when the exam time came, I better have
learned it by-heart already. There was no way around it. My school, the neighbourhood with other kids everybody basically reinforced the need-to-learn-multiplication-tables-by-heart ™. So I would
spend days and months learning it by-heart.
One onezaar one
One twozaar two
Eight onezaar eight
Eight twozaar sixteen
Eight threezaar twenty-four
“I was mostly in college when I realised that it is not “eight eightzaar sixty-four”. It actually is “eight eights are sixty four.” I never bothered to question “zaar” as a child. Because
questioning anything was a big big big big risk. What if my questions pissed off my teachers?”
As a kid, I did not like it at all. Now as an adult I know that multiplication tables are useful. Although I hardly use tables beyond 10 for mental math. Calculators are always around so performing
multiplication is not an issue.
My son turned 8 this year. If he was in a regular school, he would be in 3rd standard. Many of his friends in the neighbourhood are ramping up on math-tables. So Advay would come home asking about it
and slightly wanting to learn it.
Nandini & I tried the standard way to get him to learn tables. Advay did not like that approach. He continued to enjoy multiplying each time and writing the answer rather than just memorising it. For
him – “multiplication must be performed, not memorised.” We had to come up with a reason, a purpose for having him memorise math tables. So, we came up with one idea.
I setup a puzzle for him at home. All over the house, I put up post-it-notes with numbers that show up in the math-tables for 3, 5 & 7. The post-it-notes were pasted on doors, window frames, switch
boards, desks, chairs tub-seats, kitchen racks, washing machine, fridge etc. I then put up a fill in the blanks on the green board at home.
With this, I setup a treasure hunt game for him. The game was to go around, pick up post-it-notes and paste them next to the appropriate place on the green board. Advay simply loved the idea.
He ran around the house and harvested the post-it-notes and properly pasted them next to the numbers. He simply loved the whole game.
While he was able to quickly “hunt” for numbers against 3 and 5 tables, he was a bit slow with 7 tables. We told him – “Ok, tomorrow we will setup treasure hunt for 6, 8 & 11. If you memorise them
before the game begins, you will be able to finish it quickly.”
Suddenly, it seemed like Advay found his purpose to memorise the tables. But he still did not want to give up his “multiplication must be performed, not memorised” motto. So he sat with Nandini and
worked out the math tables for 6, 8 & 11. They worked out the answers, ways to memorise them and also the reason for memorising them. He was able to recall 11 tables easily, because he identified a
pattern. 6 & 8 was a bit of struggle. While he did commit the tables to memory, he did not do it by “mugging”. The tables meant logic, math and pattern to him. It was not abstract sounds like “eight
eightzaar sixty-four.”
It appeared like he had made some progress. The time to test came. I setup the board for home once again. This time Nandini suggested Advay to not pick post-it-notes in sequence, but to pick them as
and when he found it and paste them against the right place on the board.
Whenever he found numbers that he couldn’t immediately figure out its place for on the board, he pasted them temporarily on the chair in front of the board. And they would eventually go up on the
Finally he was done with the tables.
He seems to enjoy playing this game a lot. He keeps asking me to set it up for him every evening. Due to work, I was unable to set it up for him for a few days. And I started to notice how much he
was missing playing it.
Yesterday, I setup treasure hunting game for 4, 9 & 13. But this time, I did not give him fill in the blanks in sequence. Some of them were in random order. But he still managed to complete the
treasure-hunt-puzzle properly.
The last two times he played this game, he noticed something interesting. He said – “Appa, some numbers show up in many tables, right?”
Now that we have 90 post it notes with numbers for 3, 4, 5, 6, 7, 8, 9, 11 & 13 tables; we began observing patterns.
• Why does 30 show up in 3, 5 and 6 tables?
• How come many numbers from 3 tables also show up in 6 and 9 tables?
• Oh, 3×5 and 5×3 are the same?
• How come 3, 7, 11 & 13 shows up only once?
I asked Advay to paste all post-it-notes in a collage on the board.
Next, I told him to arrange numbers such that all 24s went into one group, all 35s into one group and so on.
From the board I then asked him to take away notes that had numbers showing up only once. A lot of clutter disappeared from the board.
Then I explained to him that 6 is both 3×2 and 6×1. And that 9 is both 3×3 and 9×1. The idea was to figure out the tables in which a post-it-note-number would show up among the tables we played a
puzzle game (which was: 3, 4, 5, 6, 7, 8, 9, 11 & 13). So, even though 6 is 2×3, we won’t write that on the board because we never played treasure-hunting game for 2 tables.
After solving for numbers 6 and 9, I placed two 27s below it and asked Advay to figure out the tables in which 27 would show up. He got it and started writing out the answer.
And slowly, he figured it out for all notes.
Finally it was all done.
While playing the game he figured that if AxB is C, then BxA is also C. For many numbers he seemed to intuitively know which tables they are likely to show up in.
Let’s talk about results: Has he become perfect in math tables? Not yet. He has gotten better for sure. But more importantly, he has genuinely begun to enjoy multiplication in general & learning the
tables in particular. To Nandini & me: that counts! | {"url":"https://www.prashanthudupa.com/index.php/multiplication-tables/","timestamp":"2024-11-04T02:11:04Z","content_type":"text/html","content_length":"83339","record_id":"<urn:uuid:1fcbbca9-434b-4ad6-9d7a-0389b0420010>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00281.warc.gz"} |
The Law of Requisite Variety
Control or regulation is most fundamentally formulated as a reduction of variety: perturbations with high variety affect the system's internal state, which should be kept as close as possible to the
goal state, and therefore exhibit a low variety. So in a sense control prevents the transmission of variety from environment to system. This is the opposite of information transmission, where the
purpose is to maximally conserve variety.
In active (feedforward and/or feedback) regulation, each disturbance D will have to be compensated by an appropriate counteraction from the regulator R. If R would react in the same way to two
different disturbances, then the result would be two different values for the essential variables, and thus imperfect regulation. This means that if we wish to completely block the effect of D, the
regulator must be able to produce at least as many counteractions as there are disturbances in D. Therefore, the variety of R must be at least as great as the variety of D. If we moreover take into
account the constant reduction of variety K due to buffering, the principle can be stated more precisely as:
V(E) ³ V(D) - V(R) - K
Ashby has called this principle the law of requisite variety: in active regulation only variety can destroy variety. It leads to the somewhat counterintuitive observation that the regulator must have
a sufficiently large variety of actions in order to ensure a sufficiently small variety of outcomes in the essential variables E. This principle has important implications for practical situations:
since the variety of perturbations a system can potentially be confronted with is unlimited, we should always try maximize its internal variety (or diversity), so as to be optimally prepared for any
foreseeable or unforeseeable contigency.
Some Comments
Ashby's Law can be seen as an application of the principle of selective variety. However, a frequently cited stronger formulation of Ashby's Law, "the variety in the control system must be equal to
or larger than the variety of the perturbations in order to achieve control", which ignores the constant factor K, does not hold in general. Indeed, the underlying "only variety can destroy variety"
assumption is in contradiction with the principle of asymmetric transitions which implies that spontaneous decrease of variety is possible (which is precisely what buffering does). For example, a
bacterium searching for food and avoiding poisons has a minimal variety of only two actions: increase or decrease the rate of random movements. Yet, it is capable to cope with a quite complex
environment, with many different types of perturbations and opportunities. Its blind "transitions" are normally sufficient to find a favourable situation, thus escaping all dangers.
Ashby's law is perhaps the most famous (and some would say the only successful) principle of cybernetics recognized by the whole Cybernetics and Systems Science community. The Law has many forms, but
it is very simple and common sensical: a model system or controller can only model or control something to the extent that it has sufficient internal variety to represent it. For example, in order to
make a choice between two alternatives, the controller must be able to represent at least two possibilities, and thus one distinction. From an alternative perspective, the quantity of variety that
the model system or controller possesses provides an upper bound for the quantity of variety that can be controlled or modeled.
Requisite Variety has had a number of uses over the years , and there are a number of alternative formulations. Variety can be quantified according to different distributions, for example
probabilistic entropies and possibilistic nonspecificities. Under a stochastic formulation, there is a particularly interesting isomorphism between the LRV, the 2nd Law of Thermodynamics, and Shannon
's 10th Theorem .
See also: Dictionary: LAW OF REQUISITE VARIETY
Reference: Heylighen F. (1992): "Principles
of Systems and Cybernetics: an evolutionary perspective
", in: Cybernetics and Systems '92, R. Trappl (ed.), (World Science, Singapore), p. 3-10. | {"url":"http://pespmc1.vub.ac.be/REQVAR.html","timestamp":"2024-11-03T03:15:15Z","content_type":"text/html","content_length":"9213","record_id":"<urn:uuid:caf8d002-619f-4f0f-9080-426d2e27904d>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00247.warc.gz"} |
┃ │ Thomas Heath ┃
┃ │ ┃
┃ │ (5 Oct 1861 - 16 Mar 1940) ┃
┃ │ ┃
┃ │ ┃
Science Quotes by Thomas Heath (2 quotes)
One feature which will probably most impress the mathematician accustomed to the rapidity and directness secured by the generality of modern methods is the deliberation with which Archimedes
approaches the solution of any one of his main problems. Yet this very characteristic, with its incidental effects, is calculated to excite the more admiration because the method suggests the tactics
of some great strategist who foresees everything, eliminates everything not immediately conducive to the execution of his plan, masters every position in its order, and then suddenly (when the very
elaboration of the scheme has almost obscured, in the mind of the spectator, its ultimate object) strikes the final blow. Thus we read in Archimedes proposition after proposition the bearing of which
is not immediately obvious but which we find infallibly used later on; and we are led by such easy stages that the difficulties of the original problem, as presented at the outset, are scarcely
appreciated. As Plutarch says: “It is not possible to find in geometry more difficult and troublesome questions, or more simple and lucid explanations.” But it is decidedly a rhetorical exaggeration
when Plutarch goes on to say that we are deceived by the easiness of the successive steps into the belief that anyone could have discovered them for himself. On the contrary, the studied simplicity
and the perfect finish of the treatises involve at the same time an element of mystery. Though each step depends on the preceding ones, we are left in the dark as to how they were suggested to
Archimedes. There is, in fact, much truth in a remark by Wallis to the effect that he seems “as it were of set purpose to have covered up the traces of his investigation as if he had grudged
posterity the secret of his method of inquiry while he wished to extort from them assent to his results.” Wallis adds with equal reason that not only Archimedes but nearly all the ancients so hid
away from posterity their method of Analysis (though it is certain that they had one) that more modern mathematicians found it easier to invent a new Analysis than to seek out the old.
— Thomas Heath
The treatises [of Archimedes] are without exception, monuments of mathematical exposition; the gradual revelation of the plan of attack, the masterly ordering of the propositions, the stern
elimination of everything not immediately relevant to the purpose, the finish of the whole, are so impressive in their perfection as to create a feeling akin to awe in the mind of the reader.
— Thomas Heath | {"url":"https://todayinsci.com/H/Heath_Thomas/HeathThomas-Quotations.htm","timestamp":"2024-11-03T00:05:09Z","content_type":"text/html","content_length":"81778","record_id":"<urn:uuid:cd673b55-7ff7-455d-ae92-cd8c1b9d1647>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00587.warc.gz"} |
1 Square Meter to Square Terameter
Square Meter [m2] Output
1 square meter in ankanam is equal to 0.14949875578764
1 square meter in aana is equal to 0.031450432189072
1 square meter in acre is equal to 0.00024710516301528
1 square meter in arpent is equal to 0.00029249202856357
1 square meter in are is equal to 0.01
1 square meter in barn is equal to 1e+28
1 square meter in bigha [assam] is equal to 0.00074749377893818
1 square meter in bigha [west bengal] is equal to 0.00074749377893818
1 square meter in bigha [uttar pradesh] is equal to 0.00039866334876703
1 square meter in bigha [madhya pradesh] is equal to 0.00089699253472581
1 square meter in bigha [rajasthan] is equal to 0.00039536861034746
1 square meter in bigha [bihar] is equal to 0.00039544123500036
1 square meter in bigha [gujrat] is equal to 0.00061776345366791
1 square meter in bigha [himachal pradesh] is equal to 0.0012355269073358
1 square meter in bigha [nepal] is equal to 0.00014765309213594
1 square meter in biswa [uttar pradesh] is equal to 0.0079732669753405
1 square meter in bovate is equal to 0.000016666666666667
1 square meter in bunder is equal to 0.0001
1 square meter in caballeria is equal to 0.0000022222222222222
1 square meter in caballeria [cuba] is equal to 0.000007451564828614
1 square meter in caballeria [spain] is equal to 0.0000025
1 square meter in carreau is equal to 0.000077519379844961
1 square meter in carucate is equal to 0.0000020576131687243
1 square meter in cawnie is equal to 0.00018518518518519
1 square meter in cent is equal to 0.024710516301528
1 square meter in centiare is equal to 1
1 square meter in circular foot is equal to 13.71
1 square meter in circular inch is equal to 1973.52
1 square meter in cong is equal to 0.001
1 square meter in cover is equal to 0.00037064492216457
1 square meter in cuerda is equal to 0.00025445292620865
1 square meter in chatak is equal to 0.23919800926022
1 square meter in decimal is equal to 0.024710516301528
1 square meter in dekare is equal to 0.0010000006597004
1 square meter in dismil is equal to 0.024710516301528
1 square meter in dhur [tripura] is equal to 2.99
1 square meter in dhur [nepal] is equal to 0.059061236854374
1 square meter in dunam is equal to 0.001
1 square meter in drone is equal to 0.00003893196765303
1 square meter in fanega is equal to 0.00015552099533437
1 square meter in farthingdale is equal to 0.00098814229249012
1 square meter in feddan is equal to 0.00023990792525755
1 square meter in ganda is equal to 0.01245822964897
1 square meter in gaj is equal to 1.2
1 square meter in gajam is equal to 1.2
1 square meter in guntha is equal to 0.0098842152586866
1 square meter in ghumaon is equal to 0.00024710538146717
1 square meter in ground is equal to 0.0044849626736291
1 square meter in hacienda is equal to 1.1160714285714e-8
1 square meter in hectare is equal to 0.0001
1 square meter in hide is equal to 0.0000020576131687243
1 square meter in hout is equal to 0.00070359937931723
1 square meter in hundred is equal to 2.0576131687243e-8
1 square meter in jerib is equal to 0.00049466500076791
1 square meter in jutro is equal to 0.0001737619461338
1 square meter in katha [bangladesh] is equal to 0.014949875578764
1 square meter in kanal is equal to 0.0019768430517373
1 square meter in kani is equal to 0.00062291148244848
1 square meter in kara is equal to 0.049832918595878
1 square meter in kappland is equal to 0.0064825619084662
1 square meter in killa is equal to 0.00024710538146717
1 square meter in kranta is equal to 0.14949875578764
1 square meter in kuli is equal to 0.074749377893818
1 square meter in kuncham is equal to 0.0024710538146717
1 square meter in lecha is equal to 0.074749377893818
1 square meter in labor is equal to 0.0000013950025009895
1 square meter in legua is equal to 5.580010003958e-8
1 square meter in manzana [argentina] is equal to 0.0001
1 square meter in manzana [costa rica] is equal to 0.00014308280488084
1 square meter in marla is equal to 0.039536861034746
1 square meter in morgen [germany] is equal to 0.0004
1 square meter in morgen [south africa] is equal to 0.00011672697560406
1 square meter in mu is equal to 0.0014999999925
1 square meter in murabba is equal to 0.000009884206520611
1 square meter in mutthi is equal to 0.079732669753405
1 square meter in ngarn is equal to 0.0025
1 square meter in nali is equal to 0.0049832918595878
1 square meter in oxgang is equal to 0.000016666666666667
1 square meter in paisa is equal to 0.12580540458988
1 square meter in perche is equal to 0.029249202856357
1 square meter in parappu is equal to 0.0039536826082444
1 square meter in pyong is equal to 0.30248033877798
1 square meter in rai is equal to 0.000625
1 square meter in rood is equal to 0.00098842152586866
1 square meter in ropani is equal to 0.001965652011817
1 square meter in satak is equal to 0.024710516301528
1 square meter in section is equal to 3.8610215854245e-7
1 square meter in sitio is equal to 5.5555555555556e-8
1 square meter in square is equal to 0.1076391041671
1 square meter in square angstrom is equal to 100000000000000000000
1 square meter in square astronomical units is equal to 4.4683704831421e-23
1 square meter in square attometer is equal to 1e+36
1 square meter in square bicron is equal to 1e+24
1 square meter in square centimeter is equal to 10000
1 square meter in square chain is equal to 0.0024710436922533
1 square meter in square cubit is equal to 4.78
1 square meter in square decimeter is equal to 100
1 square meter in square dekameter is equal to 0.01
1 square meter in square digit is equal to 2755.56
1 square meter in square exameter is equal to 1e-36
1 square meter in square fathom is equal to 0.29899751157527
1 square meter in square femtometer is equal to 1e+30
1 square meter in square fermi is equal to 1e+30
1 square meter in square feet is equal to 10.76
1 square meter in square furlong is equal to 0.000024710516301528
1 square meter in square gigameter is equal to 1e-18
1 square meter in square hectometer is equal to 0.0001
1 square meter in square inch is equal to 1550
1 square meter in square league is equal to 4.290006866585e-8
1 square meter in square light year is equal to 1.1172498908139e-32
1 square meter in square kilometer is equal to 0.000001
1 square meter in square megameter is equal to 1e-12
1 square meter in square microinch is equal to 1550001732660300
1 square meter in square micrometer is equal to 1000000000000
1 square meter in square micromicron is equal to 1e+24
1 square meter in square micron is equal to 1000000000000
1 square meter in square mil is equal to 1550003100.01
1 square meter in square mile is equal to 3.8610215854245e-7
1 square meter in square millimeter is equal to 1000000
1 square meter in square nanometer is equal to 1000000000000000000
1 square meter in square nautical league is equal to 3.2394816622014e-8
1 square meter in square nautical mile is equal to 2.9155309240537e-7
1 square meter in square paris foot is equal to 9.48
1 square meter in square parsec is equal to 1.0502647575668e-33
1 square meter in perch is equal to 0.039536861034746
1 square meter in square perche is equal to 0.01958018322827
1 square meter in square petameter is equal to 1e-30
1 square meter in square picometer is equal to 1e+24
1 square meter in square pole is equal to 0.039536861034746
1 square meter in square rod is equal to 0.039536708845746
1 square meter in square terameter is equal to 1e-24
1 square meter in square thou is equal to 1550003100.01
1 square meter in square yard is equal to 1.2
1 square meter in square yoctometer is equal to 1e+48
1 square meter in square yottameter is equal to 1e-48
1 square meter in stang is equal to 0.00036913990402362
1 square meter in stremma is equal to 0.001
1 square meter in sarsai is equal to 0.35583174931272
1 square meter in tarea is equal to 0.0015903307888041
1 square meter in tatami is equal to 0.60499727751225
1 square meter in tonde land is equal to 0.00018129079042785
1 square meter in tsubo is equal to 0.30249863875613
1 square meter in township is equal to 1.0725050478094e-8
1 square meter in tunnland is equal to 0.00020257677659833
1 square meter in vaar is equal to 1.2
1 square meter in virgate is equal to 0.0000083333333333333
1 square meter in veli is equal to 0.0001245822964897
1 square meter in pari is equal to 0.000098842152586866
1 square meter in sangam is equal to 0.00039536861034746
1 square meter in kottah [bangladesh] is equal to 0.014949875578764
1 square meter in gunta is equal to 0.0098842152586866
1 square meter in point is equal to 0.024710731022028
1 square meter in lourak is equal to 0.00019768430517373
1 square meter in loukhai is equal to 0.00079073722069493
1 square meter in loushal is equal to 0.0015814744413899
1 square meter in tong is equal to 0.0031629488827797
1 square meter in kuzhi is equal to 0.074749377893818
1 square meter in chadara is equal to 0.1076391041671
1 square meter in veesam is equal to 1.2
1 square meter in lacham is equal to 0.0039536826082444
1 square meter in katha [nepal] is equal to 0.0029530618427187
1 square meter in katha [assam] is equal to 0.0037374688946909
1 square meter in katha [bihar] is equal to 0.0079088247000071
1 square meter in dhur [bihar] is equal to 0.15817649400014
1 square meter in dhurki is equal to 3.16 | {"url":"https://hextobinary.com/unit/area/from/m2/to/sqterameter/1","timestamp":"2024-11-03T06:37:04Z","content_type":"text/html","content_length":"128376","record_id":"<urn:uuid:e8cb0964-72c9-40f1-9d44-7b25c2566d09>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00746.warc.gz"} |
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the cutter eventually becomestheregular ballend a typical variable pitch end mill, these pitch angles between two successive cutting edges are variable as exhibited in Fig. 2 (b). (2)
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For selected spindle speed ranges and given lead/tilt angles, the optimal pitch angles of variable pitch ballend cutter are determined with the proposed optimization approach. The experimental
results show the average stability of the customized optimal variable pitch ballend cutter is significantly higher than that of the regular ballend ...
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It was found that the ball mill consumed kWh/t energy to reduce the F 80 feed size of µm to P 80 product size of µm while stirred mill consumed kWh/t of energy to produce the product size of µm.
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CERAMIC LINED BALL MILL. Ball Mills can be supplied with either ceramic or rubber linings for wet or dry grinding, for continuous or batch type operation, in sizes from 15″ x 21″ to 8′ x 12′.
High density ceramic linings of uniform hardness male possible thinner linings and greater and more effective grinding volume.
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A numerical dynamicmechanical model of a planetary ballmill is developed to study the dependence of process efficiency on milling parameters like ball size and number, jar geometry and velocity
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To examine the effect of the kinetic energy dose upon the WSP yield obtained from the MCD of H 2 SO 4impregnated beechwood, a total number (n) of 47 experiments were performed in a planetary mill
operating under varied conditions,, different ball diameters (420 mm), ball counts (4391), rotational speeds (400800 rpm), and milling ...
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Cavitation experiments were conducted on a VMC1000P CNC highspeed machining center with a maximum spindle speed of 15,000 rpm. The feed rate can be varied in the range of 120,000 mm/min. A BT40
tool holder was used in the experiment, which is matched with the central water discharge spindle vertical machining center.
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Statistical experiment design was applied to optimize wet and dry milling of clinoptilolite zeolite by utilizing a planetary ball mill. For obtaining appropriate milling conditions with respect
to crystallinity retention and powder size distribution different milling parameters consisting of dry and wet milling time and rotational speed, ball to powder ratio in dry state and ball to
powder ...
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In Section 3, the Mill Load Identification Method for Ball milling Process Based on Grinding Signal Xiaoli Wang* Kaixuan Sun* He Zhang* Wei Xiong** Chunhua Yang* *School of Automation, Central
... After many experiments, set the optimal parameters: m = 4, ï ´ =1. Taking the normal working condition as an example, correlation and PE of the ...
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The cutting edge trajectory for micro milling is determined by a theoretical and empirical coupled method, based on the process kinematics, tool runout and stochastic tool wear. The proposed
surface generation model is validated with milling experiments of Al6061 under various machining conditions.
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We carried out a detailed study on the effect of particle load and ball load on grinding kinetics. by carrying out experiments at two different mill speeds (55 and 70 % critical) and four levels
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Mill Speed as a Manipulated Variable for Ball Mill Grinding Control Author links open overlay,, Show more Add to Mendeley Share Cite https:////S(17)642654Get rights and content Abstract
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Product Detail: QXQM series planetary ball mill has four ball grinding tanks installed on one disc. When the turn disc rotates, the mill pots revolve on their own axis and make 360degree
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Mill Speed is one variable that can often be easily changed with a variable frequency drive (VFD). The starting point for mill speed calculations is the critical speed. Critical speed (CS) is the
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A novel characteristic variable of fill level has been proposed, in order to reduce the influence of various factors on measuring the fill level and improve the measurement accuracy of the fill
level. ... FFT waveform of vibration signal of the mill shell in the second experiment (mill diameter = m, length = m, mill speed = rpm ...
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Quantum Nanostructures (QDs): An Overview. D. Sumanth Kumar, ... Mahesh, in Synthesis of Inorganic Nanomaterials, 2018 Ball Milling. A ball mill is a type of grinder used to grind and blend bulk
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The 5axis ballend milling experiment ... As shown in Fig. 14, with the design variable space setting and constraints, ... Surface texture generation using highfeed milling with spindle speed
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This work consisted on experimental batch grinding tests with a 30 x 30 cm ball mill in which operational variables were altered. The change of these parameters resulted in direct variation on
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... A ball mill is a type of grinder widely utilized in the process of mechanochemical catalytic degradation. It consists of one or more rotating cylinders partially filled with grinding balls...
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vibratory milling apparatus is much greater than that of planetary ball mill system; therefore, less time may be needed for the particle size reduction in the vibratory ball mills. Here, the
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Laboratory mill configuration. Experiments were carried out using the Wits pilot mill. This mill is fitted with twelve equally spaced trapezoidal lifters and is driven by a kW variable speed
motor mounted on a mill 1 gives some specifications of the mill and the operating conditions as defined for the experiments.
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Experiments were performed varying the rotation speeds in 15, 31 and 47 rpm and the clinker and grinding media filling degree in, respectively,, 5 and % and 2, 4 and 6% of the drum volume. Using
DEM, the numbers and intensity of the collisions between the grinding media and the mill walls were evaluated, along with the drum flow regime.
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In this section, simulations and experiments are performed to validate the proposed model. The cutter used here is a ballend mill with an 8 mm diameter, 4 flutes and 48 mm overhang length. The
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The 21 Liter (5 gallon) 911METALLURGY 911MPE21BM dual function Laboratory Rod Mill / Ball Mill is designed to meet the industrial requirements to grind coal, cement and a wide variety of ores.
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collector. A ...
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S. Rosenkranz, S. BreitungFaes, A. Kwade Add to Mendeley https://// Get rights and content Abstract Planetary ball mills feature attractive properties, like the possibility of dry or wet
operation, straightforward handling, cleanability and moderate costs.
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as ball lling, feed rate, mill speed, and ball size. Austin, ... variablespeed, ... The rougher and recleaner flotation experiments were conducted using Denvertype mechanically agitated cells ...
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BALL MILL EXPERIMENT MANUAL title of experiment: ball mill (variable speed) objective: to study the operation of ball mill and to calculate the efficiency of
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1. Fill the container with small metal balls. Most people prefer to use steel balls, but lead balls and even marbles can be used for your grinding. Use balls with a diameter between ½" (13 mm)
and ¾" (19 mm) inside the mill. The number of balls is going to be dependent on the exact size of your drum.
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The drive of the ball mill consists of an induction electric motor, twostage gearbox and the pinion gear. This pinion gear is mated with the girth gear mounted on circumference of the ball mill
shell and in that way it transfers the motion from the drive to the shell, see Figure 1, taken from [1]. Figure 1. Mechanical system of a ball mill ...
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The effectiveness of the designed variable pitch cutter was validated by laboratory and industrial experiments. Recently, Zhan et al. [231] established a five−axis variable−pitch milling model
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Chemical Reaction Engineering GATE-2019 - Insight into Chemical Engineering
Chemical Reaction Engineering GATE-2019
Q 1: The desired liquid-phase reaction
is accompanied by an undesired side reaction
Four isothermal reactor schemes (CSTR: ideal Continuous-Stirred Tank Reactor; PFR: ideal Plug Flow Reactor) for processing equal molar feed rates of D and E are shown in the figure. Each scheme is
designed for the same conversion. The scheme that gives the most favorable product distribution is:
Q 2: For a first-order reaction in a porous spherical catalyst pellet, diffusional effects are most likely to lower the observed rate of reaction for
Q 3: The elementary, irreversible, liquid-phase, parallel reactions $2A\rightarrow D$ and $2A\rightarrow U$, take place in an isothermal non-ideal reactor. The C-curve measured in a tracer experiment
is shown in the figure, where C(t) is the concentration of the tracer in g/m^3 at the reactor exit at time t (in min).
The rate constants are k[1] = 0.2 Litre/(mol min) and k[2] = 0.3 Litre/(mol min). Pure A is fed to the reactor at a concentration of 2 mol/Litre. Using the segregated model, the percentage conversion
in the reactor is _________ (rounded off to the nearest integer).
Q 4: A first-order irreversible liquid phase reaction $A\rightarrow B\;(k=0.1\;min^{-1})$ is carried out under isothermal, steady-state conditions in the following reactor arrangement comprising an
ideal CSTR (Continuous-Stirred Tank Reactor) and two ideal PFRs (Plug Flow Reactors).
From the information in the figure, the volume of the CSTR (in Litres) is ________ (rounded off to the nearest integer).
Q 5: The elementary liquid-phase irreversible reactions
take place in an isothermal ideal CSTR (Continuous-Stirred Tank Reactor). Pure A is fed to the reactor at a concentration of 2 mol/Litre. For the residence time that maximizes the exit concentration
of B, the percentage yield of B, defined as $\left(\frac{net\;formation\;rate\;of\;B}{consumption\;rate\;of\;A}\times100\right)$, is _______ (rounded off to the nearest integer).
Q 6: The elementary irreversible gas-phase reaction $A\rightarrow B+C$ is carried out adiabatically in an ideal CSTR (Continuous-Stirred Tank Reactor) operating at 10 atm. Pure A enters the CSTR at a
flow rate of 10 mol/s and a temperature of 450 K. Assume A, B, and C to be ideal gases. The specific heat capacity at constant pressure $\left(C_{p_i}\right)$ and heat of formation $\left(H_i^o\
right)$, of component i (i = A, B, C), are:
C_{P_A}=30\frac J{molK};\;C_{P_B}=10\frac J{molK};\;C_{P_C}=20\frac J{molK} H_A^0=-90\frac{kJ}{mol};\;H_B^0=-54\frac{kJ}{mol};\;H_C^0=-45\frac{kJ}{mol}
The reaction rate constant $k(per\;second)=0.133exp\left[\frac ER\left(\frac1{450}-\frac1T\right)\right]$, where E = 31.4 kJ/mol and universal gas constant R = 0.082 L atm/(mol K) = 8.314 J/(mol K).
The shaft work may be neglected in the analysis, and specific heat capacities do not vary with temperature. All heats of formation are referenced to 273 K. The reactor volume (in Litres) for 75%
conversion is _______ (rounded off to the nearest integer).
Q 7: Consider the reactor-separator-recycle process operating under steady-state conditions as shown in the figure.
The reactor is an ideal Continuous-Stirred Tank Reactor (CSTR), where the reaction $A+B\rightarrow C$ occurs. Assume that there is no impurity in the product and recycle streams. Other relevant
information are provided in the figure. The mole fraction of B (x[B]) in the reactor that minimizes the recycle rate is ___________ (rounded off to two decimal places). | {"url":"https://chelearning.com/chemical-reaction-engineering-gate-2019/","timestamp":"2024-11-02T12:56:26Z","content_type":"text/html","content_length":"122546","record_id":"<urn:uuid:2d259e71-2677-4c9d-95dd-939b3a3e9bc8>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00084.warc.gz"} |
Calculus in Context
Download Full Text (8.1 MB)
Download Calculus I: Chapters 1-6, plus front matter and full index (2.6 MB)
Download Calculus II: Chapters 7-12, plus front matter and full index (5.6 MB)
Download Prefaces and Table of Contents (106 KB)
Download Chapter 1: A context for calculus: pages 1-60 (409 KB)
Download Chapter 2: Successive approximations: pages 61-100 (334 KB)
Download Chapter 3: The derivative: pages 101-178 (539 KB)
Download Chapter 4: Differential equations: pages 179-272 (640 KB)
Download Chapter 5: Techniques of differentiation: pages 275-336 (510 KB)
Download Chapter 6: The integral: pages 337-418 (500 KB)
Download Chapter 7: Periodicity: pages 419-460 (1017 KB)
Download Chapter 8: Dynamical systems: pages 461-510 (1.4 MB)
Download Chapter 9: Functions of several variables: pages 511-592 (1.9 MB)
Download Chapter 10: Series and approximations: pages 593-680 (578 KB)
Download Chapter 11: Techniques of integration: pages 681-768 (485 KB)
Download Chapter 12: Case studies: pages 769-834 (491 KB)
Download Index: pages 835-845 (51 KB)
Download Supplementary: Handbook for Instructors (530 KB)
Download Supplementary: QuickBasic (836 KB)
Designing the curriculum
We believe that calculus can be for students what it was for Euler and the Bernoullis: a language and a tool for exploring the whole fabric of science. We also believe that much of the mathematical
depth and vitality of calculus lies in connections to other sciences. The mathematical questions that arise are compelling in part because the answers matter to other disciplines. We began our work
with a "clean slate," not by asking what parts of the traditional course to include or discard. Our starting points are thus our summary of what calculus is really about. Our curricular goals are
what we aim to convey about the subject in the course. Our functional goals describe the attitudes and behaviors we hope our students will adopt in using calculus to approach scientific and
mathematical questions.
