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Converting Fractions to Mixed Numbers
Learn About Converting Fractions to Mixed Numbers With Example Problems And Interactive Exercises
You may recall the example below from a previous lesson.
Example 1
In example 1, we used circles to help us solve the problem. Now look at the next example.
Example 2: At a birthday party, there are 19 cupcakes to be shared equally among 11 guests. What part of the cupcakes will each guest get?
Analysis: We need to divide 19 cupcakes by 11 equal parts. It would be time-consuming to use circles or other shapes to help us solve this problem. Therefore, we need an arithmetic method.
Step 1: Look at the fraction nineteen-elevenths below. Recall that the fraction bar means to divide the numerator by the denominator. This is shown in step 2.
Step 2:
Step 3:
Example 3:
Step 1:
Step 2:
Analysis: We need to divide 37 into 10 equal parts.
Step 1:
Step 2:
Example 5:
Analysis: We need to divide 37 into 13 equal parts.
Step 1:
Step 2:
In each of the examples above, we converted a fraction to a mixed number through long division of its numerator and denominator. Look at example 6 below. What is wrong with this problem?
Example 6:
Analysis: In the fraction seven-eighths, the numerator is less than the denominator. Therefore, seven-eighths is a proper fraction less than 1. We know from a previous lesson that a mixed number is
greater than 1.
Answer: Seven-eighths cannot be written as a mixed number because it is a proper fraction.
Example 7: Can these fractions be written as mixed numbers? Explain why or why not.
Analysis: In each fraction above, the numerator is equal to the denominator. Therefore, each of these fractions is an improper fraction equal to 1. But a mixed number is greater than 1.
Answer: These fractions cannot be written as mixed numbers since each is an improper fraction equal to 1.
After reading examples 6 and 7, you may be wondering: Which types of fractions can be written as mixed numbers? To answer this question, let’s review an important chart from a previous lesson.
Comparison of numerator and denominator: If the numerator < denominator, then the fraction < 1
Type of Fraction: proper fraction
Write As: proper fraction
Comparison of numerator and denominator: If the numerator = denominator, then the fraction = 1.
Type of Fraction: improper fraction
Write As: whole number
Comparison of numerator and denominator: If the numerator > denominator, then the fraction > 1.
Type of Fraction: improper fraction
Write As: mixed number
The answer to the question is: Only an improper fraction greater than 1 can be written to a mixed number.
Summary: We can convert an improper fraction greater than one to a mixed number through long division of its numerator and denominator.
In Exercises 1 through 5, click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is
correct or incorrect. To start over, click CLEAR. Note: To write the mixed number four and two-thirds, enter 4, a space, and then 2/3 into the form.
1. Write eleven-fifths as a mixed number.
2. Write eleven-fourths as a mixed number.
3. Write thirteen-ninths as a mixed number.
4. On field day, there are 23 pies to share equally among 7 classes. What part of the pies will each class get?
5. A teacher gives her class a spelling test worth 35 points. If there are 8 words graded equally, then how many points is each word worth? | {"url":"https://mathgoodies.com/lessons/fractions_to_mixed/","timestamp":"2024-11-05T15:27:59Z","content_type":"text/html","content_length":"51022","record_id":"<urn:uuid:a69514da-1f86-4bf3-8a11-c4505cd636e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00560.warc.gz"} |
LINKED LIST AND ITS TYPES | DSA - CoinCodeCap
LINKED LIST AND ITS TYPES | DSA
Linked list is the linear collection of data structure, where the data is not stored sequentially inside the computer memory but linked with each other by the the help of address. Each element is
called a node, contains references to the next node. this way flexibility is provided to a structure.
types of linked list
• singly linked list
• doubly linked list
• circular linked list
Singly linked list
SINGLY linked list is the linear collection of data item called nodes, where each node has
been divided into two parts
1. DATA
2. NEXT
The data part of the node collects the actual data, and the link part contains the address to the next node which has to be traversed.
Here, the head address is 4800 which determines where the data is being started. On the first node the data is 10 and the address to the next node is given 4900. This process is repeated till the
data ends and the address is null on the last node.
• Create a new node with the given value.
• Point the next of this new node to the current head of the list.
• Update the head to this new node.
2.INSERT ITEM AT THE ENDING OF SINGLY LINKED LIST
• Create a new node with the given value.
• Traverse the list to find the last node.
• Point the next of the last node to the new node.
• Create a new node with the given value.
• If the position is 0, update the new node’s next to the current head and update the head to the new node.
• Otherwise, traverse the list to find the node just before the desired position.
• Update the new node’s next to the next of the node found in step 3.
• Update the next of the node found in step 3 to the new node.
• Check if the list is empty (i.e., the head is NULL).
• If it is, there’s nothing to delete.
• If the list is not empty, update the head to the next node.
• Free the memory allocated for the original head node.
5.DELETE FROM THE END
• Check if the list is empty (i.e., the head is NULL).
• If it is, there’s nothing to delete.
• If the list is not empty, update the head to the next node.
• free the memory allocated for the original head node.
• Check if the list is empty (i.e., the head is NULL).
• If it is, there’s nothing to delete.
• If the position is 0, delete the node at the beginning.
• Otherwise, traverse the list to find the node just before the desired position.
• Update the next pointer of the node found in step 3 to the next node of the node to be deleted.
• Free the memory allocated for the node to be deleted.
Doubly linked list
Doubly linked list is a linear collection of data items called nodes, where each node has been divided into 3 parts
1. Previous
2. Data
3. Next
Previous section of the node contains address to the previous data node. Data section contains the actual data and next part contains address to the next node.
Following is the example: –
Here, the head address is 100 and the previous section is null. This means this is the starting point of doubly linked list. The next section of the 100 node is 200 which means 200 is the next node.
1.INSERT AN ITEM IN THE BEGINING OF DOUBLY LINKED LIST
• Create a new node.
• Assign the new node’s data to the given value.
• Set the new node’s next to the current head.Set the new node’s prev to NULL.
• If the list is not empty, set the current head’s prev to the new node.
• Update the head to point to the new node.
2.INSERT AN ITEM IN THE END OF DOUBLY LINKED LIST
• Create a new node.
• Assign the new node’s data to the given value.
• Set the new node’s next to NULL.
• If the list is empty, set the head to the new node.
• Traverse to the end of the list.Set the last node’s next to the new node.
• Set the new node’s prev to the last node.
3.INSERT AN ITEM AT ANY POSITION OF DOUBLY LINKED LIST
• Validate the position.
• Create a new node.
• Assign the new node’s data to the given value.
• If inserting at the beginning, update head and return.
• Traverse to the node before the desired position.
• Set the new node’s next to the current node’s next.
• Set the new node’s prev to the current node.
• If not inserting at the end, set the next node’s prev to the new node.
• Update the current node’s next to the new node.
• Check if the list is empty.
• Store the head in a temporary variable.
• Update the head to the next node.
• If the list is not empty after the update, set the new head’s prev to NULL.
• Free the temporary node.
5.DELETE FROM THE END
• Check if the list is empty.
• Traverse to the last node.
• Update the previous node’s next to NULL.
• Free the last node.
• Validate the position.
• Check if the list is empty.
• If deleting the first node, update head and return.
• Traverse to the node at the given position.
• Update the previous node’s next to the current node’s next.
• If not deleting the last node, update the next node’s prev to the current node’s prev.
• Free the current node.
Circular linked list
It is the variation of singly and doubly linked list where first node point to the last node and last node point to the first. It is used when we want traversing of data numbers of times without
relying initializing the start pointer, as well as we can visit all the nodes from any node.
• CIRCULAR SINGLY LINKED LIST
• CIRCULAR DOUBLY LINKED LIST
Circular singly linked list
If a last node of singly linked list hold the address of start node, then its called circular singly linked list.
Following is an example:-
Here, end of the list can’t be seen as there is no null value in address of the node.
1.INSERT AN ITEM IN THE BEGINING
• Create a new node.
• Assign the new node’s data to the given value.
• If the list is empty, set the new node’s next to point to itself.
• If the list is not empty, find the last node and set its next to the new node, and update the new node’s next to point to the head.
• Update the head to point to the new node.
2.INSERT ITEM AT THE ENDING
• Create a new node.
• Assign the new node’s data to the given value.
• If the list is empty, set the new node’s next to point to itself and update the head.
• If the list is not empty, traverse to the last node and update its next to point to the new node, and set the new node’s next to point to the head.
3.INSERT AN ITEM AT ANY POSITION
• Validate the position.
• Create a new node.
• Assign the new node’s data to the given value.
• If inserting at the beginning, update the head and return.
• Traverse to the node before the desired position.
• Update the new node’s next to the current node’s next.
• Update the current node’s next to the new node.
• Check if the list is empty.
• If there’s only one node, free it and set the head to NULL.
• If the list has more than one node, find the last node.
• Set the last node’s next to the second node.Free the head node.
• Update the head to point to the second node.
5.DELETE FROM THE END
• Check if the list is empty.
• If there’s only one node, free it and set the head to NULL.
• If the list has more than one node, traverse to the second last node.
• Update the second last node’s next to point to the head.
• Free the last node.
• Validate the position.
• Check if the list is empty.
• If deleting the first node, handle separately.
• Traverse to the node just before the position to be deleted.
• If the node to be deleted is not within bounds, return an error.
• Update the previous node’s next to skip the node to be deleted.
• Free the node at the given position.
Circular doubly linked list
If last node of doubly linked list hold the address of first node, and first node hold the Adrian of last node called circular doubly linked list.
Following is an example:-
Here, no null value can be seen in either previous or next part of the node and hence it doesn’t end.
1.INSERT AN ITEM IN THE BEGINING
• Create a new node.
• Assign the new node’s data to the given value.
• If the list is empty, set the new node’s next and prev to point to itself.
• If the list is not empty, find the last node and set its next to the new node, and the new node’s next to point to the head.
• Update the head’s prev to the new node.
• Set the new node’s prev to the last node.
• Update the head to point to the new node.
2.INSERT ITEM AT THE ENDING
• Create a new node.
• Assign the new node’s data to the given value.
• If the list is empty, set the new node’s next to point to itself and update the head.
• If the list is not empty, traverse to the last node and update its next to point to the new node, and set the new node’s next to point to the head.
3.INSERT AN ITEM AT ANY POSITION
• Validate the position.
• Create a new node.
• Assign the new node’s data to the given value.
• If inserting at the beginning, update the head and return.
• Traverse to the node before the desired position.
• Update the new node’s next to the current node’s next.
• Update the current node’s next to the new node.
• Check if the list is empty.
• If there’s only one node, free it and set the head to NULL.
• If the list has more than one node, find the last node.
• Set the last node’s next to the second node.
• Update the second node’s prev to the last node.
• Free the head node.
• Update the head to point to the second node.
5.DELETE FROM THE END
• Check if the list is empty.
• If there’s only one node, free it and set the head to NULL.
• If the list has more than one node, traverse to the second last node.
• Set the second last node’s next to the head.
• Update the head’s prev to the second last node.
• Free the last node.
• Validate the position.
• Check if the list is empty.
• If deleting the first node, handle separately.
• Traverse to the node at the given position.
• Update the previous node’s next to the current node’s next.
• Update the next node’s prev to the current node’s prev.
• Free the current node.
ARRAY LINKED LIST
Array is a collection of homogeneous (similar) datatype.ย Linked list is a collection of nodesย
Array elements are stored in continuous memory locations Linked list elements can be stored anywhere in the memory.
Array work with static data structures. Linked list works with dynamic data structure.
Array elements are independent.ย ย Linked list elements are dependent to each other. | {"url":"https://coincodecap.com/linked-list","timestamp":"2024-11-07T13:28:06Z","content_type":"text/html","content_length":"510464","record_id":"<urn:uuid:c105d4ab-4d96-4843-93bc-1d5726df3ca5>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00350.warc.gz"} |
The interior of charged black holes and the problem of uniqueness in general relativity
We consider a spherically symmetric, double characteristic initial value problem for the (real) Einstein-Maxwell-scalar field equations. On the initial outgoing characteristic, the data is assumed to
satisfy the Price law decay widely believed to hold on an event horizon arising from the collapse of an asymptotically flat Cauchy surface. We establish that the heuristic mass inflation scenario put
forth by Israel and Poisson is mathematically correct in the context of this initial value problem. In particular, the maximal future development has a future boundary over which the space-time is
extendible as a C ^0 metric but along which the Hawking mass blows up identically; thus, the space-time is inextendible as a C ^1 metric. In view of recent results of the author in collaboration with
I. Rodnianski, which rigorously establish the validity of Price's law as an upper bound for the decay of scalar field hair, the C ^0 extendibility result applies to the collapse of complete,
asymptotically flat, spacelike initial data where the scalar field is compactly supported. This shows that under Christodoulou's C ^0 formulation, the strong cosmic censorship conjecture is false for
this system.
All Science Journal Classification (ASJC) codes
• General Mathematics
• Applied Mathematics
Dive into the research topics of 'The interior of charged black holes and the problem of uniqueness in general relativity'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/the-interior-of-charged-black-holes-and-the-problem-of-uniqueness","timestamp":"2024-11-13T22:03:59Z","content_type":"text/html","content_length":"50207","record_id":"<urn:uuid:1f73248c-416a-4d0e-8e2e-675cda5fd3f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00413.warc.gz"} |
The Stacks project
Remark 39.8.3. Any group scheme over a field of characteristic $0$ is reduced, see [I, Theorem 1.1 and I, Corollary 3.9, and II, Theorem 2.4, Perrin-thesis] and also [Proposition 4.2.8, Perrin]. This
was a question raised in [page 80, Oort]. We have seen in Lemma 39.8.2 that this holds when the group scheme is locally of finite type.
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BFE.dev #8. can you shuffle() an array?
BFE.dev is like a LeetCode for Front End developers. I’m using it to practice my skills.
This article is about the coding problem BFE.dev#8. can you shuffle() an array?
The goal is to shuffle an array, seems pretty easy. Since shuffle is choose one of the all possible permutations, so we can choose the item one by one randomly from rest possible positions.
function shuffle(arr) {
for (let i = 0; i < arr.length; i++) {
const j = i + Math.floor(Math.random() * (arr.length - i))
;[arr[i], arr[j]] = [arr[j], arr[i]]
Code Not working
function shuffle(arr) {
for (let i = 0; i < arr.length; i++) {
const j = Math.floor(Math.random() * arr.length)
;[arr[i], arr[j]] = [arr[j], arr[i]]
Above code looks working but actually not. It loops all the positions and randomly swaps with another.
Suppose we have an array of [1,2,3,4].
Let's look at number 1. The first step is to swap 1 with all 4 positions, chance is high that it is moved to the positions other than itself and later it is traversed and swapped again.
Now let's look at number 4. It is last number to be traversed, and it might be also swapped before its turn and never be traversed again.
So 1 and 4 have different chances of be swapped, 4 is obviously lower, so the result array is not randomly shuffled.
Here is my video explaining: https://www.youtube.com/watch?v=FpKnR7RQaHM
Hope it helps, you can have a try at here
Top comments (0)
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://dev.to/jser_zanp/bfe-dev-8-can-you-shuffle-an-array-5hlo","timestamp":"2024-11-10T21:05:49Z","content_type":"text/html","content_length":"66420","record_id":"<urn:uuid:b586ff5f-0ad0-4b2f-a843-b49a84b86412>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00366.warc.gz"} |
Re: Interaction of Remove and Global variables in a Module
• To: mathgroup at smc.vnet.net
• Subject: [mg120101] Re: Interaction of Remove and Global variables in a Module
• From: Leonid Shifrin <lshifr at gmail.com>
• Date: Fri, 8 Jul 2011 04:53:55 -0400 (EDT)
• References: <201107071128.HAA15173@smc.vnet.net>
The problem you encountered is actually more subtle. Remember that when you
use Remove (as compared to Clear or ClearAll), the symbol is completely removed
from the system. This is a pretty disruptive operation. Now, what should the system do if the symbol you are removing is referenced by some other symbols?
Keeping it there unchanged would mean that the symbol has not been really removed
from the system. The solution used in Mathematica is to change a reference to a
(say "a") to Removed[a]. In practice, this means that, even when you
the symbol, those definitions that were involving it are still "spoiled" and
can not
be used. IMO, this is a very sensible design, but this is what leads to the
that puzzled you. Have a look:
f[b_]:=Module[{t},Removed[a][t_]=b t^2;]
What you have to do in your approach is to re-run the definition for "f", to
be able to
use it. This is however pretty error-prone. You can cure this by calling
Clear or ClearAll
instead of Remove, but even this approach I don't consider a good practice.
Even though
you use "a" as a function, calling Clear or Remove on it means that in a
way, you use
it as a global variable. In this post:
there is a lengthy discussion why making implicit dependencies on global
variables is a
bad practice.
Here are a few ways out. What you seem to want is to generate a function
with embedded
parameters (a closure). One way is to generate a pure function and return
In[31]:= Clear[f];
a = f[3];
a = f[5];
Out[34]= 3 t^2
Out[36]= 4 t^2
Out[39]= 5 t^2
Another way is to explicitly pass to "f" the symbol to which you want to
the definition:
In[40]:= Clear[ff,a];
Out[43]= 3 t^2
Out[45]= 4 t^2
Out[48]= 5 t^2
By making the function name an explicit parameter to "ff", you make the
situation above impossible to happen.
On Thu, Jul 7, 2011 at 3:28 PM, blamm64 <blamm64 at charter.net> wrote:
> This is what I get for querying SetDelayed
> In[1]:= ?SetDelayed
> lhs:=rhs assigns rhs to be the delayed value of lhs. rhs is maintained
> in an unevaluated form. When lhs appears, it is replaced by rhs,
> evaluated afresh each time. >>
> Note particularly the above reads AFRESH EACH time. It appears then
> the following is inconsistent behavior based on the above description:
> In[2]:= f[b_]:=Module[{t},a[t_]=b*t^2;]
> In[3]:= a[t]
> Out[3]= a[t]
> In[4]:= f[3]
> In[5]:= a[t]//InputForm
> Out[5]//InputForm=
> 3*t^2
> In[6]:= f[5]
> In[7]:= a[t]//InputForm
> Out[7]//InputForm=
> 5*t^2
> In[8]:= Remove[a]
> In[9]:= f[4]
> In[10]:= a[t]//InputForm
> Out[10]//InputForm=
> a[t]
> Apparently AFRESH is not an accurate description of how SetDelayed
> operates in this case, or I am missing something about this particular
> interaction of Module, Remove, and global variables inside Modules.