Starting Points
• Calculus is fundamentally a way of dealing with functional relationships that occur in scientific and mathematical contexts. The techniques of calculus must be subordinate to an overall view of
the questions that give rise to these relationships.
• Technology radically enlarges the range of questions we can explore and the ways we can answer them. Computers and graphing calculators are much more than tools for teaching the traditional
• The concept of a dynamical system is central to science. Therefore, differential equations belong at the center of calculus, and technology makes this possible at the introductory level.
• The process of successive approximation is a key tool of calculus, even when the outcome of the process--the limit--cannot be explicitly given in closed form.
Curricular Goals
• Develop calculus in the context of scientific and mathematical questions.
• Treat systems of differential equations as fundamental objects of study.
• Construct and analyze mathematical models.
• Use the method of successive approximations to define and solve problems.
• Develop geometric visualization with hand-drawn and computer graphics.
• Give numerical methods a more central role.
Functional Goals
• Encourage collaborative work.
• Enable students to use calculus as a language and a tool.
• Make students comfortable tackling large, messy, ill-defined problems.
• Foster an experimental attitude towards mathematics.
• Help students appreciate the value of approximate solutions.
• Teach students that understanding grows out of working on problems.
Impact of Technology
• Differential equations can now be solved numerically, so they can take their rightful place in the introductory calculus course.
• The ability to handle data and perform many computations makes exploring messy, real-world problems possible.
• Since we can now deal with credible models, the role of modelling becomes much more central to the subject.
The text illustrates how we have pursued the curricular goals. Each goal is addressed within the first chapter which begins with questions about describing and analyzing the spread of a contagious
disease. A model is built: a model which is actually a system of coupled non-linear differential equations. We then begin a numerical exploration on those equations, and the door is opened to a
solution by successive approximations. Our implementation of the functional goals is also evident. The text has many more words than the traditional calculus book--it is a book to be read. The
exercises make unusual demands on students. Most are not just variants of examples that have been worked in the text. In fact, the text has rather few "template'' examples.
Shifts in Emphasis
It will also become apparent to you that the text reflects substantial shifts in emphasis in comparison to the traditional course. Here are some of the most striking:
How the emphasis shifts:
increase: concepts, geometry, graphs, brute force, numerical solutions
decrease: techniques, algebra, formulas, elegance, closed-form solutions
Since we all value elegance, let us explain what we mean by "brute force." Euler's method is a good example. It is a general method of wide applicability. Of course when we use it to solve a
differential equation like y'(t) = t, we are using a sledgehammer to crack a peanut. But at least the sledgehammer works. Moreover, it works with coconuts (like y' = y(1 - y/10)), and it will even
knock down a house (like y' = cos^2(t)). Students also see the elegant special methods that can be invoked to solve y' = t and y' = y(1 - y/10) (separation of variables and partial fractions are
discussed in chapter 11), but they understand that they are fortunate indeed when a real problem will succumb to such methods.
The Five College Calculus Project (1995, 2008)
There are software programs are available at no charge for use with this text.
Sarah-Marie Belcastro has produced a collection of Mathematica notebooks to accompany the text. These are written using Version 6 of Mathematica.
Calculus in Context is the product of the Five College Calculus Project. Besides the introductory calculus text, the product includes computer software and a Handbook for Instructors described.
Support for the different aspects of Calculus in Context has come from several sources.
Primary funding for curriculum development and dissemination was provided by the National Science Foundation in grants DMS-14004 (1988-95) and DUE-9153301 (1991-97), awarded to Five Colleges, Inc.
Other curriculum development funding has been provided by NECUSE (New England Consortium for Undergraduate Science Education, funded by the Pew Charitable Trusts) to Smith College (1989) and Mount
Holyoke College (1990). Five Colleges, Inc. also provided start-up funds.
Equipment and software for computer classrooms has been funded by NSF grants in the ILI program: USE-8951485 to Smith College and DUE/EHR-9551919 to Mount Holyoke College. The Hewlett-Packard
Corporation contributed equipment to Mount Holyoke and Smith Colleges, and other equipment was contributed to Mount Holyoke College by IBM and the Sloan Foundation.
Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect those of the National Science Foundation.
Original version available at: http://www.math.smith.edu/~callahan/intromine.html
Recommended Citation
Callahan, James; Cox, David; Hoffman, Kenneth; O'Shea, Donal; Pollatsek, Harriet; and Senechal, Lester, "Calculus in Context" (2008). Open Educational Resources: Textbooks, Smith College,
Northampton, MA.
© Five Colleges, Inc | {"url":"https://scholarworks.smith.edu/textbooks/2/","timestamp":"2024-11-13T07:45:18Z","content_type":"text/html","content_length":"57992","record_id":"<urn:uuid:c7a75390-0b14-4f5b-988b-b27b783617bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00709.warc.gz"} |
College Physics: OpenStax
Chapter 13 Temperature, Kinetic Theory, and the Gas Laws
• Define temperature.
• Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales.
• Define thermal equilibrium.
• State the zeroth law of thermodynamics.
The concept of temperature has evolved from the common concepts of hot and cold. Human perception of what feels hot or cold is a relative one. For example, if you place one hand in hot water and the
other in cold water, and then place both hands in tepid water, the tepid water will feel cool to the hand that was in hot water, and warm to the one that was in cold water. The scientific definition
of temperature is less ambiguous than your senses of hot and cold. Temperature is operationally defined to be what we measure with a thermometer. (Many physical quantities are defined solely in terms
of how they are measured. We shall see later how temperature is related to the kinetic energies of atoms and molecules, a more physical explanation.) Two accurate thermometers, one placed in hot
water and the other in cold water, will show the hot water to have a higher temperature. If they are then placed in the tepid water, both will give identical readings (within measurement
uncertainties). In this section, we discuss temperature, its measurement by thermometers, and its relationship to thermal equilibrium. Again, temperature is the quantity measured by a thermometer.
On a cold winter morning, the wood on a porch feels warmer than the metal of your bike. The wood and bicycle are in thermal equilibrium with the outside air, and are thus the same temperature. They
feel different because of the difference in the way that they conduct heat away from your skin. The metal conducts heat away from your body faster than the wood does (see more about conductivity in
Chapter 14.5 Conduction). This is just one example demonstrating that the human sense of hot and cold is not determined by temperature alone.
Another factor that affects our perception of temperature is humidity. Most people feel much hotter on hot, humid days than on hot, dry days. This is because on humid days, sweat does not evaporate
from the skin as efficiently as it does on dry days. It is the evaporation of sweat (or water from a sprinkler or pool) that cools us off.
Any physical property that depends on temperature, and whose response to temperature is reproducible, can be used as the basis of a thermometer. Because many physical properties depend on
temperature, the variety of thermometers is remarkable. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer, the old
mercury thermometer, and the bimetallic strip (Figure 1). Other properties used to measure temperature include electrical resistance and color, as shown in Figure 2, and the emission of infrared
radiation, as shown in Figure 3.
Figure 1. The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher
temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right.
Figure 2. Each of the six squares on this plastic (liquid crystal) thermometer contains a film of a different heat-sensitive liquid crystal material. Below 95ºF, all six squares are black. When the
plastic thermometer is exposed to temperature that increases to 95ºF, the first liquid crystal square changes color. When the temperature increases above 96.8ºF the second liquid crystal square also
changes color, and so forth. (credit: Arkrishna, Wikimedia Commons)
Figure 3. Fireman Jason Ormand uses a pyrometer to check the temperature of an aircraft carrier’s ventilation system. Infrared radiation (whose emission varies with temperature) from the vent is
measured and a temperature readout is quickly produced. Infrared measurements are also frequently used as a measure of body temperature. These modern thermometers, placed in the ear canal, are more
accurate than alcohol thermometers placed under the tongue or in the armpit. (credit: Lamel J. Hinton/U.S. Navy)
Temperature Scales
Thermometers are used to measure temperature according to well-defined scales of measurement, which use pre-defined reference points to help compare quantities. The three most common temperature
scales are the Fahrenheit, Celsius, and Kelvin scales. A temperature scale can be created by identifying two easily reproducible temperatures. The freezing and boiling temperatures of water at
standard atmospheric pressure are commonly used.
The Celsius scale (which replaced the slightly different centigrade scale) has the freezing point of water at[latex]\boldsymbol{0^{\circ}\textbf{C}}[/latex]and the boiling point at[latex]\boldsymbol
{100^{\circ}\textbf{C}}.[/latex]Its unit is the degree Celsius[latex]\boldsymbol{(^{\circ}\textbf{C})}.[/latex]On the Fahrenheit scale (still the most frequently used in the United States), the
freezing point of water is at[latex]\boldsymbol{32^{\circ}\textbf{F}}[/latex]and the boiling point is at[latex]\boldsymbol{212^{\circ}\textbf{F}}.[/latex]The unit of temperature on this scale is the
degree Fahrenheit[latex]\boldsymbol{(^{\circ}\textbf{F}).}[/latex]Note that a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Only 100
Celsius degrees span the same range as 180 Fahrenheit degrees, thus one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale[latex]\boldsymbol{180/100=9/5}.[/latex]
The Kelvin scale is the temperature scale that is commonly used in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero. The
official temperature unit on this scale is the kelvin, which is abbreviated K, and is not accompanied by a degree sign. The freezing and boiling points of water are 273.15 K and 373.15 K,
respectively. Thus, the magnitude of temperature differences is the same in units of kelvins and degrees Celsius. Unlike other temperature scales, the Kelvin scale is an absolute scale. It is used
extensively in scientific work because a number of physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The kelvin is the SI unit used in scientific
Figure 4. Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales, rounded to the nearest degree. The relative sizes of the scales are also shown.
The relationships between the three common temperature scales is shown in Figure 4. Temperatures on these scales can be converted using the equations in Table 1.
To convert from . . . Use this equation . . . Also written as . . .
Celsius to Fahrenheit [latex]\boldsymbol{T(^{\circ}\textbf{F})=\frac{9}{5}T(^{\circ}\textbf{C})+32}[/latex] [latex]\boldsymbol{T_{^{\circ}\textbf{F}}=\frac{9}{5}T_{^{\circ}\textbf{C}}+32}[/latex]
Fahrenheit to Celsius [latex]\boldsymbol{T(^{\circ}\textbf{C})=\frac{5}{9}T(^{\circ}\textbf{F})-32}[/latex] [latex]\boldsymbol{T_{^{\circ}\textbf{C}}=\frac{5}{9}(T_{^{\circ}\textbf{F}}-32)}[/latex]
Celsius to Kelvin [latex]\boldsymbol{T(\textbf{K})=T(^{\circ}\textbf{C})+273.15}[/latex] [latex]\boldsymbol{T_{\textbf{K}}=T_{^{\circ}\textbf{C}}+273.15}[/latex]
Kelvin to Celsius [latex]\boldsymbol{T(^{\circ}\textbf{C})=T(\textbf{K})-273.15}[/latex] [latex]\boldsymbol{T_{^{\circ}\textbf{C}}=T_{\textbf{K}}-273.15}[/latex]
Fahrenheit to Kelvin [latex]\boldsymbol{T(\textbf{K})=\frac{5}{9}(T(^{\circ}\textbf{F})-32)+273.15}[/latex] [latex]\boldsymbol{T_{\textbf{K}}=\frac{5}{9}(T_{^{\circ}\textbf{F}}-32)+273.15}[/latex]
Kelvin to Fahrenheit [latex]\boldsymbol{T(^{\circ}\textbf{F})=\frac{9}{5}T(\textbf{K})-273.15)+32}[/latex] [latex]\boldsymbol{T_{^\textbf{F}}=\frac{9}{5}(T_{\textbf{K}}-273.15)+32}[/latex]
Table 1. Temperature Conversions
Notice that the conversions between Fahrenheit and Kelvin look quite complicated. In fact, they are simple combinations of the conversions between Fahrenheit and Celsius, and the conversions between
Celsius and Kelvin.
Example 1: Converting between Temperature Scales: Room Temperature
“Room temperature” is generally defined to be[latex]\boldsymbol{25^{\circ}\textbf{C}}.[/latex](a) What is room temperature in[latex]\boldsymbol{^{\circ}\textbf{F}}?[/latex](b) What is it in K?
To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values.
Solution for (a)
1. Choose the right equation. To convert from[latex]\boldsymbol{^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{^{\circ}\textbf{F}},[/latex]use the equation
2. Plug the known value into the equation and solve:
Solution for (b)
1. Choose the right equation. To convert from[latex]\boldsymbol{^{\circ}\textbf{C}}[/latex]to K, use the equation
2. Plug the known value into the equation and solve:
Example 2: Converting between Temperature Scales: the Reaumur Scale
The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur temperature scale, the freezing point of water is[latex]\boldsymbol{0^{\circ}\
textbf{R}}[/latex]and the boiling temperature is[latex]\boldsymbol{80^{\circ}\textbf{R}}.[/latex]If “room temperature” is[latex]\boldsymbol{25^{\circ}\textbf{C}}[/latex]on the Celsius scale, what is
it on the Reaumur scale?
To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is[latex]\boldsymbol{80^{\
circ}\textbf{R}}.[/latex]On the Celsius scale it is[latex]\boldsymbol{100^{\circ}\textbf{C}}.[/latex]Therefore[latex]\boldsymbol{100^{\circ}\textbf{C}=80^{\circ}\textbf{R}}.[/latex]Both scales start
at[latex]\boldsymbol{0^{\circ}}[/latex]for freezing, so we can derive a simple formula to convert between temperatures on the two scales.
1. Derive a formula to convert from one scale to the other:
2. Plug the known value into the equation and solve:
Temperature Ranges in the Universe
Figure 6 shows the wide range of temperatures found in the universe. Human beings have been known to survive with body temperatures within a small range, from[latex]\boldsymbol{24^{\circ}\textbf{C}}
[/latex]to[latex]\boldsymbol{44^{\circ}\textbf{C}\:(75^{\circ}\textbf{F}}[/latex]to[latex]\boldsymbol{111^{\circ}\textbf{F})}.[/latex]The average normal body temperature is usually given as[latex]\
boldsymbol{37.0^{\circ}\textbf{C}}[/latex]([latex]\boldsymbol{98.6^{\circ}\textbf{F}}[/latex]), and variations in this temperature can indicate a medical condition: a fever, an infection, a tumor, or
circulatory problems (see Figure 5).
Figure 5. This image of radiation from a person’s body (an infrared thermograph) shows the location of temperature abnormalities in the upper body. Dark blue corresponds to cold areas and red to
white corresponds to hot areas. An elevated temperature might be an indication of malignant tissue (a cancerous tumor in the breast, for example), while a depressed temperature might be due to a
decline in blood flow from a clot. In this case, the abnormalities are caused by a condition called hyperhidrosis. (credit: Porcelina81, Wikimedia Commons)
The lowest temperatures ever recorded have been measured during laboratory experiments:[latex]\boldsymbol{4.5\times10^{-10}\textbf{ K}}[/latex]at the Massachusetts Institute of Technology (USA), and
[latex]\boldsymbol{1.0\times10^{-10}\textbf{ K}}[/latex]at Helsinki University of Technology (Finland). In comparison, the coldest recorded place on Earth’s surface is Vostok, Antarctica at 183 K
[latex]\boldsymbol{(-89^{\circ}\textbf{C})},[/latex]and the coldest place (outside the lab) known in the universe is the Boomerang Nebula, with a temperature of 1 K.
Figure 6. Each increment on this logarithmic scale indicates an increase by a factor of ten, and thus illustrates the tremendous range of temperatures in nature. Note that zero on a logarithmic scale
would occur off the bottom of the page at infinity.
What is absolute zero? Absolute zero is the temperature at which all molecular motion has ceased. The concept of absolute zero arises from the behavior of gases. Figure 7 shows how the pressure of
gases at a constant volume decreases as temperature decreases. Various scientists have noted that the pressures of gases extrapolate to zero at the same temperature,[latex]\boldsymbol{-273.15^{\circ}
\textbf{C}}.[/latex]This extrapolation implies that there is a lowest temperature. This temperature is called absolute zero. Today we know that most gases first liquefy and then freeze, and it is not
actually possible to reach absolute zero. The numerical value of absolute zero temperature is[latex]\boldsymbol{-273.15^{\circ}\textbf{C}}[/latex]or 0 K.
Figure 7. Graph of pressure versus temperature for various gases kept at a constant volume. Note that all of the graphs extrapolate to zero pressure at the same temperature.
Thermal Equilibrium and the Zeroth Law of Thermodynamics
Thermometers actually take their own temperature, not the temperature of the object they are measuring. This raises the question of how we can be certain that a thermometer measures the temperature
of the object with which it is in contact. It is based on the fact that any two systems placed in thermal contact (meaning heat transfer can occur between them) will reach the same temperature. That
is, heat will flow from the hotter object to the cooler one until they have exactly the same temperature. The objects are then in thermal equilibrium, and no further changes will occur. The systems
interact and change because their temperatures differ, and the changes stop once their temperatures are the same. Thus, if enough time is allowed for this transfer of heat to run its course, the
temperature a thermometer registers does represent the system with which it is in thermal equilibrium. Thermal equilibrium is established when two bodies are in contact with each other and can freely
exchange energy.
Furthermore, experimentation has shown that if two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system C, then A is also in thermal
equilibrium with C. This conclusion may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics.
If two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C.
This law was postulated in the 1930s, after the first and second laws of thermodynamics had been developed and named. It is called the zeroth law because it comes logically before the first and
second laws (discussed in Chapter 15 Thermodynamics). An example of this law in action is seen in babies in incubators: babies in incubators normally have very few clothes on, so to an observer they
look as if they may not be warm enough. However, the temperature of the air, the cot, and the baby is the same, because they are in thermal equilibrium, which is accomplished by maintaining air
temperature to keep the baby comfortable.
Check Your Understanding
1: Does the temperature of a body depend on its size?
Section Summary
• Temperature is the quantity measured by a thermometer.
• Temperature is related to the average kinetic energy of atoms and molecules in a system.
• Absolute zero is the temperature at which there is no molecular motion.
• There are three main temperature scales: Celsius, Fahrenheit, and Kelvin.
• Temperatures on one scale can be converted to temperatures on another scale using the following equations:
• Systems are in thermal equilibrium when they have the same temperature.
• Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy.
• The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in
thermal equilibrium with C.
Conceptual Questions
1: What does it mean to say that two systems are in thermal equilibrium?
2: Give an example of a physical property that varies with temperature and describe how it is used to measure temperature.
3: When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
4: If you add boiling water to a cup at room temperature, what would you expect the final equilibrium temperature of the unit to be? You will need to include the surroundings as part of the system.
Consider the zeroth law of thermodynamics.
Problems & Exercises
1: What is the Fahrenheit temperature of a person with a[latex]\boldsymbol{39.0^{\circ}\textbf{C}}[/latex]fever?
2: Frost damage to most plants occurs at temperatures of[latex]\boldsymbol{28.0^{\circ}\textbf{F}}[/latex]or lower. What is this temperature on the Kelvin scale?
3: To conserve energy, room temperatures are kept at[latex]\boldsymbol{68.0^{\circ}\textbf{F}}[/latex]in the winter and[latex]\boldsymbol{78.0^{\circ}\textbf{F}}[/latex]in the summer. What are these
temperatures on the Celsius scale?
4: A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit temperature? What is this on the Celsius scale?
5: The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale?
6: One of the hottest temperatures ever recorded on the surface of Earth was[latex]\boldsymbol{134^{\circ}\textbf{F}}[/latex]in Death Valley, CA. What is this temperature in Celsius degrees? What is
this temperature in Kelvin?
7: (a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when there is a[latex]\boldsymbol{40.0^
{\circ}\textbf{F}}[/latex]decrease in temperature? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees.
8: (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value?
the quantity measured by a thermometer
Celsius scale
temperature scale in which the freezing point of water is[latex]\boldsymbol{0^{\circ}\textbf{C}}[/latex]and the boiling point of water is[latex]\boldsymbol{100^{\circ}\textbf{C}}[/latex]
degree Celsius
unit on the Celsius temperature scale
Fahrenheit scale
temperature scale in which the freezing point of water is[latex]\boldsymbol{32^{\circ}\textbf{F}}[/latex]and the boiling point of water is[latex]\boldsymbol{212^{\circ}\textbf{F}}[/latex]
degree Fahrenheit
unit on the Fahrenheit temperature scale
Kelvin scale
temperature scale in which 0 K is the lowest possible temperature, representing absolute zero
absolute zero
the lowest possible temperature; the temperature at which all molecular motion ceases
thermal equilibrium
the condition in which heat no longer flows between two objects that are in contact; the two objects have the same temperature
zeroth law of thermodynamics
law that states that if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object
Check Your Understanding
1: No, the system can be divided into smaller parts each of which is at the same temperature. We say that the temperature is an intensive quantity. Intensive quantities are independent of size.
Problems & Exercises
[latex]\begin{array}{lcl} \boldsymbol{\Delta{T}(^{\circ}\textbf{F})} & \boldsymbol{=} & \boldsymbol{T_2(^{\circ}\textbf{F})-T_1(^{\circ}\textbf{F})} \\ {} & \boldsymbol{=} & \boldsymbol{\frac{9}{5}
T_2(^{\circ}\textbf{C})+32.0^{\circ}-(\frac{9}{5}T_1(^{\circ}\textbf{C})+32.0^{\circ})} \\ {} & \boldsymbol{=} & \boldsymbol{\frac{9}{5}T_2(^{\circ}\textbf{C})-T_1(^{\circ}\textbf{C})=\frac{9}{5}\
Delta{T}(^{\circ}\textbf{C})} \end{array}[/latex] | {"url":"https://pressbooks.bccampus.ca/collegephysics/chapter/temperature/","timestamp":"2024-11-11T07:46:59Z","content_type":"text/html","content_length":"181796","record_id":"<urn:uuid:933fd6da-0048-4286-a3e8-8560f8ef09e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00462.warc.gz"} |
Cell Intervals: Rules of ThumbCell Intervals: Rules of Thumb
Cell Intervals: Rules of Thumb
When constructing a histogram (a graph in which the classes are identified on the horizontal axis and the class frequencies are shown on the vertical axis by the heights of the vertical bars) it is
important to choose an appropriate number of Cell Intervals (the grouping you will use to generate the classes for the histogram).
When constructing a histogram (a graph in which the classes are identified on the horizontal axis and the class frequencies are shown on the vertical axis by the heights of the vertical bars) it is
important to choose an appropriate number of Cell Intervals (the grouping you will use to generate the classes for the histogram). As a rule of thumb, the number of cell intervals “K” can be chosen
to equal the smallest whole number that makes 2K greater than the total number of measurements (n). However, this is not a hard and fast rule, so you may want to try a few different options (say, K –
1 intervals or K + 1 intervals) as well. Your interval length can then be computed to be (largest value – smallest value) divided by K.
Example: Suppose we have a group of data from a test administered to new employees to evaluate their understanding of a new financial product. The highest score possible on the test is 40. The
results from 30 employees show a low score of 17 and a high score of 39.
To find the number of cell intervals to use, we would choose the value of K that makes 2K just greater than n = 30. Thus K would equal 5 (since 25 = 32). Thus we want 5 cell intervals and the width
of our cell intervals is approximately:
Always “round up” to the next whole number. Notice in the example above that quotient came out to 4.4. Normally, we would round down to four, but if we did that we would only have four in each cell –
which would be insufficient to cover the whole range of numbers 17 to 39. See below.
17 to 20 would be the first cell.
21 to 24 would be the second cell.
25 to 28 would be the third cell.
29 to 32 would be the fourth cell.
33 to 36 would be the final cell, which would make us short on the last cell because we needed values to cover through that number 39. So, always round up to the whole number.
Let’s use five cells (intervals) with five values in each cell. It would look like this:
17 to 21
22 to 26
27 to 31
32 to 36
37 to 41
You could have chosen to use 10 intervals instead of five and if you would have chosen 10 it would have looked like this:
17 – 19
20 – 22
23 – 25
26 – 28
29 – 31
32 – 34
35 – 37
38 – 40
Notice that the last range goes to 40 which is one higher than 39, but that’s okay because eventually if we were to continue to collect data we might eventually see a 40.
So, we gave you two examples – one with five intervals, and the other example with six intervals in it. So, how do you know when to choose five, when to choose six, or when to choose seven, eight, or
nine? There is no hard and fast rule on this, but here is a “rule of thumb” list.
Now we count the number of scores that fall into each cell and we are ready to generate the frequency distribution or histogram. See below. | {"url":"https://www.sixsigmadaily.com/cell-intervals-rules-of-thumb/","timestamp":"2024-11-03T16:54:16Z","content_type":"text/html","content_length":"77557","record_id":"<urn:uuid:60daf3a0-94fe-4d78-994f-65636ad0a594>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00652.warc.gz"} |
A probabilistic algorithm for MEG source reconstruction
We present a novel algorithm for source localization based on probabilistic modeling of stimulus-evoked MEG/EEG data. This algorithm localizes multiple dipoles with the computational complexity
equivalent to a single dipole scan, and is therefore more efficient than traditional multidipole fitting procedures. The algorithm assumes that the activity of multiple dipolar sources can be
characterized by a linear combination of known temporal basis functions with unknown coefficients. We model the sensor data as arising from activity in each voxel of interest, plus background
activity. We estimate temporal basis functions from the data using a probabilistic algorithm called partitioned-factor analysis, previously developed in our lab. We model background activity outside
the voxel of interest as an unknown linear mixture of unobserved background factors plus diagonal sensor noise. We use an Expectation-Maximization algorithm to calculate MAP estimates of unknown
basis function coefficients, background mixing matrix, sensor noise covariance and the likelihood of a dipole in each voxel of interest. In simulations, the algorithm is able to accurately localize
several simultaneously-active dipoles, at SNRs typical for averaged MEG data. The algorithm performs well even in configurations that include deep sources and highly correlated sources, and thus is
superior to MUSIC and beamforming techniques which are sensitive to correlated sources. The algorithm also correctly localizes real somatosensory and auditory evoked fields to the postcentral sulcus
and lower bank of the lateral sulcus, respectively.
Conference 4th IEEE Sensor Array and Multichannel Signal Processing Workshop Proceedings, SAM 2006
Country/Territory United States
City Waltham, MA
Period 12/07/06 → 14/07/06
Dive into the research topics of 'A probabilistic algorithm for MEG source reconstruction'. Together they form a unique fingerprint. | {"url":"https://research-test.aston.ac.uk/en/publications/a-probabilistic-algorithm-for-meg-source-reconstruction","timestamp":"2024-11-13T13:01:59Z","content_type":"text/html","content_length":"59233","record_id":"<urn:uuid:aefe1cb8-052c-4221-94c9-91739fb676a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00602.warc.gz"} |
Science:Math Exam Resources/Courses/MATH101/April 2017/Question 02 (b)
MATH101 April 2017
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) (i) • Q2 (c) (ii) • Q2 (c) (iii) • Q2 (c) (iv) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b)
• Q8 • Q9 • Q10 • Q11 (a) • Q11 (b) • Q11 (c) •
Question 02 (b)
Which of the following substitutions is most helpful in evaluating the integral ${\displaystyle \int _{2}^{4}{\frac {dx}{x^{2}{\sqrt {x^{2}+2x+10}}}}}$?
F: ${\displaystyle x={\sqrt {10}}\tan u}$
G: ${\displaystyle u=x^{2}+2x+10}$
H: ${\displaystyle u=x^{2}}$
J: ${\displaystyle x=3\sec u-1}$
K: ${\displaystyle x=3\tan u-1}$
L: ${\displaystyle x={\sqrt {10}}\sec u}$
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is
correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Use the identity ${\displaystyle \sec ^{2}u=\tan ^{2}u+1}$ to simplify the inside of the square root.
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
• If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you
are stuck or if you want to check your work.
• If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem
and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
We want an expression that simplifies the one on the square root.
Note that ${\displaystyle x^{2}+2x+10=(x+1)^{2}+3^{2}}$, thus one of the expressions with a -1 on it will help us get rid of terms. The standard trigonometric identity with ${\displaystyle \tan }$
and ${\displaystyle \sec }$ is ${\displaystyle \tan {^{2}}x+1=\sec {^{2}}x}$. Note that if we substitute ${\displaystyle x=3\tan u-1}$ the expression on the inside of the square root becomes
${\displaystyle (3\tan u-1+1)^{2}+3^{2}=3^{2}(\tan {^{2}}u+1)=(3\sec u)^{2}}$ and this greatly simplifies the expression on the square root.
Answer: ${\displaystyle \color {blue}K:x=3\tan u-1}$ | {"url":"https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH101/April_2017/Question_02_(b)","timestamp":"2024-11-09T13:39:39Z","content_type":"text/html","content_length":"54410","record_id":"<urn:uuid:80a04414-c240-49fb-95cb-c9a8f667bca9>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00357.warc.gz"} |
Quadratic Forms of Random Variables
Quadratic Forms and Transformation
Let \(A = \{a_{ij}\}\) be an \(n\times n\) matrix. A quadratic function of \(n\) variables \(x = (x_1,\ldots, x_n)’\) is defined as
f(x) = x’ A x = \sum_{i,j} a_{ij} x_i x_j.
Without loss of generality, assume \(A\) is symmetric; otherwise replace \(A\) by \((A+A’)/2\).
Since \(A\) is symmetric, it has spectral decomposition
A = Q’ \Lambda Q.
\(\Lambda\) is diagonal and the diagonal elements \(\lambda_1, \ldots, \lambda_n\) are eigenvalues of \(A\). \(Q = (q_1, \ldots, q_n)\) is a orthogonal matrix with the eigenvectors \(q_i\) as
Let \(y = Q’x = Q^{-1} x\). Then we have
f (x) = x’A x = x’ Q \Lambda Q’ x = y’ \Lambda y = \sum_{i} \lambda_i y_i^2 =\sum_{i} ||q_i’ x||^2 .
Random Variables
Let \(X= (X_1,\ldots, X_n)’\) be a random vector, with expectation \(\mu\) and covariance matrix \(\Sigma\):
\mu = E[X] = (E[X_1], \ldots, E[X_n])
\Sigma = E[(X-\mu) (X-\mu)’]
The covariance matrix \(\Sigma\) is symmetric and positive semi-definite. This is because, for any vector \(b\) and \(Y= b’X\),
0 \leq Var[Y] = Var[b’X] = b’ \Sigma b.
Let \(A\) be a symmetric matrix, and define random variable \(Y = X’AX\). Then,
E[Y] = E[X’A X] = tr(E[X’A X]) = E[ tr(X’A X) ] = E[ tr(A X X’) ]
= tr(A E[ X X’ ]) = tr(A (\Sigma + \mu \mu’)) = tr(A\Sigma) + \mu’A\mu | {"url":"http://theoryapp.com/quadratic-forms-of-random-variables/","timestamp":"2024-11-12T03:11:31Z","content_type":"text/html","content_length":"97415","record_id":"<urn:uuid:fd0ce73f-52bf-4a27-b0c2-ff710d7588c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00432.warc.gz"} |
Binary Decimal And Hexadecimal
Existing Tutorials
Again, there are existing tutorials that touch on this topic.
Sigma's Learn Assembly in 28 days, day3.
Cobb's z80 Assembly, Data and Numbers.
If you still don't understand number bases, read on.
The Binary Base
The most basic counting system you can get, it is based on just 1's and 0's. Each 1 or 0 is a bit, and 8 bits make up a byte. Counting in binary is related to counting in decimal. Here's how: You
keep counting in the units digit (farthest right) until there are no more number symbols to count. Then you increment the next digit and start over again. Here's an example:
binary 0,1,10,11,100,101,110
decimal 0,1,2,3,4,5,6
The Decimal Base
Hopefully you know how this base works. It's what you use everyday. The digits are 0,1,2,3,4,5,6,7,8, and 9.
The Hexadecimal Base
Counting in ones and zeros can be a pain, and converting between decimal and binary is no better. So, to help sort ease the pain of programmers we'll use the hexadecimal base. This base uses the
digits 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E, and F. Again, counting similar binary and decimal.
Conversions of Binary, Decimal, and hexadecimal
Binary to Decimal
The easier of the two, what you do is take each digit times 2 to the power of the column it is in and add it all together. Please note that the farthest right digit is the 0th digit, not the 1st
digit. Here's and example:
Binary Number 1001
Conversion to Decimal $1*2^3 + 0*2^2 + 0*2^1 + 1*2^0=9$
Decimal to Binary
Converting back to binary is slightly more complicated. What you have to do is figure out the largest power of 2 ($2^n$) that is smaller or equal to than your number. Write the nth digit as a 1 and
continue. You then subtract the power of 2 from the number and repeat until the decimal number equals 0 (take the result of the subtraction as the new number). If you skip over a power of two, don't
forget to write a 0.
Some Useful Powers of 2 $2^0=1$,$2^1=2$,$2^2=4$,$2^3=8$,$2^4=16$,$2^5=32$,$2^6=64$,$2^7=128$,$2^8=256$,$2^9=512$,$2^10=1024$,$2^11=2048$,$2^12=4096$,$2^13=8192$,$2^14=16384$,$2^15=32768$
Decimal Number 121
Conversion to Binary
1. The largest power of 2 that works is $2^6=64$.
2. $121-64= 57$. Binary number so far: 10000000
3. Next Power of 2 that works is $2^5=32$.
4. $57-32=25$. Binary number so far: 11000000
5. Next Power of 2 that works is $2^4=16$.
6. $25-16=9$. Binary number so far: 11100000
7. Next Power of 2 is $2^3=8$.
8. $9-8=1$. Binary number so far: 11110000
9. Next Power of 2 smaller that works is $2^0=1$.
10. $1-1=0$. You're done. Don't forget to write your zeros for the powers of 2 you skipped.
Final answer: 11110001
Binary to Hexadecimal
A rather simple task, though slightly harder than converting binary to decimal. You do the same thing except make sure that when adding you don't go onto the next digit until the previous digit is
larger than F (15 in decimal).
Binary number 11110011
Conversion to Hexadecimal $1*2^7+1*2^6+1*2^5+1*2^4+0*2^3+0*2^2+1*2^1+1*2^0=F3$
You'll probably be asking "how is that easier than converting binary to decimal?"
Here's your answer: If you haven't notice yet, but each digit in hexadecimal is represented by 4 digits in binary. What does this mean? It means that you only have to compute hexadecimal a nibble (4
bits) at a time. So:
Previous number split up into nibbles 1111 0011
First nibble (right one) 0011=3
Second nibble 1111=F
Hexadecimal to Binary
Another simple task, again we'll use the notion of nibbles to help us. Take each hexadecimal digit and convert it to decimal (A=10,B=11,C=12,D=13,E=14,F=15). Then convert that number to binary.
Hexadecimal Number E7
First digit 7
Decimal equivalence 7
Binary conversion 7->0111
Second digit E
Decimal equivalence 14
Binary conversion 14->1110
Final answer 11100111
Hexadecimal to Decimal
Simple. Just take the decimal equivalence of the hexadecimal digit, times it by $16^n$ where n equals the location the digit is in (0 being the farthest right), and add them all up.
Hexadecimal Number 3BA
Decimal Conversion $3*16^2+11*16^1+10*16*0=954$
Decimal to Hexadecimal
Similar to converting from binary to hexadecimal. Just don't forget you need to add up to F (15) before going onto the next digit.
Decimal Number 68
Hexadecimal Conversion $6*A^1+8*A^0=44$
Setting up Windows Calculator to convert for you
If you didn't already know, you can use Windows Calculator to convert to certain bases. Here's how you set it up:
• Go to view
• Change to Scientific mode (as opposed to the default "Standard" mode)
• Choose the base of the number you want to convert by selecting either hex, dec, oct, or bin.
• Determine how large of a number you're going to input, and adjust the panel next to it accordingly:
byte: Fits in 8 bits
word: Fits in 16 bits
Dword: Fits in 32 bits
Qword: Fits in 64 bits
• Type in your number
• Click on the base you want to convert to.
• The number will be changed to the equivalent number for that base
Common notation
How do you distinguish one number base from other base numbers? The compiler has 2 ways:
Denote before
Use a symbol to denote the base of the number.
binary %
hexadecimal $
Note that decimal has no pre-denote form.
Denote after
Use a symbol after to denote the base of the number.
binary b
decimal d
hexadecimal h
Note that if you're using this method with hexadecimal, it is important that the first digit is not a letter (A,B,C,D,E, or F). If necessary, add a 0 to the front.
Just to save you some trouble, if you don't put a denotation for a number the compiler will assume that it is a decimal number.
Here are some questions to make sure that you really understand this. If not, perhaps you should re-read this section.
Convert the following numbers
Don't use a base converter, but a calculator is alright if you need to multiply certain things.
Binary to Decimal
1. 1110
2. 11001001
3. 110101011000
4. 1111
5. 10001000
Binary to Hexadecimal
1. 1001
2. 10101010
3. 10011101
4. 1100
5. 10000110
Decimal to Binary
1. 10
2. 36
3. 57
4. 184
5. 376
Decimal to Hexadecimal
1. 91
2. 100
3. 852
4. 4533
5. 424
Hexadecimal to Binary
1. 10
2. AB
3. 8F
4. 3E
5. B7
Hexadecimal to Decimal
1. 4
2. F8
3. EB
4. D1
5. 97
Identify the following numbers as binary, decimal, or hexadecimal according the the denotation.
1. %10001110
2. 1011
3. $100101
4. 10h
5. 1111b
6. 11
*7. AEh
8. 54d
9. %10010010011
10. 100000000001h
*What's wrong with this?
For solutions, see this page.
It takes a lot of practice to get good at converting between bases. Although you could just use Windows calculator to convert between bases, it's still useful to know how it's done by hand. If used
correctly, bases and converting between bases will make programming as well as understanding other people's code a lot easier.
page revision: 16, last edited: 27 Jun 2007 19:59 | {"url":"http://z80-heaven.wikidot.com/binary-decimal-and-hexadecimal","timestamp":"2024-11-07T20:45:49Z","content_type":"application/xhtml+xml","content_length":"43411","record_id":"<urn:uuid:db9fa685-a905-47ca-85e7-cec4ee3545a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00443.warc.gz"} |
Winning Blackjack - Odds and Favors - Agen Poker 99 Top
Winning Blackjack – Odds and Favors
Whether you play blackjack for thrill and the excitement of the winning streak, or you play blackjack simply to make a profit, blackjack odds and blackjack advantages are an important consideration.
Assuming the house has an edge in the game of blackjack, putting the odds in your favor is the sure way to a win. Odds and probabilities When a person sits down at a blackjack table to play
blackjack, one of the first things that will be scrutinized is the odds. How do you work them out? Essentially, you have to work them out mathematically so that you will have a number you can the
work out your hand versus the dealer’s up-cards. Firstly, you need to work out the dealer’s hand contains two cards of equal value, face cards or 10 or above. In order to determine this, you will
have to calculate the probability of making a “blackjack” hand on the spot. This is simply the probability that the dealer will go “blackjack” positive. In addition, you will have to calculate the
probability of the dealer going bust. This is more complicated, as the following calculations require some complicated Bee Math. However, you will certainly need to calculate these basic
probabilities of cards turning over.
Winning Blackjack – Odds and Favors
This is simply the probability that the dealer will go “blackjack”
Working out the blackjack odds
The blackjack odds are best understood by looking at the probabilities of cards coming out of the deck. Let’s say that there are four cards of one rank and three cards of another rank (for example,
Ace of spades, King of spades, Queen of spades, Jack of spades). Firstly, you need to determine the probability of any ace or king being made. In this example, there is a one in four chance of this
happening. Similarly, you can calculate the chances of a queen being made – one in nine chances, again according to the cards in the deck. Next, you need to determine the chances of any other card in
the deck being made. In this example, there is a one in four chance of this happening. Finally, you need to determine the probability of making an out hand on the flop, which is the next step in the
calculation of the odds.
Calculating the Odds
The next step in blackjack odds is to calculate the odds of making your hand in the fraction of a second that the dealer will make his hand. This is the kind of instant Odds calculation that you
would use, assuming you had super-peed calculation powers. The simplest way to explain this is as follows: given two cards of equal value, the odds of the next card is the ratio of the numerator of
the numerator of the total number of cards remaining in the deck. In our example of four cards of one rank and three cards of another, there are three cards of Ace, two cards of King, one card of
Queen and one card of Jack. Jack of spades is the highest card, and thus the chances of being dealt a second Ace or King are greater than those of getting a Jack. In this situation, you will have to
multiply your outs by four, giving you the total number of cards that will help you – 2 outs to make a winning hand, 4 outs to win and the rest to losses.
Calculating pot odds
When you are deciding whether to call a bet, calculate the odds of you winning the pot, using the pot odds. You need to be able to know the odds of making your hand in the fraction of a second, so
that you know if you can call in the end. Keep in mind that this may not be the best way to decide whether to play a hand or not, as this does not take into account other bets that may be still in
the game.
placing a bet in a game of poker is the act of putting money into the community pot of the game. The amount depends on the judgment of the judges whether it is bet to play or not. Play starts the
first or second or third and the following player can place a bet as long as the goalkeeper has not placed a bet in the previous round.
In conclusion, knowing the blackjack odds, pot odds and payoff options, you can choose the best methods to problem solve and effectively solve the blackjack with the least effort. | {"url":"https://agenpoker99.top/winning-blackjack-odds-and-favors/","timestamp":"2024-11-07T02:25:26Z","content_type":"text/html","content_length":"42413","record_id":"<urn:uuid:6b80c630-8339-482c-a93a-ae391f070e4d>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00009.warc.gz"} |
Functional, extendible arrays.
Functional, extendible arrays. Arrays can have fixed size, or can grow automatically as needed. A default value is used for entries that have not been explicitly set.
Arrays uses zero-based indexing. This is a deliberate design choice and differs from other Erlang data structures, for example, tuples.
Unless specified by the user when the array is created, the default value is the atom undefined. There is no difference between an unset entry and an entry that has been explicitly set to the same
value as the default one (compare reset/2). If you need to differentiate between unset and set entries, ensure that the default value cannot be confused with the values of set entries.
The array never shrinks automatically. If an index I has been used to set an entry successfully, all indices in the range [0,I] stay accessible unless the array size is explicitly changed by calling
Create a fixed-size array with entries 0-9 set to undefined:
A0 = array:new(10).
10 = array:size(A0).
Create an extendible array and set entry 17 to true, causing the array to grow automatically:
A1 = array:set(17, true, array:new()).
18 = array:size(A1).
Read back a stored value:
true = array:get(17, A1).
Accessing an unset entry returns default value:
undefined = array:get(3, A1)
Accessing an entry beyond the last set entry also returns the default value, if the array does not have fixed size:
undefined = array:get(18, A1).
"Sparse" functions ignore default-valued entries:
A2 = array:set(4, false, A1).
[{4, false}, {17, true}] = array:sparse_to_orddict(A2).
An extendible array can be made fixed-size later:
A3 = array:fix(A2).
A fixed-size array does not grow automatically and does not allow accesses beyond the last set entry:
{'EXIT',{badarg,_}} = (catch array:set(18, true, A3)).
{'EXIT',{badarg,_}} = (catch array:get(18, A3)).
A functional, extendible array. The representation is not documented and is subject to change without notice. Notice that arrays cannot be directly compared for equality.
array_indx() = integer() >= 0
array_opt() =
{fixed, boolean()} |
fixed |
{default, Type :: term()} |
{size, N :: integer() >= 0} |
(N :: integer() >= 0)
default(Array :: array(Type)) -> Value :: Type
Gets the value used for uninitialized entries.
See also new/2.
fix(Array :: array(Type)) -> array(Type)
Fixes the array size. This prevents it from growing automatically upon insertion.
See also set/3 and relax/1.
foldl(Function, InitialAcc :: A, Array :: array(Type)) -> B
• Function =
fun((Index :: array_indx(), Value :: Type, Acc :: A) -> B)
Folds the array elements using the specified function and initial accumulator value. The elements are visited in order from the lowest index to the highest. If Function is not a function, the call
fails with reason badarg.
See also foldr/3, map/2, sparse_foldl/3.
foldr(Function, InitialAcc :: A, Array :: array(Type)) -> B
• Function =
fun((Index :: array_indx(), Value :: Type, Acc :: A) -> B)
Folds the array elements right-to-left using the specified function and initial accumulator value. The elements are visited in order from the highest index to the lowest. If Function is not a
function, the call fails with reason badarg.
See also foldl/3, map/2.
from_list(List :: [Value :: Type]) -> array(Type)
from_list(List :: [Value :: Type], Default :: term()) ->
Converts a list to an extendible array. Default is used as the value for uninitialized entries of the array. If List is not a proper list, the call fails with reason badarg.
See also new/2, to_list/1.
from_orddict(Orddict :: indx_pairs(Value :: Type),
Default :: Type) ->
Converts an ordered list of pairs {Index, Value} to a corresponding extendible array. Default is used as the value for uninitialized entries of the array. If Orddict is not a proper, ordered list of
pairs whose first elements are non-negative integers, the call fails with reason badarg.
See also new/2, to_orddict/1.
get(I :: array_indx(), Array :: array(Type)) -> Value :: Type
Gets the value of entry I. If I is not a non-negative integer, or if the array has fixed size and I is larger than the maximum index, the call fails with reason badarg.
If the array does not have fixed size, the default value for any index I greater than size(Array)-1 is returned.
See also set/3.
is_array(X :: term()) -> boolean()
Returns true if X is an array, otherwise false. Notice that the check is only shallow, as there is no guarantee that X is a well-formed array representation even if this function returns true.
is_fix(Array :: array()) -> boolean()
Checks if the array has fixed size. Returns true if the array is fixed, otherwise false.
See also fix/1.
map(Function, Array :: array(Type1)) -> array(Type2)
Maps the specified function onto each array element. The elements are visited in order from the lowest index to the highest. If Function is not a function, the call fails with reason badarg.
See also foldl/3, foldr/3, sparse_map/2.
Creates a new, extendible array with initial size zero.
See also new/1, new/2.
Creates a new array according to the specified otions. By default, the array is extendible and has initial size zero. Array indices start at 0.
Options is a single term or a list of terms, selected from the following:
N::integer() >= 0 or {size, N::integer() >= 0}
Specifies the initial array size; this also implies {fixed, true}. If N is not a non-negative integer, the call fails with reason badarg.
fixed or {fixed, true}
Creates a fixed-size array. See also fix/1.
{fixed, false}
Creates an extendible (non-fixed-size) array.
{default, Value}
Sets the default value for the array to Value.
Options are processed in the order they occur in the list, that is, later options have higher precedence.
The default value is used as the value of uninitialized entries, and cannot be changed once the array has been created.
creates a fixed-size array of size 100.
creates an empty, extendible array whose default value is 0.
creates an extendible array with initial size 10 whose default value is -1.
See also fix/1, from_list/2, get/2, new/0, new/2, set/3.
new(Size :: integer() >= 0, Options :: array_opts()) -> array()
Creates a new array according to the specified size and options. If Size is not a non-negative integer, the call fails with reason badarg. By default, the array has fixed size. Notice that any
size specifications in Options override parameter Size.
If Options is a list, this is equivalent to new([{size, Size} | Options], otherwise it is equivalent to new([{size, Size} | [Options]]. However, using this function directly is more efficient.
array:new(100, {default,0})
creates a fixed-size array of size 100, whose default value is 0.
See also new/1.
relax(Array :: array(Type)) -> array(Type)
Makes the array resizable. (Reverses the effects of fix/1.)
See also fix/1.
Resets entry I to the default value for the array. If the value of entry I is the default value, the array is returned unchanged. Reset never changes the array size. Shrinking can be done
explicitly by calling resize/2.
If I is not a non-negative integer, or if the array has fixed size and I is larger than the maximum index, the call fails with reason badarg; compare set/3
See also new/2, set/3.
resize(Array :: array(Type)) -> array(Type)
Changes the array size to that reported by sparse_size/1. If the specified array has fixed size, also the resulting array has fixed size.
See also resize/2, sparse_size/1.
resize(Size :: integer() >= 0, Array :: array(Type)) ->
Change the array size. If Size is not a non-negative integer, the call fails with reason badarg. If the specified array has fixed size, also the resulting array has fixed size.
set(I :: array_indx(), Value :: Type, Array :: array(Type)) ->
Sets entry I of the array to Value. If I is not a non-negative integer, or if the array has fixed size and I is larger than the maximum index, the call fails with reason badarg.
If the array does not have fixed size, and I is greater than size(Array)-1, the array grows to size I+1.
See also get/2, reset/2.
size(Array :: array()) -> integer() >= 0
Gets the number of entries in the array. Entries are numbered from 0 to size(Array)-1. Hence, this is also the index of the first entry that is guaranteed to not have been previously set.
See also set/3, sparse_size/1.
sparse_foldl(Function, InitialAcc :: A, Array :: array(Type)) -> B
□ Function =
fun((Index :: array_indx(), Value :: Type, Acc :: A) -> B)
Folds the array elements using the specified function and initial accumulator value, skipping default-valued entries. The elements are visited in order from the lowest index to the highest. If
Function is not a function, the call fails with reason badarg.
See also foldl/3, sparse_foldr/3.
sparse_foldr(Function, InitialAcc :: A, Array :: array(Type)) -> B
□ Function =
fun((Index :: array_indx(), Value :: Type, Acc :: A) -> B)
Folds the array elements right-to-left using the specified function and initial accumulator value, skipping default-valued entries. The elements are visited in order from the highest index to the
lowest. If Function is not a function, the call fails with reason badarg.
See also foldr/3, sparse_foldl/3.
sparse_map(Function, Array :: array(Type1)) -> array(Type2)
Maps the specified function onto each array element, skipping default-valued entries. The elements are visited in order from the lowest index to the highest. If Function is not a function, the
call fails with reason badarg.
See also map/2.
sparse_size(Array :: array()) -> integer() >= 0
Gets the number of entries in the array up until the last non-default-valued entry. That is, returns I+1 if I is the last non-default-valued entry in the array, or zero if no such entry exists.
See also resize/1, size/1.
sparse_to_list(Array :: array(Type)) -> [Value :: Type]
Converts the array to a list, skipping default-valued entries.
See also to_list/1.
sparse_to_orddict(Array :: array(Type)) ->
indx_pairs(Value :: Type)
Converts the array to an ordered list of pairs {Index, Value}, skipping default-valued entries.
See also to_orddict/1.
to_list(Array :: array(Type)) -> [Value :: Type] | {"url":"https://www.erldocs.com/22.0.7/stdlib/array","timestamp":"2024-11-10T02:19:33Z","content_type":"text/html","content_length":"28738","record_id":"<urn:uuid:0c1ba17a-5971-4f5b-9544-0bfd571be52d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00071.warc.gz"} |
Linear Mixed Model with Censoring Options • Genstat Knowledge Base 2024
Selects information to be printed by a linear mixed model with censoring analysis and controls certain aspects of the method used.
This specifies which items of output are to be produced by the analysis.
Model Description of the model fitted by the analysis
Variance components Estimates of variance parameters
Estimated effects Estimates of regression coefficients
Predicted means Predicted means
Residual checks Uses the VCHECK procedure to check the residuals for outliers and variance stability
Summary of censoring Gives a summary of fitting the predicted values to the censored observations
Stratum variances Estimates of approximate stratum variances
Variance-covariance matrix Variance-covariance matrix for the variance parameters
Deviance The residual deviance
Wald tests Wald Tests for fixed model terms and accompanying F-statistics (if selected)
Missing value estimates Estimates of values missing from the input
Monitoring Monitoring information at each iteration
Akaike information coefficient Akaike Information Coefficient to assess the random model
Schwarz information coefficient Schwarz Information Coefficient to assess the random model
Use full likelihood for AIC/SIC The information coefficients are calculated using the full log-likelihood from the VKEEP procedure, otherwise the residual log-likelihood is used. See the VAIC
procedure for details.
This specifies which graphics are produced by the analysis.
Residual plots This gives a default residual plot from the fitted model
Means plots This gives default plot of the predicted means against the factors in the model
Censored predictions This plots the estimated y values against the observed y values
More control over the residual and mean plots can be obtained by using the Further Output dialog.
Standard errors
Tables of means and effects are accompanied by estimates of standard errors. You can choose whether Genstat computes standard errors or standard errors of differences (SEDs) for the tables.
Approximate least significant differences (LSDs) for the predicted means of the fixed terms specified in the Model terms for effects and means field can be computed by selecting LSDs. These are
calculated using the approximate numbers of residual degrees of freedom printed by the analysis in the d.d.f column in the table of tests for fixed tests (produced by selecting the Wald tests display
option). The degrees of freedom are relevant for assessing the fixed term as a whole, and may vary over the contrasts amongst the means of the term. So the LSDs should be used with caution. If you
are interested in a specific comparison, you should set up a 2-level factor to fit this explicitly in the analysis. The significance level for LSDs can be specified as a percentage (default 5) in the
accompanying field.
Method for calculating F-statistics
This controls whether Wald tests for fixed effects are accompanied with approximate F statistics and corresponding numbers of residual degrees of freedom. The computations, using the method devised
by Kenward & Roger (1997), can be time consuming with large or complicated models. So, the default setting automatic, can be used to allow Genstat to assess the model itself and decide automatically
whether to do the computations and which method to use. The other settings allow you to control what to do yourself:
none No F statistics are produced
algebraic F statistics are calculated using algebraic derivatives (which may involve large matrix calculations)
numerical F statistics are calculated using numerical derivatives (which require an extra evaluation of the mixed model equations for every variance parameter).
A variate of weights can be supplied to give varying influence of each unit on the fit of the model. This would usually correspond to a known pattern of variance of the observations, where the
weights would be the reciprocals of the variances. A variate that specifies the weights in the analysis can be selected from the drop down list or you can type its name into the field. If the
y-variate has been specified, only variates that match its length will be displayed in the list. The variates in the list are sorted in the standard available-data order.
Censoring convergence tolerance (%)
This controls the convergence criterion of the E-M algorithm in the TOBIT procedure. When the changes between cycles in Means and variance Components are both less than the given percentages, then
the algorithm will stop.
Estimate missing data values
This specifies whether predictions are formed from the fitted model for missing values of the y-variate; alternatively any units with missing values in the y-variate are excluded from the analysis.
Include units with missing factor values
This specifies whether data units with missing values in any of the factors in the fixed or random models are included in the analysis. Units with missing y values are always excluded from the
Estimate constant term
Specifies whether a constant term is included in the fixed model.
Covariates centred to zero mean
Specifies whether covariates are centred to zero mean during the analysis. This applies to all covariates in the model. If covariates are centred, tables of predicted means are based on the mean
covariate value, otherwise zero for each covariate.
Optimization method
The AI (Average Information) method is the standard optimization method for REML in Genstat. It uses sparse matrices and is particularly recommended for large datasets and/or complex models. An
alternative method is Fisher scoring which, if selected, also allows an absorbing factor to be specified in the model. This can be used to reduce the time or space requirements when fitting large
models with many parameters. A more detailed discussion of the use and choice of absorbing factors can be found in the Genstat 5 Reference Manual.
Maximum iterations
This specifies the maximum number of iterations to use to optimize the REML likelihood and in the TOBIT model E-M algorithm.
See also | {"url":"https://genstat.kb.vsni.co.uk/knowledge-base/lmmcensored-options/","timestamp":"2024-11-02T17:44:15Z","content_type":"text/html","content_length":"48018","record_id":"<urn:uuid:24dd4304-6ac8-41d4-a32b-6f5acc853633>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00620.warc.gz"} |
An improvement in the value of the objective function per unit increase in the right-hand side of a constraint is the
An improvement in the value of the objective function per unit increase in the right-hand side of a constraint is the
Solution 1
The improvement in the value of the objective function per unit increase in the right-hand side of a constraint is known as the Shadow Price or Dual Value.
Here are the steps to understand this concept:
1. In linear programming, the objective function is what you are trying to maximize or minimiz Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem.
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Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem.
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Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions. | {"url":"https://knowee.ai/questions/37993590-an-improvement-in-the-value-of-the-objective-function-per-unit","timestamp":"2024-11-11T21:47:08Z","content_type":"text/html","content_length":"364527","record_id":"<urn:uuid:e0eeaf1d-d654-48ed-99d2-d8c7d8ea958f>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00026.warc.gz"} |
[Previous: AMFs for Highway Segments] [Table Of Contents] [Next: Examples]
7. AMFs for Intersections
The procedures for determining the values of the AMFs for intersections (AMF[10] through AMF[14]) are described in this section. The values of the geometric and traffic control variables needed to
determine the AMFs are found in Intersection Geometric and Traffic Control Data. In all cases, the values for the AMFs for minor leg YIELD-controlled intersections are the same as for minor
STOP-controlled intersections.
7.1 Intersection Skew Angle (AMF[10])
The AMF for intersection skew angle (AMF[10]) is defined as follows:
For a three-leg intersection with minor-road STOP-control, the value of AMF[10] is:
• SKEW = absolute value of the intersection skew angle (degrees) expressed as the absolute value of the difference between 90 degrees and the actual angle between the major legs and minor legs of
the intersection. This absolute value is always within the range from 0 to 90 degrees (i.e., for nonzero skew angles, always measure the acute rather than the obtuse angle).
For a four-leg intersection with minor-road STOP-control, the value of AMF[10] is:
• AMF[10] = exp ( 0.0054 SKEW ) (A-15)
If the skew angle for a four-leg intersection with minor-road STOP-control differs for the two minor legs, AMF[10] is computed separately for each minor leg and the results averaged.
For a four-leg signalized intersection or an all-way STOP-controlled intersection, the value of AMF[10] is 1.00.
7.2 Intersection Traffic Control (AMF[11])
The value of AMF[11] for an intersection with all-way STOP control is 0.53. For all other intersections, the value of AMF[11] is 1.00.
7.3 Intersection Left-Turn Lanes (AMF[12])
The values of the AMF for left-turn lanes (AMF[12]) on one or more major legs to an intersection is specified in Table 10., Accident Modification Factors for Installation of Left-turn Lanes on the
Major Legs to Intersections. If there are no left-turn lanes on any major legs to the intersection, the value of AMF[12] is 1.00. These AMFs are based upon the research documented in Report No.
FHWA-RD-02-089, Safety Effectiveness of Intersection Left- and Right-Turn Lanes. The values in this table are from the judgement of an expert panel combining results from several sources.
Table 10.: Accident Modification Factors for Installation of Left-turn Lanes on the Major Legs to Intersections
│Intersection type │Intersection traffic control│One approach (leg) on which left-turn lanes installed│Both approaches (legs) on which left-turn lanes installed│
│Three-leg intersection│STOP control │0.56 │N/A │
│Three-leg intersection│Signal control │0.85 │N/A │
│Four-leg intersection │STOP control │0.72 │0.52 │
│Four-leg intersection │Signal control │0.82 │0.67 │
Note that three-leg, signal-control intersections are not modeled by CPM.
7.4 Intersection Right-Turn Lanes (AMF[13])
The value of the AMF for right-turn lanes (AMF[13]) on one or more major legs to an intersection is specified in Table 11., Accident Modification Factors for Installation of Right-turn Lanes on the
Major Legs to Intersections. If there are no right-turn lanes on any major leg to the intersection, the value of AMF[13] is 1.00. These AMFs are based upon the research documented in Report No.
FHWA-RD-02-089, Safety Effectiveness of Intersection Left- and Right-Turn Lanes.
Table 11.: Accident Modification Factors for Installation of Right-turn Lanes on the Major Legs to Intersections
│ Intersection type │Intersection traffic control│One approach (leg) on which right-turn lanes are installed│Both approaches (legs) on which right-turn lanes are installed│
│Three-leg intersection│ STOP control │ 0.86 │ N/A │
│Three-leg intersection│ Signal control │ 0.96 │ N/A │
│Four-leg intersection │ STOP control │ 0.86 │ 0.74 │
│Four-leg intersection │ Signal control │ 0.96 │ 0.92 │
Note that three-leg, signal-control intersections are not modeled by CPM.
7.5 Intersection Sight Distance (AMF[14])
The value of the AMF for llimited intersection sight distance (AMF[14]) at three- and four-leg intersections with minor leg STOP control is specified in Table 12., Accident Modification Factors for
Intersection Sight Distance Limitations in Quadrants of Three-Leg and Four-Leg Intersections with Minor STOP Control. For four-leg signalized intersections and all-way STOP-controlled intersections,
the value of AMF[14] is 1.00.
Table 12.: Accident Modification Factors for Intersection Sight Distance Limitations in Quadrants of
Three-Leg and Four-Leg Intersections with Minor STOP Control
│Number of quadrants with limited intersection sight distance │Accident Modification Factor (AMF[14])│
│0 │1.00 │
│1 │1.05 │
│2 │1.10 │
│3 │1.15 │
│4 │1.20 │
[Previous: AMFs for Highway Segments] [Top] [Next: Examples] | {"url":"https://ihsdm.org/public/html/user/cpm/cpm_em.7.html","timestamp":"2024-11-06T21:25:43Z","content_type":"text/html","content_length":"19925","record_id":"<urn:uuid:a5669999-d1f6-443a-a3e0-e27b95450b15>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00484.warc.gz"} |
Gaps and missing numbers finder
This online calculator helps you find gaps and missing numbers in an integer sequence.
Disclaimer: This calculator was made for a specific purpose to find gaps in a continuous number sequence. If you are looking for math problem solving, then it is probably about arithmetic sequence or
geometric sequence - please look at Arithmetic sequence calculator and problems solver or Geometric sequence calculator and problems solver
Let's suppose you have a text file of consecutive numbers, such as below, with each number on its own line:
However, there are some missing numbers and gaps. In the example above, 5 is missing, and there are two gaps: between 8 and 11 and also between 14 and 17.
Of course, in this example, it is easy to see the problems just by looking at numbers, but if we were talking about thousands of numbers, it would be too cumbersome to locate the gaps by eye. That's
why I've created this calculator. You can paste in a list of numbers, and it will display all the missing numbers and gaps it was able to find in the sequence, bounded by the first and the last line
of the list.
I have also added a couple of options:
• You can set the "Collapse gaps" checkbox to output gaps in a collapsed form, in other words, taking 8 - 11 as an example, the 8 is the start of the gap and 11 is the end. This option is turned on
by default. If you uncheck it, you will get a list of all numbers in a gap, i.e. 8, 9, 10, 11.
• You can set the "Use regular expression to parse number" checkbox to parse the line of your text file with regular expression.
The reason for the second option is that your list is unlikely to include just numbers. Your text file will probably include additional text, something like:
1 Paper beats stone
2 Scissors beat paper
3 Stone beats scissors
For that case, you can use a regular expression to extract the number from the line and then check the sequence. You need to set the "Use regular expression to parse number" checkbox and enter a
regular expression and match group, which will be used to extract the number.
Default regular expression matches the line where the number is placed at the beginning of the line.
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PLANETCALC, Gaps and missing numbers finder | {"url":"https://pt.planetcalc.com/7471/?thanks=1","timestamp":"2024-11-10T05:19:36Z","content_type":"text/html","content_length":"36930","record_id":"<urn:uuid:181517e3-cce5-451e-8799-594ee3c0d7b6>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00401.warc.gz"} |
Computing Coherence of Two Signals
Hey there!
I've been looking into the math of coherence of signals, and I'm not quite sure how to compute it... Is there a phase variable in the Processing code, or even a coherence one? And in case there
isn't, does anyone know any pointers as to how I can compute it?
Any help would really be appreciated...
I should phrase it differently:
I've looked throught all the processing code, everything I could find seemed to be the FFT, so we have the Complex Discrete Fourier Transform. Is that sufficient for getting the coherence of two
Thanks a lot for the link!