> However, if I go back, after executing the last line above (<a> has
> been Removed), and place the cursor in the input line where <f> is
> defined and hit Enter, which I thought would be identical to just
> evaluating <f> AFRESH again, and then execute the <f[4]> line again,
> then the global <a> definition is re-constituted.
> The documentation for Remove reads the name is no longer recognized by
> Mathematica. My understanding is that if the same name is defined
> AFRESH, it will once again be recognized.
> So if anyone would let me know what I am missing, regarding why the
> definition of <a> is not created AFRESH each time <f> is evaluated, I
> would appreciate it.
> Please don't construe the definition of <f> as my way of
> 'parameterizing' a function definition, I just use that definition to
> convey the apparent inconsistency.
> -Brian L. | {"url":"http://forums.wolfram.com/mathgroup/archive/2011/Jul/msg00161.html","timestamp":"2024-11-05T05:44:09Z","content_type":"text/html","content_length":"34807","record_id":"<urn:uuid:4bd6c12a-4d34-451b-92f6-e93aaad0008b>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00739.warc.gz"} |
Implement HDL-Optimized CORDIC-Based Square Root for Positive Real Numbers
This example shows two implementations of HDL-optimized CORDIC square root in the CORDIC Square Root Resource-Shared block and the CORDIC Square Root Fully-Pipelined block. The first implementation
has a resource-shared architecture that is optimized for low hardware utilization. The second implementation has a fully-pipelined architecture that is optimized for high throughput. The core
algorithm of both blocks uses CORDIC in hyperbolic vectoring mode to compute the approximation of square root (see Compute Square Root Using CORDIC). This CORDIC-based algorithm is different from the
Simulink® Sqrt block, which uses bisection and Newton-Raphson methods. The algorithm in the HDL-optimized CORDIC square root blocks require only iterative shift-add operations.
The input data for this example is a real non-negative scalar. Both blocks use the AMBA AXI handshake protocol at the input and output interfaces.
Generate HDL Code from Resource-Shared Architecture and Fully-Pipelined Architecture
The CORDIC Square Root Resource-Shared block and the CORDIC Square Root Fully-Pipelined block use different architectures for HDL code generation.
Resources-Shared Architecture
The CORDIC Square Root Resource-Shared block uses a resource-shared architecture. In the generated hardware design, this architecture prioritizes the hardware utilization constraint over throughput.
The block has one CORDIC kernel, which is reused for all CORDIC shift-and-add iterations. A controller based on a MATLAB Function block controls the AMBA AXI handshake process and CORDIC workflow.
This controller is a Moore machine. The state register and internal counter of the controller are modeled by persistent variables. For guidelines, see Initialize Persistent Variables in MATLAB
Functions. The control logic and algorithm data path are intentionally separated for better maintainability and reusability.
Assuming that the upstream block and downstream block are always ready, the block timing diagram is as shown.
The first u input is u1, u2 is the second u input, y1 is the square root of u1, y2 is the square root of u2, and so on.
After a successful input data transaction, this block starts computing the square root approximation. During the computation, the block ignores all other input data. Once the computation is done, the
block holds the result at output port and asserts its validOut signal. After a successful output data transaction, the block becomes ready again. The block asserts its ready signal and waits for the
input handshake signal.
Fully-Pipelined Architecture
The CORDIC Square Root Fully-Pipelined block uses a fully-pipelined architecture. In the generated hardware design, this architecture prioritizes throughput over hardware utilization constraints.
Each CORDIC shift-and-add iteration is performed by a dedicated CORDIC kernel, so the block can accept new data on any clock cycle if its downstream block is available. The cascaded CORDIC structure
is implemented by a for-each subsystem with a Unit Delay Enabled Synchronous block, Selector block, and Mux block. The pipeline registers in the data path are modeled by Unit Delay Enabled
Synchronous block so the downstream ready signal can control the data flow. A Unit Delay Enabled Resettable Synchronous block models the pipeline registers in the data path. The downstream ready
signal controls the valid signal flow, and the restart signal resets the pipeline registers along the valid signal path.
Assuming that the upstream block and downstream block are always ready, the block timing diagram is as shown.
The first u input is u1, u2 is the second u input, y1 is the square root of u1, y2 is the square root of u2, and so on.
After a successful input data transaction, this block computes the square root approximation of the input data. Because of its fully-pipelined nature, the block is able to accept input data on any
cycle, including on consecutive cycles. Once the computation is done, the block holds the result at output port, asserts its validOut signal, and waits for the downstream handshake signal. The ready
signal is a direct feedthrough of the readyIn signal for back-pressure propagation. If the downstream block is not ready, this block also pauses accordingly.
Define Simulation Parameters
Specify the number of input samples and data type.
Specify the data type as Fixed, Single, or Double.
For fixed-point data type, specify the word length and fraction length.
wordLength = 16;
FractionLength = 10;
Define the maximum CORDIC shift value. In fixed point, this value cannot exceed wordLength - 1.
switch lower(DT)
case 'fixed'
maximumShiftValue = wordLength - 1;
case 'single'
maximumShiftValue = 23;
case 'double'
maximumShiftValue = 52;
maximumShiftValue = 52;
Generate Nonnegative Input u
u = abs(randn(1,numSamples));
Cast to Selected Data Type
switch lower(DT)
case 'fixed'
u = cast(u,'like',fi([],1,wordLength,FractionLength));
case 'single'
u = single(u);
case 'double'
u = double(u);
u = double(u);
Configure Model Workspace and Run Simulation
model = 'CORDICSquareRootHDLOptimizedModel';
out = sim(model);
Verify Output Solutions
Compare fixed-point results with built-in floating point results.
yBuiltIn = sqrt(double(u))'
yBuiltIn = 3×1
yShared =
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 16
FractionLength: 10
yPipelined = out.yPipelined(1:numSamples)
yPipelined =
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 16
FractionLength: 10
Verify that the CORDIC Square Root Resource-Shared block and the CORDIC Square Root Fully-Pipelined block return identical results in fixed point.
if strcmpi(DT,'fixed')
yShared == yPipelined %#ok
ans = 3x1 logical array
Block Latency Equations
The block latency is the number of clock cycles between a successful input and when the corresponding output becomes valid.
The CORDIC-based square root approximation has two main steps: normalization and CORDIC shift-add iterations. Thus, the total latency is normalization latency plus the latency of the CORDIC shift-add
The latency of normalization is determined by the input word length.
if isfi(u) && isfixed(u)
% If input is fi, normalization latency is nextpow2(u.WordLength)+1.
normLatency = nextpow2(u.WordLength)+1;
% If input is floating point, normalization latency is 0.
normLatency = 0;
The number of CORDIC iterations is determined by the CORDIC maximum shift value. This example uses wordLength - 1 for best precision.
sequence = fixed.cordic.hyperbolic.shiftSequence(maximumShiftValue);
numOfCORDICIterations = length(sequence);
CORDIC Square Root Resource-Shared Block
For the resource-shared architecture, the CORDIC shift-add iterations latency is the number of CORDIC iterations plus two.
blockLatencySharedSim = normLatency + numOfCORDICIterations + 2 %#ok
blockLatencySharedSim =
After a successful output, the resource-shared block becomes ready in the next clock cycle.
CORDIC Square Root Fully-Pipelined Block
For the fully-pipelined architecture, the CORDIC shift-add iterations latency is the number of CORDIC iterations plus one.
blockLatencyPipelined = normLatency + numOfCORDICIterations + 1
blockLatencyPipelined =
When the downstream block is available, the CORDIC Square Root Fully-Pipelined block can accept new data on any clock cycle, so it is always ready.
Benchmark Block Latency from Simulation
To verify the block latency equations, log the ready/valid handshake signals to measure the block latency in simulation.
validInHistoryShared = out.validInShared;
readyHistoryShared = out.readyShared;
validOutHistoryShared = out.validOutShared;
readyInHistoryShared = out.readyInShared;
validInHistoryPipelined = out.validInPipelined;
readyHistoryPipelined = out.readyPipelined;
validOutHistoryPipelined = out.validOutPipelined;
readyInHistoryPipelined = out.readyInPipelined;
CORDIC Square Root Resource-Shared Block Latency
Find the data transaction time for the CORDIC Square Root Resource-Shared block.
tDataInShared = find(validInHistoryShared & readyHistoryShared == 1);
tDataOutShared = find(validOutHistoryShared & readyInHistoryShared == 1);
Find the rising edge of the ready signal.
tReadyShared = find(diff(readyHistoryShared) == 1) + 1;
Compute the block latency from a successful input to the output.
blockLatencySharedSim = tDataOutShared - tDataInShared(1:numSamples)
blockLatencySharedSim = 3×1
Compute the block latency from a successful output to when the block becomes ready again.
readyLatencySharedSim = tReadyShared - tDataOutShared
readyLatencySharedSim = 3×1
CORDIC Square Root Fully-Pipelined Block Latency
Find the data transaction time.
tDataInPipelined = find(validInHistoryPipelined & readyHistoryPipelined == 1);
tDataOutPipelined = find(validOutHistoryPipelined & readyInHistoryPipelined == 1);
Compute the block latency from a successful into to the corresponding output.
blockLatencyPipelinedSim = tDataOutPipelined(1:numSamples) - tDataInPipelined(1:numSamples)
blockLatencyPipelinedSim = 3×1
Hardware Resource Utilization
Both blocks in this example support HDL code generation using the Simulink® HDL Workflow Advisor. For an example, see HDL Code Generation and FPGA Synthesis from Simulink Model (HDL Coder)(HDL Coder)
and Implement Digital Downconverter for FPGA (HDL Coder) (DSP HDL Toolbox).
This example data was generated by synthesizing the block on a Xilinx® Zynq®-7000 SoC ZC702 Evaluation Kit. The synthesis tool was Vivado® v2022.1 (win64).
These parameters were used for synthesis:
• Input data type: sfix16_En10
• maximumShiftValue: 15 (WordLength - 1)
• Target frequency: 200 MHz
This table shows the post-place-and-route resource utilization results for the CORDIC Square Root Resource-Shared block.
Resource Usage Available Utilization (%)
Slice LUTs 445 53200 0.84
Slice Registers 92 106400 0.09
DSPs 0 220 0.00
Block RAM Tile 0 140 0.00
This table shows the timing summary for the CORDIC Square Root Resource-Shared block.
Requirement 5 ns (200 MHz)
Data Path Delay 5.337 ns
Slack -0.312 ns
Clock Frequency 184.71 MHz
This table shows the post-place-and-route resource utilization results for the CORDIC Square Root Fully-Pipelined block.
Resource Usage Available Utilization (%)
Slice LUTs 926 53200 1.74
Slice Registers 701 106400 0.66
DSPs 0 220 0.00
Block RAM Tile 0 140 0.00
This table shows the timing summary for the CORDIC Square Root Fully-Pipelined block.
Requirement 5 ns (200 MHz)
Data Path Delay 5.016 ns
Slack 0.009 ns
Clock Frequency 200.36 MHz
Related Topics | {"url":"https://kr.mathworks.com/help/fixedpoint/ug/implement-hdl-optimized-cordic-based-square-root-for-positive-real-numbers.html","timestamp":"2024-11-06T01:28:07Z","content_type":"text/html","content_length":"92688","record_id":"<urn:uuid:23aacd91-696f-4592-abd5-39db472e144c>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00600.warc.gz"} |
Adjacency Matrix Template Excel
Adjacency Matrix Template Excel - Adjacency list or adjacency matrix. Web i have an excel spreadsheet which depicts a binary adjacency matrix in the form shown below (with about a 1000. Interior
design / adjacency diagram editor. Web to make this a little easier i have made an adjacency matrix template that you can download and manipulate to fit the needs of your projects. Web creating
adjacency matrix for any network in excel is a matter of 5 minutes.when the data is very. Web use this adjacency matrix diagram to show the relationship between 2 adjacent pairs. Web matrix diagrams
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Adjacency Matrix Interior Design Design Talk
Web creating adjacency matrix for any network in excel is a matter of 5 minutes.when the data is very. Web i have an excel spreadsheet which depicts a binary adjacency matrix in the form shown below
(with about a 1000. Web use this adjacency matrix diagram to show the relationship between 2 adjacent pairs. Web the spreadsheet must be in.
Adjacency Matrix Template Excel
Web creating adjacency matrix for any network in excel is a matter of 5 minutes.when the data is very. Web use this adjacency matrix diagram to show the relationship between 2 adjacent pairs.
Interior design / adjacency diagram editor. Web to make this a little easier i have made an adjacency matrix template that you can download and manipulate to.
How To Make An Adjacency Matrix Architecture Design Talk
Web i have an excel spreadsheet which depicts a binary adjacency matrix in the form shown below (with about a 1000. Adjacency list or adjacency matrix. Interior design / adjacency diagram editor. Web
to make this a little easier i have made an adjacency matrix template that you can download and manipulate to fit the needs of your projects. Web.
Adjacency Matrix Template Excel
Web to make this a little easier i have made an adjacency matrix template that you can download and manipulate to fit the needs of your projects. Web creating adjacency matrix for any network in
excel is a matter of 5 minutes.when the data is very. Interior design / adjacency diagram editor. Adjacency list or adjacency matrix. Web use this.
Adjacency matrix plots with R and ggplot2 Matthew Lincoln, PhD
Web the spreadsheet must be in one of the following formats: Web matrix diagrams can be useful in a wide range of applications from aligning requirements with specifications to delegating
responsibilities based on. Web to make this a little easier i have made an adjacency matrix template that you can download and manipulate to fit the needs of your projects..
Programming Matrix • Gustin Design Services
Web i have an excel spreadsheet which depicts a binary adjacency matrix in the form shown below (with about a 1000. Web use this adjacency matrix diagram to show the relationship between 2 adjacent
pairs. Web matrix diagrams can be useful in a wide range of applications from aligning requirements with specifications to delegating responsibilities based on. Web the spreadsheet.
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Adjacency Matrix Template Excel
Interior design / adjacency diagram editor. Web to make this a little easier i have made an adjacency matrix template that you can download and manipulate to fit the needs of your projects. Web use
this adjacency matrix diagram to show the relationship between 2 adjacent pairs. Web creating adjacency matrix for any network in excel is a matter of.
How to Make a Matrix on Excel Microsoft Excel Help YouTube
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and manipulate to fit the needs of your projects. Web i have an excel spreadsheet which depicts a binary adjacency matrix in the.
Adjacency Matrix Template Excel
Web creating adjacency matrix for any network in excel is a matter of 5 minutes.when the data is very. Web the spreadsheet must be in one of the following formats: Web i have an excel spreadsheet
which depicts a binary adjacency matrix in the form shown below (with about a 1000. Adjacency list or adjacency matrix. Web use this adjacency.
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Web Creating Adjacency Matrix For Any Network In Excel Is A Matter Of 5 Minutes.when The Data Is Very.
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depicts a binary adjacency matrix in the form shown below (with about a 1000.
Web Use This Adjacency Matrix Diagram To Show The Relationship Between 2 Adjacent Pairs.
Interior design / adjacency diagram editor. Adjacency list or adjacency matrix.
Related Post: | {"url":"https://www.trendysettings.com/en/adjacency-matrix-template-excel.html","timestamp":"2024-11-08T23:33:30Z","content_type":"text/html","content_length":"28414","record_id":"<urn:uuid:4af9cf8b-4f2f-4611-94cd-ec145b6ec1b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00050.warc.gz"} |
Monte Carlo methods
Generally speaking, Monte Carlo methods are algorithms that simply incorporate random sampling to make approximations. Three most common use cases include Mathematical optimization, Numerical
integration, and drawing samples from Probability distributions.
The general pattern followed by most Monte Carlo methods is as follow:
1. Define domain of possible inputs
2. Generate random values from a Probability distribution over that domain
3. Perform some analysis on the resulting samples
A common example used to introduce Monte Carlo methods and the power of random sampling for approximation is the estimation of $\pi$ using uniform random samples inside of a square. | {"url":"https://samgriesemer.com/Monte_Carlo_methods","timestamp":"2024-11-12T12:44:32Z","content_type":"text/html","content_length":"8325","record_id":"<urn:uuid:aa4f05ba-33d2-4a2b-9db2-2f92cf8f2f81>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00414.warc.gz"} |
Distance from Point to Plane Calculator
Last updated:
Distance from Point to Plane Calculator
Whenever you need to calculate the distance from a point to a plane in 3D space, Omni is here to help. If you are new to the topic of finding the minimum distance from a point to a plane, you can
read the article below, where we discuss:
• What the shortest distance from a point to a plane is;
• The distance from point to plane formula;
• How to find the distance from a point to a plane by hand; and
• Particular cases, like the distance to the xy plane from a point or the distance from any plane to the space origin.
🙋 If our distance from point to plane calculator is not exactly what you're looking for, check out our distance calculator, which covers the topic of distance in a much broader context.
What is the shortest distance from point to plane?
When someone gives us a point and a plane in 3D space, the shortest distance from one to the other is along the line perpendicular to the plane dropped from the point. In other words, it is the
magnitude of the normal vector that starts from the given point and ends at the plane.
💡 Visit Omni's vector magnitude calculator if you need a refresher on that.
An equivalent explanation that might fancy your imagination may be the following: image a ball centered at the point. We inflate the ball until its surface touches the plane. The radius of this ball
is exactly the distance from our point to the plane!
Keep this image in your head, and you'll never forget what the minimum distance from a point to a plane is!
Distance from point to plane formulas
Once we know what the perpendicular distance from a point to a plane means, let's discuss how to calculate it for a given point (a,b,c) and plane. We'll discuss two approaches: when you have the
standard form equation of your plane and when you have its normal vector and one point in the plane.
Standard form
Here we assume your plane is given by the equation Ax + By + Cz + D = 0.