What helped me in the end was the following paper:
I managed to compute the coherence of 2 FFTs in Processing, so it's in real time. If anyone is interested in the code, just PM me, or I can post it!
Luc, that's great, thanks.
Do you have a Github link? Or consider posting on Github.
Sure, you can find it here:
I am not sure whether or not I have computed it well though. The values are between 0 and 1, of course, but I have noticed the following:
When measuring frontal alpha (8-12hz) coherence in an eyes closed sample session, the values are dancing around at about .8, which is way too high it seems, doesn't it? When I compare a sample
eyes-closed session with a meditation session, my alpha coherence drops.
Will do some more measurements, but I would be really thankful if someone could look over the code in order to see if there is an obvious mistake... But I'm pretty sure it's right | {"url":"https://openbci.com/forum/index.php?p=/discussion/1115/computing-coherence-of-two-signals","timestamp":"2024-11-06T04:32:17Z","content_type":"text/html","content_length":"51980","record_id":"<urn:uuid:1a117c07-3543-43dc-85ab-39664ff75240>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00361.warc.gz"} |
0 - 1 Knapsack Problem
Use dynamic programming to implement the solution of a problem.
We'll cover the following
Problem statement
For each item, you are given its weight and its value. You want to maximize the total value of all the items you are going to put in the knapsack such that the total weight of the items is less than
the knapsack’s capacity. What is this maximum total value?
The only condition is to consider all subsets of items. There can be two cases for every item:
1. The item is included in the optimal subset.
2. The item is not included in the optimal set.
Therefore, the maximum value that can be obtained from n items is the maximum of the following two values:
1. Maximum value obtained by n - 1 items and W weight (excluding ${n^{th}}$ item)
2. Value of ${n^{th}}$ item plus maximum value obtained by n - 1 items and W minus weight of the ${n^{th}}$ item (including ${n^{th}}$ item). If the weight of the ${n^{th}}$ item is greater than W,
the ${n^{th}}$ item cannot be included and case 1 is the only possibility.
Let’s look at the recurrence relation:
• Base case: If we have explored all the items or if we have reached the maximum capacity of Knapsack,
if (n=0 or W=0)
return 0
• If the weight of the ${n^{th}}$ item is greater than the capacity of the knapsack, we cannot include this item:
if (weight[n] > W)
return solve(n-1, W)
• Otherwise, we can,
return max{solve(n-1, W), solve(n-1, W-weight[n])}
Here, the expression solve(n - 1, W) means that we have not included the item and the expression solve(n - 1, W - weight[n]) means that we have included that item in the knapsack.
If we build the recursion tree for the above relation, we can see that the property of overlapping sub-problems is satisfied. So, let’s try to solve it using dynamic programming.
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MA8352 Notes Linear Algebra and Partial Differential Equations
MA8352 Notes Linear Algebra and Partial Differential Equations Regulation 2017 Anna University
MA8352 Notes Linear Algebra and Partial Differential Equations
MA8352 Notes Linear Algebra and Partial Differential Equations Regulation 2017 Anna University free download. Linear Algebra and Partial Differential Equations Notes MA8352 pdf free download.
OBJECTIVES: MA8352 Notes Linear Algebra and Partial Differential Equations
To introduce the basic notions of groups, rings, fields which will then be used to solve related problems.
To understand the concepts of vector space, linear transformations and diagonalization.
To apply the concept of inner product spaces in orthogonalization.
To understand the procedure to solve partial differential equations.
To give an integrated approach to number theory and abstract algebra, and provide a firm basis for further reading and study in the subject.
OUTCOMES: MA8352 Notes Linear Algebra and Partial Differential Equations
Upon successful completion of the course, students should be able to:
Explain the fundamental concepts of advanced algebra and their role in modern
mathematics and applied contexts.
Demonstrate accurate and efficient use of advanced algebraic techniques.
Demonstrate their mastery by solving non – trivial problems related to the concepts and by proving simple theorems about the statements proven by the text.
Able to solve various types of partial differential equations.
Able to solve engineering problems using Fourier series.
TEXTBOOKS: MA8352 Notes Linear Algebra and Partial Differential Equations
1. Grewal B.S., “Higher Engineering Mathematics”, Khanna Publishers, New Delhi, 43rd Edition, 2014.
2. Friedberg, A.H., Insel, A.J. and Spence, L., “Linear Algebra”, Prentice Hall of India, New Delhi, 2004.
REFERENCES: MA8352 Notes Linear Algebra and Partial Differential Equations
1. Burden, R.L. and Faires, J.D, “Numerical Analysis”, 9th Edition, Cengage Learning, 2016.
2. James, G. “Advanced Modern Engineering Mathematics”, Pearson Education, 2007.
3. Kolman, B. Hill, D.R., “Introductory Linear Algebra”, Pearson Education, New Delhi, First Reprint, 2009.
4. Kumaresan, S., “Linear Algebra – A Geometric Approach”, Prentice – Hall of India, New Delhi, Reprint, 2010.
5. Lay, D.C., “Linear Algebra and its Applications”, 5th Edition, Pearson Education, 2015.
6. O‟Neil, P.V., “Advanced Engineering Mathematics”, Cengage Learning, 2007.
7. Strang, G., “Linear Algebra and its applications”, Thomson (Brooks/Cole), New Delhi, 2005.
8. Sundarapandian, V. “Numerical Linear Algebra”, Prentice Hall of India, New Delhi, 2008.
Subject name Linear Algebra and Partial Differential Equations
Semester 3
Subject Code MA8352
Regulation 2017 regulation
MA8352 Notes Linear Algebra and Partial Differential Equations Click Here to download
MA8352 Syllabus Linear Algebra and Partial Differential Equations
MA8352 Important Questions Linear Algebra and Partial Differential Equations
MA8352 Question Bank Linear Algebra and Partial Differential Equations
5 replies on “MA8352 Notes Linear Algebra and Partial Differential Equations Regulation 2017 Anna University”
Please send Sem4 important questions and notrs
sir send me text book LAPDE
202/63B RAGAVENDRA NAGER
PH: 9841186375
Can you teach me online MA 8352 – subject
sir LAPDE handwritten notes and textbook
kindly send to my mail | {"url":"https://padeepz.net/ma8352-notes-linear-algebra-and-partial-differential-equations-regulation-2017-anna-university/","timestamp":"2024-11-08T05:05:46Z","content_type":"text/html","content_length":"60223","record_id":"<urn:uuid:76165268-d5b4-4e0c-baa8-3b075c1c5866>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00469.warc.gz"} |
5th Annual Conference of IT4Innovations
Extended Abstract The main goal of this talk is to put together theoretical results on intermediate quantifiers which were proposed in several papers (see e.g. [1, 2, 3, 4]) with the Fuzzy GUHA
method [5], and to introduce a linguistic characterization of natural data using generalized intermediate quantifiers. The theory of intermediate quantifiers was introduced by Nov´ak in [3] and now
is a constituent of the theory of Fuzzy Natural Logic (FNL), which is a mathematical counterpart of the concept of Natural Logic introduced by Lakoff [6]. This theory is based on Łukasiewicz fuzzy
type theory (Ł- FTT) [4], which is one of the existing higher-order fuzzy logics. Fuzzy GUHA is a special method for automated search of association rules from numerical data. Generally, obtained
associations are in the form A s B, which means that the occurrence of A is associated with the occurrence of B, where A and B are formulae created from objects’ attributes. As proposed by H´ajek et
al. [5], the original GUHA method allowed only boolean attributes to be involved. Some parts of their approach was independently re-invented by Agrawal [7] many years later and is also known as the
mining of association rules or market basket analysis. A detailed book on the GUHA method is [8], where one can find distinct statistically approved associations between attributes of given objects.
Fuzzy GUHA is an extension of a classical GUHA method for fuzzy data. In this paper, we work with associations in the form of IF-THEN rules composed of evaluative linguistic expressions, which allow
the quantities to be characterized with vague linguistic terms such as “very small”, “big”, “medium” etc. To measure the interestingness of a rule, many numerical characteristics or indices have been
proposed (see [9, 10] for a nice overview). As a supplement to them, we try to utilize the theory of intermediate quantifiers to characterize the intensity of association, which allows us to use
linguistic characterizations such as “almost all”, “most”, “some”, or “few”. As a result, we may automatically obtain the following sentences from numerical bio-statistical data: Almost all people,
who suffer atopic tetter, live in an area affected by heavy industry and smoke, suffer from asthma. Most people who smoke and suffer from respiratory diseases also suffer from ischemic disease of
leg. In the practice, it is often the case that some data are not available e.g. due the error in measures, missing results, or if the respondent is not willing to answer or has no opinion on the
given subject. We can completely remove the cases with missing values to obtain clean data, but it can result in an excessive loss of information. Alternatively, we can handle missing values by using
fuzzy partial logics, which were proposed by Bˇehounek and Nov´ak in [11]. They provide formal apparatus for several types of missing information such as “unknown” or “undefined” (i.e. not
meaningful) value. Basically, the semantics of these logics formed by algebras of truth values is extended by a special value “”. | {"url":"https://events.it4i.cz/event/5/timetable/?view=standard_numbered_inline_minutes","timestamp":"2024-11-13T09:56:31Z","content_type":"text/html","content_length":"186760","record_id":"<urn:uuid:cec2822b-c18a-4129-bbd7-1cbd9f6b17c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00399.warc.gz"} |
exceptional structure
Added connection and reference between the Leech lattice and the octonions.
diff, v7, current
thanks, hadn’t seen that (and it seems unpublished?) BTW, we do have an entry for Yang-Hui He
diff, v6, current
Added a reference
• Yang-Hui He, John McKay, Sporadic and Exceptional, (arXiv:1505.06742)
diff, v5, current
Added some more cases, and pointed out some interrelations.
diff, v2, current
made page name singular
v1, current
Page created, but author did not leave any comments.
v1, current
I added a link to an associated talk by Wilson.
He ends with applications:
□ Perhaps this gives us a better understanding of why the Leech lattice exists.
□ Perhaps it will give us new ways to prove important properties of the Leech lattice and the Conway group.
□ Perhaps we can use octonions to simplify the construction of the Monster.
□ Perhaps it will explain the ‘2-local group’ BDI(4) which contains Co3 and looks as though it should be some kind of twist of ‘skew-symmetric 3×3 matrices over octonions’
Could one dare to hope that growing the superpoint $\mathbb{R}^{0|3}$ could make contact with these things?
added pointer to
• Piero Truini, Michael Rios, Alessio Marrani, The Magic Star of Exceptional Periodicity (arXiv:1711.07881)
prodded by today’s
• Michael Rios, Alessio Marrani, David Chester, Exceptional Super Yang-Mills in $D=27+3$ and Worldvolume M-Theory (arXiv:1906.10709)
Haven’t absorbed either yet. But if these hold water, I should.
diff, v9, current
added publication data for this item:
• Piero Truini, Michael Rios, Alessio Marrani, The Magic Star of Exceptional Periodicity, J. Phys.: Conf. Ser. 1194 (2019) 012106 [arXiv:1711.07881, doi:10.1088/1742-6596/1194/1/012106&
diff, v11, current | {"url":"https://nforum.ncatlab.org/discussion/9920/","timestamp":"2024-11-15T04:34:45Z","content_type":"application/xhtml+xml","content_length":"52506","record_id":"<urn:uuid:47fed2de-9bcb-4dc8-9de7-d07d93271c70>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00021.warc.gz"} |
Symmetry Points | Symmetry Protocol
Symmetry Rewards Points System Documentation
The Symmetry platform utilizes a comprehensive points system to incentivize and reward users and managers for their engagement and contributions. This system covers several areas of interaction
within the platform, as detailed below.
Total Value Locked (TVL) Points
Mechanism: Managers accrue points daily, calculated from the total value locked in their baskets. Points are derived by multiplying the TVL by a Basket Age Multiplier.
Basket Age Multiplier: This factor adjusts according to the basket's age, starting at 1.0 for new baskets and increasing to a maximum of 2.0 for those older than 360 days.
Example: A manager with a basket valued at $1000, using a multiplier of 1.5 for a basket created six months ago, would receive 1500 points that day.
Commencement: TVL point tracking initiated in the period between February 24th and 28th, 2024.
Holder Points
Allocation: Points are awarded daily to users based on the value of any Symmetry baskets they hold.
Exception: Points for basket managers are consolidated under TVL points to prevent double counting.
Volume Points
Allocation: Managers are granted points daily, equal to 1/20th of their basket's trade volume.
Example: A daily trade volume of $2000 awards the manager with 100 points.
Calculation Start: This points calculation began with the launch of Symmetry, based on each basket's total converted volume.
Platform Interaction Points
Allocation: Users earn points for interacting with the Symmetry platform through transactions.
Formula: The calculation formula is interaction_points = (log(max(interactions/3, 1), base=2) + 1) * 500.
Examples: Initial interaction yields 500 points; 5 interactions total approximately 870 points; 100 interactions bring about 3000 points; and after 10,000 interactions, approximately 6300 points.
Referral Points
Mechanism: Users earn a 10% share of the points collected by individuals they refer to Symmetry, provided these referred users achieve a minimum level of engagement. This engagement is defined by
either having at least 10 interactions with the Symmetry engine or amassing a minimum of 5,000 points. This requirement is designed to discourage the creation of fake accounts and ensure
meaningful participation within the platform.
Example: If a referred user earns 1000 points each in TVL, Holder, Volume, and Interaction Points in a day, the referrer gains 400 points.
Network Benefits: A 1% benefit extends to points collected from users who were referred by your referrals.
Referral Bonus: Users referred by someone receive at least 1,000 points as a referral bonus. Special referral links can provide even higher bonus points.
Bonus Points
Bonus Points: Additional points are assigned for specific activities, including early interactions within given periods or roles within the community. Users who interacted with Symmetry during
February 2024 will be awarded "SMFebruary" Bonus points. Active community members, special Discord role holders, and other contributors will be awarded various bonus points.
Level Advancement Formula: The transition of total points into levels is governed by a specific mathematical formula:
This formula ensures that as users accumulate points through their activities and contributions, their levels within the platform adjust accordingly to reflect their presence.
This points system is designed to foster active participation and reward users and creators for their contributions to the Symmetry ecosystem, ensuring a dynamic and engaging platform for all | {"url":"https://docs.symmetry.fi/introduction/symmetry-points","timestamp":"2024-11-13T09:16:59Z","content_type":"text/html","content_length":"285191","record_id":"<urn:uuid:0337b8db-f0bb-4944-95bc-f55d2730b84b>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00672.warc.gz"} |
Carnegie Learning Algebra I Student Text Volume 1 3rd Edition Chapter 3 Exercise 3.4 Linear Functions - Math Book Answers
Carnegie Learning Algebra I Student Text Volume 1 3rd Edition Chapter 3 Exercise 3.4 Linear Functions
Carnegie Learning Algebra I Student Text Volume 1 3rd Edition Chapter 3 Linear Functions
Page 196 Problem 1 Answer
Given Alexis is a flower vendor who grows and sells her own fresh-cut flowers.
She has a stand in the city, and she distributes flowers to pedestrians during the day.
She charges $ 5 for each flower, and each month she randomly gives away two flowers for free.
We have to write a linear function, a(x) to represent how much money Alexis earns each month.
Use x to represent the number of flowers she sells each month. Write the function in the simplest form.
Remember Alexis earn on x−2 flower every month.
Alexis distributes x flowers every month and each month she randomly gives away two flowers for free
So she earn on x−2 flowers
She charges $ 5 for each flower
Then her total earning is a(x)=5(x−2)
A linear function represents total earning is a(x)=5(x−2).
Page 196 Problem 2 Answer
Given Alexis is a flower vendor who grows and sells her own fresh-cut flowers.
She has a stand in the city, and she distributes flowers to pedestrians during the day.
She charges $ 5 for each flower, and each month she randomly gives away two flowers for free.
We have to answer what property did you use to write the simplified form of the function.
Distribute the x.
The function is a(x)=5(x−2)
Use distributive property a(x)=5x−10
Use the distributive property to write the simplified form of the function.
Page 196 Problem 3 Answer
Given Alexis is a flower vendor who grows and sells her own fresh-cut flowers.
She has a stand in the city, and she distributes flowers to pedestrians during the day.
She charges $ 5 for each flower, and each month she randomly gives away two flowers for free
We have to describe the function. Is it increasing or decreasing and Is it discrete or continuous, and explain our reasoning.
The function is increasing since her earing is increasing as the number of flowers she distributes.
The x must be a whole number, the graph is discrete since the domain is not all real numbers.
The function is increasing since her earing is increasing as the number of flowers she distributes and the graph is discrete since the domain is not all real numbers.
Page 197 Problem 4 Answer
Here we have to complete the table shown.
First, determine the unit of measure for each expression.
Then, describe the contextual meaning of each part of the function.
Finally, choose a term from the word box to describe the mathematical meaning of each part.
input value output value rate of change y-intercept
Expression a(x) addresses the measure of cash she procures so its unit is dollars and its numerical importance is the yield.
Expression 5 addresses the sum she procures per bloom sold so its unit is dollars per blossom and its numerical significance is pace of progress.
Expression x addresses the quantity of blossoms she circulates so its unit is a number of blossoms and its numerical importance is input.
Expression(x−2) addresses the quantity of blossoms she brings in cash from so its unit is blossoms yet it doesn’t have numerical importance (which is the reason the phone is concealed in the table in
your book).
Expression−10 is the y-block of the improved on structure a(x)=5x−10. Its unit should then be a similar unit as a(x).
Since the y-block happens when x=0,−10 is the measure of cash she loses when she conveys 0 blossoms.
That is, it addresses the measure of cash she loses from parting with 2 blossoms.
Expression 5x−2 is equivalent to a(x) so it has a similar unit, context-oriented importance, and numerical significance as a(x).
Page 197 Problem 5 Answer
Given Alexis is a flower vendor who grows and sells her own fresh-cut flowers.
She has a stand in the city, and she distributes flowers to pedestrians during the day.
She charges $ 5 for each flower, and each month she randomly gives away two flowers for free.
We have to find how much will Alexis earn in a month if she distributes 45 flowers and show our work.
Substitute x=45 in equation.
The equation for the amount she earn
Hence, she earn $215.
If she distributes 45 flowers she earn $215.
Page 198 Problem 6 Answer
Given Bashir is also a flower vendor in a different part of the same city.
He sells flowers for $3 each and gives away 4 flowers for free each month.
He also earns an extra $6 each month by selling one of his homemade bracelets.
We have to write a linear function, b(x), to represent the amount of money Bashir earns each month and be sure to simplify our function.
Remember Bashir earn on x−4 flowers and one of his homemade bracelets.
He sells x flowers and gives away 4 flowers for free each month
So he earn onx−4 flowers
He charges $3 for each flower then he earn 3(x−4) on flower
He also earns an extra $6 each month by selling one of his homemade bracelets.
then the total earing b(x)=3(x−4)+6
simplifying the equation
A linear function represents total earning is b(x)=3x−6.
Page 198 Problem 7 Answer
Given Bashir is also a flower vendor in a different part of the same city.
He sells flowers for $3 each and gives away 4 flowers for free each month.
He also earns an extra $6 each month by selling one of his homemade bracelets.
We have to explain the meaning of the rate of change and the y-intercept of each function.
In Alexis’ functiona(x)=5x−10 here rate of change 5 represents the amount of money she earns per flower and the y-intercept amount of money she lose by giving 2 flowers free.
In Bashir’s functionb(x)=3x−6 here rate of change 3 represents the amount of money he earns per flower and the y-intercept amount of money she lose by giving 4 flowers free but earns $6 from selling
In Alexis’ function here rate of change 5 represents the amount of money she earns per flower and the y-intercept amount of money she lose by giving 2 flowers free and in Bashir’s function here rate
of change 3 represents the amount of money he earns per flower and the y-intercept amount of money she lose by giving 4 flowers free but earns $6 from selling braclet.
Page 198 Problem 8 Answer
Given Bashir is also a flower vendor in a different part of the same city.
He sells flowers for $3 each and gives away 4 flowers for free each month.
He also earns an extra $6 each month by selling one of his homemade bracelets.
We have to compare the units of the: output values, input values, rate of change, andy-intercepts of both functions and what do we notice.
The output value and input value and y-intercept have the same unit of the dollar, the input values have a unit of the number of the flower and the rate of change have unit of dollar per flower both
functions have the same unit and positive rate change and negative y-intercept.
After comparing the units we notice both functions have the same unit and positive rate change and negative y-intercept.
Page 199 Problem 9 Answer
Given Bashir and Alexis decide to become business partners and combine their monthly earnings.
They will each continue to sell to their own customers in different parts of the city.
Bashir distributes twice as many flowers each month as Alexis.
We have to find at the end of the month when Alexis and Bashir combine their earnings, about how much will they will earn from each flower sold and explain our prediction.
Find the average of their both they will earn from each flower.
Alexis sells each flower at $5 and Bashir sells each flower at $3.
For combing earning find the average of their both they will earn from each flower.
Thus, 5+3/2
=8/2 =4
Hence, both combine earing is $4.
They will earn $4 from each flower sold.
Page 199 Problem 10 Answer
Given suppose in one month Alexis distributes 20 flowers.
We have to use Alexis’ function to calculate her earnings and show our work.
Substitutex=20 in the equation.
The equation for the amount she earn
Hence, she earn $90.
Page 199 Problem 11 Answer
Given suppose in one month Alexis distributes 20 flowers.
We have to use Bashir’s function to calculate her earnings and show our work.
Substitutex=40 in the equation.
The equation for the amount he earn b(x)=3x−6
Bashir distributes twice as many flowers each month as Alexis.
So substitutex=40
Hence, he earn $144.
If in one month Alexis distributes 20 flowers then from Bashir’s function his earning is $144.
Page 199 Problem 12 Answer
Given suppose in one month Alexis distributes 20 flowers.
We have to find how much money would Bashir and Alexis make together if they combined their earnings.
Find the sum of both they earn.
Alexis earn $90 and Bashir earn $144
So they both combined earning is 90+114=204
Hence, they both combined earning is $204.
Bashir and Alexis combined earnings is $204.
Page 199 Problem 13 Answer
Given suppose in one month Alexis distributes 20 flowers.
We have to use our answer to part (c) to determine the average selling price of each flower after Alexis and Bashir combined their earnings and does this match our prediction Find the ratio of their
both earnings and their sold flower.
They both sold a 20+40=60 flower And they combined earning is $204
Then the average selling price is 204/60=3.40
This is less than our prediction.
The average selling price of each flower is $3.40 it is less than our prediction.
Page 200 Problem 14 Answer
Given Nick tried to write a new function, c(x), to represent Alexis’ and Bashir’s combined earnings. He said, “I can add the two functions like this:
Madison disagreed. She said, “That’s not right. You can’t add the functions because the x-values in the two functions don’t mean the same thing, so they might be different values.”
We have to tell who’s correct -Madison or Nick and explain your reasoning.Madison is right.
The x in a(x) addresses the quantity of blossoms Alexis dispersed and the x in $b(x)$ addresses the quantity of blossoms Bashir conveyed.
Since Bashir dispersed twice however many blossoms as Alexis, the worth of x are not the equivalent.
To consolidate the capacities to make c(x), the x in b(x)=3x−6 would need to be supplanted with 2x giving b(x)=3(2x)−6=6x−6 so the x in every situation address similar worth of the quantity of
blossoms Alexis dispersed.
Madison is right x represent different values in each function so Bashir sold twice flowers as Alexis.
Page 200 Problem 15 Answer
Here we have to use our answers to Question 4 and Nick’s function to show why his function is not correct and explain our method Use Nick’s function isc(x)=8x−16
where x is the number of flowers they sell.
From Question 4 Bashir and Alexis make together earned $204 and they sell 60 flowers together.
Nick’s function isc(x)=8x−16where x is the number of flowers they sell.
So c(60)=8(60)−16
Nick’s function is not correct because c(x)≠204.
Page 200 Problem 16 Answer
Here we have to answer what does the slope represents in Nick’s function and how does this compare with your answer to Question 4, part (d)The slope of the Nick equation c(x)=8x−16 is 8 it represents
the average price per flower.
But in Question 4 , part (d) average price per flower is $4 it does not increase.
In Nick equation is 8 presents the average price per flower but in Question 4, part (d) average price per flower is $4 it does not increase.
Page 201 Problem 17 Answer
Given Nick could actually add the two functions together.
However, he did not recognize that the input values were different for Alexis and Bashir.
To add two functions together, you must ensure the input values represent the same thing in both functions.
A model can be used to represent the input values Let x represent the total number of flowers Alexis and Bashir distribute each month.
The model shows that Bashir distributes twice as many flowers as Alexis each month and that together the number of flowers adds up to x.
We have to write an expression to represent Alexis’ share of the total flowers distributed.
Then write an expression to represent Bashir’s share of the total flowers distributed.
Let $x$ represent the total number of flowers distributed.
If x is the total number of flowers distributed.
From the model Bashir’s share of the total flowers distributed is 2/3x.And Alexis’ share of the total flowers distributed is 1/3x.
An expression to represent Alexis’ share is 1/3x and Bashir’s share is 2/3x.
Page 201 Problem 18 Answer
Given Nick could actually add the two functions together. However, he did not recognize that the input values were different for Alexis and Bashir.
To add two functions together, you must ensure the input values represent the same thing in both functions.
A model can be used to represent the input values.
We have to rewrite Alexis’ and Bashir’s functions so they show each person’s share of the total earnings.
Then, add the functions to determine a new function, c(x), that describes the combined amount of money Alexis and Bashir earn each month and show our work.
Replace x with in the Alexis function with1/3x and with 2/3x in the Bashir equation.
Then add both equations.
The Alexis function a(x)=5x−10
Replace x with 1/3x in the Alexis function right side
The Bashir equation b(x)=3x−6
Replace x with 2/3x in the Bashir equation right side
Now add both equations for the total earnings.
Alexis’ person’s share equation is a(x)=5/3x−10, Bashir person’s share equation is b(x)=6/3x−6 and the combined equationc(x)=3.67x−16.
Page 201 Problem 19 Answer
Given Nick could actually add the two functions together.
However, he did not recognize that the input values were different for Alexis and Bashir.
To add two functions together, you must ensure the input values represent the same thing in both functions.
A model can be used to represent the input values.
Let x represent the total number of flowers Alexis and Bashir distribute each month.
The model shows that Bashir distributes twice as many flowers as Alexis each month and that together the number of flowers adds up to x.
We have to answer what does the slope of the new function mean and what does the y-intercept of the new function meanThe new function is c(x)=3.67x−16
where the slope of the equation is 3.67 it represents the average rate per flower and the y-intercept is -16 it represents the total amount of flower they in give free.
The slope 3.67 it represents the average rate per flower and the y-intercept is -16 it represents the total amount of flower they give free.
Page 202 Problem 20 Answer
Given Alexis and Bashir decide to partner with an investor.
The investor will supply money for equipment, flower seeds, and other materials. In return, the investor will receive $ 0.50 for every flower distributed.
We have to answer In this case, why was it possible to determine a new function without rewriting d(x).
In both functions, x represents the total number of flowers they distribute so they have the same input values that’s why it was not we can write a new function without rewriting d(x).
Both functions have the same input values that’s why it was not we can write a new function without rewriting d(x).
Page 202 Problem 21 Answer
Given Alexis and Bashir decide to partner with an investor.
The investor will supply money for equipment, flower seeds, and other materials.
In return, the investor will receive $ 0.50 for every flower distributed We have to think about each problem situation and compare the functions t(x) and c(x) and what do we notice.
The functions t(x)=3.17x−16 and c(x)=3.67x−16.
They both have a y-intercept means the amount flower they give free it’s not affected by the amount.
The slope t(x) is 0.05 less than c(x)because the investor has a share 0.05 per flower.
They both have a y-intercept and The slope t(x) is 0.05 less than c(x).
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Trigonometric Ratios MCQ Quiz [PDF] Questions Answers | Trigonometric Ratios MCQs App e-Book Download: Test 6
Class 8 Math MCQs - Chapter 6
Trigonometric Ratios Multiple Choice Questions (MCQ) PDF Download - 6
The Trigonometric ratios Multiple Choice Questions (MCQ Quiz) with Answers PDF (Trigonometric Ratios MCQ PDF e-Book) download Ch. 6-6 to learn Grade 8 Math Course. Study Solving Right Angled
Triangles Multiple Choice Questions and Answers (MCQs) PDF, Trigonometric Ratios Quiz Questions and Answers for homeschool certification courses. The Trigonometric Ratios MCQ App Download: Free
educational app for solving right angled triangles, angles and trigonometrical ratio, trigonometrical ratios, applications of trigonometry test prep for online study.
The MCQs: Consider a right angle triangle ABC, if PQ = y, QR = x, PR = 102.6 and angle of A is 67.8° then the values of x and y respectively are; "Trigonometric Ratios" App (Android, iOS) with
answers: 85.027, 40.258; 94.993, 38.766; 68.237, 12.365; 98.657, 23.854; for homeschool certification courses. Practice Solving Right Angled Triangles MCQ Questions, download Apple Book (Free Sample)
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Trigonometric Ratios MCQs with Answers PDF Download: Quiz 6
MCQ 26:
Consider a right angle triangle ABC, if PQ = y, QR = x, PR = 102.6 and angle of A is 67.8° then the values of x and y respectively are
1. 94.993, 38.766
2. 85.027, 40.258
3. 68.237, 12.365
4. 98.657, 23.854
MCQ 27:
If tan A is 0.573 then the value of angle A in a right angle triangle is
1. 42.43°
2. 34.86°
3. 31.53°
4. 29.81°
MCQ 28:
The tan P of triangle PQR with respect to P is calculated as
1. PQ/PR
2. PR/PQ
3. QR/PR
4. QR/PQ
MCQ 29:
If AB = 23.8 and AC = 38.6 then the angle of A in a right angle triangle is
1. 72.25°
2. 51.93°
3. 62.82°
4. 48.65°
MCQ 30:
A house is built at the top of cliff. From the foot of cliff at a distance of 75 m, the angle of elevation of the top of the house is 51° and the angle of elevation at the top of the cliff 41°. The
height of the house is
1. 38.25 m
2. 36 m
3. 32.42 m
4. 27.42 m
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XGBoost Configure fit() "feature_weights" Parameter
XGBoost allows you to assign different selection probabilities to features when using the colsample_bytree or colsample_bylevel parameters.
This can be useful when you know certain features are more informative than others and want the model to focus on them.
Here’s how you can train an XGBoost model with feature weights using the scikit-learn API.
from sklearn.datasets import make_regression
from sklearn.model_selection import train_test_split
from xgboost import XGBRegressor
from sklearn.metrics import mean_squared_error
import numpy as np
# Generate synthetic regression dataset
X, y = make_regression(n_samples=1000, n_features=10, n_informative=5, noise=0.1, random_state=42)
# Split data into train and test sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
# Initialize XGBRegressor
model = XGBRegressor(n_estimators=100, learning_rate=0.1, colsample_bytree=0.8, colsample_bylevel=0.8, random_state=42)
# Create feature_weights array
feature_weights = np.zeros(X_train.shape[1])
feature_weights[:5] = 1 # Higher weights for informative features
# Fit model with feature_weights
model.fit(X_train, y_train, feature_weights=feature_weights)
# Make predictions and evaluate performance
y_pred = model.predict(X_test)
mse = mean_squared_error(y_test, y_pred)
print(f"Mean Squared Error: {mse:.4f}")
In this example:
1. We generate a synthetic regression dataset using make_regression with 10 features, 5 of which are informative.
2. We initialize an XGBRegressor with colsample_bytree=0.8 and colsample_bylevel=0.8, which means that 80% of features will be randomly selected at each tree and level.
3. We create a feature_weights array that assigns a weight of 1 to the first 5 features (which are informative) and 0 to the rest. This gives the informative features a higher probability of being
selected during colsampling.
4. We fit the model using the feature_weights parameter to pass our weights.
5. Finally, we make predictions on the test set and evaluate the model’s performance using mean squared error.
By assigning higher weights to informative features, we can guide XGBoost to focus on them during training, potentially improving the model’s performance. | {"url":"https://xgboosting.com/xgboost-configure-fit-feature_weights-parameter/","timestamp":"2024-11-13T06:21:29Z","content_type":"text/html","content_length":"9236","record_id":"<urn:uuid:179ed154-2645-467c-b97f-1aa1f531cdb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00515.warc.gz"} |
Numerical investigation of a heat-generating fluid with regard solidification for different cooling regimes on the boundaries (preprint IBRAE-1995-06)
Preprint IBRAE-95-06
Aksenova A.E., Chudanov V.V., Vabishchevich P.N.
The quasisteady natural convection of a heat-generating fluid with regard to solidification for the different cooling regimes on the boundaries is investigated numerically in present work. The main
goal of this paper is investigation of different cooling regimes influence on the distribution of heat flux on the cavity boundaries.
To solve unsteady problems of hydrodynamics with heat conductivity, new efficient numerical method, inspected at the wide set of test problem, was employed. The obtained numerical results were
verified on reability via conducting of calculations on the sequence of the fine grids.
These results may be used for construction of simplified models describing the behaviour of a molten heat-generating masses (corium) in different retention system(core catcher).
Bibliographical reference
Aksenova A.E., Chudanov V.V., Vabishchevich P.N. NUMERICAL INVESTIGATION OF A HEAT-GENERATING FLUID WITH REGARD SOLIDIFICATION FOR DIFFERENT COOLING REGIMES ON THE BOUNDARIES (in Russian). –
Preprint ¹ IBRAE-95-06. Moscow: Nuclear Safety Institute of the Russian Academy of Sciences (IBRAE RAN), 1995. – 21 p. – Refs.: 17 items
Program system RCS: organization of calculations for modeling of physics problems (preprint IBRAE-1995-05)
Preprint IBRAE-95-05
Varenkov V.V., Pervichko V.A., Popkov A.G., Chudanov V.V.
This paper describes a program system for organization of problem solving in field of mathematical physics. Principles of manipulations with calculation modules and data are considered. System
structure and types of handled objects are described. An example demonstrates the organization of computing environment.
Bibliographical reference
Varenkov V.V., Pervichko V.A., Popkov A.G., Chudanov V.V. PROGRAM SYSTEM RCS: ORGANIZATION OF CALCULATIONS FOR MODELING OF PHYSICS PROBLEMS (in Russian). Preprint IBRAE-95-05. Moscow. Nuclear Safety
Institute. 1995. 16 p. – Refs.: 6 items.
Numerical investigation of the heat exchange in a heat-generating fluid with regard to solidification (preprint IBRAE-1995-04)
Preprint IBRAE-95-04
Aksenova A.E., Chudanov V.V., Vabishchevich P.N.
The quasisteady natural convection of a heat-generating fluid with phase changes in the enclosures of a square section with isothermal boundary conditions is investigated numerically in present
work. The main goal of this paper is correlation relations for heat fluxes distribution at the domain boundaries depending on Rayleigh and Ostrogradskii numbers.
To solve unsteady problems of hydrodynamics with heat conductivity, new efficient numerical method, inspected at the wide set of test problem, was employed. The obtained numerical results were
verified on reability via conducting of calculations on the sequence of the fine grids.
These results may be used for construction of simplified models describing the behaviour of a molten heat-generating masses (corium) in different retention system(core catcher).
Bibliographical reference
Aksenova A.E., Chudanov V.V., Vabishchevich P.N. NUMERICAL INVESTIGATION OF THE SOME SINGULARITIES OF A HEAT-GENERATING FLUID BEHAVIOR WITH REGARD TO SOLIDIFICATION (in Russian). – Preprint ¹
IBRAE-95-04. Moscow: Nuclear Safety Institute of the Russian Academy of Sciences (IBRAE RAN), 1995. – 21 p. – Refs.: 16 items
Numerical simulation the buoyancy driven flow of the stratified heat-generating fluid (preprint IBRAE-1995-03)
Preprint IBRAE-95-03
Aksenova A.E., Chudanov V.V., Strizhov V.F., Vabishchevich P.N.
The 2D natural convection of a heat-generating fluid in a stratified corium for the different regimes of cooling on the upper boundary in a geometric domain similar to the vessel lower head is
investigated numerically in present work. This paper deals to the analysis of the corium nonhomogeneity influence on the heat flux distribution on the lateral, lower and upper surfaces in depending
on the cooling regimes at the upper boundary (Biot number) and different aspect ratio. To solve unsteady problems of hydrodynamics with heat conductivity, new efficient numerical method, inspected
at the wide set of test problem, was employed. These results may be used for construction of simplified models describing the behaviour of a molten heat-generating masses (corium) in different
retention system(core catcher).
Bibliographical reference
Aksenova A.E., Chudanov V.V., Strizhov V.F., Vabishchevich P.N. NUMERICAL SIMULATION THE BUOYANCY DRIVEN FLOW OF THE STRATIFIED HEAT-GENERATING FLUID (in Russian). – Preprint ¹ IBRAE-95-03. Moscow:
Nuclear Safety Institute of the Russian Academy of Sciences (IBRAE RAN), 1995. – 18 p. – Refs.: 10 items
Numerical investigation of the heat exchange in a heat-generating fluid with melting and solidification depending on Ostrogradskii and Rayleigh numbers (preprint IBRAE-1995-01)
Preprint IBRAE-95-01
Aksenova A.E., Chudanov V.V., Vabishchevich P.N.
The quasisteady natural convection of a heat-generating fluid with phase changes in the enclosures of a square section with isothermal boundary conditions is investigated numerically in present
work. The main goal of this paper is correlation relations for heat fluxes distribution at the domain boundaries depending on Rayleigh and Ostrogradskii numbers.
To solve unsteady problems of hydrodynamics with heat conductivity, new efficient numerical method, inspected at the wide set of test problem, was employed. The obtained numerical results were
verified on reability via conducting of calculations on the sequence of the fine grids.
These results may be used for construction of simplified models describing the behaviour of a molten heat-generating masses (corium) in different retention system(core catcher).
Bibliographical reference
Aksenova A.E., Chudanov V.V., Vabishchevich P.N. NUMERICAL INVESTIGATION OF THE HEAT EXCHANGE IN A HEAT-GENERATING FLUID WITH MELTING AND SOLIDIFICATION DEPENDING ON OSTROGRADSKII AND RAYLEIGH
NUMBERS. – Preprint ¹ IBRAE-95-01. Moscow: Nuclear Safety Institute of the Russian Academy of Sciences (IBRAE RAN), 1995. – 25 p. – Refs.: 21 items
The interactive preprocessor for data generation in application to mathematical physics problems (preprint IBRAE NSI-32-94)
Preprint IBRAE ¹ NSI-32-94
Aksenova À.Å., Varenkov V.V., Pervichko V.A., Popkov A.G., Chudanov V.V.
In this paper an interactive preprocessor program intended for data generation in application to mathematical physics problems is described. The process of generation of computational regions, grids
and boundary conditions with help of package is considered.
Bibliographical reference
Aksenova À.Å., Varenkov V.V., Pervichko V.A., Popkov A.G., Chudanov V.V. THE INTERACTIVE PREPROCESSOR FOR DATA GENERATION IN APPLICATION TO MATHEMATICAL PHYSICS PROBLEMS (in Russian). Preprint
NSI-32-94. Moscow: Nuclear Safety Institute, 1994. 17 p. – Refs.: 7 items.
Models Of Radionuclides Transport In Atmosphere From Integrated Software Package «NOSTRADAMUS» (preprint IBRAE NSI-31-94)
Preprint IBRAE NSI-31-94
Arutunjan R.V., Bolshov L.A., Belikov V.V., Belikova G.V., Goloviznin V.M., Fokin A.L, Kanevsky M.F., Peschany S.E., Semenov V.N., Shilkova S.V, Sorokovikova O.S., Starodubtseva L.P.
The characteristics of the radionuclides transport model imbedded into the NOSTRADAMUS integrated software package of the Nuclear Safety Institute are described.
Bibliographical reference
Arutunjan R.V., Bolshov L.A., Belikov V.V., Belikova G.V., Goloviznin V.M., Fokin A.L, Kanevsky M.F., Peschany S.E., Semenov V.N., Shilkova S.V, Sorokovikova O.S., Starodubtseva L.P. MODELS OF
RADIONUCLIDES TRANSPORT IN ATMOSPHERE FROM INTEGRATED SOFTWARE PACKAGE «NOSTRADAMUS». Preprint NSI-31-94. Moscow. Nuclear Safety Institute. September 1994. 23 p. – Refs.: 33 items
Numerical simulation of the natural convection in porous medium (preprint IBRAE NSI-33-94)
Preprint IBRAE NSI-33-94
Aksenova A.E., Chudanov V.V., Vabishchevich P.N.
A numerical algorithm is described in the work for 2D convection problem of fluid in porous medium in the stream function — vorticity — temperature formulation. Numerical implementation of the
operator-splitting scheme used in calculations is based on the solution of elliptic grid problems at every time-level. The algorithm validation is conducted on the benchmark solutions for
buoyancy-driven flow of fluid in a square cavity with side walls of different temperatures.
Bibliographical reference
Aksenova A.E., Chudanov V.V.,Vabishchevich P.N. NUMERICAL SIMULATION OF THE NATURAL CONVECTION IN POROUS MEDIUM. – Preprint ¹ ¹ NSI-33-94. Moscow: Nuclear Safety Institute of the Russian Academy of
Sciences (IBRAE RAN), 1994. – 22 p. – Refs.: 12 items
Comparative Analysis of Spatial Interpollation Methods by Using Chernobyl Fallout Data (in Russian) (preprint IBRAE NSI-26-94)
Preprint IBRAE NSI-26-94
Arutyunyan R.V., Bolshov L.A., Kanevsky Ì.F., Linge I.I., Savel’eva E.A.
The spatial data analysis of the Chernobyl fallout by using different methods is presented. Results of the cross-validation and prediction maps are described. Several recommendations for the
selection of the model and model-dependent parameters are discussed.
Bibliographical reference
Arutyunyan R.V., Bolshov L.A., Kanevsky Ì.F., Linge I.I., Savel’eva E.A. COMPARATIVE ANALYSIS OF SPATIAL INTERPOLLATION METHODS BY USING CHERNOBYL FALLOUT DATA. (in Russian). Preprint NSI-26-94.
Moscow: Nuclear Safety Institute, August 1994. 43 p. – Refs.: 22 items
Sample Geostatistical Analysis of Chernobyl Fallout (preprint IBRAE NSI-25-94)
Preprint IBRAE NSI-25-94
Arutyunyan R.V., Bolshov L.A., Demyanov V.V., Kanevsky Ì.F., Linge I.I.
One of the approaches to the spatial data analysis is presented in the work. The following methods were sequentially applied to the Chenobyl data: cell declustering, moving window statistics,
structural analysis, cross-validation and kriging. Anysotropical case was under special concern, as well as comparison between isotropic and anisotropic modelling. Ordinary kriging results were
compared with classical inverse distance squared interpolation. All used methods are fully and clearly described.
Bibliographical reference
Arutyunyan R.V., Bolshov L.A., Demyanov V.V., Kanevsky Ì.F., Linge I.I. SAMPLE GEOSTATISTICAL ANALYSIS OF CHERNOBYL FALLOUT (in Russian). Preprint NSI-25-94. Moscow. Nuclear Safety Institute. 1994.
47 p. – Refs.: 16 items | {"url":"http://en.ibrae.ac.ru/pubs/153/page/17/","timestamp":"2024-11-12T17:35:12Z","content_type":"text/html","content_length":"23876","record_id":"<urn:uuid:797ee049-bdec-42e0-80eb-dfccbd4c033c>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00537.warc.gz"} |
TI-Basic Developer
The iPart() Command
Returns the integer part of a number.
Menu Location
• Press 2nd MATH to enter the MATH popup menu.
• Press 1 to enter the Number submenu.
• Press 4 to select iPart(.
This command works on all calculators.
1 byte
The iPart() command returns the integer part of a number (removing all the digits after the decimal). Another way of thinking about it is it rounds a number towards 0: positive numbers get rounded
down to an integer, and negative numbers get rounded up to an integer.
There are several other rounding commands available, which work in subtly different ways:
• ceiling() always rounds up to the next higher integer.
• floor() always rounds down to the next lower integer. int() does the same thing as floor().
• round() rounds to any given place value, including to an integer; it rounds up or down, whichever is nearest.
However, iPart() is the only one that has a counterpart fPart() which returns the fractional part of a number. This follows the rule that iPart(x)+fPart(x) always equals x.
Using iPart() on the result of a division — iPart(x/y) — is useful so often that there's a specific command, intDiv(), for doing so.
iPart() can also be applied to complex numbers, lists, and matrices, rounding everything that there is to round in each of them.
{-3 3}
Related Commands | {"url":"http://tibasicdev.wikidot.com/68k:ipart","timestamp":"2024-11-10T08:10:29Z","content_type":"application/xhtml+xml","content_length":"27619","record_id":"<urn:uuid:33ef2df1-f0bc-4b28-bff3-8ceb22a8ae86>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00619.warc.gz"} |
Simulate Bates
Simulate Bates, Heston, and CIR sample paths by quadratic-exponential discretization scheme
Since R2020a
[Paths,Times,Z] = simByQuadExp(MDL,NPeriods) simulates NTrials sample paths of a Heston model driven by two Brownian motion sources of risk, or a CIR model driven by one Brownian motion source of
risk. Both Heston and Bates models approximate continuous-time stochastic processes by a quadratic-exponential discretization scheme. The simByQuadExp simulation derives directly from the stochastic
differential equation of motion; the discrete-time process approaches the true continuous-time process only in the limit as DeltaTime approaches zero.
[Paths,Times,Z] = simByQuadExp(___,Name,Value) specifies options using one or more name-value pair arguments in addition to the input arguments in the previous syntax.
[Paths,Times,Z,N] = simByQuadExp(MDL,NPeriods) simulates NTrials sample paths of a Bates model driven by two Brownian motion sources of risk, approximating continuous-time stochastic processes by a
quadratic-exponential discretization scheme. The simByQuadExp simulation derives directly from the stochastic differential equation of motion; the discrete-time process approaches the true
continuous-time process only in the limit as DeltaTime approaches zero.
[Paths,Times,Z,N] = simByQuadExp(___,Name,Value) specifies options using one or more name-value pair arguments in addition to the input arguments in the previous syntax.
You can perform quasi-Monte Carlo simulations using the name-value arguments for MonteCarloMethod, QuasiSequence, and BrownianMotionMethod. For more information, see Quasi-Monte Carlo Simulation.
Simulate Bates Sample Paths by Quadratic-Exponential Discretization Scheme
Create a bates object.
AssetPrice = 80;
Return = 0.03;
JumpMean = 0.02;
JumpVol = 0.08;
JumpFreq = 0.1;
V0 = 0.04;
Level = 0.05;
Speed = 1.0;
Volatility = 0.2;
Rho = -0.7;
StartState = [AssetPrice;V0];
Correlation = [1 Rho;Rho 1];
batesObj = bates(Return, Speed, Level, Volatility,...
JumpFreq, JumpMean, JumpVol,'startstate',StartState,...
batesObj =
Class BATES: Bates Bivariate Stochastic Volatility
Dimensions: State = 2, Brownian = 2
StartTime: 0
StartState: 2x1 double array
Correlation: 2x2 double array
Drift: drift rate function F(t,X(t))
Diffusion: diffusion rate function G(t,X(t))
Simulation: simulation method/function simByEuler
Return: 0.03
Speed: 1
Level: 0.05
Volatility: 0.2
JumpFreq: 0.1
JumpMean: 0.02
JumpVol: 0.08
Use simByQuadExp to simulate NTrials sample paths directly from the stochastic differential equation of motion; the discrete-time process approaches the true continuous-time process only in the limit
as DeltaTimes approaches zero.
NPeriods = 2;
[Paths,Times,Z,N] = simByQuadExp(batesObj,NPeriods)
Paths = 3×2
80.0000 0.0400
64.3377 0.1063
31.5703 0.1009
Z = 2×2
0.5377 1.8339
-2.2588 0.8622
The output Paths is returned as a (NPeriods + 1)-by-NVars-by-NTrials three-dimensional time-series array.
Quasi-Monte Carlo Simulation Using Bates Model
This example shows how to use simByQuadExp with a Bates model to perform a quasi-Monte Carlo simulation. Quasi-Monte Carlo simulation is a Monte Carlo simulation that uses quasi-random sequences
instead pseudo random numbers.
Define the parameters for the bates object.
AssetPrice = 80;
Return = 0.03;
JumpMean = 0.02;
JumpVol = 0.08;
JumpFreq = 0.1;
V0 = 0.04;
Level = 0.05;
Speed = 1.0;
Volatility = 0.2;
Rho = -0.7;
StartState = [AssetPrice;V0];
Correlation = [1 Rho;Rho 1];
Create a bates object.
Bates = bates(Return, Speed, Level, Volatility, ...
JumpFreq, JumpMean, JumpVol,'startstate',StartState, ...
Bates =
Class BATES: Bates Bivariate Stochastic Volatility
Dimensions: State = 2, Brownian = 2
StartTime: 0
StartState: 2x1 double array
Correlation: 2x2 double array
Drift: drift rate function F(t,X(t))
Diffusion: diffusion rate function G(t,X(t))
Simulation: simulation method/function simByEuler
Return: 0.03
Speed: 1
Level: 0.05
Volatility: 0.2
JumpFreq: 0.1
JumpMean: 0.02
JumpVol: 0.08
Perform a quasi-Monte Carlo simulation by using simByQuadExp with the optional name-value arguments for 'MonteCarloMethod','QuasiSequence', and 'BrownianMotionMethod'.
[paths,time,z] = simByQuadExp(Bates,10,'ntrials',4096,'montecarlomethod','quasi','quasisequence','sobol','BrownianMotionMethod','brownian-bridge');
Input Arguments
MDL — Stochastic differential equation model
bates object | heston object | cir object
Stochastic differential equation model, specified as a bates, heston, or cir object. You can create these objects using bates, heston, or cir.
Data Types: object
NPeriods — Number of simulation periods
positive scalar integer
Number of simulation periods, specified as a positive scalar integer. The value of NPeriods determines the number of rows of the simulated output series.
Data Types: double
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: [Paths,Times,Z,N] = simByQuadExp(bates_obj,NPeriods,'DeltaTime',dt)
NTrials — Simulated trials (sample paths) of NPeriods observations each
1 (single path of correlated state variables) (default) | positive scalar integer
Simulated trials (sample paths) of NPeriods observations each, specified as the comma-separated pair consisting of 'NTrials' and a positive scalar integer.
Data Types: double
DeltaTime — Positive time increments between observations
1 (default) | scalar | column vector
Positive time increments between observations, specified as the comma-separated pair consisting of 'DeltaTime' and a scalar or a NPeriods-by-1 column vector.
DeltaTime represents the familiar dt found in stochastic differential equations, and determines the times at which the simulated paths of the output state variables are reported.
Data Types: double
NSteps — Number of intermediate time steps within each time increment dt (specified as DeltaTime)
1 (indicating no intermediate evaluation) (default) | positive scalar integer
Number of intermediate time steps within each time increment dt (specified as DeltaTime), specified as the comma-separated pair consisting of 'NSteps' and a positive scalar integer.
The simByQuadExp function partitions each time increment dt into NSteps subintervals of length dt/NSteps, and refines the simulation by evaluating the simulated state vector at NSteps − 1
intermediate points. Although simByQuadExp does not report the output state vector at these intermediate points, the refinement improves accuracy by allowing the simulation to more closely
approximate the underlying continuous-time process.
Data Types: double
MonteCarloMethod — Monte Carlo method to simulate stochastic processes
"standard" (default) | string with values "standard", "quasi", or "randomized-quasi" | character vector with values 'standard', 'quasi', or 'randomized-quasi'
Monte Carlo method to simulate stochastic processes, specified as the comma-separated pair consisting of 'MonteCarloMethod' and a string or character vector with one of the following values:
• "standard" — Monte Carlo using pseudo random numbers.
• "quasi" — Quasi-Monte Carlo using low-discrepancy sequences.
• "randomized-quasi" — Randomized quasi-Monte Carlo.
If you specify an input noise process (see Z and N), simByQuadExp ignores the value of MonteCarloMethod.
Data Types: string | char
QuasiSequence — Low discrepancy sequence to drive stochastic processes
"sobol" (default) | string with value "sobol" | character vector with value 'sobol'
Low discrepancy sequence to drive the stochastic processes, specified as the comma-separated pair consisting of 'QuasiSequence' and a string or character vector with one of the following values:
• "sobol" — Quasi-random low-discrepancy sequences that use a base of two to form successively finer uniform partitions of the unit interval and then reorder the coordinates in each dimension
If MonteCarloMethod option is not specified or specified as"standard", QuasiSequence is ignored.
Data Types: string | char
BrownianMotionMethod — Brownian motion construction method
"standard" (default) | string with value "brownian-bridge" or "principal-components" | character vector with value 'brownian-bridge' or 'principal-components'
Brownian motion construction method, specified as the comma-separated pair consisting of 'BrownianMotionMethod' and a string or character vector with one of the following values:
• "standard" — The Brownian motion path is found by taking the cumulative sum of the Gaussian variates.
• "brownian-bridge" — The last step of the Brownian motion path is calculated first, followed by any order between steps until all steps have been determined.
• "principal-components" — The Brownian motion path is calculated by minimizing the approximation error.
If an input noise process is specified using the Z input argument, BrownianMotionMethod is ignored.
The starting point for a Monte Carlo simulation is the construction of a Brownian motion sample path (or Wiener path). Such paths are built from a set of independent Gaussian variates, using either
standard discretization, Brownian-bridge construction, or principal components construction.
Both standard discretization and Brownian-bridge construction share the same variance and therefore the same resulting convergence when used with the MonteCarloMethod using pseudo random numbers.
However, the performance differs between the two when the MonteCarloMethod option "quasi" is introduced, with faster convergence seen for "brownian-bridge" construction option and the fastest
convergence when using the "principal-components" construction option.
Data Types: string | char
Antithetic — Flag to use antithetic sampling to generate the Gaussian random variates
false (no antithetic sampling) (default) | logical with values true or false
Flag to use antithetic sampling to generate the Gaussian random variates that drive the Brownian motion vector (Wiener processes), specified as the comma-separated pair consisting of 'Antithetic' and
a scalar numeric or logical 1 (true) or 0 (false).
When you specify true, simByQuadExp performs sampling such that all primary and antithetic paths are simulated and stored in successive matching pairs:
• Odd trials (1,3,5,...) correspond to the primary Gaussian paths.
• Even trials (2,4,6,...) are the matching antithetic paths of each pair derived by negating the Gaussian draws of the corresponding primary (odd) trial.
If you specify an input noise process (see Z), simByEuler ignores the value of Antithetic.
Data Types: logical
Z — Direct specification of dependent random noise process for generating Brownian motion vector
generates correlated Gaussian variates based on the Correlation member of the heston, bates, or cir object (default) | function | three-dimensional array of dependent random variates
Direct specification of the dependent random noise process for generating the Brownian motion vector (Wiener process) that drives the simulation, specified as the comma-separated pair consisting of
'Z' and a function or an (NPeriods ⨉ NSteps)-by-NBrowns-by-NTrials three-dimensional array of dependent random variates.
If you specify Z as a function, it must return an NBrowns-by-1 column vector, and you must call it with two inputs:
• A real-valued scalar observation time t
• An NVars-by-1 state vector X[t]
Data Types: double | function
N — Dependent random counting process for generating number of jumps
random numbers from Poisson distribution with parameter JumpFreq from a bates object (default) | three-dimensional array | function
Dependent random counting process for generating the number of jumps, specified as the comma-separated pair consisting of 'N' and a function or an (NPeriods ⨉ NSteps) -by-NJumps-by-NTrials
three-dimensional array of dependent random variates. If you specify a function, N must return an NJumps-by-1 column vector, and you must call it with two inputs: a real-valued scalar observation
time t followed by an NVars-by-1 state vector X[t].
The N name-value pair argument is supported only when you use a bates object for the MDL input argument.
Data Types: double | function
StorePaths — Flag that indicates how Paths is stored and returned
true (default) | logical with values true or false
Flag that indicates how the output array Paths is stored and returned, specified as the comma-separated pair consisting of 'StorePaths' and a scalar numeric or logical 1 (true) or 0 (false).
If StorePaths is true (the default value) or is unspecified, simByQuadExp returns Paths as a three-dimensional time-series array.
If StorePaths is false (logical 0), simByQuadExp returns Paths as an empty matrix.
Data Types: logical
Processes — Sequence of end-of-period processes or state vector adjustments
simByQuadExp makes no adjustments and performs no processing (default) | function | cell array of functions
Sequence of end-of-period processes or state vector adjustments, specified as the comma-separated pair consisting of 'Processes' and a function or cell array of functions of the form
The simByQuadExp function runs processing functions at each interpolation time. The functions must accept the current interpolation time t, and the current state vector X[t] and return a state vector
that can be an adjustment to the input state.
If you specify more than one processing function, simByQuadExp invokes the functions in the order in which they appear in the cell array. You can use this argument to specify boundary conditions,
prevent negative prices, accumulate statistics, plot graphs, and more.
The end-of-period Processes argument allows you to terminate a given trial early. At the end of each time step, simByQuadExp tests the state vector X[t] for an all-NaN condition. Thus, to signal an
early termination of a given trial, all elements of the state vector X[t] must be NaN. This test enables you to define a Processes function to signal early termination of a trial, and offers
significant performance benefits in some situations (for example, pricing down-and-out barrier options).
Data Types: cell | function
Output Arguments
Paths — Simulated paths of correlated state variables
Simulated paths of correlated state variables for a heston, bates, or cir model, returned as a (NPeriods + 1)-by-NVars-by-NTrials three-dimensional time series array.
For a given trial, each row of Paths is the transpose of the state vector X[t] at time t. When StorePaths is set to false, simByQuadExp returns Paths as an empty matrix.
Times — Observation times associated with simulated paths
column vector
Observation times for a heston, bates, or cir model associated with the simulated paths, returned as a (NPeriods + 1)-by-1 column vector. Each element of Times is associated with the corresponding
row of Paths.
Z — Dependent random variates for generating Brownian motion vector
Dependent random variates for a heston, bates, or cir model for generating the Brownian motion vector (Wiener processes) that drive the simulation, returned as an (NPeriods ⨉ NSteps)
-by-NBrowns-by-NTrials three-dimensional time-series array.
N — Dependent random variates for generating jump counting process vector
Dependent random variates for a bates model for generating the jump counting process vector, returned as a (NPeriods ⨉ NSteps)-by-NJumps-by-NTrials three-dimensional time-series array.
More About
Antithetic Sampling
Simulation methods allow you to specify a popular variance reduction technique called antithetic sampling.
This technique attempts to replace one sequence of random observations with another of the same expected value, but smaller variance. In a typical Monte Carlo simulation, each sample path is
independent and represents an independent trial. However, antithetic sampling generates sample paths in pairs. The first path of the pair is referred to as the primary path, and the second as the
antithetic path. Any given pair is independent of any other pair, but the two paths within each pair are highly correlated. Antithetic sampling literature often recommends averaging the discounted
payoffs of each pair, effectively halving the number of Monte Carlo trials.
This technique attempts to reduce variance by inducing negative dependence between paired input samples, ideally resulting in negative dependence between paired output samples. The greater the extent
of negative dependence, the more effective antithetic sampling is.
In the Heston stochastic volatility model, the asset value process and volatility process are defined as
$\begin{array}{l}dS\left(t\right)=\gamma \left(t\right)S\left(t\right)dt+\sqrt{V\left(t\right)}S\left(t\right)d{W}_{S}\left(t\right)\\ dV\left(t\right)=\kappa \left(\theta -V\left(t\right)\right)dt+\
sigma \sqrt{V\left(t\right)}d{W}_{V}\left(t\right)\end{array}$
• γ is the continuous risk-free rate.
• θ is a long-term variance level.
• κ is the mean reversion speed for the variance.
• σ is the volatility of volatility.
You can simulate any vector-valued CIR process of the form
• X[t] is an NVars-by-1 state vector of process variables.
• S is an NVars-by-NVars matrix of mean reversion speeds (the rate of mean reversion).
• L is an NVars-by-1 vector of mean reversion levels (long-run mean or level).
• D is an NVars-by-NVars diagonal matrix, where each element along the main diagonal is the square root of the corresponding element of the state vector.
• V is an NVars-by-NBrowns instantaneous volatility rate matrix.
• dW[t] is an NBrowns-by-1 Brownian motion vector.
Bates models are bivariate composite models. Each Bates model consists of two coupled univariate models.
• A geometric Brownian motion (gbm) model with a stochastic volatility function and jumps that is expressed as follows.
This model usually corresponds to a price process whose volatility (variance rate) is governed by the second univariate model.
• A Cox-Ingersoll-Ross (cir) square root diffusion model that is expressed as follows.
This model describes the evolution of the variance rate of the coupled Bates price process.
[1] Andersen, Leif. “Simple and Efficient Simulation of the Heston Stochastic Volatility Model.” The Journal of Computational Finance 11, no. 3 (March 2008): 1–42.
[2] Broadie, M., and O. Kaya. “Exact Simulation of Option Greeks under Stochastic Volatility and Jump Diffusion Models.” In Proceedings of the 2004 Winter Simulation Conference, 2004., 2:535–43.
Washington, D.C.: IEEE, 2004.
[3] Broadie, Mark, and Özgür Kaya. “Exact Simulation of Stochastic Volatility and Other Affine Jump Diffusion Processes.” Operations Research 54, no. 2 (April 2006): 217–31.
Version History
Introduced in R2020a
R2022b: Perform Brownian bridge and principal components construction
Perform Brownian bridge and principal components construction using the name-value argument BrownianMotionMethod.
R2022a: Perform Quasi-Monte Carlo simulation
Perform Quasi-Monte Carlo simulation using the name-value arguments MonteCarloMethod and QuasiSequence. | {"url":"https://es.mathworks.com/help/finance/bates.simbyquadexp.html","timestamp":"2024-11-04T08:30:05Z","content_type":"text/html","content_length":"154657","record_id":"<urn:uuid:b7fdf37d-69b3-4d2a-8785-4784b2205675>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00297.warc.gz"} |
Jun 25, 2014
Reaction score
10 years of service
Revisiting expressions
In the section Introduction to programming, we had defined an expression as “A mathematical entity that evaluates to a value”. However, the term mathematical entity is somewhat vague. More precisely,
an expression is a combination of literals, variables, operators, and functions that evaluates to a value.
A literal is simply a number, such as 5, or 3.14159. When we talk about the expression “3 + 4″, both 3 and 4 are literals. Literals always evaluate to themselves.
You have already seen variables and functions. Variables evaluate to the values they hold. Functions evaluate to produce a value of the function’s return type. Because functions that return void do
not have return values, they are usually not part of expressions.
Literals, variables, and functions are all known as operands. Operands are the objects of an expression that are acted upon. Operands supply the data that the expression works with.
The last piece of the expressions puzzle is operators. Operators tell how to combine the operands to produce a new result. For example, in the expression “3 + 4″, the + is the plus operator. The +
operator tells how to combine the operands 3 and 4 to produce a new value (7).
You are likely already quite familiar with standard arithmetic operators, including addition (+), subtraction (-), multiplication (*), and division (/). Assignment (=) is an operator as well.
Operators come in two types:
Unary operators act on one operand. An example of a unary operator is the – operator. In the expression -5, the – operator is only being applied to one operand (5) to produce a new value (-5).
Binary operators act on two operands (known as left and right). An example of a binary operator is the + operator. In the expression 3 + 4, the + operator is working with a left operand (3) and a
right operand (4) to produce a new value (7).
Note that some operators have more than one meaning. For example, the – operator has two contexts. It can be used in unary form to invert a number’s sign (eg. -5), or it can be used in binary form to
do arithmetic subtraction (eg. 4 – 3).
This is just the tip of the iceberg in terms of operators. We will take an in-depth look at operators in more detail in a future section. | {"url":"https://katz.to/threads/1-5.851/","timestamp":"2024-11-12T10:15:51Z","content_type":"text/html","content_length":"74867","record_id":"<urn:uuid:1240d32d-fa70-46ac-b96d-4dd7787926c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00260.warc.gz"} |
Neural Networks: Theory and Implementation (NNTI, Winter 2024/2025)
This is is the version for CS, systems engineering, wirtschaftsinformatik, etc. No CoLi students permitted in this edition. The corresponding LSF entry .