The distance between this plane and your point (a,b,c) is.
$\footnotesize L = \frac{|A \cdot a + B \cdot b + C \cdot c + D|}{\sqrt{A^2 + B^ 2 + C^2}}$
• L — Distance;
• A, B, C, and D — Coefficients of the standard form plane equation; and
• a, b, and c — Coordinates of your point.
What to do if the denominator is zero, you wonder? How do I calculate the distance? Well, you don't. Recall that the condition for Ax + By + Cz + D = 0 to describe a plane in 3D space is that A, B,
and C must not all be zero. This translates into A^2 + B^2 + C^2 > 0. So, if it happens that you get zero in the denominator, it means your plane equation is wrong.
Normal vector and a point
Here we assume your plane is given by the normal vector n = [A, B, C] and point p = (x, y, z) belonging to the plane.
To get the formula for distance between this plane (given by n and p) to your point (a,b,c), you first need to derive the standard form plane equation and then apply the formula we've given above. To
save you some time, we've done it for you — here's the final formula, ready for you to apply directly to your data:
$\footnotesize L = \frac{|A (a\!-\!x) + B (b\!-\!y) + C (c\!-\!z)|}{\sqrt{A^2 + B^ 2 + C^2}}$
In this version:
• L is the distance;
• A, B, C are the coefficients of the normal vector, n;
• x, y, and z are the coordinates of the point p belonging to the plane; and
• a, b, and c are the coordinates of the point from which you calculate distance.
The condition that A^2 + B^2 + C^2 > 0 corresponds to the fact that the magnitude of the normal vector cannot be zero.
🙋 Need a refresher on vectors? Try our vector calculator!
And that's how we find the distance from a point to a plane. As you can see, these formulas are not very hard, but they are not the simplest as well. Fortunately, our distance from point to plane
calculator can perform the calculations for you!
How to use this perpendicular distance from point to plane calculator?
When you want to use our tool to determine the shortest distance from point to plane, you need to:
1. Input the point's coordinates into the fields a, b, and c.
2. Decide on how you want to input the plane. You can choose between
□ Standard form equation; and
□ Normal vector and one point from the plane.
3. Whatever you choose, input the data into the fields of our distance from point to plane calculator.
4. Our calculator will display the result immediately.
How do I compute the distance from a point to a plane?
To determine the distance from a point to a plane:
1. Write down the standard form equation of your plane. It should be in the form Ax + By + Cz + D = 0.
2. Calculate A^2 + B^2 + C^2. If it equals zero, your plane equation is wrong.
3. Write down the coordinates of the point (a, b, c) from which you want to calculate distance.
4. Calculate |A×a + B×b + C×c + D|.
5. Divide it by the square root of the value from Step 2. This is the distance you're looking for!
How do I find the distance from a point to the xy plane?
The distance from the point (a, b, c) to the xy plane is equal to the absolute value of the last coordinate, i.e., to |c|. To see how this follows from the general formula for the distance from a
point to a plane, you need to plug in A = B = D = 0 and C = 1, which gives |C×c| / √(C^2) = |C × c| / |C| = |c|.
What is the distance from point (1,1,1) to x+y=0 plane?
The distance is √2 ≈ 1.41. To get this result, we apply the formula distance = |A×a + B×b + C×c|/√(A^2 + B^2 + C^2) with A=B=1 and C=D=0 and a = b = c = 1. We obtain distance = |1×1 + 1×1 + 0×1| / √
(1^2 + 1^2) = |2| / √2 = √2.
What is the distance from plane to space origin?
To compute the distance from the plane Ax + By + Cz + D = 0 to the point (0,0,0):
1. Compute A^2 + B^2 + C^2.
2. Take the square root of the number from Step 1.
3. Compute the absolute value of D and divide it by the number from Step 2.
4. That's it! You can verify your result with an online distance from point to plane calculator. | {"url":"https://www.omnicalculator.com/math/distance-from-point-to-plane","timestamp":"2024-11-06T01:31:19Z","content_type":"text/html","content_length":"491384","record_id":"<urn:uuid:714389ba-eb08-4f0c-b544-645b1c9bb777>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00123.warc.gz"} |
Talk:Draw a rotating cube - Rosetta CodeTalk:Draw a rotating cube
The task description currently ask that "[The cube] should be oriented so that 2 of the vertices are pointing up and down."
I can think of two ways of interpreting this specification.
One interpretation has an edge of the cube "at the top" (with the two vertices on that edge "at the top") and another edge "at the bottom".
The other interpretation has one vertex "at the top" and another vertex "at the bottom".
So I guess the issue with "2 of the vertices pointing up and down" is that a vertex that is pointing up would not be pointing down, nor vice versa. So did this mean "2 each pointing up and 2 others
pointing down" or did this mean "for the set of all vertices pointing up and down, there can be only two of them".
I hope this makes sense. --Rdm (talk) 19:13, 4 May 2015 (UTC)
It should have 1 at the top, and one at the bottom. I'll change the description. 12Me21 (talk) 21:03, 4 May 2015 (UTC)
The wording "other vertex" is confusing, because there are 5 of them.
Maybe "the cube should rotate around an axis that goes thru one vertex, the center of the cube, and the opposing vertex", and "that axis should be oriented up-down" Hajo (talk) 12:28, 5 May
2015 (UTC)
That's good. 12Me21 (talk) 12:05, 7 May 2015 (UTC) | {"url":"https://rosettacode.org/wiki/Talk:Draw_a_rotating_cube","timestamp":"2024-11-03T18:41:20Z","content_type":"text/html","content_length":"43101","record_id":"<urn:uuid:c26e8f07-934f-4332-bd36-8519ae5dc20d>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00428.warc.gz"} |
Ants in action
What have ants to do with finding best solutions? Well, ants can find the shortest way to the food being almost blind. How? Because of their cooperation and smell. They leave pheromone when they
move. When deciding where to go, they choose the direction where the pheromone smell is the strongest.
In a computer algorithm, each simulated ant also has to choose its direction of movement. It takes into account two factors: intensities of pheromones (maximized) and distances to next locations
(minimized). While moving, each ant leaves the pheromone.
Although each ant is independent, it unconsciously affects the behavior of other ants. This is how the pheromone gradually marks the globally optimal (shortest) path.
The program below (click and choose your language) demonstrates the ant algorithm. Green lines correspond to the intensity of pheromone, and the black line is the shortest path found so far.
The ant algorithm can be used to solve various optimization problems, not just the problem of finding the shortest path (called the Traveling Salesman Problem). | {"url":"http://en.alife.pl/opt/e/aa","timestamp":"2024-11-13T04:29:55Z","content_type":"application/xhtml+xml","content_length":"12641","record_id":"<urn:uuid:587d6c7d-cfec-420a-812b-803035ce7905>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00004.warc.gz"} |
Chapter 9.9: Fitting Exponential Models to Data
Learning Objectives
In this section, you will:
• Build an exponential model from data.
• Build a logarithmic model from data.
• Build a logistic model from data.
In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra
to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With
regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.
Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is
typically used to indicate that the equation or function approximates a real-world situation.
We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing
their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key
features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways
regression is used to model real-world phenomena.
Building an Exponential Model from Data
As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a
cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or
decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential
regression, so let’s review exponential growth and decay.
Recall that exponential functions have the form
• The initial value of the model is
□ If
□ If exponential decay. As x-axis. In other words, the outputs never become equal to or less than zero.
As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable
Exponential Regression
Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer
to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form,
Note that:
How To
Given a set of data, perform exponential regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
a. Clear any existing data from the lists.
b. List the input values in the L1 column.
c. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
a. Use ZOOM [9] to adjust axes to fit the data.
b. Verify the data follow an exponential pattern.
3. Find the equation that models the data.
a. Select “ExpReg” from the STAT then CALC menu.
b. Use the values returned for a and b to record the model,
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Using Exponential Regression to Fit a Model to Data
In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level
(BAC) with the risk of being in an accident. (Figure) shows results from the study^[1] . The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person
with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.
BAC 0 0.01 0.03 0.05 0.07 0.09
Relative Risk of Crashing 1 1.03 1.06 1.38 2.09 3.54
BAC 0.11 0.13 0.15 0.17 0.19 0.21
Relative Risk of Crashing 6.41 12.6 22.1 39.05 65.32 99.78
a. Let
b. After 6 drinks, a person weighing 160 pounds will have a BAC of about
Show Solution
a. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the
exponential pattern shown in (Figure):
Figure 1.
Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,
Converting from scientific notation, we have:
Notice that (Figure):
Figure 2.
b. Use the model to estimate the risk associated with a BAC of
If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.
Try It
(Figure) shows a recent graduate’s credit card balance each month after graduation.
Month 1 2 3 4 5 6 7 8
Debt ($) 620.00 761.88 899.80 1039.93 1270.63 1589.04 1851.31 2154.92
a. Use exponential regression to fit a model to these data.
b. If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?
Show Solution
a. The exponential regression model that fits these data is
b. If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?
No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a
model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).
Building a Logarithmic Model from Data
Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods,
and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or
decrease that helps us determine whether a logarithmic model is best.
Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we
can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on
graphing utilities,
• All input values,
• The point
• If
• If
Logarithmic Regression
Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic
function to a set of data points. This returns an equation of the form,
Note that
• all input values,
• when
• when
How To
Given a set of data, perform logarithmic regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
a. Clear any existing data from the lists.
b. List the input values in the L1 column.
c. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
a. Use ZOOM [9] to adjust axes to fit the data.
b. Verify the data follow a logarithmic pattern.
3. Find the equation that models the data.
a. Select “LnReg” from the STAT then CALC menu.
b. Use the values returned for a and b to record the model,
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Using Logarithmic Regression to Fit a Model to Data
Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.
(Figure) shows the average life expectancies, in years, of Americans from 1900–2010^[2] .
Year 1900 1910 1920 1930 1940 1950
Life Expectancy(Years) 47.3 50.0 54.1 59.7 62.9 68.2
Year 1960 1970 1980 1990 2000 2010
Life Expectancy(Years) 69.7 70.8 73.7 75.4 76.8 78.7
a. Let
b. Use the model to predict the average American life expectancy for the year 2030.
Show Solution
a. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the
scatterplot follows a logarithmic pattern as shown in (Figure):
Figure 3.
Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model,
Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in (Figure):
Figure 4.
b. To predict the life expectancy of an American in the year 2030, substitute
If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030.
Try It
Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. (Figure) shows the number of games sold, in thousands, from the years 2000–2010.
Year 2000 2001 2002 2003 2004 2005
Number Sold (thousands) 142 149 154 155 159 161
Year 2006 2007 2008 2009 2010 –
Number Sold (thousands) 163 164 164 166 167 –
a. Let
b. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.
Show Solution
a. The logarithmic regression model that fits these data is
b. If sales continue at this rate, about 171,000 games will be sold in the year 2015.
Building a Logistic Model from Data
Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and
the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living
space or nutrients.
It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population
growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:
Recall that:
• when
is the limiting value, sometimes called the carrying capacity, of the model.
Logistic Regression
Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a
logistic function to a set of data points. This returns an equation of the form
Note that
• The initial value of the model is
• Output values for the model grow closer and closer to
How To
Given a set of data, perform logistic regression using a graphing utility.
1. Use the STAT then EDIT menu to enter given data.
a. Clear any existing data from the lists.
b. List the input values in the L1 column.
c. List the output values in the L2 column.
2. Graph and observe a scatter plot of the data using the STATPLOT feature.
a. Use ZOOM [9] to adjust axes to fit the data.
b. Verify the data follow a logistic pattern.
3. Find the equation that models the data.
a. Select “Logistic” from the STAT then CALC menu.
b. Use the values returned for
4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Using Logistic Regression to Fit a Model to Data
Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. (Figure) shows the percentage of Americans with cellular service
between the years 1995 and 2012^[3] .
Year Americans with Cellular Service (%) Year Americans with Cellular Service (%)
1995 12.69 2004 62.852
1996 16.35 2005 68.63
1997 20.29 2006 76.64
1998 25.08 2007 82.47
1999 30.81 2008 85.68
2000 38.75 2009 89.14
2001 45.00 2010 91.86
2002 49.16 2011 95.28
2003 55.15 2012 98.17
a. Let
b. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
c. Discuss the value returned for the upper limit,
Show Solution
a. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot
follows a logistic pattern as shown in (Figure):
Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,
Next, graph the model in the same window as shown in (Figure) the scatterplot to verify it is a good fit:
b. To approximate the percentage of Americans with cellular service in the year 2013, substitute
According to the model, about 98.8% of Americans had cellular service in 2013.
c. The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a
population have cellular service?) If the model were exact, the limiting value would be
Try It
(Figure) shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.
Year Seal Population (Thousands) Year Seal Population (Thousands)
1997 3.493 2005 19.590
1998 5.282 2006 21.955
1999 6.357 2007 22.862
2000 9.201 2008 23.869
2001 11.224 2009 24.243
2002 12.964 2010 24.344
2003 16.226 2011 24.919
2004 18.137 2012 25.108
a. Let
b. Use the model to predict the seal population for the year 2020.
c. To the nearest whole number, what is the limiting value of this model?
Show Solution
a. The logistic regression model that fits these data is
b. If the population continues to grow at this rate, there will be about
c. To the nearest whole number, the carrying capacity is 25,657.
Access this online resource for additional instruction and practice with exponential function models.
Visit this website for additional practice questions from Learningpod.
Key Concepts
• Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and
closer to zero.
• We use the command “ExpReg” on a graphing utility to fit function of the form (Figure).
• Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
• We use the command “LnReg” on a graphing utility to fit a function of the form (Figure).
• Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
• We use the command “Logistic” on a graphing utility to fit a function of the form (Figure).
Section Exercises
1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.
Show Solution
Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic
model best describes populations.
2. What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense?
3. What is regression analysis? Describe the process of performing regression analysis on a graphing utility.
Show Solution
Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT
then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined,
select the appropriate regression analysis command from the STAT then CALC menu.
4. What might a scatterplot of data points look like if it were best described by a logarithmic model?
5. What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?
Show Solution
The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.
For the following exercises, match the given function of best fit with the appropriate scatterplot in (Figure) through (Figure). Answer using the letter beneath the matching graph.
11. To the nearest whole number, what is the initial value of a population modeled by the logistic equation
Show Solution
12. Rewrite the exponential model
13. A logarithmic model is given by the equation
Show Solution
14. A logistic model is given by the equation t does
15. What is the y-intercept on the graph of the logistic model given in the previous exercise?
Show Solution
For the following exercises, use this scenario: The population
16. Graph the population model to show the population over a span of
17. What was the initial population of koi?
Show Solution
18. How many koi will the pond have after one and a half years?
19. How many months will it take before there are
Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity.
For the following exercises, use this scenario: The population
20. Graph the population model to show the population over a span of
21. What was the initial population of wolves transported to the habitat?
Show Solution
22. How many wolves will the habitat have after
23. How many years will it take before there are
Show Solution
about 5.4 years.
24. Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.
For the following exercises, refer to (Figure).
x f(x)
25. Use a graphing calculator to create a scatter diagram of the data.
26. Use the regression feature to find an exponential function that best fits the data in the table.
27. Write the exponential function as an exponential equation with base
Show Solution
28. Graph the exponential equation on the scatter diagram.
29. Use the intersect feature to find the value of
For the following exercises, refer to (Figure).
x f(x)
30. Use a graphing calculator to create a scatter diagram of the data.
31. Use the regression feature to find an exponential function that best fits the data in the table.
Show Solution
32. Write the exponential function as an exponential equation with base
33. Graph the exponential equation on the scatter diagram.
34. Use the intersect feature to find the value of
For the following exercises, refer to (Figure).
x f(x)
1 5.1
2 6.3
3 7.3
4 7.7
5 8.1
6 8.6
35. Use a graphing calculator to create a scatter diagram of the data.
36. Use the LOGarithm option of the REGression feature to find a logarithmic function of the form
37. Use the logarithmic function to find the value of the function when
Show Solution
38. Graph the logarithmic equation on the scatter diagram.
39. Use the intersect feature to find the value of
For the following exercises, refer to (Figure).
x f(x)
1 7.5
3 5.2
4 4.3
5 3.9
6 3.4
7 3.1
8 2.9
40. Use a graphing calculator to create a scatter diagram of the data.
41. Use the LOGarithm option of the REGression feature to find a logarithmic function of the form
Show Solution
42. Use the logarithmic function to find the value of the function when
43. Graph the logarithmic equation on the scatter diagram.
44. Use the intersect feature to find the value of
For the following exercises, refer to (Figure).
x f(x)
1 8.7
2 12.3
3 15.4
4 18.5
5 20.7
6 22.5
7 23.3
9 24.6
10 24.8
45. Use a graphing calculator to create a scatter diagram of the data.
46. Use the LOGISTIC regression option to find a logistic growth model of the form
47. Graph the logistic equation on the scatter diagram.
48. To the nearest whole number, what is the predicted carrying capacity of the model?
49. Use the intersect feature to find the value of
For the following exercises, refer to (Figure).
2 28.6
4 52.8
5 70.3
7 99.9
8 112.5
10 125.8
11 127.9
15 135.1
17 135.9
50. Use a graphing calculator to create a scatter diagram of the data.
51. Use the LOGISTIC regression option to find a logistic growth model of the form
Show Solution
52. Graph the logistic equation on the scatter diagram.
53. To the nearest whole number, what is the predicted carrying capacity of the model?
54. Use the intersect feature to find the value of
55. Recall that the general form of a logistic equation for a population is given by
Show Solution
Working with the left side of the equation, we see that it can be rewritten as
Working with the right side of the equation we show that it can also be rewritten as
56. Use a graphing utility to find an exponential regression formula
57. Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary.
Show Solution
First rewrite the exponential with base e:
58. Find the inverse function
59. Use the result from the previous exercise to graph the logistic model
Show Solution
The graph of y-intercept at (0, 4) and horizontal asymptotes at y = 0 and y = 20. The graph of x– intercept at (4, 0) and vertical asymptotes at x = 0 and x = 20.