Starts: 22.10.24
Location: HS1 in E25 (large lecture hall math)
Exam: tbd
Registration for participation: please register here
Registration for the exam has to be done separately. Please check LSF and the corresponding deadline.
There will be four tutorials groups. Details to follow.
1. Linear Algebra and Principal Component Analysis (PCA)
2. Numerical Computation
3. Machine Learning Basics
4. Deep Feedforwad Neural Networks
5. Regularization for Deep Learning
6. Optimization for Deep Learning
7. Convolutional Neural Networks
8. Sequence Modelling: Recurrent and Recursive Neural Networks
Text Books:
Neural Networks and Deep Learning by Charu C. Aggarwal
Deep Learning by Aaron Courville, Ian Goodfellow, Joshua Bengio
Geometry of Deep Learning, Jong Chul Ye
Deep Learning Architectures: A Mathematical Approach, Ovidiu Calin | {"url":"https://www.lsv.uni-saarland.de/neural-networks-theory-and-implementation-nnti-winter-2024-2025/","timestamp":"2024-11-11T03:20:33Z","content_type":"text/html","content_length":"93492","record_id":"<urn:uuid:b434191e-a995-4028-9276-faabd6f30b77>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00869.warc.gz"} |
Math in Books: Who's #1?
May 9, 2012
In an earlier post, I closed by hinting at the mathematics of ranking. In this modern era, the topic is particularly relevant: the ranking algorithms are hard at work whenever you type something
into a search engine, rate a movie on Netflix, or look at a product on Amazon. It's also a popular area of study among sports enthusiasts, for whom accurate rankings of the relative strengths of
teams can make all the difference in a fantasy league or a betting pool.
Because of all of these accessible applications, it should come as no surprise that the mathematics of ranking is the subject of a new book, titled Who's #1? The Science of Rating and Ranking.
Written by applied mathematicians Amy N. Langville and Carl D. Meyer, the book tackles a variety of methods used to extract ratings or rankings given some collection of input data.
This book is chock full of information. The first part of the book discusses several different algorithms used to construct rankings. There are plenty of applications of the type I mentioned above,
though the authors seem to be particularly fond of sports. They keep one running example from the 2005 NCAA football season running throughout; viewed through the lens of this common example, it's
easier to compare and contrast the ranking and rating methods they discuss.
They also devote a good amount of space to tweaking general ranking methods to suit your particular needs. Want a football ranking system that weighs late season games more heavily than early season
games? Want to understand what you should do to break ties in your data (if anything)? The authors illustrate a variety of alterations that can be made to the basic framework they provide. These
changes allow for a countless supply of ways to rank a given collection of data, each with its own strengths and weaknesses.
In the last part of the book, the authors focus on the ranked lists obtained from the methods they've previously discussed. How can you best aggregate information from a number of ranked lists to
create one super ranked list? This is a question faced by metasearch engines, for example. Also, how can you measure the difference between two ranked lists, and is it possible to objectively
declare one ranked list "better" than another? These are the types of questions addressed in the latter part of the book.
Examples are sprinkled liberally throughout, and the authors provide plenty of references for data junkies looking to apply the methods discussed to data that interests them. I will give one such
example in a follow-up post, focused on my favorite baseball team, to give you a better sense of what this book is all about. Here's Google's explanation of how internet search works, if that's more
up your alley.
Some caveats: a bit of mathematical sophistication will help if you really want to appreciate this book. If the only matrix you're familiar with is the one with Keanu Reeves in it, for example, you
may be in trouble. But if you've taken a Linear Algebra course or two, you should have no trouble following the methods described in the text. And if you're a teacher looking for a way to make an
undergraduate course more relevant, this book may be able to help. Do you have students who are in to sports, for example? It's no coincidence the book was released in time for March Madness -
since ESPN offers $10,000 to the most accurate bracket each year, the right application of the methods in this book could bring in a tidy profit.
My one complaint is the book's discussion of Arrow's Impossibility Theorem, a result from social choice theory most commonly affiliated with voting systems. Since a voting rule is really just a rule
to aggregate a large collection of rankings (i.e. ballots), it arises fairly naturally in their discussions a couple of times. I've discussed this theorem before, so won't spend much time on it
here, but the theorem essentially states that any reasonably constructed voting system can't satisfy a set of common-sense criteria that one would like any voting system to satisfy. The theorem,
however, only applies to ranked voting systems, i.e. systems in which the voters provide a ranking of the candidates (e.g. Candidate A in 1st place, Candidate B in 2nd place, Candidate C in 3rd
place). What the authors fail to mention, however, is that Arrow's theorem need not apply to rated voting systems, i.e. systems in which the votes provide a rating of the candidates (e.g. Candidate
A 100 points, Candidate B 20 points, Candidate C 0 points). Since the authors discuss the importance of the distinction between ratings and rankings, it's unfortunate that they overlook this point
when discussing Arrow's theorem.
Experts in the field may also be upset if their favorite method for ranking is not included, but the authors do a good job of providing honorable mentions in their epilogue. There's plenty of
material that they don't discuss, but references are given throughout, so the interested reader can quite easily further pursue his or her own interests.
All in all, it's quite an interesting read, so if you're curious, I'd encourage you to check it out!
Some final remarks:
1. In the interest of full disclosure, this post is based on a review copy of the book I received from Princeton University Press.
2. If you're interested in purchasing a copy of this book (or other books I've discussed on the site), please consider purchasing through the Amazon widget now displayed on the right hand side.
This isn't a big money making operation here, but if I could cover the annual site maintenance costs on the back of you building your library, it seems like a win-win!
Psst ... did you know I have a brand new website full of interactive stories? You can check it out here!
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Can a 6 sided shape be a parallelogram?
Can a 6 sided shape be a parallelogram?
A parallelogon must have an even number of sides and opposite sides must be equal in length and parallel (hence the name). A less obvious corollary is that all parallelogons have either four or six
sides; a four-sided parallelogon is called a parallelogram.
How many parallelogram are in a hexagon?
A hexagon has 3 parallelograms – Math Central.
What is a hexagon classified as?
A hexagon is an example of a polygon, or a shape with many sides. Hex is a Greek prefix which means ‘six. ‘ A regular hexagon has six sides that are all congruent, or equal in measurement. A regular
hexagon is convex, meaning that the points of the hexagon all point outward.
Why hexagon is not a quadrilateral?
In geometry, a hexagon (from Greek ἕξ, hex, meaning , and γωνία, gonía, meaning ) is a six-sided polygon or 6-gon. The total of the internal angles of any simple (non-self-intersecting) hexagon is
720°. A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners).
What is next to hexagon?
List of n-gons by Greek numerical prefixes
Sides Names
5 pentagon
6 hexagon
7 heptagon septagon
8 octagon
How many triangles are in a hexagon?
six equilateral triangles
A regular hexagon can be dissected into six equilateral triangles by adding a center point. This pattern repeats within the regular triangular tiling.
How many parallelogram are there in a figure?
Detailed Solution. Hence, ’18’ is the correct answer.
What is special about a hexagon?
Mathematically, the hexagon has 6 sides – what makes this particular shape so interesting is that the hexagonal shape best fills a plane with equal size units and leaves no wasted space. Hexagonal
packing also minimises the perimeter for a given area because of its 120-degree angles.
Is a hexagon a convex or concave?
For example, a regular hexagon is also a convex polygon because all of the interior angles equal 120°, which is less than 180°.
Do all hexagons have 6 vertices?
A hexagon has six straight sides and six vertices (corners). It has six angles inside it that add up to 720°.
What are the 4 types of parallelograms?
Types of Parallelograms
• Rhombus (or diamond, rhomb, or lozenge) — A parallelogram with four congruent sides.
• Rectangle — A parallelogram with four congruent interior angles.
• Square — A parallelogram with four congruent sides and four congruent interior angles.
Is a hexagon a regular shape?
No. It is a Hexagon. But a Hexagon is ANY shape composed of 6 intersecting lines. A regular hexagon is a 6 sided shape where ALL lines are the same length and ALL angles are equal in size.
Is a hexagon an irregular polygon?
A six-sided figure, also known as a hexagon, is a polygon commonly found in geometry. Hexagons can either be regular or irregular depending in the length of each side. Finding the perimeter of a
hexagon is relatively easy and requires only simple addition or multiplication.
Is hexagon a regular polygon?
A hexagon is a six sided polygon. The sides of a hexagon are straight line segments. A hexagon is a planar figure, a figure that exists in a plane. A hexagon can be concave or convex. If a convex
hexagon is equilateral (the sides are the same length), then the hexagon is a regular hexagon.
How many parallel sides does a regular hexagon have?
a regular hexagon has 3 pairs of parallel sides. a regular octagon has 4 pairs of parallel sides. those are the sides that are opposite the central angles. | {"url":"https://yoursageinformation.com/can-a-6-sided-shape-be-a-parallelogram/","timestamp":"2024-11-13T07:50:41Z","content_type":"text/html","content_length":"68595","record_id":"<urn:uuid:d643ed04-dbee-49bf-a69a-cc4b5b79ba74>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00815.warc.gz"} |
Detecting Outliers
10.1.7 Detecting Outliers
An outlier is statistically an observation which is numerically distant from the rest of the data. Origin provides methods and tools to help finding and testing for outliers.
To determine whether there is an outlier in a data set from repeated measurements, tools for Grubbs test and Dixon's Q test are available, and it's also possible to roughly visualize the outlier
using the Q-Q plot.
To detect an outlier from regression, you may use the standardized residuals.
Once you’ve statistically determined if a point is an outlier, you can then mask the point using the Regional Mask Tool button, , on the Tools toolbar.
Grubbs Test
For a series of repeated measured data listed in a column, in order to detect if there is an outlier or not with Grubbs Test:
1. Select from menu Statistics:Descriptive Statistics:Grubbs Test to open the grubbs dialog.
2. Select the input data range, significance level and other settings, click OK.
1. Open the Command Window.
2. Call the X-Function grubbs directly.
The result will be output to both Result Log and Command Window, a report sheet will also be generated and if Outlier Plot is selected, a worksheet with plot data will also be generated:
ox The value of the suspected point
index Row index of suspected point
gstat The calculated g value from suspected point
critical The critical g value at the specified significance level
pval The p value for the test
sig sig=1 means there is an outlier, sig=0 means there is no outlier
conclusion A statement of conclusion indicating the statistical result.
rd The worksheet range to put the plot data for outlier plot, if the Outlier Plot option is selected.
rt The worksheet range to put the report table.
Dixon's Q Test
For a series of repeated measured data listed in a column (Sample size from 3 to 10), in order to detect if there is an outlier or not with Dixon's Q Test:
1. Select from menu Statistics:Descriptive Statistics:Dixon's Q Test to open the qtest dialog.
2. Select the input data range, significance level and click OK.
1. Open the Command Window.
2. Call the X-Function qtest directly.
The result will be output to both Result Log and Command Window:
ox The value of the suspected point
index Row index of suspected point
qstat The calculated Q value from suspected point
critical The critical Q value at the specified significance level
sig sig=1 means there is an outlier, sig=0 means there is no outlier
conclusion A statement of conclusion indicating the statistical result.
rd The worksheet range to put the plot data for outlier plot, if the Outlier Plot option is selected..
rt The worksheet range to put the report table.
Detect Outlier with Residual Plot
You can perform a regression (Linear, Polynomial or Nonlinear Curve Fitting), and then use the standardized residuals to determine which data points are outliers.
The following short tutorial will show you how to make use of residual plot to detect outlier:
1. Start with a new workbook and import the file \Samples\Curve Fitting\Outlier.dat
2. Click and select the second column and use the menu item Plot : Symbol : Scatter to create a scatter plot.
3. With the graph active, use the menu item Analysis : Fitting : Linear Fit to bring up the Linear Fit dialog. Note that if you have used the Linear Fit dialog before, there will be a fly-out menu
and you need to select the Open Dialog... sub menu.
4. Expand the Fit Options tree node in the dialog, and uncheck the Apparent Fit check box.
5. Expand the Residual Analysis tree node in the dialog, and check the Standardized check box.
6. Change the Recalculate drop-down at the top of the dialog to Auto and press the OK button at the bottom of the dialog. The dialog will close and linear regression will be performed on the data.
7. Select the FitLinearCurves1 result sheet in the data workbook and scroll to the right side to view the Standardized Residual column. You will note that the value in row 6 in this columns is
-2.54889, hence this data point is an outlier:
8. Make the graph active and then click and hold down the mouse left button on the Regional Mask Tool button in the Tools toolbar. Select the Mask Points on Active Plot submenu which will be the
first item in the fly-out menu:
9. With the above submenu selected, go to the graph and click on the 6th data point to mask the point. This changes the input data to the linear fit operation and the auto update mechanism will
trigger. The linear fit will be repeated with this particular masked point left out. The fit curve in the graph and the pasted parameters will automatically update. Your result graph should then
look like below: | {"url":"https://d2mvzyuse3lwjc.cloudfront.net/doc/Origin-Help/Detect-Outlier","timestamp":"2024-11-13T09:18:14Z","content_type":"text/html","content_length":"131102","record_id":"<urn:uuid:189a3ee6-582e-42ee-9bd9-421f002e110f>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00328.warc.gz"} |
FREE Fraction Of a Number Activity Using Google Slides
Get 150+ Free Math Worksheets!
This hands-on fraction of a number activity uses google slides to allow children to manipulate objects to find the answer. Perfect for spring.
Does it feel like spring where you are? Spring came a little early for us this year, and we are enjoying the spring-like temperatures and seeing the flowers start to bloom…though we are not enjoying
the pollen or the allergies that come with it.
Today, I have a fun fraction of a number activity that has a spring theme.
Prep – Work
I’m really enjoying creating google slides. It is a fun way to get in hands-on work without pulling out tons of stuff. Though there is a big need for that, so please don’t cut that out of your
lessons. But once in a while, it is nice to be able to manipulate objects without all the mess. And this fraction of a number activity does just that.
Prep-work is simple.
1. Download the slide by clicking on the large blueish button at the end of the post.
2. Next, check your email and you will get a link to the slides.
3. When the slide comes up, you will need to save it so you and your children can move the objects.
4. Finally, teach the lesson and have fun with it.
Fraction Of a Number Activity:
This video with take you through how to use this activity. Click here to get it.
The biggest thing to remember is you are teaching your children that fractions are equal groups, and this fraction of a number activity drives that home. They get to make groups based on a fraction
and then sort the flowers to make equal groups.
Once they are done with that they may enjoy some of these paper activities.
I hope you enjoy!
You’ve Got This,
After this activity you may enjoy some pencil and paper activities:
This game is a fun way to practice this skill.
Or solve some word problems for Fraction of A Number | {"url":"https://youvegotthismath.com/fraction-of-a-number-activity/","timestamp":"2024-11-05T02:34:36Z","content_type":"text/html","content_length":"331678","record_id":"<urn:uuid:67ed469a-fe73-4846-8acf-b2d91c850594>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00287.warc.gz"} |
Kiwi Shading Data
kiwishade {DAAG} R Documentation
Kiwi Shading Data
The kiwishade data frame has 48 rows and 4 columns. The data are from a designed experiment that compared different kiwifruit shading treatments. There are four vines in each plot, and four plots
(one for each of four treatments: none, Aug2Dec, Dec2Feb, and Feb2May) in each of three blocks (locations: west, north, east). Each plot has the same number of vines, each block has the same number
of plots, with each treatment occurring the same number of times.
This data frame contains the following columns:
Total yield (in kg)
a factor with levels east.Aug2Dec, east.Dec2Feb, east.Feb2May, east.none, north.Aug2Dec, north.Dec2Feb, north.Feb2May, north.none, west.Aug2Dec, west.Dec2Feb, west.Feb2May, west.none
a factor indicating the location of the plot with levels east, north, west
a factor representing the period for which the experimenter placed shading over the vines; with levels: none no shading, Aug2Dec August - December, Dec2Feb December - February, Feb2May February -
The northernmost plots were grouped together because they were similarly affected by shading from the sun in the north. For the remaining two blocks shelter effects, whether from the west or from the
east, were thought more important.
Snelgar, W.P., Manson. P.J., Martin, P.J. 1992. Influence of time of shading on flowering and yield of kiwifruit vines. Journal of Horticultural Science 67: 481-487.
Maindonald J H 1992. Statistical design, analysis and presentation issues. New Zealand Journal of Agricultural Research 35: 121-141.
print("Data Summary - Example 2.2.1")
kiwimeans <- aggregate(yield, by=list(block, shade), mean)
names(kiwimeans) <- c("block","shade","meanyield")
print("Multilevel Design - Example 9.3")
kiwishade.aov <- aov(yield ~ shade+Error(block/shade),data=kiwishade)
sapply(split(yield, shade), mean)
kiwi.table <- t(sapply(split(yield, plot), as.vector))
kiwi.means <- sapply(split(yield, plot), mean)
kiwi.means.table <- matrix(rep(kiwi.means,4), nrow=12, ncol=4)
kiwi.summary <- data.frame(kiwi.means, kiwi.table-kiwi.means.table)
names(kiwi.summary)<- c("Mean", "Vine 1", "Vine 2", "Vine 3", "Vine 4")
mean(kiwi.means) # the grand mean (only for balanced design)
if(require(lme4, quietly=TRUE)) {
kiwishade.lmer <- lmer(yield ~ shade + (1|block) + (1|block:plot),
## block:shade is an alternative to block:plot
## Residuals and estimated effects
xyplot(residuals(kiwishade.lmer) ~ fitted(kiwishade.lmer)|block,
data=kiwishade, groups=shade,
layout=c(3,1), par.strip.text=list(cex=1.0),
xlab="Fitted values (Treatment + block + plot effects)",
ylab="Residuals", pch=1:4, grid=TRUE,
scales=list(x=list(alternating=FALSE), tck=0.5),
key=list(space="top", points=list(pch=1:4),
ploteff <- ranef(kiwishade.lmer, drop=TRUE)[[1]]
qqmath(ploteff, xlab="Normal quantiles", ylab="Plot effect estimates",
version 1.25.6 | {"url":"https://search.r-project.org/CRAN/refmans/DAAG/html/kiwishade.html","timestamp":"2024-11-04T04:21:36Z","content_type":"text/html","content_length":"5434","record_id":"<urn:uuid:8dbd101f-0a0c-487c-a574-68cc91f93f75>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00413.warc.gz"} |
NEPSE SHORT TERM TREND - Page 2
· March 5, 2022, 2:13 pm
Thankyou sir for your guidance.
I would like to share my analysis on similar basis.
This would be a Short term analysis for end of WAVE C
Part 1
we assume that wave C can retrace upto 2.236 of A (ie.2459)
Part 2: Subminuttee: 5th has extended
As the 5th wave has already extended, during the 5th wave extension it can be expected to extend atleast 1.618. & Considering the part 1 analysis, 5th is going for 1.618. Such we can say it has
possibility to reach even 2370.
Part 3: Micro degree (currently on making of 3rd or 5th)
3rd will extend upto 1.618 of 1 =ie. 2510
4th will retrace upto 0.236 = 2550 (from wave 2-3)
5th will extend upto 1 of 1st. i.e (2447)
Part 4
Considering the past swing a swing bottom is due somewhere around end of next week: ~10March
And if we draw a channel wave C it will intersect at 2440 on this day.
Analyzing on 4 parts
we can predict that the final bottom can be expected at range (min2370 to max 2459) (diff = 89pts), somewhere around 10-3-2022
In a more likely situation is it will seek its bottom in range of 2440 to 2459
With a min expected drop: 3.5% (2459)
& max expected drop: 6.9% (2370)
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Last edited on March 5, 2022, 4:18 pm by Gunjan shrestha
· March 12, 2022, 3:17 pm
A bounce t6towards 2670 to 2730 is possible in both cases whether it is preferred or alternative.
Hydro seems finishing 2ns wave with equality of wave wave A and C.
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· March 12, 2022, 3:18 pm
A bounce towards 2670 to 2730 is possible in both cases whether it is preferred or alternative.
Hydro seems finishing 2nd wave with equality of wave wave A and C.
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Last edited on March 12, 2022, 3:19 pm by bishnu.p.basyal@gmail.com
· March 14, 2022, 11:25 am
Both Possibilities are open. With Red counting post correction towards 2600 NEPSE will rise aggressively by breaking channel and other alternative says it may move slowly enclosing within the
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· April 6, 2022, 12:43 pm
Whether it is preferred or alternative. a bounce towards 2700 is highly probable. Rising above today's high of 2458 will confirm the same.
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Last edited on April 6, 2022, 8:03 pm by bishnu.p.basyal@gmail.com
· April 6, 2022, 8:06 pm
A story telling of two charts on different time frames:
A bottom leading to November 3 to December 13 & Recent fall from March 14 to April 6 made s
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Last edited on April 6, 2022, 8:14 pm by bishnu.p.basyal@gmail.com
· May 1, 2022, 1:06 am
There is slight change in the cycle wave. The count has been changed to Intermediate Degree Wave (1) at 3227 and current decline can be considered as Intermediate Degree Wave (2). It does not change
ultimate target of NESPE as primary Degree Wave 3 which can be estimated above 6000 till 2024 AD but modification in Wave Degree only. Main Reason for change in Wave Degree is because of time spent
for current decline. Minor Degree wave generally takes about 3-4 months only while Intermediate Degree Wave can take longer time say about 8-12 months as well.
Here are some Price time Calculations with suggests Either Wave (2) has already been made or May take a bit longer which varies from another 3 to 13 days. Whatever may be the situation. An Explosive
and Surprising Wave (3) is about to start soon.
Wave a of Wave W was 35 (+-1 of 34) days affair.
Wave c of Wave W was 25 days affair.
Wave a of Wave Y was 33 (+-1 of 34) days affair.
Wave c of Wave Y took exactly 25 days.
Wave W took 71 days. So, if time equality applies then within next 3 days Wave Y may complete if fall continues from Monday onwards.
Level of 2090 and 50% retracement from 1149-3227 & Wave W equals Wave Y as well in log scale.
Level of 2237 in arithmetic Scale & 2290 in Log Scale equals wave A with Wave C as alternative count.
Thus, in Nutshell, Wave (2) bottom has already been made at 2287 on 21st of April 2022 or it may drift lower in the range of 2240 to 2090 within next 3 days (to match 71 days) or 13 days which
matches with 61.8% Fibonacci time for Wave (2) of Entire Wave (1).
Click for thumbs down.0Click for thumbs up.0
· May 1, 2022, 1:07 am
There is slight change in the cycle wave. The count has been changed to Intermediate Degree Wave (1) at 3227 and current decline can be considered as Intermediate Degree Wave (2). It does not
change ultimate target of NESPE as primary Degree Wave 3 which can be estimated above 6000 till 2024 AD but modification in Wave Degree only. Main Reason for change in Wave Degree is because of
time spent for current decline. Minor Degree wave generally takes about 3-4 months only while Intermediate Degree Wave can take longer time say about 8-12 months as well.
Here are some Price time Calculations with suggests Either Wave (2) has already been made or May take a bit longer which varies from another 3 to 13 days. Whatever may be the situation. An
Explosive and Surprising Wave (3) is about to start soon.
Wave a of Wave W was 35 (+-1 of 34) days affair.
Wave c of Wave W was 25 days affair.
Wave a of Wave Y was 33 (+-1 of 34) days affair.
Wave c of Wave Y took exactly 25 days.
Wave W took 71 days. So, if time equality applies then within next 3 days Wave Y may complete if fall continues from Monday onwards.
Level of 2090 and 50% retracement from 1149-3227 & Wave W equals Wave Y as well in log scale.
Level of 2237 in arithmetic Scale & 2290 in Log Scale equals wave A with Wave C as alternative count.
Thus, in Nutshell, Wave (2) bottom has already been made at 2287 on 21st of April 2022 or it may drift lower in the range of 2240 to 2090 within next 3 days (to match 71 days) or 13 days which
matches with 61.8% Fibonacci time for Wave (2) of Entire Wave (1).
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Tanya - MATLAB Central
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URGENT... Matlab System::FatalException!!!
Hi all,, I have SERIOUS problem with my Matlab. I m going to do classification by using SIFT featrues (vl_sift function from VLF...
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Eigen vector in SVD??
Im going to compute the eigen value and eigen vector from my Matrix data fro the classification. The rows represent the diff...
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PCA calculation for classification?
To do PCA, we have to compute the covariance matrix from our input data and then eigen decomposition is performed in that covari...
10 years ago | 1 answer | 0
Calculation of Principal Component Analysis
I am strugling with PCA stuff. So for example I have : Data=100*3 substractdata=data-mean (the size will be s...
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Proof that the computed Eigen Vectors is righ?t
Hii all.. I am working in PCA to do Dimensionally Reduction. I already extract the eigen values and eigen vectors from the r...
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Suggestion about method for recognition based on machine learning?
I am going to develop a traffic sign recognition based on machine learning (SVM). I just have to work at the recognition part (N...
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Plot eigen Vector e1 & e2 in matlab?
How to be able plot eigen vector in Matlab an make it sure have the exact position as a new axes of our data?? here is my cod...
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PCA generated initial matrix in gaussian and ellipse?
I have to do PCA in Matlab for object recognition. For the input matrix / data, it should be fulfill the condition (to be dis...
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How to get the number of significant Eigen Value?
Im working in Matlab to compute the PCA. I already compute the Eigen Value and the Eigen Vector. I used this matlab function ...
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URGENT!!! Matlab Symbolic Variable Error???
Im tring to substract the diagonal values with *eigval* and store the new value in the matrix *Diagonal* : CovarianceMatrix...
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Matrix Calculation in MATLAB
Could someone help me solve this problem in Matlab.. Suppose I have this Matriks A=[2-x 5 2 3-x ] So, it...
10 years ago | 2 answers | 0
How to calculate determinant in PCA?
Im going to program PCA, but for that, I have to calculate the Eigen Vector and Eigen Value. My question is in calculate the ...
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What is Surface Contamination - Definition
What is Surface Contamination – Definition
Surface contamination means that radioactive material has been deposited on surfaces (such as walls, floors). It may be loosely deposited, much like ordinary dust, or it may be quite firmly fixed by
chemical reaction. Radiation Dosimetry
Radioactive contamination consist of radioactive material, that generate ionizing radiation. It is the source of radiation, not radiation itself.
Surface Contamination
Surface contamination means that radioactive material has been deposited on surfaces (such as walls, floors). It may be loosely deposited, much like ordinary dust, or it may be quite firmly fixed by
chemical reaction. This distinction is important, and we classify surface contamination on the basis of how easily it can be removed:
• Free Contamination. In the case of free contamination (or loose contamination), the radioactive material can be spread. This is surface contamination that can easily be removed with simple
decontamination methods. For example, if dust particles containing various radioisotopes land on the person’s skin or garments, we can clean it up or remove clothes. Once a person has been
decontaminated, all of the particulate radioactivity sources are eliminated, and the individual is no longer contaminated. Free contamination is also a more serious hazard than fixed
contamination, because dust particles can become airborne and they can be easily ingested. This leads to an internal exposure by radioactive contaminants. Although almost all contaminants are
beta radioactive with accompanying gamma emission, but there is also the possibility of alpha contamination in any nuclear fuel handling areas.
• Fixed Contamination. In the case of fixed contamination, the radioactive material cannot be spread, since it is chemically or mechanically bound to structures. It cannot be removed by normal
cleaning methods. Fixed contamination is a less serious hazard than free contamination, it cannot be re-suspended or transferred to skin. Therefore the hazard is usually an external one only. On
the other hand, it depends on the level of contamination. Although almost all contaminants are beta radioactive with accompanying gamma emission, but there is also the possibility of alpha
contamination in any nuclear fuel handling areas. Unless the level of contamination is very severe, the gamma radiation dose rate will be small and external exposure will be significant only in
contact with, or very close to, the contaminated surfaces. Since beta particles are less penetrating than gamma rays, the beta dose rate can be high only at contact. A value of 1 mSv/h at contact
for a contamination level of 400 – 500 Bq/cm^2 is fairly representative.
Calculation of Shielded Dose Rate in Sieverts from Contaminated Surface
Assume a surface, which is contamined by 1.0 Ci of ^137Cs. Assume that this contaminant can be aproximated by the point isotropic source which contains 1.0 Ci of ^137Cs, which has a half-life of 30.2
years. Note that the relationship between half-life and the amount of a radionuclide required to give an activity of one curie is shown below. This amount of material can be calculated using λ, which
is the decay constant of certain nuclide:
About 94.6 percent decays by beta emission to a metastable nuclear isomer of barium: barium-137m. The main photon peak of Ba-137m is 662 keV. For this calculation, assume that all decays go through
this channel.
Calculate the primary photon dose rate, in sieverts per hour (Sv.h^-1), at the outer surface of a 5 cm thick lead shield. Then calculate the equivalent and effective dose rates for two cases.
1. Assume that this external radiation field penetrates uniformly through the whole body. That means: Calculate the effective whole-body dose rate.
2. Assume that this external radiation field penetrates only lungs and the other organs are completely shielded. That means: Calculate the effective dose rate.
Note that, primary photon dose rate neglects all secondary particles. Assume that the effective distance of the source from the dose point is 10 cm. We shall also assume that the dose point is soft
tissue and it can reasonably be simulated by water and we use the mass energy absorption coefficient for water.
See also: Gamma Ray Attenuation
See also: Shielding of Gamma Rays
The primary photon dose rate is attenuated exponentially, and the dose rate from primary photons, taking account of the shield, is given by:
As can be seen, we do not account for the buildup of secondary radiation. If secondary particles are produced or if the primary radiation changes its energy or direction, then the effective
attenuation will be much less. This assumption generally underestimates the true dose rate, especially for thick shields and when the dose point is close to the shield surface, but this assumption
simplifies all calculations. For this case the true dose rate (with the buildup of secondary radiation) will be more than two times higher.
To calculate the absorbed dose rate, we have to use in the formula:
• k = 5.76 x 10^-7
• S = 3.7 x 10^10 s^-1
• E = 0.662 MeV
• μ[t]/ρ = ^ 0.0326 cm^2/g (values are available at NIST)
• μ = 1.289 cm^-1 (values are available at NIST)
• D = 5 cm
• r = 10 cm
The resulting absorbed dose rate in grays per hour is then:
1) Uniform irradiation
Since the radiation weighting factor for gamma rays is equal to one and we have assumed the uniform radiation field (the tissue weighting factor is also equal to unity), we can directly calculate the
equivalent dose rate and the effective dose rate (E = H[T]) from the absorbed dose rate as:
2) Partial irradiation
In this case we assume a partial irradiation of lungs only. Thus, we have to use the tissue weighting factor, which is equal to w[T] = 0.12. The radiation weighting factor for gamma rays is equal to
one. As a result, we can calculate the effective dose rate as:
Note that, if one part of the body (e.g.,the lungs) receives a radiation dose, it represents a risk for a particularly damaging effect (e.g., lung cancer). If the same dose is given to another organ
it represents a different risk factor.
If we want to account for the buildup of secondary radiation, then we have to include the buildup factor. The extended formula for the dose rate is then:
We hope, this article, Surface Contamination, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information
about radiation and dosimeters. | {"url":"https://www.radiation-dosimetry.org/what-is-surface-contamination-definition/","timestamp":"2024-11-03T13:57:24Z","content_type":"text/html","content_length":"469113","record_id":"<urn:uuid:74905c54-7165-42ba-bc05-8e7cada80db0>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00305.warc.gz"} |
Lesson 13
Divide Whole Numbers by Unit Fractions
Warm-up: Notice and Wonder: Quilt (10 minutes)
The purpose of this warm-up is for students to describe the rectangles in the representation of a quilt, which will be useful when students divide strips of paper into unit fraction sized pieces in a
later activity. While students may notice and wonder many things about this image, the variety of lengths and colors of fabric strips is the important discussion point.
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
• “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Share and record responses.
Student Facing
What do you notice? What do you wonder?
Activity Synthesis
• “These pictures show women from Gee's Bend, Alabama, who have been making quilts for generations. How would you describe the quilt they are working on?” (It is colorful. There are rectangles.