Chapter Review Exercises
1. Determine whether the function
Show Solution
exponential decay; The growth factor,
2. The population of a herd of deer is represented by the function
3. Find an exponential equation that passes through the points
Show Solution
4. Determine whether (Figure) could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.
x 1 2 3 4
f(x) 3 0.9 0.27 0.081
5. A retirement account is opened with an initial deposit of $8,500 and earns
Show Solution
6. Hsu-Mei wants to save $5,000 for a down payment on a car. To the nearest dollar, how much will she need to invest in an account now with
Show Solution
continuous decay; the growth rate is negative.
8. Suppose an investment account is opened with an initial deposit of
9. Graph the function y-intercept.
Show Solution
domain: all real numbers; range: all real numbers strictly greater than zero; y-intercept: (0, 3.5);
10. Graph the function y-axis on the same axes, and give the y-intercept.
11. The graph of y-axis and stretched vertically by a factor of y-intercept, domain, and range.
Show Solution
12. The graph below shows transformations of the graph of
25. State the domain, vertical asymptote, and end behavior of the function
32.Use properties of logarithms to expand
33.Use properties of logarithms to expand
Show Solution
34. Condense the expression
35. Condense the expression
Show Solution
40. Use logarithms to find the exact solution for no solution.
41. Use logarithms to find the exact solution for no solution.
Show Solution
no solution
42. Find the exact solution for no solution.
43. Find the exact solution for no solution.
Show Solution
no solution
44. Find the exact solution for no solution.
45. Find the exact solution for no solution.
Show Solution
46. Use the definition of a logarithm to solve.
47. Use the definition of a logarithm to find the exact solution for
Show Solution
48. Use the one-to-one property of logarithms to find an exact solution for no solution.
49. Use the one-to-one property of logarithms to find an exact solution for no solution.
Show Solution
50. The formula for measuring sound intensity in decibels
51. The population of a city is modeled by the equation
52. Find the inverse function
53. Find the inverse function
Show Solution
For the following exercises, use this scenario: A doctor prescribes
54. To the nearest minute, what is the half-life of the drug?
55. Write an exponential model representing the amount of the drug remaining in the patient’s system after
Show Solution
For the following exercises, use this scenario: A soup with an internal temperature of
56. Use Newton’s Law of Cooling to write a formula that models this situation.
57. How many minutes will it take the soup to cool to
For the following exercises, use this scenario: The equation
58. How many people started the rumor?
59. To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity?
60. What is the carrying capacity?
For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a
function that is linear, exponential, or logarithmic.
x f(x)
1 3.05
2 4.42
3 6.4
4 9.28
5 13.46
6 19.52
7 28.3
8 41.04
9 59.5
10 86.28
Show Solution
x f(x)
0.5 18.05
3 15.33
5 14.55
7 14.04
10 13.5
12 13.22
13 13.1
15 12.88
17 12.69
20 12.45
63. Find a formula for an exponential equation that goes through the points e.
Show Solution
64. What is the carrying capacity for a population modeled by the logistic equation
65. The population of a culture of bacteria is modeled by the logistic equation
For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best
described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
x f(x)
1 409.4
2 260.7
3 170.4
4 110.6
6 44.7
7 32.4
8 19.5
9 12.7
10 8.1
x f(x)
0.15 36.21
0.25 28.88
0.5 24.39
0.75 18.28
1 16.5
1.5 12.99
2 9.91
2.25 8.57
2.75 7.23
3 5.99
3.5 4.81
Show Solution
x f(x)
2 22.6
4 44.2
5 62.1
7 96.9
8 113.4
10 133.4
11 137.6
15 148.4
17 149.3
1. •Source: Indiana University Center for Studies of Law in Action, 2007 ↵
2. •Source: Center for Disease Control and Prevention, 2013 ↵
3. •Source: The World Bank, 2013 ↵ | {"url":"https://ecampusontario.pressbooks.pub/sccmathtechmath1/chapter/fitting-exponential-models-to-data/","timestamp":"2024-11-03T00:23:00Z","content_type":"text/html","content_length":"306134","record_id":"<urn:uuid:f14b8474-f993-4ef2-bccd-b9c70122a759>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00798.warc.gz"} |
Slides from David Babbel's Fixed Indexed Annuity Study - Recent Historical Evidence | Annuity Digest
In the last slide (46), how is the performance of a 9 year fixed indexed annuity measured over a span of 14 years? Is the 9 year annualized return simply applied to the 14 year period?
No. The calculation is for only nine years, or through 10/2008, whichever comes first. The way to read the chart is to consider the horizontal axis. The data point above 1995 shows the average rate
of return over the ensuing 9 years, from 1/95 thru 12/2001. The data point above 1996 shows the average rate of return over the ensuing 9 years, from 1/96 thru 12/2002. etc.
Once the starting date gets to 1/2002, there are less than 9 years in our average. For example, the point above 2002 shows the average rate of return from 1/2002 thru 10/2008, almost 9 years. The
point above 2003 shows the average rate of return from 1/2003 thru 10/2008, almost 8 years. And so forth.
Thank you very much for your research and in depth education of the widely misunderstood and criticized by the brokerage industry investment - fixed annuity.
Professor David Babel just proved what I have experienced in reality with my clients since 1999.
Good luck with your work.
The data for 2009 will be interesting.
The study is loaded and misleading.
It compares annuity returns with investment strategies that only a numbskull would use.
Try this: A 65-year-old male living in Wisconsin who invested $100,000 in an annuity would receive $626 a month ($7,512 a year) for life. If he started in 1995 and lived to age 79 he would have
collected $105,168 on his annuity and when he died he would get nothing; the annuity company would get what was left of the principal.
Now let's see what would happen if this person instead placed his $100,000 in Vanguard's Wellesley Income fund. He would receive the same $7,512 a year as the annuitant -- but by age 79 he would have
$141,246 in Wellesley to leave to his heirs.
Investors are constantly told to avoid high-commission, high-fee financial instruments.
If an annuity company can pay a salesperson a 5% commission for selling one of these things and then extract ongoing fees of 1.5% to run it you know it's a bad deal that can be creamed by an
appropriate alternative investment.
What, exactly, is misleading and loaded about the study? You make no points in support of this comment.
You also do nothing to substantiate the annuity rate in your calculation. What type of annuity is this? Are these current rates or are they from 1995?
There are annuities that satisfy bequest motives. Were you using calculations for an immediate life annuity?
How are you calculating the Wellington return? The price of this fund on January 3, 1995 was $19.53 and the price the week or December 2008 was $23.83.
What marginal tax rate are you assuming for the Wellington fund interest/dividend yield and were those net return incorporated into your calculation?
How does the investor receive "the same $7,512 per year as the annuitant" when the funds are invested in the Wellington fund?
Are you saying that the total return on the Wellington fund is precisely the same as the return on the annuity rate that seems to have been picked out of thin air?
What guarantee does the Wellington fund provide? What if the person invested the week of September 17, 2007 when the fund price as around 35 and was forced to liquidate all or a portion of their
assets in 2008 (a loss in excess of 35%)?
I find it ironic that this type of completely unsubstantiated (and comic at times) criticism is in response to an academic study that so successfully exposed the unsubstantiated myths (equity indexed
annuities are poor investments)that exist in the financial industry and media.
I would say that the quality of the analysis in the original comments by Stephen are at the complete opposite end of the spectrum compared to the analysis by Babbel--night and day. | {"url":"https://annuitydigest.com/blog/tom/slides-david-babbels-fixed-indexed-annuity-study-recent-historical-evidence","timestamp":"2024-11-08T02:07:59Z","content_type":"application/xhtml+xml","content_length":"71946","record_id":"<urn:uuid:75250a18-d006-4138-99d0-63a356515ba0>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00288.warc.gz"} |
Puzzles and Riddles that Improve Math Ability
This is Day 11 of 31 Days of Math Learning Success. Follow all the days here and check out others that are writing for 31 days here.
“The solution often turns out more beautiful than the puzzle.”
~Richard Dawkins
“Did you always like math?” I was asked once.
It was a strange question, because I’ve never really liked math.
After pondering it, I realized that I’ve done well in math because I’ve always liked puzzles!
Puzzles & Riddles Instead of Math
Puzzles and riddles get your thinking juices flowing. And math is all about thinking.
To improve your math skills, start doing more puzzles and riddles.
Note: “puzzles” are not limited to jigsaw puzzles. In fact, I don’t even think about that kind when I use the word puzzle. Although I’m sure they have a benefit on thinking and processing too.
Here is a list of some of my favorite puzzles and riddles, along with some great places to find them:
Some puzzle/riddle collections:
You can even invent your own puzzle or create a maze!
Have fun!
Are you enjoying the 31 Days of Math Learning Success? Share it on Twitter, Facebook and Pinterest!
This post may contain affiliate links. When you use them, you support us so we can continue to provide free content!
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://mathfour.com/games/puzzles-and-riddles-that-improve-math-ability","timestamp":"2024-11-08T02:46:37Z","content_type":"text/html","content_length":"38733","record_id":"<urn:uuid:b1e6128c-ed2e-4ff2-a0c3-000a6d1dc0de>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00652.warc.gz"} |
Wavelet Time Scattering with GPU Acceleration
Wavelet Time Scattering with GPU Acceleration — Spoken Digit Recognition
This example shows how to accelerate the computation of wavelet scattering features using gpuArray (Parallel Computing Toolbox). You must have Parallel Computing Toolbox™ and a supported GPU. See GPU
Computing Requirements (Parallel Computing Toolbox) for details. This example uses an NVIDIA Titan V GPU with compute capability 7.0. The section of the example that computes the scattering transform
provides the option to use the GPU or CPU if you wish to compare GPU vs CPU performance.
This example reproduces the exclusively CPU version of the scattering transform found in Spoken Digit Recognition with Wavelet Scattering and Deep Learning.
Clone or download the Free Spoken Digit Dataset (FSDD), available at https://github.com/Jakobovski/free-spoken-digit-dataset. FSDD is an open data set, which means that it can grow over time. This
example uses the version committed on 08/20/2020 which consists of 3000 recordings of the English digits 0 through 9 obtained from six speakers. The data is sampled at 8000 Hz.
Use audioDatastore to manage data access and ensure random division of the recordings into training and test sets. Set the location property to the location of the FSDD recordings folder on your
computer. In this example, the data is stored in a folder under tempdir.
location = fullfile(tempdir,'free-spoken-digit-dataset','recordings');
ads = audioDatastore(location);
The helper function, helpergenLabels, defined at the end of this example, creates a categorical array of labels from the FSDD files. List the classes and the number of examples in each class.
ads.Labels = helpergenLabels(ads);
The FSDD dataset consists of 10 balanced classes with 300 recordings each. The recordings in the FSDD are not of equal duration. Read through the FSDD files and construct a histogram of the signal
LenSig = zeros(numel(ads.Files),1);
nr = 1;
while hasdata(ads)
digit = read(ads);
LenSig(nr) = numel(digit);
nr = nr+1;
grid on
xlabel('Signal Length (Samples)')
The histogram shows that the distribution of recording lengths is positively skewed. For classification, this example uses a common signal length of 8192 samples. The value 8192, a conservative
choice, ensures that truncating longer recordings does not affect (cut off) the speech content. If the signal is greater than 8192 samples, or 1.024 seconds, in length, the recording is truncated to
8192 samples. If the signal is less than 8192 samples in length, the signal is symmetrically prepended and appended with zeros out to a length of 8192 samples.
Wavelet Time Scattering
Create a wavelet time scattering network using an invariant scale of 0.22 seconds. Because the feature vectors will be created by averaging the scattering transform over all time samples, set the
OversamplingFactor to 2. Setting the value to 2 will result in a four-fold increase in the number of scattering coefficients for each path with respect to the critically downsampled value.
sn = waveletScattering('SignalLength',8192,'InvarianceScale',0.22,...
The settings of the scattering network results in 326 paths. You can verify this with the following code.
[~,npaths] = paths(sn);
Split the FSDD into training and test sets. Allocate 80% of the data to the training set and retain 20% for the test set. The training data is for training the classifier based on the scattering
transform. The test data is for assessing the model's ability to generalize to unseen data.
rng default;
ads = shuffle(ads);
[adsTrain,adsTest] = splitEachLabel(ads,0.8);
Form a 8192-by-2400 matrix where each column is a spoken-digit recording. The helper function helperReadSPData truncates or pads the data to length 8192 and normalizes each record by its maximum
value. The helper function casts the data to single precision.
Xtrain = [];
scatds_Train = transform(adsTrain,@(x)helperReadSPData(x));
while hasdata(scatds_Train)
smat = read(scatds_Train);
Xtrain = cat(2,Xtrain,smat);
Repeat the process for the held-out test set. The resulting matrix is 8192-by-600.
Xtest = [];
scatds_Test = transform(adsTest,@(x)helperReadSPData(x));
while hasdata(scatds_Test)
smat = read(scatds_Test);
Xtest = cat(2,Xtest,smat);
Apply the scattering transform to the training and test sets. Move both the training and test data sets to the GPU using gpuArray. The use of gpuArray with a CUDA-enabled NVIDIA GPU provides a
significant acceleration. With this scattering network, batch size, and GPU, the GPU implementation computes the scattering features approximately 15 times faster than the CPU version. If you do not
wish to use the GPU, set useGPU to false. You can also alternate the value of useGPU to compare GPU vs CPU performance.
useGPU = true;
if useGPU
Xtrain = gpuArray(Xtrain);
Strain = sn.featureMatrix(Xtrain);
Xtrain = gather(Xtrain);
Xtest = gpuArray(Xtest);
Stest = sn.featureMatrix(Xtest);
Xtest = gather(Xtest);
Strain = sn.featureMatrix(Xtrain);
Stest = sn.featureMatrix(Xtest);
Obtain the scattering features for the training and test sets.
TrainFeatures = Strain(2:end,:,:);
TrainFeatures = squeeze(mean(TrainFeatures,2))';
TestFeatures = Stest(2:end,:,:);
TestFeatures = squeeze(mean(TestFeatures,2))';
This example uses a support vector machine (SVM) classifier with a quadratic polynomial kernel. Fit the SVM model to the scattering features.
template = templateSVM(...
'KernelFunction', 'polynomial', ...
'PolynomialOrder', 2, ...
'KernelScale', 'auto', ...
'BoxConstraint', 1, ...
'Standardize', true);
classificationSVM = fitcecoc(...
TrainFeatures, ...
adsTrain.Labels, ...
'Learners', template, ...
'Coding', 'onevsone', ...
'ClassNames', categorical({'0'; '1'; '2'; '3'; '4'; '5'; '6'; '7'; '8'; '9'}));
Use k-fold cross-validation to predict the generalization accuracy of the model. Split the training set into five groups for cross-validation.
partitionedModel = crossval(classificationSVM, 'KFold', 5);
[validationPredictions, validationScores] = kfoldPredict(partitionedModel);
validationAccuracy = (1 - kfoldLoss(partitionedModel, 'LossFun', 'ClassifError'))*100
validationAccuracy = 97.2500
The estimated generalization accuracy is approximately 97%. Now use the SVM model to predict the held-out test set.
predLabels = predict(classificationSVM,TestFeatures);
testAccuracy = sum(predLabels==adsTest.Labels)/numel(predLabels)*100
The accuracy is also approximately 97% on the held-out test set.
Summarize the performance of the model on the test set with a confusion chart. Display the precision and recall for each class by using column and row summaries. The table at the bottom of the
confusion chart shows the precision values for each class. The table to the right of the confusion chart shows the recall values.
figure('Units','normalized','Position',[0.2 0.2 0.5 0.5]);
ccscat = confusionchart(adsTest.Labels,predLabels);
ccscat.Title = 'Wavelet Scattering Classification';
ccscat.ColumnSummary = 'column-normalized';
ccscat.RowSummary = 'row-normalized';
As a final example, read the first two records from the dataset, calculate the scattering features, and predict the spoken digit using the SVM trained with scattering features.
sig1 = helperReadSPData(read(ads));
scat1 = sn.featureMatrix(sig1);
scat1 = mean(scat1(2:end,:),2)';
plab1 = predict(classificationSVM,scat1);
Read the next record and predict the digit.
sig2 = helperReadSPData(read(ads));
scat2 = sn.featureMatrix(sig2);
scat2 = mean(scat2(2:end,:),2)';
plab2 = predict(classificationSVM,scat2);
t = 0:1/8000:(8192*1/8000)-1/8000;
plot(t,[sig1 sig2])
grid on
axis tight
title('Spoken Digit Prediction - GPU')
The following helper functions are used in this example.
helpergenLabels — generates labels based on the file names in the FSDD.
function Labels = helpergenLabels(ads)
% This function is only for use in Wavelet Toolbox examples. It may be
% changed or removed in a future release.
tmp = cell(numel(ads.Files),1);
expression = "[0-9]+_";
for nf = 1:numel(ads.Files)
idx = regexp(ads.Files{nf},expression);
tmp{nf} = ads.Files{nf}(idx);
Labels = categorical(tmp);
helperReadSPData — Ensures that each spoken-digit recording is 8192 samples long.
function x = helperReadSPData(x)
% This function is only for use Wavelet Toolbox examples. It may change or
% be removed in a future release.
N = numel(x);
if N > 8192
x = x(1:8192);
elseif N < 8192
pad = 8192-N;
prepad = floor(pad/2);
postpad = ceil(pad/2);
x = [zeros(prepad,1) ; x ; zeros(postpad,1)];
x = single(x./max(abs(x)));
See Also
Related Examples
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A wooden packaging box of dimension 30 cm by 50 cm by 85 cm has a mass of 70 ke. Determine the: (1) maximum pressure; (ii) minimum pressure it can exert on a horizontal surface. (5 marks)craft1 electrical June/July 2020 - Question & AnswersA wooden packaging box of dimension 30 cm by 50 cm by 85 cm has a mass of 70 ke. Determine the: (1) maximum pressure; (ii) minimum pressure it can exert on a horizontal surface. (5 marks)craft1 electrical June/July 2020
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A wooden packaging box of dimension 30 cm by 50 cm by 85 cm has a mass of 70 ke.
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Free Games
Just wondering what free games people have found on the internet that they would recommend to play.
We now bring you live footage from the World Championship Staring Final.
Ancient Domains of Mystery. Haven't beaten it in over 1000 attempts.