There are different colored pieces of fabric.)
• Consider showing students examples of abstract or improvised quilts by Gee’s Bend Quiltmakers from the website of Souls Grown Deep.
Activity 1: Paper Strips (20 minutes)
The purpose of this activity is for students to solve problems about dividing a whole number by a unit fraction in a way that makes sense to them. The context of quilt making is used so students can
visualize a strip of paper that is a whole number length being cut into fractional sized pieces. As students describe how the problems are similar and different, listen for the authentic language
they use to describe division. The paper strip, or tape, is a helpful diagram to use when dividing a whole number by a unit fraction because students recognize important relationships between the
divisor, dividend, and quotient (MP7). For example, if the length of the strip stays the same, but the size of the piece gets smaller, then the number of pieces will get bigger.
This activity uses MLR2 Collect and Display. Advances: Conversing, Reading, Writing.
• Groups of 2
• Refer to the picture from the warm up.
• “If the blue strip of fabric under the woman’s chin is 1 meter long, about how long is the short gray strip next to it?” (\(\frac{1}{6}\) meter)
• 1–2 minutes: independent think time
• 8–10 minutes: partner work time
MLR2 Collect and Display
• Circulate, listen for, and collect the language students use to describe what was the same and different about the strategies they used to determine the number of pieces of paper for each color.
Listen for:
□ The size of the piece changed.
□ The pieces were shorter.
□ There were more pieces.
□ I made more cuts.
• Record students’ words and phrases on a visual display and update it throughout the lesson.
Student Facing
Below are diagrams that show strips of different colored paper. Each strip is 2 feet long. The paper strips will be cut into different sized pieces.
1. The red strip will be cut into pieces that are \(\frac{1}{2}\) foot long. How many pieces will there be?
2. The green strip will be cut into pieces that are \(\frac{1}{3}\) foot long. How many pieces will there be?
3. The yellow strip will be cut into pieces that are \(\frac{1}{4}\) foot long. How many pieces will there be?
4. Describe what was the same about the problems you solved. Describe what was different.
Advancing Student Thinking
Students may not immediately make a connection with the situation and division, since the number of pieces that results (the quotient) is greater than the number of feet being divided (the dividend).
Consider asking, “How does this situation represent division?” Allow students to recognize that the cuts are similar to partitions in a diagram, or sharing. They may also recognize that the quotient
represents the number of groups, while the fraction being divided is the size of each group. Have them write a division equation for each situation, if it helps. They will write division equations in
the next activity.
Activity Synthesis
• “Are there any other words or phrases that are important to include on our display?”
• As students share responses, update the display by adding (or replacing) language, diagrams, or annotations.
• Remind students to borrow language from the display as needed.
• Display:
\(2 \div \frac {1}{2} = 4\)
\(2 \div \frac {1}{3} = 6\)
\(2 \div \frac {1}{4} = 8\)
• “How do these equations represent the problems about the paper strips?” (The 2 is for 2 feet of paper, and the fractions show the size of the pieces that the paper is being cut into. The 4, 6,
and 8 are the number of pieces of each color of paper.)
Activity 2: More Paper Strips (15 minutes)
The purpose of this activity is for students to represent division of a whole number by a unit fraction with diagrams and equations. The context is the same as the previous activity so students can
use a tape diagram to solve the problem, if they choose. In the previous activity, students recognized that when the length of paper stays the same and the size of the piece gets smaller, there are
more pieces of paper. In this activity, students will consider what happens when the length of the paper changes, but the size of the pieces stays the same.
Representation: Access for Perception. Provide access to strips of paper for students to cut and fold. Ask students to identify correspondences between the number of pieces/folds and the fraction
they represent.
Supports accessibility for: Conceptual Processing, Memory
• 1–2 minutes: independent think time
• 6–8 minutes: partner work time
• Monitor for students who:
□ determine that Kiran will have 12 pieces of paper
□ can explain how the equation \(2 \div \frac{1}{6} = 12\) represents the yellow strip of paper being cut into \(\frac16\)-foot strips
□ describe the equation \(3 \div \frac {1}{6} = 18\) as representing a 3 foot strip of paper being cut into 18 pieces that are each \(\frac {1}{6}\) of a foot long
Student Facing
Kiran has a yellow strip of paper that is 2 feet long. He wants to cut the strip into pieces that are
foot long.
1. How many pieces will Kiran have? Explain or show your reasoning.
2. Write a division equation to represent the situation.
3. Describe how the equation \(3 \div \frac{1}{6} = 18\) represents a strip of paper that is 3 feet long being cut into equal-sized pieces.
Advancing Student Thinking
If students do not write an equation that represents the situation, show them \(2 \div \frac{1}{6}\) and ask, “How does this expression represent the situation?”
Activity Synthesis
• Ask previously identified students to share their solutions.
• Display: \(2 \div \frac {1}{6} = 12\)
• “How does this equation represent the yellow strip of paper?” (The strip of paper is 2 feet long and it is cut into pieces that are \(\frac {1}{6}\) of a foot long so there will be 12 pieces.)
• Display: \(3 \div \frac {1}{6} = 18\)
• “How does this equation represent a different strip of paper being cut into equal sized pieces?” (A 3 foot piece of paper is cut into pieces that are \(\frac {1}{6}\) of a foot long so there are
18 pieces.)
• “Why is the quotient larger than the dividend in both of these equations?” (Because you are cutting a whole number length into fractional sized pieces, so there will be more pieces than when you
Lesson Synthesis
"Today, we solved problems about cutting strips of paper into small pieces. We wrote equations to represent dividing a whole number by a unit fraction.”
\(2 \div \frac {1}{2} = 4\)
\(2 \div \frac {1}{3} = 6\)
\(2 \div \frac {1}{4} = 8\)
\(2 \div \frac {1}{6} = 12\)
“These are some of the equations we discussed today. Why is the quotient getting larger in each equation?” (Because the size of the piece is getting smaller, so there will be more pieces.)
Display: \(3 \div \frac {1}{6} = 18\)
“Here is another equation we discussed. In this equation, the size of the piece is the same as the equation above it. Why is the quotient larger than when 2 is divided by \(\frac16\)?” (3 is being
divided into smaller pieces, instead of 2, so you get more pieces.)
“We are going to learn more about the relationships between the numbers in division equations with unit fractions in the next lesson.”
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Linear Pricing
Procurement Glossary & Terminologies
Linear Pricing
A charting scale known as a linear (arithmetic) pricing scale is plotted with absolute values evenly spaced apart on the vertical y-axis and is majorly employed by traders. No matter the asset’s
price level, when the change occurs, each unit change is represented by the same vertical distance on the chart.
A comparison can be made between the logarithmic and linear scales. Depending on the type of price scale used to examine the data, different traders may read a stock chart differently.
How Linear Price Scales Work
The financial sector frequently uses two types of price scale charts: linear and logarithmic (log) pricing scales. Technical analysts can utilise either type of chart. The charts are all typically
produced by software automation. Since they rely on static units that reflect absolute values, linear pricing scale charts are easier to create by hand.
Since the unit value movements in logarithmic charts are often expressed in percentages rather than constants, they typically need sophisticated chart programming. The exact x-axis dates will be used
for linear and logarithmic chart charting.
An arithmetic chart is another name for a linear price scale. The chart’s linear price scale does not scale or represent movement with its % change. With each change in unit value equivalent to a
constant unit value, the linear price scale shows changes in the price level. Linear pricing scales are simpler to draw manually because each value change on the grid is constant.
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Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion
Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 1 Laws of Motion Notes, Textbook Exercise Important Questions and Answers.
Maharashtra State Board Class 9 Science Solutions Chapter 1 Laws of Motion
Class 9 Science Chapter 1 Laws of Motion Textbook Questions and Answers
1. Match the first column with appropriate entries in the second and third columns and remake the table.
2. Clarify the differences
A. Distance and displacement
Distance Displacement
(i) Distance is the length of the actual path travelled by an object. (i) Displacement is the minimum distance between the starting and finishing points.
(ii) It is a scalar quantity. (ii) It is a vector quantity.
(iii) It is either equal to or greater than displacement. (iii) It is either equal to or less than distance.
(iv) Distance travelled is always positive. (iv) Displacement may be positive or negative or zero.
B. Uniform and non-uniform motion.
Uniform motion Non-uniform motion
(i) If an object covers equal distances in equal intervals of time it is said to be in uniform (i) If an object moves unequal distances in equal intervals of time, its motion is said to be
motion. nonuniform.
(ii) Distance – time graph for uniform motion is a straight line. (ii) Distance – time graph for non-uniform motion is not a straight line.
(iii) In uniform motion, acceleration is zero. (iii) In non-uniform motion acceleration is non-zero.
3. Complete the following table.
4. Complete the sentences and explain them.
a. The minimum distance between the start and finish points of the motion of an object is called the ……….. of the object.
b. Deceleration is ………………………. acceleration
c. When an object is in uniform circular motion, its ………………………. changes at every point.
d. During collision ………………………. remains constant.
e. The working of a rocket depends on Newton’s ………………………. law of motion.
5. Give scientific reasons.
a. When an object falls freely to the ground, its acceleration is uniform.
• When the body falls freely to the ground, there are equal changes in velocity of the body in equal intervals of time.
• Thus the acceleration of the body is constant, and it possesses uniform acceleration.
b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
• Action and reaction forces act on different bodies.
• They don’t act on the same body, hence they cannot cancel each other’s effect.
• Hence, even though the magnitudes of action force and reaction force are equal, they do not cancel each other.
c. It is easier to stop a tennis ball as compared to a cricket ball, when both are traveling with the same velocity.
• Momentum of an object depends on its mass as well as its velocity.
• Cricket ball is heavier than a tennis ball. Although they are thrown with the same velocity, cricket ball has more momentum than a tennis ball.
• The force required to stop a cricket ball is more than a tennis ball.
• Hence it is easier to stop a tennis ball than a cricket ball moving with same velocity.
d. The velocity of an object at rest is considered to be uniform.
• When a body is at rest there is no change in velocity.
• A body with constant velocity is said to be in uniform motion.
• Hence, the state of rest is an example of uniform motion.
6. Take 5 examples from your surroundings and give an explanation based on Newton’s laws of motion.
7. Solve the following examples.
a) An object moves 18 m in the first 3 s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed? (Ans: 6 m/s)
Total distance (d) = 18 + 22 + 14 = 54 m
Total time taken (t) = 3 + 3 + 3 = 9 sec
To find:
Average speed = ?
The object moves with an average speed of 6 m/s.
b) An object of mass 16 kg is moving with an acceleration of 3 m/s^2. Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration? (Ans: 48
N, 2 m/s^2)
The force acting on the 1 body is 48 N and the acceleration of the 2” body is 2 m/s^2
c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both
move together. Determine their velocity. (Ans: 0.15 m/s)
The plank embedded with the bullet moves with a velocity of 0.15 m/s.
d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed? (Ans: 2.25 m/s^2)
Total distance (d) = 100 + 80 + 45 = 225 m
Total time taken (t) = 40 + 40 + 20 = 100 sec
To find:
Average speed =?
The person swims with an average speed of 2.25 m/s.
Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers
(i) Who will take less time to reach the school and why?
Prashant will take less time as the path followed by him is the shortest.
(a) Every morning, Swaralee walks round the edge of a circular field having a radius of 100 m. As shown in figure (a), if she starts from the point A and takes one round, how much distance has she
walked and what is her displacement?
Radius (r) = 100 m
Distance covered = Circumference of the circle
= 2 nr
= 2 x 3.14 x 100
= 628 m
Displacement = 0 m (Shortest distance between initial and final position is zero)
(b) If a car, starting from point P, goes to point Q (see figure 1.9) and then returns to point P, how much distance has it travelled and what is its displacement?
Distance covered = PQ + QP
= 360 + 360
= 720 m
Displacement = 0 m (The shortest distance between initial and final position is zero)
Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers
(A) Choose and write the correct option:
Laws Of Motion Class 9 Questions And Answers Maharashtra Board Question 1.
The displacement that occurs in unit time is called …………….. .
(a) displacement
(b) distance
(c) velocity
(d) acceleration
(c) velocity
Laws Of Motion Class 9 Maharashtra Board Exercise Answers Question 2
The unit of velocity in the SI system is …………….. .
(a) cm/s
(b) rn/s^2
(c) um/s^2
(d) rn/s
(d) m/s
Laws Of Motion Class 9 Maharashtra Board Question 3.
v^2 = u^2 + 2as is the relation between and …………….. .
(a) speed and velocity
(b) distance and acceleration
(c) displacement and velocity
(d) speed and distance
(c) displacement and velocity
Class 9 Science Notes Chapter 1 Laws Of Motion Question 4.
…………….. is the relation between displacement and time.
(a) v = u + at
(b) v^2 = u^2 + 2as
(c) s = ut + 1/2 at^2
(d) v = u + 2as
(c) s = ut + 1/2 at^2
Class 9 Science Chapter 1 Laws Of Motion Question Answer Question 5.
The force necessary to cause an acceleration of 1 m/s^2 in an object of mass 1 kg is called …………….. .
(a) 1 dyne
(b) 1 m/s
(c) 1 Newton
(d) 1 cm/s
(c) 1 Newton.
9th Science Chapter 1 Laws Of Motion Question 6.
Even if the displacement of an object is zero, the actual distance traversed by it …………….. .
(a) may not be zero.
(b) will be zero
(c) will be constant
(d) will be infinity
(a) may not be zero
Question 7.
If the velocity changes by equal amounts in equal time intervals, the object is said to be in …………….. .
(a) uniform acceleration
(b) uniform velocity
(c) non-uniform acceleration
(d) non-uniform motion
(a) uniform acceleration
Question 8.
If an object is moving with a uniform velocity …………….. .
(a) its speed remains the same, but direction of motion changes
(b) its speed changes but direction of motion is same
(c) its speed and direction both change
(d) its speed and direction both remain the same
(d) its speed and direction both remain the same
Question 9.
is an example of positive acceleration.
(a) A stone is thrown vertically upwards
(b) A stone falls freely towards the earth
(c) Brakes are applied by the truck driver
(d) The train arriving at the station
(b) a stone falls freely towards the earth
Question 10.
An object continues to remain at rest or in a state of uniform motion along a straight line unless an …………….. acts on it.
(a) internal imbalanced force
(b) external unbalanced force
(c) internal balanced force
(d) external balanced force
(b) external unbalanced force
Question 11.
The …………….. is proportional to the applied force and it occurs in the direction of the force.
(a) change of momentum
(b) rate of change of velocity
(c) change of velocity
(d) rate of change of momentum
(d) rate of change of momentum
Question 12.
…………….. is a relative concept.
(a) Motion
(b) Direction
(c) Power
(d) Acceleration
(a) Motion
Question 13.
A body is said to be in motion if it changes its …………….. with respect to its surroundings.
(a) position
(b) direction
(c) speed
(d) force
(a) position
Question 14.
A body is said to be at …………….. if it does not change its position with respect to its surroundings.
(a) Motion
(b) Rest
(c) Gravity
(d) Force
(b) Rest
Question 15.
…………….. is the length of the actual path travelled by an object in motion while going from one point to another.
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
(a) Distance
Question 16.
The distance covered by a body in unit time is called its …………….. .
(a) velocity
(b) speed
(c) displacement
(d) rest
(b) speed
Question 17.
S.I. unit of speed is …………….. and in C.G.S unit it is …………….. .
(a) m/s and cm/s
(b) km/s and cm/s
(c) m/s and mm/s
(d) m/s and nm/s
(a) m/s , cm/s
Question 18.
The distance travelled in a particular direction by an object in unit time is called its …………….. .
(a) velocity
(b) speed
(c) displacement
(d) rest
(a) velocity
Question 19.
Units of speed and velocity are the
(a) Same
(b) Different
(c) Greater than each other
(d) Unequal
(a) same
Question 20.
…………….. . is related to distance, while …………….. is related to displacement.
(a) Gravity and magnetism
(b) Speed and force
(c) Speed and velocity
(d) Motion and rest
(c) Speed, velocity
Question 21.
If an object covers equal distances in equal time intervals, it is said to be moving with …………….. speed.
(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
(a) uniform
Question 22.
If an object covers unequal distances in equal time Intervals, it is said to be moving with speed.
(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
(b) non uniform
Question 23.
The rate of change of velocity is called
(a) Speed
(b) Acceleration
(c) Velocity
(d) Rest
(b) acceleration
Question 24.
Speed of light in dry air is …………….. m/s.
(a) 3 x 10^7
(b) 3 x 10^8
(c) 3 x 10^9
(d) 3 x 10^3
(b) 3 x 10^8
Question 25.
When velocity of a body increases, its acceleration is …………….. .
(a) Negative
(b) Zero
(c) Positive
(d) Equal
(c) positive
Question 26.
When velocity of a body decreases, its acceleration is …………….. .
(a) Negative
(b) Zero
(c) Positive
(d) Equal
(a) negative
Question 27.
Negative acceleration is also called or
(a) Deceleration or retardation
(b) Deceleration or acceleration
(c) acceleration or retardation
(d) Zero
(a) deceleration or retardation
Question 28.
In case of motion, object travels equal …………….. in equal intervals of time.
(a) Uniform, distance
(b) Non-Uniform, distance
(c) Uniform, displacement
(d) Uniform, displacement
(a) uniform, distances
Question 29.
Motion of an object was studied by …………….. .
(a) Sir Albert Einstein
(b) Sir Thomas Edison
(c) Sir Isaac Newton
(d) Sir Ravindranath Tagore
(c) Sir Issac Newton
Question 30.
When an object moves in a circular path with uniform speed, its motion is …………….. motion.
(a) Non uniform circular
(b) Random circular
(c) Uniform circular
(d) Uniform linear
(c) uniform circular
Question 31.
When a coin moves along a circular path, the direction of its motion at every point is …………….. .
(a) Circular
(b) Concave
(c) Tangential
(d) Convex
(c) tangential
Question 32.
For all uniformly accelerated motions, the velocity-time graph is a …………….. .
(a) Curved line
(b) Straight line
(c) Negative line
(d) Positive line
(b) straight line
Question 33.
In the distance-time graph, the slope of the straight line indicates …………….. .
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
(b) velocity
Question 34.
The first equation of motion gives relation between …………….. and time.
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
(b) velocity
Question 35.
Newton’s first law explains the phenomenon of
(a) Rest
(b) Inertia
(c) Speed
(d) Velocity
(b) inertia
Question 36.
…………….. cause a change in the state of an object at rest or in uniform motion.
(a) Balanced forces
(b) Zero forces
(c) Unbalanced forces
(d) None of them
(c) Unbalanced forces
Question 37.
To describe an object’s momentum, we must specify its …………….. and …………….. .
(a) Mass and displacement
(b) Mass and direction
(c) Mass and velocity
(d) Mass and acceleration
(c) mass and velocity
Question 38.
…………….. is the product of mass and velocity of an object.
(a) Speed
(b) Acceleration
(c) Momentum
(d) Force
(c) Momentum
Question 39.
The rate of change of momentum is proportional to the applied …………….. .
(a) Balanced force
(b) Unbalanced force
(c) Mass
(d) Velocity
(b) unbalanced force
Question 40.
S.I. unit of momentum is
(a) kg cm/s
(b) kg m/s
(c) gm/s
(d) m/s
(b) kg m/s
Question 41.
…………….. is always conserved in a collision.
(a) Force
(b) Power
(c) Speed
(d) Total momentum
(d) Total momentum
Question 42.
When a bullet is fired from the gun, the gun moves in backward direction. This motion is called as …………….. .
(a) Momentum
(b) Velocity
(c) Acceleration
(d) Recoil
(d) Recoil
Question 43.
In CGS system, the unit of force is …………….. .
(а) Newton
(b) Watt
(c) Horse power
(d) Dyne
(d) Dyne.
(B) 1. Find the odd man out:
Question 1.
Displacement, Force, Momentum, Mass
Question 2.
Speed, Power, Energy, Acceleration
Question 3.
Newton’s 1st law, Newton’s 2nd law, Newton’s 3rd law, Kepler’s laws of motion
Newton’s 3rd law
(B) 2. Find out the correlation
Question 1.
Speed zero: Body at rest :: Negative acceleration : Retardation
Question 2.
Displacement : Vector quantity :: Distance : Scalar quantity
Vector quantity
Question 3.
When body comes to rest at the end of the motion : Final velocity is zero :: When body is at rest at the starting of motion : Initial velocity is zero
Initial velocity is zero
Question 4.
Uniform circular motion: Displacement is zero :: Uniform velocity : Acceleration is zero
Displacement is zero
Question 5.
Inertia : Newton’s 1st law :: Rate of change of momentum : Newton’s 2nd law
Newton’s 2nd law
Question 6.
Balanced force : Body at rest :: Force equation : Mass x acceleration
Body at rest
(B) 3. Distinguish between:
Question 1.
Positive acceleration and Negative acceleration
Positive acceleration Negative acceleration
(i) When the velocity of a body increases, acceleration is said to be positive acceleration. (i) When the velocity of a body decreases, acceleration is said to be negative acceleration.
Question 2.
Scalar quantity and Vector quantity
Scalar quantity Vector quantity
(i) Scalar quantities are physical quantities having magnitude only. (i) Vector quantities are physical quantities having both magnitude and direction.
Question 3.
Balanced force and Unbalanced force
Balanced force Unbalanced force
(i) Balanced force keeps the body at rest. (i) Balanced force keeps the body at rest.
(B) 4. State whether the following statements are true or false:
Question 1.
The velocity of a body is given by the distance covered by it in unit time in a given direction.
Question 2.
Displacement is a scalar quantity.
Question 3.
Uniform acceleration means that the body is moving with a uniform velocity.
Question 4.
The direction of acceleration can be opposite to that of velocity.
Question 5.
Work is a vector quantity.
Question 6.
Displacement is always greater than distance.
Question 7.
The distance and displacement are equal only if, motion is along a straight path.
Question 8.
If an object experiences acceleration, a force is acting on it.
Question 9.
A train pulling out from a station is in uniform motion.
Question 10.
If a bus in motion is suddenly stopped, the passengers fall backwards.
Question 11.
If a single force is acting on an object, it will always accelerate.
Question 12.
In circular motion, direction of motion is tangential.
Question 13.
The inertia of a body is measured in terms of its mass.
(B) 5. Name the following:
Question 1.
The scientist who summarized motion in a set of equations of motion.
Isaac Newton
Question 2.
Motion of an object along a circular path with uniform speed.
Uniform circular motion
Question 3.
What is the backward motion of the gun called?
Question 4.
The motion in which the object covers equal distance in equal intervals of time.
Uniform motion
Question 5.
S. I. unit of acceleration.
Question 6.
CGS unit of momentum
g cm/s
(B) 6. Answer the following in one sentence:
Question 1.
When is acceleration said to be positive?
When the velocity of a body increases, acceleration is said to be positive acceleration.
Question 2.
What is negative acceleration?
When the velocity of a body decreases, acceleration is said to be negative acceleration.
Question 3.
What is the direction of velocity of an object performing uniform circular motion?
The direction of velocity is along the tangential direction to its position.
Question 4.
Give the mathematical expression used to determine velocity of an object moving with uniform circular motion.
is the expression used to determine velocity of a body moving with uniform circular motion.
Question 5.
What kind of force keeps the body at rest?
Balanced force keeps the body at rest.
Question 6.
Which law of motion gives the measure of force?
Newton’s second law of motion gives the measure of force.
Question 7.
What are vectors and scalars?
Scalars are physical quantities having magnitude only whereas, vectors are physical quantities having both magnitude and direction.
Question 8.
Which of the quantities distance, speed, velocity, time and displacement are scalars and which are vectors?
Distance, speed and time are scalars displacement and velocity are vectors.
Give formula:
Question 1.
Force =
Mass x Acceleration = ma
Question 2.
Final velocity (v) =
Initial Velocity + (Acceleration x Time) = u + at
Question 3.
Displacement (s) =
Question 4.
Final velocity^2 (v^2) =
Initial Velocity^2 + 2 x Acceleration x Displacement = u^2 + 2as
Question 5.
velocity of an object moving with uniform circular motion =
Give scientific reasons:
Question 5.
Motion is relative.
• The motion of an object depends on the observer, hence a body may appear moving for one person and at the same time at rest for another one.
• Hence, motion is relative.
Question 6.
Newton’s first law of motion is called as law of inertia.
Heavier objects offer more inertia.
• Inertia is related to the mass of the object.
• As mass is the quantity of matter in a body, we need to exert more force to push a heavier body.
• Hence heavier objects offer more inertia.
• As the same property is described by Newton’s first law of motion, it is called as Law of Inertia.
Question 7.
The launching of a rocket is based on Newton’s third law of motion.
• Newton’s third law of motion states that ‘Every action force has an equal and opposite reaction force which acts simultaneously.’
• When the fuel in a rocket is ignited, it bums as a result of chemical reaction.
• The exhaust gases escape with a great force in the backward direction.
• It exerts an equal and opposite reaction force on the rocket, due to which the rocket moves in the forward direction.
• Thus, the principle of launching of rocket is based on Newton’s third law of motion.
Solve the following numerical:
Question 1.
An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity?
Total distance travelled = 400 m
Total displacement = 0, as he returns to his original position.
Total time = 25 seconds.
To find:
Average speed = ?
Average velocity = ?
The athlete runs at an average speed of 16 m/s and velocity 0 m/s.
Question 2.
An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s^2 before taking off. How much distance would it have covered on the runway?
The distance covered on the runway is 1440 m.
Question 3.
A kangaroo can jump 2.5 m vertically. What must be the initial velocity of the kangaroo?
a = 9.8 m/s^2, s = 2.5 m, v = 0,
To find:
u = ?
v^2 = u^2 + 2as
v^2 = u^2 + 2as
0^2 = u^2 + 2 x (-9.8) (2.5) : (Negative sign is used as the acceleration is in the direction opposite to that of velocity.)
0= u^2 – 49 u^2
= 49 u
= 7 m/s
The initial velocity of the kangaroo must be 7 m/s.
Question 4.
A motorboat starts from rest and moves with uniform acceleration, if it attains the velocity of 15 m/s in 5s, calculate the acceleration and the distance travelled in that time.
Initial velocity, u = 0
Final velocity, v = 15 m/s time, t = 5 s.
To find:
Acceleration (a) = ?
Distance (s) = ?
From the first equation of motion
From the second equation of motion, the distance covered will be
The acceleration is 3 m/s^2 and distance travelled is 37.5 m.
Question 5.
The mass of a cannon is 500 kg and it recoils with a speed of 0.25 m/s. What is the momentum of the cannon?
mass of the cannon = 50 kg
recoil speed = 0.25 m/s
To find:
Momentum = ?
Momentum = m x v
Momentum = m x v
= 500 x 0.25
= 125 kg m/s
The momentum of cannon is 125 kg m/s
Answer the following in short:
Question 1.
Explain the three different ways to change the velocity.
As velocity is related to speed and direction, it can be changed by :
• Changing the speed while keeping the direction constant.
• Changing direction while keeping speed constant.
• Changing both speed as well as direction of motion.
Question 2.
Explain what is positive, negative and zero acceleration.
• Positive Acceleration: When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity.
• Negative Acceleration: When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the
direction of velocity.
• Zero Acceleration: If the velocity of the object does not change with time, it has zero acceleration.
Question 3.
What inference do we draw from the velocity-time graph for a uniformly accelerated motion?
• From velocity-time graph we can infer whether velocity changes by equal amounts in equal intervals of time or not.
• Thus, for all uniformly accelerated motion, the velocity – time graph is a straight line and slope of the line gives the acceleration.
• For non-uniformly accelerated motion, velocity-time graph can have any shape according to variation in velocity with respect to time.
Question 4.
State the three equations of motion and give the relationship explained by them.
• v = u + at: This is the relation between velocity and time.
• s = ut + 1/2 at^2 : This is the relation between displacement and time
• v^2 = u^2 + 2as : This is the relation between displacement and velocity.
Question 5.
What are the implications of Newton’s Third Law of motion?
• Action and reaction are terms that express force.
• These forces act in pairs. One force cannot exist by itself.
• Action and reaction forces act simultaneously.
• Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other’s effect.
Question 6.
Explain recoil and recoil velocity. Derive its expression.
• Let us consider the example of a bullet fired from a gun. When a bullet of mass nij is fired from a gun of mass m[2], its velocity becomes v[2], and its momentum becomes m^. Before firing the
bullet, both the gun and the bullet are at rest. Hence, total initial momentum is zero.
• According to Newton’s third law of motion, the total final momentum also has to be zero. Thus, the forward-moving bullet causes the gun to move backward after firing.
• This backward motion of the gun is called its recoil. The velocity of recoil, v[2] is such that,
Complete the flow chart:
(1) Types of force and their effects
(2) Newton’s laws
Distinguish between:
Question 1.
Speed and velocity.
Speed Velocity
(i) Speed is the distance covered by a body in unit time. (i) The displacement that occurs in unit time is called velocity.
(ii) It is a scalar quantity. (ii) It is a vector quantity.
(iii) Speed is equal to or greater than velocity. (iii) Velocity is equal to or less than speed.
(iv) Speed = distance/time (iv) Velocity = displacement/time
(v) It is always positive or zero but never negative. (v) It may be positive, Zero or negative.
Question 2.
Balanced force and unbalanced force.
Balanced force Unbalanced force
(i) Two equal forces applied on a body in the opposite direction. (i) Two unequal forces applied on a body.
(ii) This force does not change the state of rest or the state of uniform motion (ii) This force can change the state of rest or the state of uniform motion of a body in a straight line.
Give examples:
Question 1.
Scalar quantities
Time, Volume, Speed, Mass, Temperature, Distance, Entropy, Energy, and Work
Question 2.
Vector quantities
Acceleration, Velocity, Momentum, Force, and Weight
Answer the following questions:
Observe the figure and answer the questions:
(a) Measure the distance between points A and B in different ways as shown in figure (I).
Distances measured may be of different lengths depending on the path taken.
(b) Now measure the distance along the dotted line. Which distance is correct according to you and why?
Dotted line shows the shortest way of reaching from A to B.
(c) Observe the following figures. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path?
If we increase the number of sides of the polygon and make it infinite, then we will have to change the direction an infinite number of times. The shape of the path thus obtained will be a circle.
Observe the figure and answer the questions
Question 1.
What will be the effect on the velocity of the motorcycle if its speed increases or decreases, but its direction remains unchanged?
If the speed is increased the velocity of the motorcycle will increase and if the speed is decreased the velocity of the motorcycle will decrease
Question 2.
In case of a turning on the road, will the velocity and speed be same?
As speed is scalar quantity while velocity is vector quantity so by turning velocity will change while speed remains same
Question 3.
If, on a turning, we change the direction as well as the speed of the motorcycle, what will be the effect on its velocity?
Its velocity will change because velocity depends on speed as well as direction and here both speed and direction are changed.
Question 1.
A person travels a distance of 72 km in 4 hours. Calculate average speed in m/s.
Given :
Total distance (d)
= 72 km
= 72×1000
= 72000 m
Total time taken (t)
= 4 hours
= 4 x 3600 (v lhr = 3600 sec)
= 14400 s
To find:
Average speed = ?
The person travels with average speed of 5 m/s
Question 2.
balls have masses of 50 gm and 100 gm and they are moving along the same line in the same direction with velocities of 3 m/s and 1.5 m/s respectively. They collide with each other and after the
collision, the first ball moves with a velocity of 2.5 m/s. Calculate the velocity of the other ball after collision.
Final velocity of second ball after collision is 1.75 m/s.
Write laws and explain write implications:
Question 1.
Newton’s third law of motion
‘Every action force has an equal and opposite reaction force which acts simultaneously’.
• Action and reaction are terms that express force.
• These forces act in pairs. One force cannot exist by itself.
• Action and reaction forces act simultaneously.
• Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other’s effect.
Question 2.
Explain Newton’s second law of motion and derive the formula.
Newton’s second law explains about change in momentum. It states that The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the
Suppose an object of mass ‘m’ has an initial velocity ‘u. When a force ‘F’ is applied in direction of its velocity for time ‘t’, its velocity becomes ‘y’. Then, the total initial momentum of the body
= ‘mu’. Its final momentum after time t = ‘mv’.