I was hoping to find some free RPGs, but it seems most of the freeware are RPGs with zero to little graphics.
I'm not a graphics whore by any means, but I'd atleast like them to resemble those found in the IE games.
"Console exclusive is such a harsh word." - Darque
"Console exclusive is two words Darque." - Nartwak (in response to Darque's observation)
Well, I have been playing System Shock 2 for the last hour or so and once I reached the actual game and got past the main medical room, I had 4 scares within 3 minutes. The first involved hearing a
loud gun shot, while no one was around me. The second was when someone just started talking and I looked around and saw a "ghost." The third and fourth both involved my first experience with hybrids.
I took the ghosts advice and went into the area he came from. Then I heard someone whispering, "Flesh...flesh...," which rather freaked me out. Again looking around, this deformed half-human thing
with a large tube of flesh between his head and shoulder came out of a room and jumped at me. Of course, I prompty beat him to death with my all-powerful wrench. Then, I left the area for a moment to
just recheck everything (I hadnt expected an attack so soon) when all the previous room had only dead bodies and maybe a box or two. I noticed a computer help poster on the wall I had missed before
and clicked on it. While reading it, I again heard strange voices, basically to the effect of, "I see you...Im coming to eat you..." Naturally enough, I was again freaked out. I tried to exit the
inventory screen quickly but by the time I did, the hybrid had reached me and was attacking. I tried to get my bearing but it was dark and I couldnt seem to find him. And so I died.
It is an impressive game that can amuse, entertain, scare and kill you within the first half-hour (the other half-hour was introduction and prologue).
And I find it kind of funny
I find it kind of sad
The dreams in which I'm dying
Are the best I've ever had
I was hoping to find some free RPGs, but it seems most of the freeware are RPGs with zero to little graphics.
I'm not a graphics whore by any means, but I'd atleast like them to resemble those found in the IE games.
Well, if nothing else games like Nethack, Moria (and its *band descendants), and ADoM are living pieces of RPG history, and you can enjoy them in the vein of RPG appreciation even if they aren't
strictly your bag.
I play Nethack until I'm sick of it every few years and I try to win a few games of ADoM every few months, get enraged when my characters die and swear it off for another stretch. They're just
distillations of good gameplay, in large part because they can't hide behind any other features.
I keep meaning to try this one, but I haven't had time/inclination...
Maybe someone else can, or already has? Someone who doesn't hate this time of game by default, that is...
this site is pretty interesting to wade through - it's where i got the above link from. There are A LOT of games there, from terrible graphics to not too bad graphics, with user ratings and such.
Some games listed are no longer free tho, and some are outdated (non-existant). But mostly, a fun browse.
“Things are as they are. Looking out into the universe at night, we make no comparisons between right and wrong stars, nor between well and badly arranged constellations.” – Alan Watts
If you don't mind a rather hokey interface, you can't get a much better storyline than Circuit's Edge. The author, George Alec Effinger went through some crazy stuff in his life, and as a consequence
this grittier-than-a-grit-and-glasspaper-smoothie story hits hard. It's also a good detective plot, although strictly linear in progression. Solutions to problems require you to immerse yourslef in
the world and the character. No obvious purple key to purple door rubbish.
"It wasn't lies. It was just... bull****"."
-Elwood Blues
tarna's dead; processing... complete. Disappointed by Universe. RIP Hades/Sand/etc. Here's hoping your next alt has a harp.
Is the actual full game because I've been looking for this game awhile now?
Yep, it's the full monty. I think the movies are missing though unfortunately, although this doesn't affect the start of the game too badly since the intro movies aren't too great and most of the
narrative is told through in-game radio messages and logs. I never quite finished SS2 so I haven't seen the end cinematic, anyone want to shed some light on the importance of the end movie?
Of course, if you're really desperate, you can always try the electronic mule.
We now bring you live footage from the World Championship Staring Final.
The movies aren't missing, they're packed using |33+ warez methods, IIRC.
Anything that has to do with being a soldier or commander on the front lines of battle, action is more my style
The movies aren't missing, they're packed using |33+ warez methods, IIRC.
So, is there a way to get them to play while you're playing the game or do I have ot watch them separately?
I heard Wasteland is a freebie. Anyone know if it's true and where to get it?
"Console exclusive is such a harsh word." - Darque
"Console exclusive is two words Darque." - Nartwak (in response to Darque's observation)
The movies aren't missing, they're packed using |33+ warez methods, IIRC.
So, is there a way to get them to play while you're playing the game or do I have ot watch them separately?
I just downloaded it; the movies are packed and require DivX5+ codex, which (v.6) is a free download from divx.com.
Is System Shock worth playing first? (System Shock 2 looks and sounds awesome!)
I heard Wasteland is a freebie. Anyone know if it's true and where to get it?
You can download Wasteland at Underdogs as well, they have a host of old RPGs. Check the System Shock 2 link.
Also Enemy Territory is an excellent team based FPS that plays like a tug of war with teams trying to achieve objectives while the other team stops them. It's only a couple of years old, so it's not
as antiquated as most other free games..
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Class 8
Maths Revision Worksheet
Ques 1. Find the greatest valu... | Filo
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Class 8 Maths Revision Worksheet Ques 1. Find the greatest values of and so that the even number is divisible by both 3 and 5 . Ques. Find the value of in each of the following:| Ques.3. By what
number should we multiply to get Ques. A) find the additive inverse of B) find the multiplicative inverse of Ques. Find the ratio of Ques6. Find the value of A and B in the following:
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Class 8 Maths Revision Worksheet Ques 1. Find the greatest values of and so that the even number is divisible by both 3 and 5 . Ques. Find the value of in each of the following:| Ques.3.
Question By what number should we multiply to get Ques. A) find the additive inverse of B) find the multiplicative inverse of Ques. Find the ratio of Ques6. Find the value of A and B in the
Text following:
Updated On Jul 17, 2023
Topic All topics
Subject Mathematics
Class Class 12
Answer Type Video solution: 1
Upvotes 111
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CPM Homework Help
On graph paper, plot $ABCD$ if $A(0,3),B(2,5),C(6,3)$, and $D(4,1)$.
a. Rotate $ABCD\ 90^\circ$ clockwise ($↻$) about the origin to form $A'B'C'D'$. Name the coordinates of $B'$.
b. Translate $A'B'C'D'$ up $8$ units and left $7$ units to form $A''B''C''D''$. Name the coordinates of $C''$.
c. After rotating $ABCD$ $180^\circ$ to form $A'''B'''C'''D'''$, Arah noticed that the image lined up exactly with the original. In this case, about what point was $ABCD$ rotated? How did you locate
Use the eTool below to solve each part.
Click the link at right for the full version of the eTool. Int2 3-97 HW eTool | {"url":"https://homework.cpm.org/category/CCI_CT/textbook/int2/chapter/3/lesson/3.2.3/problem/3-97","timestamp":"2024-11-03T16:30:27Z","content_type":"text/html","content_length":"39697","record_id":"<urn:uuid:9d967fd5-e2ea-4fd7-9edc-8b432428470e>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00067.warc.gz"} |
History of the Golden Ratio - Golden Ratio in Nature | Math Ratios
History of the Golden Ratio in Nature
The golden ratio, a mathematical constant approximately equal to 1.618, has been identified in various aspects of the natural world. From the arrangement of petals on a flower to the spiral of a
seashell, the golden ratio holds a significant place in natural phenomena.
The Golden Ratio and Fibonacci
The understanding of the golden ratio in nature is often attributed to Leonardo of Pisa, more commonly known as Fibonacci. In his book "Liber Abaci," he introduced a sequence of numbers, now known as
the Fibonacci sequence. As the sequence progresses, the ratio of successive terms approximates the golden ratio.
F(n+1) / F(n) ≈ Φ (for large n)
Observing the Golden Ratio in Nature
The golden ratio's appearance in nature remained largely anecdotal until the 19th and early 20th centuries when mathematicians and scientists began to formally document these instances.
In botany, the golden ratio is often seen in the arrangement of leaves, branches, and petals, a phenomenon known as phyllotaxis. The spirals seen in pinecones, pineapples, and sunflowers also exhibit
patterns that correlate with the Fibonacci sequence and the golden ratio.
The golden ratio has also been observed in the physical proportions of animals. For example, the body lengths of ants and various sections of the human finger have proportions that approximate the
golden ratio.
The Golden Ratio, Mathematics, and Science
Mathematicians and scientists have explored the golden ratio's presence in nature extensively. Biologists, physicists, chemists, and mathematicians alike have found the golden ratio a fascinating
phenomenon to study.
Famous mathematician Johannes Kepler, best known for his laws of planetary motion, described the golden ratio as a "precious jewel" due to its frequent appearance in geometry.
In the 20th century, physicist Roger Penrose further extended the golden ratio's reach into the realm of quantum physics. He used the golden ratio to create a tiling pattern, known as Penrose tiling,
that exhibits fivefold symmetry and aperiodicity - a pattern that never repeats.
The Golden Ratio and the Human Perception of Beauty
The golden ratio's appeal extends beyond science and mathematics. Many believe it plays a role in human perceptions of beauty and completeness, often referenced in the context of art, architecture,
and even facial proportions. However, this subject remains a topic of much debate among psychologists and mathematicians.
Controversies and Misconceptions
While the golden ratio's presence in nature and its link to aesthetics is intriguing, it's essential to approach these topics critically. Some claims about the ubiquity of the golden ratio have been
debunked or exaggerated. Despite this, the golden ratio continues to captivate researchers, artists, and mathematicians alike due to its unique properties and pervasive nature.
The golden ratio's influence spans across various fields, intertwining mathematics, nature, art, and science in its unique way. Its discovery and further investigation by mathematicians such as
Fibonacci and Kepler have shed light on its intriguing properties and potential connections to the natural world. Despite controversies and misconceptions, the golden ratio continues to be a
fascinating topic of study, bearing testimony to the beautiful interplay between mathematics and nature.
The Golden Ratio Tutorials
If you found this ratio information useful then you will likely enjoy the other ratio lessons and tutorials in this section: | {"url":"https://www.mathratios.com/tutorial/golden-ratio-in-math-nature.html","timestamp":"2024-11-08T17:09:25Z","content_type":"text/html","content_length":"10248","record_id":"<urn:uuid:2607763a-b91f-44ed-a448-3fe054448adf>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00487.warc.gz"} |
Elliptical Coordinates in Double Integrals - Application
Elliptical Coordinates (Double Integrals)
Krystian Karczyński
Founder and General Manager of eTrapez.
Graduate of Mathematics at Poznan University of Technology. Mathematics tutor with many years of experience. Creator of the first eTrapez Courses, which have gained immense popularity among students
He lives in Szczecin, Poland. He enjoys walks in the woods, beaches and kayaking.
There are times in life when the region of integration in a double integral is an ellipse….
What do we do then?
Elliptical Coordinates
A neat method of solving this is usually to use the so-called elliptical coordinates. It’s something like polar coordinates, the mechanism works in a similar way, only you substitute different things
for x and y, and the Jacobian is different. The interpretation of ‘r’ is also different. So, to sum up, if you know how to switch to polar coordinates (which is usually done when the region of
integration is a circle), you’ll easily get the hang of elliptical coordinates too.
So we have the integral: and the region of integration bounded by an ellipse centered at the origin, with the equation: . Let’s make sure the right side of the ellipse equation is 1, alright? If, for
example, it’s 9, you can easily make it 1 by dividing both sides of the equation by 9.
The region of integration drawn looks like this:
What a and b mean is clear from the drawing. Be careful, because if the denominator in the ellipse equation under is, for example, 9, it means that , for obvious reasons, right?
Now with such a “clean” situation, we move to elliptical coordinates, substituting:
Meaning of Variables in Elliptical Coordinates
The angle means exactly the same as in polar coordinates, and means something different. In basic problems with an ellipse given by a neat equation , simply assume that ranges from zero to one (in
more complex cases, substitute and into the ellipse equation and calculate the upper limit of r).
The Jacobian in elliptical coordinates is equal to .
Remembering the Jacobian, we switch to the integral in elliptical coordinates:
where the variables and are bounded: in the range from zero to one, and depending on whether we are talking about the whole ellipse, half of it, or, for example, a quarter – just like in polar
Just take it and calculate.
Calculate the integral , where D is the ellipse with the equation: .
Following the above scheme, we substitute:
We take the region of integration:
And calculate the integral:
Which is, of course, just a formality by now 🙂
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algebra book Archives - Math Research of Victor Porton
A new book for mathematicians and programmers published: Axiomatic Theory of Formulas or Algebraic Theory of Formulas. The book is an undergraduate level but contains a new theory. Get it: PAPERBACK
E-BOOK From the preface: This new mathematical theory developed by the book author researches the properties of mathematical formulas (aka expressions). Naturally this theory […] | {"url":"https://math.portonvictor.org/tag/algebra-book/","timestamp":"2024-11-08T07:49:46Z","content_type":"text/html","content_length":"92324","record_id":"<urn:uuid:b9032ab8-d83b-4c97-9401-c33715527353>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00488.warc.gz"} |
Celsius to Fahrenheit - Formula, Examples | C to F Conversion - [[company name]] [[target location]], [[stateabr]]
Celsius to Fahrenheit Formula, Examples | C to F Conversion
Whenever you watch the weather channel or any other news programs that discusses the weather, they will also speak about moisture level, precipitation, and temperature. The metric system is utilized
by majority of the world to assess these in numericals.
Still, we use the rarely adopted imperial system in the United States. In the imperial system, the temperature is assessed in degrees Fahrenheit (°F). In comparison, in the metric system, it is
calculated in degrees Celsius (°C).
You may decide on a voyage to a country that utilizes the metric system, or you might need to convert Celsius to Fahrenheit for a academic assignment. Regardless of the reason, learning how to make
this conversion is essential.
This blog takes into account how to convert Celsius (also called centigrade) to Fahrenheit without turning to a temperature conversion diagram. And we promise, this will be not only for math class
but additionally for realistic scenarios.
What Is the Conversion Formula for Celsius to Fahrenheit?
The Celsius and Fahrenheit levels are employed for assessing temperature. The elementary difference between the two temperature scales is that the scientists who created them chose distinct starting
Celsius utilizes water’s boiling and freezing points as citation points, and Fahrenheit uses a salt water mixture's freezing and boiling points.
In other words, 0 °C is the temperature at which water freezes, while 100 °C is the temperature at which it boils. On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F.
Celsius to Fahrenheit transformation is made practicable with a easy formula.
If we know temperature data in Celsius, we can convert it to Fahrenheit by using the following formula:
Fahrenheit = (Celsius * 9/5) + 32
Let’s test this equation by converting the boiling point of water from Celsius to Fahrenheit. The boiling point of water is 100 °C so we could plug this unit into our equation like so:
Fahrenheit = (100 °C * 9/5) + 32
When we work out this equation for Fahrenheit, we find the result is 212 °F, which is what we anticipated.
At the moment let’s attempt to converting a temp which isn’t moderately simple to retain, like 50 °C. We place this temperature into our equation and find the result same as we did before:
Fahrenheit = (50 °C * 9/5) + 32
Fahrenheit = 90 + 32
When we solve the formula, we find the result is 122 °F.
We could also employ this formula to change from Fahrenheit to Celsius. We only have to solve the equation for Celsius. This will give us the following formula:
Celsius = (Fahrenheit – 32) * 5/9
So, let’s try transforming the freezing point of water from Fahrenheit to Celsius. Bear in mind, the freezing point of water is 32 °F thus we can plug this unit into our formula:
Celsius = (32 °F – 32) * 5/9
Celsius = 0 * 5/9
When we work out this formula, the result will be 0 degrees Celsius is equal to 32 degrees Fahrenheit, precisely as we predicted.
How to Convert from Celsius to Fahrenheit
At the moment that we have this details, let’s get to figuring it out and practice some conversions. Easily follow these steps!
Steps to Convert from Celsius to Fahrenheit
1. Gather the temp in Celsius which you want to convert.
2. Put the Celsius value into the formula.
Fahrenheit = (Celsius * 9/5) + 32
3. Work out the formula.
4. The answer will be the temperature in Fahrenheit!
These are the quintessential steps. However, if you desire to check your work, it is handy to bear in mind the freezing and boiling points of water to do a sanity check. This step is elective but
double-checking your result is always a good thinking.
Example 1
Let’s place this knowledge into use by following the steps last explained with an example.
Exercise: Convert 23 C to F
1. The temperature we need to convert is 23 degrees Celsius.
2. We place this unit into the equation, giving us:
Fahrenheit = (23 * 9/5) + 32
Fahrenheit = 41.4 +32
3. When we figure out the equation, the solution is 73.4 degrees Fahrenheit.
Good! Now we are aware that 23 degrees in the Celsius scale results in a calm day in late spring.
Example 2
Now let’s try one more example: converting a temp which isn’t as straightforward to remember.
Exercise: Convert 37 C to F
1. The temp we want to convert is 37 degrees Celsius.
2. We put this number into the equation, providing us:
Fahrenheit = (37 * 9/5) + 32
3. When we work out the equation, we discover that the answer is 98.6 degrees in Fahrenheit temp. This refers to the normal body temperature!
That’s all you need to know! These are the quick and simple steps to convert temps from Celsius to Fahrenheit. Just keep in mind\bear in mind the equation and plug in the numbers accordingly.
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Main Ideas
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• When two quantities vary at a constant rate with respect to each other, the graph of their relationship forms a straight line.
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4-3 Relations Objectives Students will be able to: 1) Represent relations as sets of ordered pairs, tables, mappings, and graphs 2) Find the inverse of. - ppt download
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[Solved] 1. A population of wild hogs have heights | SolutionInn
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1. A population of wild hogs have heights that are normally distributed with a mean of 27 with a standard deviation of 8.5. A recent
1. A population of wild hogs have heights that are normally distributed with a mean of 27" with a standard deviation of 8.5". A recent Fish and Game trapping program captured 42 of these hogs and
found a sample mean height of 24" and a sample standard deviation of 4.7". Does this data give good evidence that the population of hogs have heights less than 27". Use an alpha of 5%.
a. What would the hypothesis look like for this situation?(A)
b. What is the test statistic and p value for this test? (C)
Test Stat =_________ P-Value =___________
2. Based on a P-Value of 3% and a alpha of 1% which decision would you choose? Circle or highlight the correct choice. (B)
Reject the null in favor of the alternative.
Fail to reject the null in favor of the alternative.
3. You are testing the claim that the mean weight for a production line of parts is less than 34g. You reject the null. What does this outcome mean in terms of the problem?(D)
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Paper 1, Section II, E
Let $\left(X_{n}\right)_{n \geqslant 0}$ be a Markov chain.
(a) What does it mean to say that a state $i$ is positive recurrent? How is this property related to the equilibrium probability $\pi_{i}$ ? You do not need to give a full proof, but you should
carefully state any theorems you use.
(b) What is a communicating class? Prove that if states $i$ and $j$ are in the same communicating class and $i$ is positive recurrent then $j$ is positive recurrent also.
A frog is in a pond with an infinite number of lily pads, numbered $1,2,3, \ldots$ She hops from pad to pad in the following manner: if she happens to be on pad $i$ at a given time, she hops to one
of pads $(1,2, \ldots, i, i+1)$ with equal probability.
(c) Find the equilibrium distribution of the corresponding Markov chain.
(d) Now suppose the frog starts on pad $k$ and stops when she returns to it. Show that the expected number of times the frog hops is $e(k-1)$ ! where $e=2.718 \ldots$ What is the expected number of
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7e Electricity V - SPM Physics
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1. Question 1 of 10
1 point(s)
When a 5Ω resistor is connected to the terminal of a cell, the current in the resistor is 5A. When the resistor is replaced by another resistor with resistance 30Ω, the current becomes 1A. Find
the internal resistance of the cell
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2. Question 2 of 10
1 point(s)
An experiment was carried out to determine the e.m.f. and the internal resistance of a dry cell. The result was plotted on a graph as shown in the diagram below. Find the internal resistance of
the cell.
[Back] [Check] [Next]
3. Question 3 of 10
1 point(s)
An experiment was carried out to determine the e.m.f. and the internal resistance of a dry cell. The result was plotted on a graph as shown in the diagram below. What is the e.m.f of the cell?
[Back] [Check] [Next]
4. Question 4 of 10
1 point(s)
A current of 0.50A flows through a 100 Ω resistor. Find the power dissipated from the resistor?
[Back] [Check] [Next]
5. Question 5 of 10
1 point(s)
A bulb rated 240 V, 18 W is operated from a 240V power source. Find the current flowing through the bulb.
[Back] [Check] [Next]
6. Question 6 of 10
1 point(s)
3 light bulbs of resistance 1Ω, 2Ω and 3Ω are connected in series to a power source of 240V. Find the power of the light bulb of resistance 2Ω.
[Back] [Check] [Next]
7. Question 7 of 10
1 point(s)
The figure above shows that an ideal battery is connected in parallel to two resistors with resistance 2 Ω and 4 Ω. Find the power dissipated in the 2 Ω resistor.
[Back] [Check] [Next]
8. Question 8 of 10
1 point(s)
A 2000W heater is used to heat 500 cm3 of water. Find the minimum time needed to heat the water from 25 to 100 °C. (Specific Heat Capacity of water = 4200Jkg-1oC-1)
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9. Question 9 of 10
1 point(s)
An electric heater of power 3600W is used to heat water for 15 minutes. If the cost of 1 unit (kilowatt-hour) energy consumption is 24 cent, what is the cost of this heating process?
[Back] [Check] [Next]
10. Question 10 of 10
1 point(s)
An electric motor raises a mass of 20kg to a height of 5m in 10s. Given that the electric motor is connected to a power source of 80V and current 2A. Find the efficiency of the motor.
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Perturbative renormalization of lattice N = 4 super Yang-Mills theory
We consider N = 4 super Yang-Mills theory on a four-dimensional lattice. The lattice formulation under consideration retains one exact supersymmetry at non-zero lattice spacing. We show that this
feature combined with gauge invariance and the large point group symmetry of the lattice theory ensures that the only counterterms that appear at any order in perturbation theory correspond to
renormalizations of existing terms in the bare lattice action. In particular we find that no mass terms are generated at any finite order of perturbation theory. We calculate these renormalizations
by examining the fermion and auxiliary boson self energies at one loop and find that they all exhibit a common logarithmic divergence which can be absorbed by a single wavefunction renormalization.
This finding implies that at one loop only a fine tuning of the finite parts is required to regain full supersymmetry in the continuum limit.
• Extended supersymmetry
• Lattice gauge field theories
ASJC Scopus subject areas
• Nuclear and High Energy Physics
Dive into the research topics of 'Perturbative renormalization of lattice N = 4 super Yang-Mills theory'. Together they form a unique fingerprint. | {"url":"https://experts.syr.edu/en/publications/perturbative-renormalization-of-lattice-n-4-super-yang-mills-theo","timestamp":"2024-11-06T11:10:33Z","content_type":"text/html","content_length":"49726","record_id":"<urn:uuid:53ddcdd8-798c-4f3d-b76a-6d70737bb3f3>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00312.warc.gz"} |
Cryptography: Attacking OCB2 – react0r blog
Recently I read the Trail of Bits Crypto 2019 Takeaways blog post. After studying the reference literature, I wanted to implement the attack described in Iwata's Plaintext Recovery Attack of OCB2*
[0]. This blog goes over and discusses the implementation.
From the post:
>The Best Paper award went to the Cryptanalysis of OCB2: Attacks on Authenticity and Confidentiality. This amalgamation of three papers released last fall demonstrates attacks against OCB2, an
ISO-standard authenticated encryption scheme, for forging signatures on arbitrary messages and full plaintext recovery. This result is especially shocking because all three versions of OCB have been
standardized, and were thought to have airtight security proofs. Fortunately, OCB1 and OCB3 are not vulnerable to this attack because it relies on details specific to how OCB2 applies the XEX* mode
of operation.
Likewise, the following passage from the winning paper [1] motivated me:
>For our most relevant attacks we have C code that breaks the OCB2 reference implementation [...]
A block cipher encrypts plaintext in fixed-size \(n\)-bit blocks. For example, \(n = 128\) using AES-128. For plaintext exceeding \(n\) bits, the simplest approach is to partition the plaintext into
\(n\)-bit blocks and encrypt them one by one [2]. This method is called the electronic code book (ECB) block cipher mode of operation. In contrast, offset code book 2 (OCB2) is another mode of
operation. Precisely, OCB2 [3] is a symmetric-key nonce-based authenticated encryption with associated data (AEAD) block cipher mode. OCB2 appeared at ASIACRYPT 2004 and is standardized(\(\dagger\))
as ISO/IEC 19772:2009.
Algorithms of OCB2
Given a plaintext message (\(M\)), associated data (\(A\)), nonce (\(N\)), and key (\(K\)), the OCB2 encryption function returns a ciphertext (\(C\)) and authentication tag (\(T\)). Vice versa, given
a ciphertext, associated data, nonce, and key, the OCB2 decryption function produces plaintext. If the authentication tag is not valid for the ciphertext, associated data, nonce, and key, then the
decryption function indicates failure \((\bot)\).
The specifications of the OCB2 encryption and decryption algorithms above are derived from [3]. Note that the details of the parellelizable message authentication code (PMAC) and length functions are
not relevant to the attacks.
Encryption Decryption
Algorithm \(OCB2.E_{K}(N, A, M)\) Algorithm \(OCB2.D_{K}(N, A, C, T)\)
1. \(L \gets E(N)\) 1. \(L \gets E(N)\)
2. For \(i = 1\) to \(m − 1\): 2. For \(i =1\) to \(m - 1\):
3. \(\hspace{0.42cm} C[i] \gets 2^{i}L \oplus E(2^{i}L \oplus M[i])\) 3. \(\hspace{0.42cm} M[i] \gets 2^i L \oplus E^{-1}(2^iL \oplus C[i])\)
4. \( Pad \gets E(2^{m}L \oplus len(M[m]))\) 4. \( Pad \gets E(2^{m} L \oplus len(C[m]))\)
5. \( C[m] \gets M[m] \oplus msb_{|M[m]|}(Pad)\) 5. \( M[m] \gets C[m] \oplus msb_{|C[m]|}(Pad)\)
6. \( \Sigma \gets C[m] || 0^∗ \oplus Pad\) 6. \( \Sigma \gets C[m] || 0^* \oplus Pad\)
7. \( \Sigma \gets M[1] \oplus ... \oplus M[m − 1] \oplus\Sigma\) 7. \( \Sigma \gets M[1] \oplus ... \oplus M[m-1] \oplus \Sigma\)
8. \( T \gets E(2^{m}3L \oplus \Sigma)\) 8. \( T^* \gets E(2^{m}3L \oplus \Sigma)\)
9. If \(A \ne \epsilon\) then \(T \gets T \oplus PMAC_{E}(A) \hspace{0.75cm}\) 9. If \(A \ne \epsilon\) then \(T^* \gets T^* \oplus PMAC_{E}(A)\)
10. \(T \gets msb_{\tau}(T)\) 10. \(T^* \gets msb_{\tau}(T^{*})\)
11. Return \((C,T)\) 11. If \(T = T^{*}\) return \(M\)
\(\hspace{0.42cm}\) 12. Else return \(\bot\) \(\hspace{0.42cm}\) [3]
The goal of OCB2 is that (\(C\)) protects the confidentiality and authenticity of (\(M\)), as well as the authenticity of (\(A\)). Confidentiality can be defined as indistinguishability from random
bits. This means an attacker cannot distinguish OCB2 output from an equal amount of random bits. In addition, the authenticity of the cipher refers to the authenticity of ciphertexts [4]. In other
words, an attacker is unable to produce valid nonce-ciphertext pairs not already obtained.
OCB2 and AES-128
To build the OCB2 and AES-128 reference implementations:
1. Obtain their source code.
$ wget https://web.cs.ucdavis.edu/~rogaway/ocb/ocb-ref/rijndael-alg-fst.{c,h}
$ wget https://web.cs.ucdavis.edu/~rogaway/ocb/ocb.{c,h}
2. Compile.
$ gcc ocb.c rijndael-alg-fst.c -o ocb
3. Execute. The following is abbreviated test vector output using the reference OCB2 encryption and decryption functions. Note that in this case the additional data is denoted by (H).
$ ./ocb
H : 000102030405060708090A0B0C0D0E0F101112131415161718191A1B1C1D1E1F2021222324252627
M : 000102030405060708090A0B0C0D0E0F101112131415161718191A1B1C1D1E1F2021222324252627
C : F75D6BC8B4DC8D66B836A2B08B32A6369F1CD3C5228D79FD6C267F5F6AA7B231C7DFB9D59951AE9C
T : 65A92715A028ACD4AE6AFF4BFAA0D396
Breaking OCB2
On October 2018, Inoue and Minematsu published a fast attack [1] that broke the authenticity of OCB2. In [2], Poettering expanded on the observations and broke the notion of confidentiality using a
theoretical approach. By November 2018, in [3], Iwata undermined OCB2 confidentiality by showing a practical plaintext recovery attack. As stated above, [4] is the amalgamation of the papers.
Attacks on Authenticity
In [1, § 4.1] Inoue and Minematsu described four attacks on authenticity. Using an adversarial game between attacker and challenger, in [2, § 3.1 (Fig. 3)] Poettering formalized their minimal attack
in relation to an adversary (\(\mathcal{A})\). Summarized below, the game assumes access to encryption (\(\mathcal{E})\) and decryption (\(\mathcal{D)}\) oracles.
Game Adversary \(\mathcal{A}^{\mathcal{E}(...),\mathcal{D}(...)}\)
1. Pick any \(N\) and \(M[2] \in \{0,1\}^{n}\)
2. Let \(M \gets (len(0^n) || M[2]) \)
3. Query \(C \gets \mathcal{E}(N,\epsilon,M)\)
4. Let \(C' \gets (C[1] \oplus len(0^n) || M[2] \oplus C[2])\)
5. Query \(M' \gets \mathcal{D}(N,\epsilon,C')\)
The game above constructs an example of an unauthentic, yet valid ciphertext \(C'\). Note that \(C' \ne C \) because they have different lengths. See steps (2) and (5) above. Thus, a successful
forgery is implied if the ciphertext is accepted by the decryption oracle. In fact, Inoue and Minematsu show in [5, § 4.1] that \(T'\) is always accepted as valid. In the context of the adversarial
game, \(Adv^{int}(\mathcal{A}) = 1\). In other words, the adversary breaks authenticity with probability \(1\) and a nonce-ciphertext pair that the adversary did not possess is obtained.
The following is output from Inoue and Minematsu's implementation found within the appendices of [1].
Test for minimal attack
Encryption query:
Nonce: 000102030405060708090A0B0C0D0E0F
Plaintext: 0000000000000000000000000000008000000000000000000000000000000000
Ciphertext: 713F06475FB9F34089FFC8DAAE1C370CCD6E6EB9ED7E4B79046246B2923F8B35
Tag: 89B544765C609A07D81D62148CD21E4D
Decryption query (forgery):
Forged Nonce (the same as encryption): 000102030405060708090A0B0C0D0E0F
Forged AD (empty):
Forged Ciphertext: 713F06475FB9F34089FFC8DAAE1C378C
Forged Tag: CD6E6EB9ED7E4B79046246B2923F8B35
Tags match: 1.
Forged Plaintext: 9FDFC45BC1A1458F12F034265483C57C
Here, the decryption oracle also reveals the block \(L=E(N)\) and the forged plaintext \(M' = 2L \oplus len(0^n)\). OCB2 treats blocks in the domain \(\mathcal{M} = \{0,1\}^n\) of its block cipher as
a finite field. Each block is a polynomial over the field \(GF(2^{128})\). To derive \(L\), note that \(E(N) = L = (M' \oplus len(0^n))/2\). To divide \((M' \oplus len(0^n))\) by \(2\) multiply it by
the inverse of the monomial \(2X \in GF(2^{128})\).
The pair (\(N\),\(L\)) is an internal block cipher input-output mapping. At least three additional input-output pairs can be gathered. Note that these mappings don't leak under normal circumstances.
The knowledge of \(L\) is leveraged to undermine OCB2 confidentiality.
Attacks on Confidentiality
In [6], Poettering extends the results of Inoue and Minematsu, and shows that OCB2 also admits a distinguishing attack. Thus, in the formal indistinguishability chosen ciphertext attack (IND-CCA)
settings, OCB2 does not achieve the confidentiality or privacy notion. In the first version of [6, § 4], Poettering notes that it remains an open question if the approach can also be used to attack
OCB2 implementations in real-world environments.
Iwata Plaintext Recovery Attack of OCB2
Iwata showed in [0] that confidentiality is also broken in a practical sense and arbitrary ciphertexts can be decrypted. The crux of the attack is the block-swapping technique introduced by Inoue and
Minematsu in [5, § 4.3]. Originally, in the context of an advanced forgery attack, Poettering notes that the technique was first used for plaintext recovery by Iwata. Both Iwata and Poettering seem
to have independently discovered plaintext recovery attacks of OCB2. The difference between their technique is the way the adversary learns \(L\).
In the following pseudo code, the attacker's goal is to recover \(M^*\). Note that it also follows the same security model as [0], assumes at least \(m=3\) blocks of plaintext, and access to
encryption and decryption oracles. Finally, let \((C^{*},T^{*})\) be the encryption of \((N^*,A^*,M^*)\).