So, the rate of change of momentum
Hence by Newtons second la of motion, 4he rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F ∝ kma (k = Constant of proportionaLity and value is 1).
∴ F = ma
Question 3.
State the law of conservation of momentum and derive the formula.
Let mass of object A and B be m[1] and m[1] respectively
Let their initial velocity be u[1] and u[2] Let their final velocity be v[1] and v[2]
We know,
P = mv
Let their initial momentum be m[1]u[1] and m[2] u[2]
Let their final momentum be m[1]v[1] and m[2]v[2
Total initial momentum = (m[1]u[1] + m[2]u[2])
Total final momentum = (m[1]v[1] + m[2]v[2])
If F[2] is the force that acts on object B,
i.e. The magnitude of total of total final momentum = the magnitude of total initial momentum
Complete the paragraph:
Question 1.
Moving Objects
‘Distance’ is the length of the actual path travelled by an object in motion while going from one point to another, whereas displacement is the minimum distance between the starting and finishing
points. The displacement that occurs in unit time is called velocity. The units of speed and velocity are the same. In the SI system, the unit is m/s while in the CGS system; it is cm/s. Speed is
related to distance while velocity is related to the displacement. If the motion is along a straight line, the values of speed and velocity are the same, otherwise they can be different. The first
scientist to measure speed as the distance /time was Galileo. The speed of sound in dry air is 343.2 m/s while the speed of light is about 3 x 108 m/s. The speed of revolution of the earth around the
sun is about 29770 m/s.
Question 2.
Types of motion
If an object covers unequal distances in equal time intervals, it is said to be moving with non-uniform speed. For example the motion of a vehicle being driven through heavy traffic. If an object
covers equal distances in equal time intervals, it is said to be moving with uniform speed. The rate of change of velocity is called acceleration. If the velocity changes by equal amounts in equal
time intervals, the object is said to be in uniform acceleration. If the velocity changes by unequal amounts in equal time intervals, the object are said to be non-uniform acceleration. The speed of
the tip is constant, but the direction of its displacement and therefore, its velocity is constantly changing. As the tip is moving along a circular path, its motion is called uniform circular
Question 3.
Newton’s laws and conservation of momentum
An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it. The rate of change of momentum is proportional to the
applied force and the change of momentum occurs in the direction of the force. Momentum has magnitude as well as direction. Its direction is the same as that of velocity. In SI system, the unit of
momentum is kg m/s, while in CGS system, it is g cm/s. If an unbalanced force applied on an object causes a change in the velocity of the object, then it also causes a change in its momentum.
The force necessary to cause a change in the momentum of an object depends upon the rate of change of momentum. Every action force has an equal and opposite reaction force which acts simultaneously.
As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the
gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket.
Answer the following in detail:
Question 1.
What is speed? State its units and types. Explain instantaneous speed and average speed.
The speed of a body is the distance travelled in unit time. The units of speed in CGS system is cm/s and in SI system is m/s.
There are two types of speed :
□ Uniform speed : When a body covers equal distance in equal intervals of time throughout its motion, it is said to have uniform speed.
□ Non-uniform or variable speed: When a body covers unequal distance in equal intervals of time it is said to have non-uniform speed.
The speed of the body at any instant is called instantaneous speed. Average speed is the ratio of total distance covered to total time taken.
Question 2.
What is velocity? State its units and types.
The velocity of a body is the distance travelled by a body in a particular direction in unit time. Thus, rate of change of displacement is called velocity.
v = s/t
where: s = displacement; t = time taken; v = velocity
(MKS unit: m/s CGS unit: cm/s)
There are two types of velocities :
□ Uniform velocity: If there is equal displacement taking place in equal intervals of time, it is uniform velocity.
□ Non-uniform velocity or variable velocity: If there is unequal displacement in equal intervals of time, it is non-uniform velocity.
Question 3.
What is acceleration? State its units and types.
(i) Acceleration is a rate of change in velocity. It is a vector quantity, \(a=\frac{v-u}{t}\)
where : v = final velocity; u = initial velocity;
a = acceleration
Units of acceleration in SI system is m/s2 and CGS system is cm/s2.
(ii) Types of acceleration: .
(a) Uniform acceleration : If the change in velocity is equal in equal intervals of time, the acceleration is uniform acceleration.
(b) Non-uniform acceleration : If the change in velocity is unequal in equal intervals of time, the acceleration is a non-uniform acceleration.
(iii) Kinds of acceleration:
Positive acceleration : When the velocity of an object goes on increasing, it is said to have Positive acceleration.
Negative acceleration : When the velocity of an object goes on decreasing, it is said to have negative acceleration or retardation or deceleration.
Zero acceleration : If the velocity of the object does not change with time, it has zero acceleration.
Question 4.
Explain Newton’s second law of motion and derive the formula.
Newton’s second law explains about change in momentum. It states that ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the
Suppose an object of mass’m’ has an initial velocity ‘u’. When a force ‘F’ is applied in the direction of its velocity for time’t’, its velocity becomes ‘v’. Then, the total initial momentum of the
body = ‘mu’. Its final momentum after time t = ‘mv’.
So, the rate of change of momentum
Hence by Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F = kma (k = Constant of proportionality and value is 1).
∴ F = ma
Question 5.
State the law of conservation of momentum and derive the formula.
(i) Let mass of objects A and B be and m[2] respectively
Let their initial velocity be m[1] and u[2]
Let their final velocity be V[1] and v[2]
(ii) We know,
P = mv
Let their initial momentum be m[1]u[1] and m[2]u[2]
Let their final momentum be m[1]v[1] and m[2]v[2]
(iii) Total initial momentum = (m[1]u[1] + m[2]u[2])
Total final momentum = (m[1]v[1] + m[2]v[2])
(iv) If F[2] is the force that acts on object B,
i.e. The magnitude of total final momentum = the magnitude of total initial momentum.
Question 6.
Obtain the equations of motion by graphical method:
(a) Equation for velocity-time relation.
Velocity-time graph: shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after
time t it reaches the point B on the graph.
The initial velocity of the object = u = OD
(b) Equation for displacement-time relation.
Suppose that an object is in uniform acceleration ‘a’ and it has covered the distance ‘s’ within time’t’. From the graph the distance covered by the object during time’t’ is given by the area of
quadrangle DOEB.
(c) Equation for displacement-velocity relation.
We can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.
This is Newton’s third equation of motion.
Read the paragraph and answer the questions:
1. Paragraph
(a) When a bat strikes a ball, the ball exerts an equal and opposite force on the bat. The force : acting on the ball projects it with high velocity, Due to the large mass of bat compared to ball,
reaction force on the bat slows down the bat’s motion.
(b) When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward
direction. This is the reaction force. Due to the large mass of the gun, it j moves only a little distance backward. This backward movement of the gun is called the recoil of the gun.
(c) In a rocket, burning fuel creates a push on the front of the rocket pushing it forward. This creates an equal and opposite push on the exhaust gas backwards.
(i) Which of Newton’s law examples are given here?
Newton’s 3rd law is stated by the above example
(ii) When a rifle is fired it is pushed back this movement is called what?
Movement of rifle getting pushed back after firing is called recoil.
(iii) What does the ball acquire after it gets velocity?
The ball acquires momentum after it gets velocity.
(iv) State newton’s 1st law of motion
An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.
(v) Which force is required to produce motion in an object?
Unbalanced force is required to produce motion in an object.
2. Paragraph 2
constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force. The second law explains how the velocity
of an object changes when it is subjected to an external force. The law defines a force to be equal to changes in momentum (mass times velocity) per change in time. Newton also developed the calculus
of mathematics, and the “changes” “expressed in j the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations
experienced by an object subjected to an external force.) For an object with a constant mass the second law states that the force F is the product of an objects mass and its acceleration a:
F = m * a
For an external applied force the change in velocity depends on the mass of the object.
A force will cause in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.
The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force
on object: A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine
(i) A chalk kept on the table remains in the position of rest until moved by the teacher. Which law of motion is followed in this situation?
Newton’s 1st law is followed in this situation e
(ii) What will happen to momentum if the mass and acceleration both are doubled?
If the mass and acceleration both are doubled then the momentum will be increased four times
(iii) What will happen to momentum if the mass and acceleration both are halved?
If the mass and acceleration both are doubled then the momentum will be decreased one four times
(iv) A moving stone filled truck collides with a moving car coming from opposite direction. Why is it observed that only the car is pushed backward?
A stone filled truck as more mass than a car hence it has more momentum thus it is observed that only the car is pushed backward
(v) What will happen to the force if the jet engines do not produce enough thrust to push the aeroplane in mid-air?
If the jet engines do not produce enough thrust to push the aeroplane in mid-air the force of aeroplane and resistance of air will become balanced thus putting the aeroplane at rest which will result
in a crash. | {"url":"https://maharashtraboardsolutions.in/maharashtra-board-class-9-science-solutions-chapter-1/","timestamp":"2024-11-02T23:59:42Z","content_type":"text/html","content_length":"130480","record_id":"<urn:uuid:8075296d-c91b-4875-818d-0115a6f0c16a>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00507.warc.gz"} |
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GitHub kuriousai ScratchDL Deep Learning framework from. Building with Python from First Principles, Deep Learning from Scratch, Seth Weidman, O'reilly media. Read "Deep Learning from Scratch
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even many software engineers. Deep Learning from Scratch: Building with Python from First Principles Seth Weidman With the resurgence of neural networks in the 2010s, deep learning has become
essential for machine learning practitioners and even many software engineers. It was mostly for me to keep the code I was writing for the book organized, but my hope is readers can clone this repo
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The Order Of Operations Three Steps A Math Worksheet From The | Order of Operation Worksheets
The Order Of Operations Three Steps A Math Worksheet From The
The Order Of Operations Three Steps A Math Worksheet From The
The Order Of Operations Three Steps A Math Worksheet From The – You may have heard of an Order Of Operations Worksheet, but what specifically is it? In addition, worksheets are an excellent means for
students to practice new skills and testimonial old ones.
What is the Order Of Operations Worksheet?
An order of operations worksheet is a sort of math worksheet that requires students to carry out math operations. These worksheets are split right into 3 main sections: subtraction, multiplication,
and also addition. They additionally consist of the examination of exponents and also parentheses. Pupils who are still finding out how to do these tasks will certainly discover this kind of
worksheet valuable.
The major purpose of an order of operations worksheet is to aid pupils learn the appropriate means to solve mathematics formulas. If a trainee does not yet comprehend the idea of order of operations,
they can review it by referring to an explanation web page. On top of that, an order of operations worksheet can be divided into several categories, based on its trouble.
An additional vital purpose of an order of operations worksheet is to show trainees just how to carry out PEMDAS operations. These worksheets start with simple troubles related to the standard rules
and also build up to more intricate problems entailing every one of the regulations. These worksheets are a great way to present young learners to the enjoyment of fixing algebraic formulas.
Why is Order of Operations Important?
One of the most important points you can discover in mathematics is the order of operations. The order of operations makes sure that the math problems you address are constant.
An order of operations worksheet is a great method to educate trainees the appropriate method to fix mathematics formulas. Before students begin utilizing this worksheet, they may need to review
concepts related to the order of operations.
An order of operations worksheet can aid pupils create their abilities in addition and also subtraction. Educators can utilize Prodigy as a simple means to set apart technique as well as supply
appealing material. Natural born player’s worksheets are an excellent way to help trainees learn more about the order of operations. Educators can start with the standard principles of division,
addition, as well as multiplication to help trainees construct their understanding of parentheses.
Worksheets On Order Of Operations
7Th Grade Math Pemdas Worksheets Rule Order Of Operations Tiktokcook
Use These Free Algebra Worksheets To Practice Your Order Of Operations
Worksheets On Order Of Operations
Worksheets On Order Of Operations provide a terrific resource for young learners. These worksheets can be conveniently tailored for specific needs. They can be discovered in three degrees of problem.
The initial degree is straightforward, needing pupils to practice using the DMAS technique on expressions containing four or even more integers or 3 operators. The second degree requires students to
utilize the PEMDAS method to streamline expressions making use of external and also inner parentheses, brackets, and also curly braces.
The Worksheets On Order Of Operations can be downloaded completely free and also can be printed out. They can then be examined utilizing addition, subtraction, multiplication, and division. Students
can additionally use these worksheets to review order of operations and using backers.
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Concrete maturity calculation - How to estimate the strength - civilMedium
Ever thought about estimating concrete strength by a calculation? There are several ways to do strength investigation in a non-destructive way. In fact, the concrete maturity method is one of the
best methods to estimate real strength.
The best gain from the maturity method is the time. You can cut out the waiting time to check the 28 days strength of laboratory specimens if you estimate the real-time strength using the maturity
method at the site. The concept was established in 1950 and got accepted in ASTM and various standards during the time.
The concrete maturity method is simply defined as the relationship of strength to the temperature & age of concrete. Some of the accepted standards of this method are ASTM C1074, ASTM C918, ACI 228 –
1R, AASHTO T325.
Different approaches of concrete maturity method
There are 3 common ways to estimate the maturity of the concrete.
Temperature – Time factor method (TTF):
This is also known as the Nurse – Saul maturity function. The first method developed using the concept and still using in the ASTM C1074 due to its simple calculation.
TTF concrete maturity method assumes a linear relationship between maturity and the strength of concrete. You can either select the datum temperature or calculate for accurate results. Typical datum
temperature is -10C because that considered as the freezing point of concrete where no longer water will involve in cement hydration. However, from 0°C to -10°C segment, the hydration is slower. It
is conservative to omit that part and consider the datum temperature at 0°C.
The following equation calculates the maturity in the time factor method.
The maturity can graphically present as the area of temperature vs. time graph
Concrete Maturity – Area below the temperature vs time curve
So it’s clear that the high curing temperature results in high maturity as well as strength gain in the concrete.
Use the maturity to calculate the strength of the concrete using the following equation
These A, B values are constant for the mix properties. Materials and their proportions will define these constants. Then maintaining the exact mix proportions at the site throughout the time is
essential. Use 15 specimens for compressive strength test and 2 samples for temperature measurements and calculate A & B. It’s always good to calculate the A, B values than using values corresponding
to target 28-day strength.
You can use Plowman’s coefficients for A & B values are in the table below for a rough estimation of strengths.
Equivalent Age method:
An exponential relationship between temperature and maturity is assuming in this method. It estimates the age of curing at a specific temperature, which gives the same maturity at different
temperatures. This allows us to convert the actual age and temperature into an equivalent curing age and predict the strength.
Weighted maturity method:
Relationship similar to Nurse – Saul method is using in this method too. Summation of hardening time and few cement factors required to calculate the maturity. However, the weighted maturity method
is not accepted in American standards.
The most crucial part of the maturity method is calibration. According to the concrete maturity concept, identical mixes having similar maturity should give you equal strengths. This allows you to do
the calibration in a laboratory.
The recommendation in the standard to use 17 number of specimens for the calibration. You can use 5 sets of ( 5 x 3 = 15) samples for the strength testing while using the remaining 2 for measuring
temperatures. Testing specimens in 1, 3, 7, 14, and 28 days is the typical way of distributing. However, some suggest 1, 2, 3, 7, and 14 days’ time because the maturity method is mostly used in early
strength evaluations.
Soon after the calibration, calculate A & B values and use any time where the same mix is using. Simple spreadsheet calculation will quickly solve the most relevant A, B coefficients.
Typical applications of the concrete maturity method
The concrete maturity method is commonly using in building constructions. It’s essential to estimate the supports and form removal time in highrise buildings. Maturity of slabs, beams, columns, and
footings are separately calculated to assess the strengths and curing time.
Pre-cast constructions are another situation where concrete maturity is using. The earliest possible from removal can save time and labor. The same method can use to estimate the curing,
transportation time from the pre-cast yard to the site.
Post-tensioning constructions widely use concrete maturity method in estimating tensioning time. It is mandatory to reach the required strength before post-tensioning.
The maturity technique is using in estimating thermal stresses and evaluating the capacity in mass concretes. Concrete dams and foundations typically have large concrete pours that need to
investigate the possibility of thermal cracks.
Pavements and bridges commonly use the maturity method to calculate traffic pass on time. It’s essential to reach strength before allowing the traffic load.
Shotcrete is a widely using ground stabilization method in the industry. It’s useful to estimate the strength of a newly placed shotcrete before you return to the site. Maturity is one of the ways
available to assess the early strength in shotcrete.
Advantages and limitations
The number one advantage of the concrete maturity method is the cost advantage. Rather than waiting for the laboratory results, this method allows us to estimate the strength quickly at any time.
Removing the forms and moving into the next layer of construction can speed up according to the results. These will cost 1000-2000 USD for the calibration, and 1-2 USD/m^2 and the saving will be
several 10 to 15 thousand dollars by cutting of delays and curing times.
Another advantage is that this method allows us to estimate the strength at any desired place. It’s just a matter of placing some thermocouples in the concrete, and you are capable of estimating
strength on the point.
Other than above, less laboratory involvement and quick estimation of strength are significant advantages in the concrete maturity method. With the record of temperature history, just a simple
calculation will estimate the strength.
Even with many plus points, the maturity method has its limitations. Understanding them before using the technique is a must.
• The method is a strength approximation technique, not the exact strength.
• Mostly accurate in early strength calculations
• Strength estimation is not as straightforward as a cube or cylindrical specimen testing. Need to arrange the thermocouples and collect readings throughout the time
• Your datum temperature can affect the maturity
• Any change in the mix can make the calibration invalid. So keeping the exact mix proportion is essential throughout the time. | {"url":"https://civilmedium.com/concrete-maturity-method-in-estimating-strengths/","timestamp":"2024-11-02T18:20:58Z","content_type":"text/html","content_length":"125068","record_id":"<urn:uuid:cc2b9de1-2c44-4a9a-8689-f83c204dd740>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00685.warc.gz"} |
How to use the MultiOutputClassifier in scikit-learn.
Using the MultiOutputClassifier
If you're interested in using the same dataset X to predict two labels, y1 and y2, then you may take a shortcut. Instead of making two pipelines, you can also use a single pipeline that contains a
copy of a model. One for each model. Scikit-Learn allows you to do this by using a MultiOutputClassifier.
Let's demonstrate this method by using the titanic dataset.
import numpy as np
import pandas as pd
from sklearn.neighbors import KNeighborsClassifier
from sklearn.linear_model import LogisticRegression
from sklearn.multioutput import MultiOutputClassifier, MultiOutputRegressor
df = pd.read_csv("https://calmcode.io/datasets/titanic.csv")
The idea is to predict the survived column and the pclass column.
labels = df[['survived', 'pclass']].values
X = df.assign(sex=lambda d: d['sex'] == 'male')[['sex', 'age', 'fare']]
We can pass the X and labels arrays into the .fit() method of our MultiOutputClassifier.
clf = MultiOutputClassifier(LogisticRegression()).fit(X, labels)
This will train a LogisticRegression for each model. Note that you're also free to train any other scikit-learn compatible classifier here. Here's another example with the KNeighborsClassifier.
clf = MultiOutputClassifier(KNeighborsClassifier()).fit(X, labels)
You can also explore the estimator probabilities. Note that you'll get two arrays as output here. One for each label.
If you'd like you can also inspect both trained models individually. They are stored in the .estimators_ property of the model.
If you're looking for a quick way to predict multiple labels from a single dataset then this meta-trick will work for you. It mainly works when you want sensible predictions but there's no need for
state of the art. If you're interested in having the most optimal pipeline out there then you may want to make two seperate pipelines instead.
Note that there's also a regressor variant of this trick if that's of interest.
If you're looking for an interesting use-case for this technique: consider confidence intervals! When your dataset has an upper/lower bound then this is something that you can predict as well. This
technique is explored in more detail in this blogpost. | {"url":"https://calmcode.io/course/scikit-meta/multi-output","timestamp":"2024-11-06T18:55:24Z","content_type":"text/html","content_length":"31454","record_id":"<urn:uuid:d74d3858-ce90-4399-bc55-2e465e4edd2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00360.warc.gz"} |
Number Comparison Made Fun: 4 Engaging Ways to Challenge Your StudentsNumber Comparison Made Fun: 4 Engaging Ways to Challenge Your Students - Teaching Perks
Number comparison is an essential math skill for 1st graders. It is important to make the learning experience challenging and engaging to keep your high level students motivated and interested.
In this blog post, we will explore four creative ways to challenge your students that have mastered the basics of greater than less than and help them extend their number comparison skills. From
number comparison riddles to number comparison with addition, these activities will encourage your students to think critically and logically. Let’s dive into these exciting activities and start
having some fun with numbers!
Number Comparison Riddles
Read 2 riddles to the students and have them figure out what the numbers are. Once they have the two mystery numbers they can write the number comparison.
Teacher: I am going to give you two riddles. You must solve both riddles and tell me which number is greater than and which is less than.
I am the sum of how many fingers two students have plus how many ears two students have.
20 + 4 = 24
I am the difference of how many students are in your class minus how many people are in your math group.
21 – 6 = 15
Students would say 24 is the greater number because 24 > 15.
Another extension would be to have students write the numbers in multiple forms.
For example, twenty-four > fifteen or 20 + 4 > 10 + 5
Number Comparison Mystery
Tell students you are thinking of a mystery number. Give them greater than/less than clues to figure out the number you are thinking of.
Example: 87
Teacher: The number I am thinking of is greater than 78 but less than 92.
Student: 81
Teacher: The number I am thinking of is more than 81 but less than 92.
Student: 89
Teacher: The number I am thinking of is fewer than 89 but greater than 86
Continue until students guess the number. You can also let students think of a mystery number and let the teacher or a partner guess their number as they give greater then/less than clues. You could
also use numbers with a hundreds place.
Number Comparison Order
Give students 5 numbers and have them put them in order from greatest to least or least to greatest. Start with 5 two digit numbers and then move on to three-digit numbers. Have students use symbols
when ordering the numbers. You can also have them explain orally or in writing why they chose to order the numbers the way they did.
Example: 43 < 49 < 58 < 61< 78
The student might say “I know that 4 tens is less than 5, 6, and 7 tens so I knew 43 and 49 would go first. I had to look at the ones place and knew that 3 was less than 4 so that told me 43 should
be first with 49 second. Then I put the tens places from the other three numbers in order and knew it should be 58, 61 and then 78.
Number Comparison With Addition
Give students two addition problems. They will use symbols to compare the problems. You can ask them to explain orally or in writing how they got their answer.
Example: 23 + 17 _ 32 + 12
Answer: 23 + 17 < 32 + 12 because 40 < 44
Since these are bigger numbers to add you may want to show students before they begin how they can use what they have learned about with place value and base ten to solve. They could draw quick
pictures and make groups of ten to find the sum. You could also do easier problems such as 7 + 9 ___ 8 + 10.
Number Comparison Conclusion
In conclusion, number comparison is an important math skill for first-grade students, and challenging them to extend their understanding beyond the basics is crucial to keeping them motivated and
engaged. The four creative activities outlined in this blog post provide exciting ways to help students think critically and logically about number comparison. By making math fun and interactive,
students are more likely to develop a love for learning and build a solid foundation for their future mathematical endeavors.
The ideas included in this post and many more just like these can be found in this place value lesson plan unit. Complete with 15 Daily lesson plans, differentiated two-sided work mats,
differentiated independent practice pages, games, higher order thinking questions to ask during lessons, challenge extensions and so much more, these lessons will have you all set so that you don’t
have to plan a thing!
Leave a Reply Cancel reply | {"url":"https://teachingperks.com/number-comparison-4-ways-to-challenge/","timestamp":"2024-11-06T05:18:36Z","content_type":"text/html","content_length":"86968","record_id":"<urn:uuid:f422f2d5-ac78-4932-b805-23bc2896aa42>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00826.warc.gz"} |
RE: st: how to deal with censoring at zero (a lot of zeroes) for a labor
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
RE: st: how to deal with censoring at zero (a lot of zeroes) for a laboratory re
From "Daniel Waxman" <[email protected]>
To <[email protected]>
Subject RE: st: how to deal with censoring at zero (a lot of zeroes) for a laboratory re
Date Mon, 6 Jun 2005 21:26:30 -0400
Point well-taken about asking for partially-informed opinions, thanks.
The cube root works, although the fit (i.e. predicted vs. observed outcomes)
is not quite as good as with the log-transformed variable. Conceptually,
I'm not sure that I understand the problem in throwing away the
zeroes--essentially I would be saying that there is substantially more
uncertainty at the zero point than at any measurable one, and therefore I
want my model to predict outcome relative to the lowest measurable
concentration rather than relative to the zero point. It does affect the
numerical results of goodness-of-fit tests--is that the conceptual problem?
But making a somewhat arbitrary choice for the zero point also has problems.
Would it be cheating to just replace the zero with the best-fit as
determined by solving the logit equation backwards? Probably...
[please feel free to censor debate if this has gone back & forth too many
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Nick Cox
Sent: Monday, June 06, 2005 8:45 AM
To: [email protected]
Subject: RE: st: how to deal with censoring at zero (a lot of zeroes) for a
laboratory re
I am always queasy when expected to approve,
or invited to disapprove, a proposed analysis.
How can anyone give a really worthwhile opinion
of what is sensible for someone's project on
this amount of information?
Nevertheless the notion of throwing away
half the data on this basis is rather alarming.
I have found cube roots often useful for non-negative
variables. This is partly empirical, partly that
zero goes to zero, but there is also an arm-waving basis
that cube roots work well for gamma distributions (cf. the
Wilson-Hilferty transformation). More generally,
powers falling towards zero in effect have the logarithm
as their limit.
[email protected]
Daniel Waxman
> Maarten, Kevin,
> Thank you very much for your replies. So for now I am just
> going give up
> trying to make distributional assumptions and to drop the half of the
> observations which are zero or non-detectable prior to log
> transforming the
> predictor and to creating the logistic model. In fact,
> whether I do this or
> change the zero to half of the lowest detectable value (i.e.
> .005) doesn't
> have much of an effect on the logistic odds ratio.
> If anybody has any objections to this (or sees how a
> statistical reviewer
> for a medical journal might have objections), please let me know.
> Daniel
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of maartenbuis
> Sent: Sunday, June 05, 2005 7:38 PM
> To: [email protected]
> Subject: Re: st: how to deal with censoring at zero (a lot of
> zeroes) for a
> laboratory re
> I am tired: The cirtical assumption behind Multiple Imputation is that
> the probability of missingness does not depend on the value of the
> missing variable itself (Missing At Random, or MAR). This is obviously
> not the case with censoring. My objection against (conditional) mean
> imputation, and my remark about selecting on the independent variables
> still hold. So, given that you have a large number of observations, I
> would just ignore the zero observations.
> Maarten
> --- In [email protected], "maartenbuis"
> <maartenbuis@y...> wrote:
> > Hi Daniel,
> >
> > It looks to me like you could use -tobit- for log(tropin) and just a
> > constant. The predicted values should give you the
> extrapolations you
> > want. (This will be the same value for all missing observations: the
> > mean of the log-normal distribution conditional on being
> less than the
> > censoring value)
> >
> > However, These are actually missing values, and apperently
> you want to
> > create imputations for them. If you just use the values you obtained
> > from -predict- you will be assuming that you are as sure about these
> > values as you are about the values you actually observed,
> and thus get
> > standard errors that are too small. If you really want to
> impute, than
> > you could have a look at -mice- (findit mice). Alternatively, you
> > could use the results from -tobit- to generate multiple imputations.
> > Mail me if you want to do that, and I can write, tonight or
> tomorrow,
> > an example for the infamous auto dataset. However, censoring on the
> > independent variable is generally much less a problem than censoring
> > on the dependent variable, so ignoring (throwing away) the censored
> > observation, should not lead to very different estimates.
> >
> > HTH,
> > Maarten
> >
> > --- "Daniel Waxman" <dan@a...> wrote:
> > > I am modeling a laboratory test (Troponin I) as an independent
> > > (continuous) predictor of in-hospital mortality in a sample of
> > > 10,000 subjects. <snip> The problem is the zero values,
> what they
> > > represent, and what to do with them. The distribution
> of results
> > > ranges from the minimal detectable level of .01 mcg/L to
> 94 mcg/L,
> > > with results markedly skewed to the left (nearly half the results
> > > are zero; 90% are < .20. results are given in increments
> > > of .01). Of course, zero is a censored value which represents a
> > > distribution of results between zero and somewhere below .01.
> > <snip.
> > > I found a method attributed to A.C. Cohen of doing
> essentially this
> > > which uses a lookup table to calculate the mean and standard
> > > deviation of an assumed log-normal distribution based upon the
> > > non-censored data and the proportion of data points that are
> > > censored, but there must be a better way to do this in Stata.
> > >
> > > Any thoughts on (1) whether it is reasonable to assume the
> > > log-normal distribution (I've played with qlognorm and
> plognorm, but
> > > it's hard to know what is good enough), and if so (2) how
> to do it?
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Multiplication Table
Multiplication Table of 12 Charts: Today internet and technology is so much advanced that it has become very easy for the students to use it for their study purpose, now there are so many courses
available that if a student is having difficulty in understanding a particular subject than he or she can easily refer to the internet and clear their doubt.
Multiplication tables is one of the main topics of a subject math’s and it is also the topic where most of the students get stuck and to help the students we are bringing one of the tables that is
the multiplication table of 12 in charted form. The students can refer to our chart and practice in order to make themselves perfect.
Times Table 2 Charts
Times Table 3 Charts
Times Table 4 Charts
Time Table 5 Charts
Times Table 6 Charts
Times Table 7 Charts
Times Table 8 Charts
Times Table 9 Charts
Times Table 10 Charts
Times Table 11 Charts
Times Table 12 Charts
Blank Times Table
12 Multiplication Times Tables Chart
For this youth, games is one of the most relaxing thing because it gives pleasure to the kids, through playing games they get a break from their daily work, but sometimes the kids are so much
addictive towards the game that they don’t pay attention to their studies and slowly-slowly their minds gets bored from studies.
So in order to help the kids so that through playing games they can also learn the multiplication table of 12, we are bringing a game which will contain the time table.
12 Times Table Games
The most interesting thing about the game is that it designed in such a manner that the kids will like the game and at the same time they can also learn the tables. This way their parents will also
not feel uncomfortable in providing their respective devices such as smartphone and Pc’s. The parents can refer to our game as they can help their child to understand the game if they find any
difficulty or are stuck at any level.
Times Table 1-10 Charts
Times Table 1-12 Charts
Times Table 1-15 Charts
Times Table 1-100 Charts
Times Table 1-20 Charts
Time Table 1-25 Charts
Times Table 1-30 Charts
Times Table 1-1000 Charts
Worksheet For Grade 2
Worksheet For Grade 3
12 Times Tables Trick
Tricks is something which makes a particular thing easier, similarly is in a subject like maths. If a student is finding difficulties in a particular question than surely he has to learn the trick in
order to solve that particular question. 12 multiplication table is also a part of maths and when it comes to tables than every student faces some problems as it is a new topic for the kid.
Therefore to help the kids so that they come to know the various tricks we are coming up with our 12 times table trick, here the kids and students will find all the tricks in detailed form which will
help them to master the table in a very quick time.
Multiplication Table 12 Worksheet
We are also proposing our worksheet which will have the 12 times table worksheet, the worksheet will be available free of cost and the users who will use this worksheet will not have to spend any
money in order to use it.
The worksheet will be very helpful for the students who are in lower grades because the worksheet will have the table in a very simplified manner and the kids will find it very interesting in using
it. The worksheet will have designs which will make the sheet more colorful and the size of the sheet is editable, therefore the users can make changes according to their choice.
Multiplication Table 12 Chart
As we mentioned before that the worksheet will have fonts and designs, apart from that the users will get the ways of how to deal with questions which are related to the table of 12. The kids and
students can carry the worksheet along with them wherever they want to and also if they want to paste the worksheet in their notebook than they can also do so.
The worksheet will have the table in a very detailed way as for one question various solutions will be their which will help the kids to understand well and clear their concept. The worksheet is
available in various forms such as pdf, word, ppt and excel too, the users can also download the worksheet and store it in their storage devices such as in phones and computers.
12 Multiplication Table Maths
The table of 12 can be printed and what the users just have to do is to download the table and after downloading it they can go to any nearby stationary where printer is available and get the table
printed. The users will surely not have much difficulty in executing our table.
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