1. To recover \(L^* = E_K(N^*)\) first perform Inoue-Minematsu Minimal Attack as per [5, § 4.1] and description above:
2. Fix any \(N \ne N^*\) and \(M[2] \in \{0,1\}^{n}\) and let \(M \gets (len(0^n) || M[2]) \)
3. Make an encryption query \(C \gets \mathcal{E}(N,\epsilon,M)\)
4. Let \(C'\gets (C[1] \oplus len(0^n) || M[2] \oplus C[2])\)
5. Make a decryption query \(M' \gets \mathcal{D}(N,\epsilon,C')\) and obtain \(L = (M' \oplus len(0^n))/2 \)
6. For \(i \in \{1,...,m-1\}\) define \((X[i],Y[i]) \in \{\{0,1\}^n\}^2\) such that \(E(X[i]) = Y[i]\) as follows:
\(\hspace{0.42cm}(X[i],Y[i]) \gets (M[i] \oplus 2^{i}L ),C[i] \oplus 2^{i}L )\)
7. Let \((X[m],Y[m]) \gets (len(M[m]) \oplus 2^{m}L, Pad)\)
8. Let \( (N'', L'') \) be one of \(m\) derived internal input-output pairs \((X[i],Y[i])\)
9. Fix any \(A''\) and \(M''[2] \in \{0,1\}^n\) and let \(M'' = (N^* \oplus 2L''||M''[2])\)
10. Make an encryption query \(C'' \gets \mathcal{E}(N'',A'',M'')\)
11. Let \(L^*\) be \(C''[1] \oplus 2L''\).
12. Modify \(C^*\) as per [5, § 4.3]. Fix indices \(j,k \in \{1,...,m^*-1\}\) such that \(C^*[j] \ne C^*[k]\).
13. Define \(C^$ = (C^$[1],...,C^$[m^*]) \) as follows:
\(\hspace{0.42cm} C^$[i] \gets C^*[i]\) for \(i \in \{1,...,m^*\} \setminus \{j,k\} \)
\(\hspace{0.42cm} C^$[j] \gets C^*[k] \oplus 2^kL^* \oplus 2^jL^*\)
\(\hspace{0.42cm} C^$[k] \gets C^*[j] \oplus 2^kL^* \oplus 2^jL^*\)
14. Make a decryption query \(M^$ \gets \mathcal{D}(N^*,A^*,C^$,T^*)\)
15. Swap modified blocks \(k\) and \(j\) of \(M^$\) to obtain goal \( M^*\) as follows:
\(\hspace{0.42cm} M^*[i] \gets M^$[i] \) for \(i \in \{1,...,m^*\} \setminus \{j,k\}\)
\(\hspace{0.42cm} M^*[j] \gets M^$[k] \oplus 2^kL^* \oplus 2^jL^*\)
\(\hspace{0.42cm} M^*[k] \gets M^$[j] \oplus 2^kL^* \oplus 2^jL^*\)
The implementation was based on the code supplied in [5] and can be found on Github. While it's straightforward, the reader is encouraged to review it. One function worth noting multiplies a block by
the inverse of the monomial \(2X \in GF(2^{128})\). To implement this, we shift the block's representative coefficient vector of bits to the right by \(1\). Polynomials are represented using the
block type. Each block is \(16\) bytes or \(128\) bits. The function is presented here for convenience.
void GF2128DivideByTwo(block A, block B) { // A = B*2^{-1}
uint8_t bottom_right = B[15] & 1; /* low bit of A */
A[0] = (B[0] >> 1) ^ (bottom_right * 0x80);
uint8_t i;
for (i = 0; i < 15; i++) {
A[i+1] = (B[i+1] >> 1) | (B[i] << 7);
if(bottom_right) {
A[15] ^= 0x43;
Instructions to apply the attack implementation are:
1. Obtain an implementation of the pseudocode above.
$ git clone https://gist.github.com/HAITI/50fb494e449531e51c172aa8e24a9cf3
2. Obtain the reference implementations and apply deltas to OCB2 exposing some internal functions.
$ cd 50fb494e449531e51c172aa8e24a9cf3/
$ wget https://web.cs.ucdavis.edu/~rogaway/ocb/ocb-ref/rijndael-alg-fst.{c,h}
$ wget https://web.cs.ucdavis.edu/~rogaway/ocb/ocb.{c,h}
$ patch < ocb.c.patch
$ patch < ocb.h.patch
3. Compile.
$ gcc -Wall ocb.c rijndael-alg-fst.c OCB2Attack.c -o OCB2Attack
4. Execute Iwata's attack using "SIXTEENBYTESHERE" as the key (K).
$ ./OCB2Attack 1 "SIXTEENBYTESHERE"
N*: F92B7D06100AD46203B990314628A28F
M*: 414141414141414141414141414141414242424242424242424242424242424243434343434343434343434343434343
Encryption Oracle
N*: F92B7D06100AD46203B990314628A28F
C*: 4C410A19DF0A0FCB5F1106946B80CEB3CEE86E9146ADC69282CA5B34B255C4317EF27F2CF5E4EF702AE740D51DC8D31F
T*: 553CA97C7AA0558CEDE15122C910D87B
Iwata Plaintext Recovery Attack
Inoue Minematsu Minimal Attack
N: 23EE09B7C678B5CE01A2F2315AB97C41
M: 0000000000000000000000000000008000000000000000000000000000000000
Encryption Oracle
C: B34D61A888966251FB16A772764F5F81D38F1949785B20465917DC23E0EA191B
T: B714E61ED1288D2D106A0A9D17E2A9DF
N': 23EE09B7C678B5CE01A2F2315AB97C41
C': B34D61A888966251FB16A772764F5F01
T': D38F1949785B20465917DC23E0EA191B
Decryption Oracle
M': B4AC4F0A0CE15B0526274F8596B1DA9F
C': B34D61A888966251FB16A772764F5F81D38F1949785B20465917DC23E0EA191B
Recover L
N'': 69589E1419C2B60A4C4E9F0B2D63B439
L'': D38F1949785B20465917DC23E0EA191B
M'': 5E354F94E0BC94EEB196287687FC903E00000000000000000000000000000000
Encryption Oracle
Recover L*
C'': DE0CB9261F981E130FFA9E02DB545F3D0B35E2EBF12CC320380DD81122038F27
L*: 79128BB4EF2E5E9FBDD526451A806D8C
C*[1]: 4C410A19DF0A0FCB5F1106946B80CEB3
C*[2]: CEE86E9146ADC69282CA5B34B255C431
C*[3]: 7EF27F2CF5E4EF702AE740D51DC8D31F
C$[i]: 7EF27F2CF5E4EF702AE740D51DC8D31F
C$[j]: D887572B244801D30E348EAAED54A99E
C$[k]: 5A2E33A3BDEFC88AD3EFD30A3481A31C
C$: D887572B244801D30E348EAAED54A99E5A2E33A3BDEFC88AD3EFD30A3481A31C7EF27F2CF5E4EF702AE740D51DC8D31F
Decryption Oracle
M$: 542D7BF820A78503CEBC97DC1D432FED572E78FB23A48600CDBF94DF1E402CEE43434343434343434343434343434343
! M*[1]: 41414141414141414141414141414141
! M*[2]: 42424242424242424242424242424242
! M*[3]: 43434343434343434343434343434343
[0] 2018 Iwata Plaintext Recovery Attack of OCB2*
[1] 2019 Inoue et. al. Cryptanalysis of OCB2: Attacks on Authenticity and Confidentiality
[2] 2001 Menezes et. al. Handbook of Applied Cryptography
[3] 2009 Rogaway Efficient Instantiations of Tweakable Blockciphers and Refinements to Modes OCB and PMAC
[4] 2014 Krovetz-Rogaway RFC 7253: The OCB Authenticated-Encryption Algorithm
[5] 2018 Inoue-Minematsu Cryptanalysis of OCB2
[6] 2018 Poettering Breaking the confidentiality of OCB2 | {"url":"https://blog.react0r.com/2019/12/06/cryptography-attacking-ocb2/","timestamp":"2024-11-05T06:47:54Z","content_type":"text/html","content_length":"63500","record_id":"<urn:uuid:8dc56e16-7ccb-46a7-958e-5904128bd0ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00036.warc.gz"} |
Skin weight optimization (matrix form) (3) - Rodolphe Vaillant's homepage
Matrix form second attempt
\nabla \mathbf s_i \sum\limits_{j \in \mathcal N(i)} (c_{ij} \mathbf S_i) \vec w_i - \nabla \mathbf s_i \sum\limits_{j \in \mathcal N(i)} (c_{ij} \mathbf S_j) \vec w_j = \nabla \mathbf s_i \vec b_i
It would be interesting to express the above in terms of the Laplacian L and skinning matrix S.
No comments | {"url":"http://mobile.rodolphe-vaillant.fr/entry/96/weight-optimization-3/jumble","timestamp":"2024-11-10T22:50:27Z","content_type":"application/xhtml+xml","content_length":"12849","record_id":"<urn:uuid:54ff508d-77f1-4dd9-bd9b-32747d11fc94>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00310.warc.gz"} |
A trapezoid is a quadrilateral with exactly 1 pair of parallel sides.
In the Polypad canvas below, create different trapezoids to explore area of trapezoids. Here is an example to get you started:
Are you able to use your examples to create a formula for the area of any trapezoid?
You may have noticed the area seems to be equal to the height multiplied by half of the sum of parallel sides a and b. Indeed, the general approach for the area of any trapezoid is
$A = 1/2(A+B)*H$.
Try and use some of the visuals below to write up a proof for this formula:
Further Reading | {"url":"https://polypad.amplify.com/lesson/trapezium-area","timestamp":"2024-11-04T18:39:54Z","content_type":"text/html","content_length":"22388","record_id":"<urn:uuid:f35bde44-b8f8-4060-ba3a-8207ec660501>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00355.warc.gz"} |
A.4 – Transportation and Accessibility | The Geography of Transport Systems
A.4 – Transportation and Accessibility
Author: Dr. Jean-Paul Rodrigue
Accessibility is a key element in transport geography and geography in general since it is a direct expression of mobility either in terms of people, freight, or information.
1. Defining Accessibility
Mobility is a choice made by users and is, therefore, a way to evaluate the impacts of infrastructure investment and related transport policies on regional development. Well-developed and efficient
transportation systems offer high accessibility levels, while less-developed ones have lower levels of accessibility. Thus, accessibility is linked with an array of economic and social opportunities,
but congestion can also have a negative impact on mobility.
Accessibility is the measure of the capacity of a location to be reached from, or to be reached by, different locations. Therefore, the capacity and the arrangement of transport infrastructure
are key elements in the determination of accessibility.
All locations are not equal because some are more accessible than others, which implies inequalities. Thus, accessibility is a proxy for spatial inequalities and remains fundamental since only a
small subset of an area is the most accessible. The notion of accessibility relies on two core concepts:
• The first is location, where the relativity of space is estimated in relation to transport infrastructures since they offer the means to support mobility. Each location has a set of referential
attributes, such as its population or level of economic activity.
• The second is distance, which is derived from the physical separation between locations. Distance can only exist when there is a possibility to link two locations through transportation. It
expresses the friction of distance, and the location with the least friction relative to others is likely to be the most accessible. The friction of distance is commonly expressed in units such
as in kilometers or in time, but variables such as cost or energy spent can also be used.
There are two spatial categories applicable to accessibility problems, which are interdependent:
• The first type is topological accessibility, which is related to measuring accessibility in a system of nodes and paths (a transportation network). It is assumed that accessibility is a
measurable attribute significant only to specific elements of a transportation system, such as terminals (airports, ports, or subway stations).
• The second type is contiguous accessibility, which involves measuring accessibility over a surface. Under such conditions, accessibility is a cumulative measure of the attributes of every
location over a predefined distance, as space is considered contiguous. It is also referred to as isochrone accessibility.
Last, accessibility is a good indicator of the underlying spatial structure since it takes into consideration location as well as the inequality conferred by distance to other locations.
Relationship between Distance and Opportunities
Topological and Contiguous Accessibility
Accessibility and Spatial Structure
Global Accessibility: Time to the Nearest Large City
2. Connectivity and Total Accessibility
The most basic accessibility measure involves network connectivity, where a network is represented as a connectivity matrix (C1), which expresses the connectivity of each node with its adjacent
nodes. The number of columns and rows in this matrix is equal to the number of nodes in the network, and a value of 1 is given for each cell where this is a connected pair and a value of 0 for each
cell where there is an unconnected pair. Simple networks and their connectivity matrices are rare. Thus, the matrix becomes exponentially more complex with the number of nodes. The summation of this
matrix provides a very basic measure of accessibility, also known as the degree of a node:
$\large C1 = \displaystyle\sum_{j}^{n} C_{ij}$
• C1 = degree of a node.
• Cij = connectivity between node i and node j (either 1 or 0).
• n = number of nodes.
The connectivity matrix does not consider all the possible indirect paths between nodes. Under such circumstances, two nodes could have the same degree but may have different accessibilities. To
consider this attribute, the Total accessibility matrix (T) is used to calculate the total number of paths in a network, including direct and indirect paths. Its calculation involves the following
$\large T = \displaystyle\sum_{k=1}^{D} Ck ewline$
$\large C1 = \displaystyle\sum_{j}^{n} C_{ij} ewline$
$\large Ck = \displaystyle\sum_{i}^{n} \displaystyle\sum_{j}^{n} c_{ij}^{1} \: \times \: c_{ji}^{k-1} (\forall k eq1)$
• D = the diameter of the network.
Thus, total accessibility would be a more comprehensive accessibility measure than network connectivity.
Creation of a Connectivity Matrix with a Link Table
Simple Connectivity Matrix
More Complex Connectivity Matrix
Total Accessibility Matrix T
3. The Shimbel Index and the Valued Graph
The main focus of measuring accessibility does not necessarily involve measuring the total number of paths between locations but rather the shortest paths between them. Even if several paths between
two locations exist, the shortest one is likely to be selected. In congested networks, the shortest path may change according to the current traffic level in each segment. Consequently, the Shimbel
index calculates the minimum number of paths necessary to connect one node with all the nodes in a defined network. The Shimbel accessibility matrix, also known as the D-Matrix, includes each
possible node pair with the shortest path.
The Shimbel index and its D-Matrix fail to consider that a topological link between two nodes may involve variable distances. Thus, it can be expanded to include the notion of distance, where value
is attributed to each link in the network. The valued graph matrix, or L-Matrix, represents such an attempt. It is very similar to the Shimbel accessibility matrix. The only difference is that
instead of showing the minimal path in each cell, it provides a minimal distance between each node of the network.
Shimbel Distance Matrix (D-Matrix)
Valued Graph Matrix (L-Matrix)
4. Geographic and Potential Accessibility
From the accessibility measure developed so far, it is possible to derive two simple and highly practical measures, defined as geographic and potential accessibility. Geographic accessibility
considers that the accessibility of a location is the summation of all distances between other locations divided by the number of locations. The lower its value, the more accessible a location is.
$\large A(G) = \displaystyle\sum_{i}^{n} \displaystyle\sum_{j}^{n} \frac {d_{ij}}{n} ewline$
$\large d_{ij} = L$
• A(G) = geographical accessibility matrix.
• dij = shortest path distance between location i and j.
• n = number of locations.
• L = valued graph matrix.
This measure (A(G)) is an adaptation of the Shimbel Index and the Valued Graph, where the most accessible place has the lowest summation of distances. Locations can be nodes in a network or cells in
a spatial matrix.
Potential accessibility is a more complex measure than geographic accessibility since it simultaneously includes the concept of distance weighted by the attributes of a location. All locations are
not equal, and thus, some are more important than others. Potential accessibility can be measured as follows:
$\large A(P) = \displaystyle\sum_{i}^{n} P_{i}+ \displaystyle\sum_{j}^{n} \frac {P_{j}}{d_{ij}}$
• A(P) = potential accessibility matrix.
• dij = friction of distance between place i and j (derived from valued graph matrix).
• Pj = attributes of place j, such as population, retailing surface, parking space, etc.
• n = number of locations.
The potential accessibility matrix is not transposable since locations do not have the same attributes, which brings the underlying notions of emissiveness and attractiveness:
• Emissiveness is the capacity to leave a location, the sum of the values of a row in the A(P) matrix.
• Attractiveness is the capacity to reach a location, the sum of the values of a column in the A(P) matrix.
Geographic Accessibility
Potential Accessibility
Although accessibility can be solved using a spreadsheet (or manually for simpler problems), Geographic Information Systems have proven to be a very useful and flexible tool to measure accessibility,
notably over a surface simplified as a matrix (raster representation). This can be done by generating a distance grid for each place and then summing all the grids to form the total summation of
distances (Shimbel) grid. The cell having the lowest value is thus the most accessible location.
Related Topics
• BTS (2001) Special Issue on Methodological Issues in Accessibility, Journal of Transportation and Statistics, Vol. 4, No. 2/3, Bureau of Transportation Statistics, Sept/Dec.
• Burns, L.D. (1979) Transportation, Temporal, and Spatial Components of Accessibility. Lexington, MA: Lexington Books.
• El-Geneidy, A.M., and D.M. Levinson (2006) Access to Destinations: Development of Accessibility Measures. Retrieved from the University of Minnesota Digital Conservancy, https://hdl.handle.net/ | {"url":"https://transportgeography.org/contents/methods/transportation-accessibility/","timestamp":"2024-11-01T19:34:00Z","content_type":"text/html","content_length":"160563","record_id":"<urn:uuid:27959fc7-dac1-48df-9613-e6176f492f9e>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00882.warc.gz"} |
Statistics, Stochastic Processes, Operations Research - DPC-MA1
Course details
Statistics, Stochastic Processes, Operations Research
DPC-MA1 FEKT DPC-MA1 Acad. year 2022/2023 Winter semester 4 credits
The course focuses on consolidating and expanding students' knowledge of probability theory, mathematical statistics and theory of selected methods of operations research. Thus it begins with a
thorough and correct introduction of probability and its basic properties. Then we define a random variable, its numerical characteristics and distribution. On this basis we then build descriptive
statistics and statistical hypothesis testing problem, the choice of the appropriate test and explanation of conclusions and findings of tests. In operational research we discuss linear programming
and its geometric and algebraic solutions, transportation and assignment problem, and an overview of the dynamic and probabilistic programming methods and inventories. In this section the
illustrative examples are taken primarily from economics. In the next the course includes an introduction to the theory of stochastic processes types. Therefore, it starts with repetition of
necessary mathematical tools (matrices, determinants, solving equations, decomposition into partial fractions, probability). Then we construct the theory of stochastic processes, where we discuss
Markovský processes and chains, both discrete and continuous. We include a basic classification of state and students learn to determine them. Great attention is paid to their asymptotic properties.
The next section introduces the award transitions between states and students learn the decision-making processes and their possible solutions. In conclusion, we mention the hidden Markov processes
and possible solutions.
Subject specific learning outcomes and competences
After completing the course the student will be able to:
• Describe the role of probability using set operations.
• Calculate basic parameters of random variables, both continuous and discrete ones.
• Define basic statistical data. List the basic statistical tests.
• Select the appropriate method for statistical processing of input data and perform statistical test.
• Explain the nature of linear programming.
• Convert a word problem into the canonical form and solve it using a suitable method.
• Perform sensitivity analysis in a geometric and algebraic way.
• Convert the specified role into its dual.
• Explain the difference between linear and nonlinear programming.
• Describe the basic properties of random processes.
• Explain the basic Markov property.
• Build an array of a Markov chain.
• Explain the procedure to calculate the square matrix.
• Perform the classification of states of Markov chains in discrete and continuous case.
• Analyze a Markov chain using the Z-transform in the discrete case and the Laplace transformation in the continuous case.
• Explain the procedure for solving decision problems.
• Describe the procedure for solving the decision-making role with alternatives.
• Discuss the differences between the Markov chain and hidden Markov chain.
The aim of the subject is to deepen and broaden students' knowledge of statistical data processing and statistical tests. To provide students with basic knowledge in the field of operations research
a teach them to use some optimization methods suitable for use in e.g. economics. Next is to provide students with a comprehensive overview of the basic concepts and results relating to the theory of
stochastic processes and especially Markov chains and processes. We show possibilities of application of the decision-making processes of various types.
Prerequisite knowledge and skills
We require knowledge at the level of bachelor's degree, i.e. students must have proficiency in working with sets (intersection, union, complement), be able to work with matrices, handle the
calculation of solving systems of linear algebraic equations using the elimination method and calculation of the matrix inverse, know graphs of elementary functions and methods of their design,
differentiate and integrate of basic functions.
• Miller, I., Miller, M.: John E. Freund's Mathematical Statistics. 8th Edition. Prentice Hall, Inc., New Jersey 2012.
• Taha, H.A.: Operations research. An Introduction. 9th Edition, Macmillan Publishing Company, New York 2013.ISBN-13: 978-0132555937
• Anděl, J.: Statistické úlohy, historky a paradoxy. Matfyzpress, MFF UK Praha, 2018.
• Zapletal, J.: Základy počtu pravděpodobnosti a matematrické statistiky. PC-DIR,VUT, Brno, 1995
• Papoulis, A., Pillai, S. U.: Probability, Random Variables and Stochastic Processes, 4th Edition, 2012. ISBN-13: 978-0071226615
• Nagy, I.: Základy bayesovského odhadování a řízení, ČVUT, Praha, 2003
• Sarma, R. D.:Basic Applied Mathematics for the Physical Sciences 3rd New edition Edition, 2017, ISBN-13: 978-8131787823
• Montgomery, D.C., Runger, G.C.: Applied Statistics and Probability for engineers. 6th Edition. John Wiley \& Sons, Inc., New York 2015.ISBN-13: 978-1118539712.
1. Classical and axiomatic definitions of probability. Conditional probability, total probability. Random variable, numerical characteristics.
2. Discrete and continuous distributions of random variables. Properties of the normal distribution. Limit theorems.
3. Statistics. Selection. Statistical processing of the material. Basic parameters and characteristics of the population selection.
4. Basic point and interval estimates. Goodness. Analysis of variance.
5. Operations Research. Linear programming. Graphic solution. Simplex method.
6. Dual role. The sensitivity analysis. The economic interpretation of linear programming.
7. Nonlinear programming.
8. Solving of problems of nonlinear programming.
9. Random processes, basic concepts, characteristics of random processes.
10. Discrete Markov chain. Homogeneous Markov chains, classification of states. Regular Markov chains, limit vector, the fundamental matrix, and the median of the first transition.
11. Absorption chain mean transit time, transit and residence. Analysis of Markov chains using Z-transform. Calculation of powers of the transition matrix.
12. Continuous time Markov chains. Classification using the Laplace transform. Poisson process. Linear growth process, linear process of extinction, linear process of growth and decline.
13. Markov decision processes. The award transitions. Asymptotic properties. Decision-making processes with alternatives. Hidden Markov processes.
I. Statistics (5 weeks)
Basic notions from probability and statistics. Statistical sets. Point and interval estimates.Testing hypotheses with parametres (not only for normal distribution). Tests of the form of distribution.
Regression analysis. Tests of good accord. Non-parametric tests.
II. Stochastic processes(4 weeks)
Deterministic and stochastic problems. Characteristics of stochastic processes. Limit, continuity, derivation and integral of a stochastic process. Markov, stationary, and ergodic processes.
Canonical and spectral division of a stochastic process.
III. Operation analysis (4 weeks)
Principles of operation analysis, linear and nonlinear programming. Dynamic programming, Bellman principle of optimality. Theory of resources. Floating averages and searching hidden periods.
Students may be awarded
Up to 100 points for the final exam, which consists of writen and oral part. Entering the written part of the exam includes theoretical and numerical task that are used to verify the orientation of
student in statistic, operation research and stochastic processes. Taking numerical task is to verify the student’s ability to apply various methods of technical and economic practice.
Teaching methods and criteria
Teaching methods depend on types of classes. They are described in Article 7 of the Study and Examination Regulations of Brno University of Technology.
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A note on discriminating Poisson processes from other point processes with stationary inter arrival times
Gusztáv Morvai; Benjamin Weiss
Kybernetika (2019)
• Volume: 55, Issue: 5, page 802-808
• ISSN: 0023-5954
We give a universal discrimination procedure for determining if a sample point drawn from an ergodic and stationary simple point process on the line with finite intensity comes from a homogeneous
Poisson process with an unknown parameter. Presented with the sample on the interval $\left[0,t\right]$ the discrimination procedure ${g}_{t}$, which is a function of the finite subsets of $\left[0,t
\right]$, will almost surely eventually stabilize on either POISSON or NOTPOISSON with the first alternative occurring if and only if the process is indeed homogeneous Poisson. The procedure is based
on a universal discrimination procedure for the independence of a discrete time series based on the observation of a sequence of outputs of this time series.
Morvai, Gusztáv, and Weiss, Benjamin. "A note on discriminating Poisson processes from other point processes with stationary inter arrival times." Kybernetika 55.5 (2019): 802-808. <http://eudml.org/
abstract = {We give a universal discrimination procedure for determining if a sample point drawn from an ergodic and stationary simple point process on the line with finite intensity comes from a
homogeneous Poisson process with an unknown parameter. Presented with the sample on the interval $[0,t]$ the discrimination procedure $g_t$, which is a function of the finite subsets of $[0,t]$, will
almost surely eventually stabilize on either POISSON or NOTPOISSON with the first alternative occurring if and only if the process is indeed homogeneous Poisson. The procedure is based on a universal
discrimination procedure for the independence of a discrete time series based on the observation of a sequence of outputs of this time series.},
author = {Morvai, Gusztáv, Weiss, Benjamin},
journal = {Kybernetika},
keywords = {Point processes},
language = {eng},
number = {5},
pages = {802-808},
publisher = {Institute of Information Theory and Automation AS CR},
title = {A note on discriminating Poisson processes from other point processes with stationary inter arrival times},
url = {http://eudml.org/doc/295069},
volume = {55},
year = {2019},
TY - JOUR
AU - Morvai, Gusztáv
AU - Weiss, Benjamin
TI - A note on discriminating Poisson processes from other point processes with stationary inter arrival times
JO - Kybernetika
PY - 2019
PB - Institute of Information Theory and Automation AS CR
VL - 55
IS - 5
SP - 802
EP - 808
AB - We give a universal discrimination procedure for determining if a sample point drawn from an ergodic and stationary simple point process on the line with finite intensity comes from a
homogeneous Poisson process with an unknown parameter. Presented with the sample on the interval $[0,t]$ the discrimination procedure $g_t$, which is a function of the finite subsets of $[0,t]$, will
almost surely eventually stabilize on either POISSON or NOTPOISSON with the first alternative occurring if and only if the process is indeed homogeneous Poisson. The procedure is based on a universal
discrimination procedure for the independence of a discrete time series based on the observation of a sequence of outputs of this time series.
LA - eng
KW - Point processes
UR - http://eudml.org/doc/295069
ER -
You must be logged in to post comments.
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Adding Three One Digit Numbers Worksheets
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download of the PDF math worksheet Teacher versions include both the question
Add 3 Single digit Numbers K5 Learning
Add 3 single digit numbers Grade 1 Addition Worksheet Find the sum 1 4 5 6 2 7 6 2 3 1 8 4 4 3 7 7 5 8 2 2 6 4 9 3 7 2 5 0 8 1 0 6 9 3 3 3 10 5 1 7 11 1 5 5 12 7 6 9 13 9 3 7 14 2 8 5 15 2 1 3 16 8 5
Adding Three One Digit Numbers A Adding Three One Digit Numbers A math worksheet The size of the PDF file is 20796 bytes Preview images of the first Print Open The Download button initiates a
download of the PDF math worksheet Teacher versions include both the question
Add 3 single digit numbers Grade 1 Addition Worksheet Find the sum 1 4 5 6 2 7 6 2 3 1 8 4 4 3 7 7 5 8 2 2 6 4 9 3 7 2 5 0 8 1 0 6 9 3 3 3 10 5 1 7 11 1 5 5 12 7 6 9 13 9 3 7 14 2 8 5 15 2 1 3 16 8 5
Number 3 Worksheets Adding Three Single Digits Additon Worksheet 2 First Grade Worksheets
3 Digit Addition Worksheets
Adding Three Digit Numbers Within One Thousand Worksheet Turtle Diary
2nd Grade Math Worksheets On Adding Three One Digit Numbers Vertically Steemit
Add Three One Digit Numbers Worksheet Turtle Diary
Add 3 digit Number To 1 digit Worksheets For Kids Online SplashLearn
Add 3 digit Number To 1 digit Worksheets For Kids Online SplashLearn
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RSICC Home Page
RSIC CODE PACKAGE CCC-131
1. NAME AND TITLE
ANTE 2: Adjoint Monte Carlo Time-Dependent Neutron Transport Code in Combinatorial Geometry.
PAT: Neutron Element Data Tape Generator (from ENDF Format.)
ADCROS: Adjoint Data Generator.
MAS: Monte Carlo Calculation.
Two versions are packaged: ANTE 2 for the CDC 6600 (A) and ANTE-BELLM for the IBM 360 (B).
Mathematical Applications Group, Inc. (MAGI), White Plains, New York.
Bell Telephone Laboratories, Whippany, New Jersey.
FORTRAN IV; CDC 6600 (A) OR IBM 360/75/91 (B).
4. NATURE OF PROBLEM SOLVED
Given a finite neutron transmitting medium (hereafter called the object or transmission object) and given a neutron detector embedded in that medium, what would be the response of the detector to an
arbitrary neutron flux field in which the object is immersed? ANTE 2 computes an importance weighting function over a surface which contains the object. Thus any arbitrary flux field may be weighted
by this function to determine the effect this field would produce upon the enclosed detector.
5. METHOD OF SOLUTION
Whereas conventional neutron Monte Carlo programs seek to generate flux distributions (which may be called the contravariant aspect) from specified sources and are usable in estimating effects upon
arbitrary (within practical limits) detectors, this program deals with the covariant aspect and seeks to generate importance or adjoint distributions for definite detectors usable in estimating
effects from arbitrary (within practical limits) sources.
There appear to be four principal features of the program.
1. Geometry. The following objects are building blocks of the geometry: sphere, right elliptical cylinder, truncated right angle cone, ellipsoid, or convex polyhedron of 4, 5, or 6 sides. These
building blocks may be combined in arbitrary fashion--unions or intersections. An elegant Boolean notation is used to describe the combinations.
2. Scattering kernels. Differential cross-section data must be analyzed to construct probability distributions of the phase coordinates of a particle prior to a collision which produces some definite
exit phase coordinates. This data preparation may be viewed as the heart of the calculation and may be usefully adapted to other Monte Carlo calculations of adjoints.
3. Standard Monte Carlo techniques are used in particle tracking and construction of histories.
4. Scoring. The transmission object is considered to be completely enclosed by a scoring surface which must be a rectangular parallelepiped or a sphere. Scoring bins represent meshes of time, energy,
polar, and azimuthal angles. The surfaces of the rectangular parallelepiped may be further subdivided to provide spatial binds. In the case of spherical enclosure, no provision is made for surface
subdivision or for recording azimuthal angles of escape. Hence, only spherically symmetric problems can be addressed with this option.
Included in the package are the following routines: PAT abstracts required cross section data from ENDF/B files; ADCROS receives cross-section data from PAT, operates on it to provide scattering
kernels and other basic data required for adjoint calculation and makes an ADCROS tape; CPROC processes geometric description provided in input and places results on ADCROS tape; and MAS performs the
calculation after the above preparations are made.
None noted.
7. TYPICAL RUNNING TIME
Estimated running time for the sample problem on the IBM 360/75: PAT, 1 minute; ADCROS, 5 minutes; and MAS, 7 seconds.
The codes are operable on the CDC 6600 or the IBM 360/75/91 computers with standard I/O, and a maximum of 9 tape units or direct access devices. Approximately 400K is required in the GO step.
ANTE 2 was designed for the FORTRAN IV CDC 6600 Operating System. It is also operable on the IBM 360/75/91 Operating System using an OS 360 FORTRAN H compiler.
10. REFERENCES
a) Included in documentation:
O. Cohen, "ANTE 2A FORTRAN Computer Code for the Solution of the Adjoint Neutron Transport Equation by the Monte Carlo Technique," DASA-2396 (January 1970).
O. Cohen and W. Guber, "ANTE-BELLMA Computer Program for the Solution of the Adjoint Neutron Time Dependent Transport Equation by the Monte Carlo Technique," MR-7003
(March 1970).
b) Background Information:
W. Guber, "The Combinatorial Geometry Technique for the Description and Computer Processing of Complex Three-Dimensional Objects," MR-7004/2 (March 1970).
11. CONTENTS OF CODE PACKAGE
Included are the referenced documents and one (1.2MB) DOS diskette which contains the source codes and sample problem input. A library of cross sections (ENDF/B-Version I) for use in the sample
problem is included in the package.
12. DATE OF ABSTRACT
August 1971; revised July 1982, February 1985. | {"url":"https://rsicc.ornl.gov/codes/ccc/ccc1/ccc-131.html","timestamp":"2024-11-11T00:26:10Z","content_type":"text/html","content_length":"7877","record_id":"<urn:uuid:279344fb-a7dd-4c40-9288-6c163f3603a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00120.warc.gz"} |
How to Add Fractions that are Mixed Numbers (like 2 & 3/4 + 3 & 5/6 ) with Free Online Tutoring in Math
How to Add Fractions that are Mixed Numbers (like 2 & 3/4 + 3 & 5/6 ) with Free Online Tutoring in Math
Unlock the secrets of adding mixed number fractions with our latest blog post, the sixth article in our comprehensive eleven-part series!
This post includes five sample problems to enhance your child’s understanding right from the start.
Adding fractions that are mixed numbers, like 2 ¾ + 3 ⅚, can be very confusing. Many children struggle with this concept, leading to gaps in their understanding. Without proper instruction on
illustrating the addition process, they often feel lost and frustrated, making it difficult for them to keep up in more advanced math classes. This confusion can undermine their confidence and impede
their academic progress.
Adding fractions that are mixed numbers is often challenging for children, leaving them with a limited understanding.
This lack of understanding can cause frustration and confusion, especially as they progress to higher levels of mathematics. Without a solid grasp of this concept, children may find it hard to
succeed in more advanced math classes, which can affect their overall academic performance and confidence.
The solution lies in learning to illustrate the addition process. In our blog post, your child will:
• Learn to Illustrate Addition: Visual models help clarify the process of adding mixed number fractions.
• Build a Strong Foundation: Step-by-step lessons ensure your child thoroughly understands each concept.
• Achieve Long-term Success: A solid grasp of fractions is crucial for future mathematical success.
Visit our website (https://www.teachersdungeon.com/) for a comprehensive educational program designed to help kids become proficient in mathematics. By mastering these concepts, your child will gain
a deeper, more concrete understanding of adding fractions that are mixed numbers, paving the way for a successful educational journey. Don’t miss out on this valuable resource—empower your child’s
learning today!
Articles within this series on Fractions:
Solving problems that deal with fractions is simple when you develop a concrete understanding. I have had incredible results with with the students in my class! The strategies taught within this
article work with children who have ADHD, Dyslexia, and other learning disabilities. Virtually every one of my students who has learned the strategies within this HOW TO DO FRACTIONS article has
passed the standards based assessment for adding, subtracting, multiplying and dividing fractions.
I have scaffold the problems in each lesson.
The first problem in this article is a “Watch Me” problem. The second is a “Work with Me” problem. All the rest are “On Your Own” problems.
*If your child needs a bit more support, they should complete the “On Your Own” problems as a “Work with Me” problem. I have a number of students with gaps in their learning and others with a
variety of learning disabilities. I have had incredible success, by having those students complete 5 to 7 problems within each lesson as a “Work with Me” problem. They play a bit of the video,
then pause it and copy, then watch a bit more, pause it and copy. My students Play – Pause – and Copy until the entire problem is solved. This is like having a personal tutor working through each
and every problem with your child. Every one of my students who has used this strategy has passed the Common Core Proficiency Exam.
How to Add Fractions that are Mixed Numbers
Online Tutoring in Math: Challenge 1
Watch Me
Red Tailed Hawk
You are on a hike when you see a Red Tailed Hawk soaring across the sky. It is absolutely beautiful! The hawk is circling round and round in two circles that form an 8. The upper portion of the
Eight is 2 & 3/4 miles around. The lower portion of the Eight is 3 & 5/6 miles around.
How far does the Red Tailed Hawk fly each time he completes his figure-8 shape in the sky?
Watch this Free Tutoring for Math Video!
Press PLAY and Watch this Free Tutoring for Math Video below. Then copy these strategies into your notes!
How to Add Fractions that are Mixed Numbers
Online Tutoring in Math: Challenge 2
Work With Me
Hungry Chipmunk
You are camping in Yellowstone National Park. You are sitting on the picnic table at your campsite eating chips when a cute, little chipmunk scampers up to you. You feed him 2 & 2/3 bags of chips.
The next day you are eating more chips when the friendly chipmunk reappears. This time you feed him 2 & 3/5 bags.
How many bags of chips did you feed your Hungry Chipmunk?
Watch this Free Tutoring for Math Video!
Gather your materials and press PLAY. We’ll solve this problem together, while you watch the math tutorial video below.
Do your children get frustrated when they make a mistake?
We all make mistakes. As a matter of fact, making mistakes is an essential part of the learning process. This is why at the end of each of the following “On Your Own” challenges I encourage
children to fix their mistakes. Finding and fixing your own mistake is the fastest way to learn.
How to Add Fractions that are Mixed Numbers
Online Tutoring in Math: Challenge 3
On Your Own
Mural Project
Your class was asked to create two murals for your school. The murals will be painted on a number of walls near your school’s gymnasium. Your teacher has chosen two art projects for the murals.
One is a drawing of Native Americans. The other is a drawing of the people from your community. You get to use 1 & 4/9 walls for one mural and 1 & 2/3 walls for the other drawing.
How many walls will be covered by these two drawings?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
How to Add Fractions that are Mixed Numbers
Online Tutoring in Math: Challenge 4
On Your Own
Pool-Slash Party
You are invited to a Pool-Splash Party at your friend’s house. Your friend has become famous, because she dips her head in the water and then flings her hair back. Her photos are all over the
internet and you can’t wait to watch her do her “hair fling thing”. In order to get from your house to her pool you will need to walk 1 & 1/2 miles to the bus stop. Next, you will need to ride the
bus another 2 & 5/8 miles before you reach her home.
How far will you travel when you go to the Pool-Splash Party?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
How to Add Fractions that are Mixed Numbers
Online Tutoring in Math: Challenge 5
On Your Own
Newspaper Reporter
You are a reporter for your school newspaper. This week’s topic is underwater exploration. You have decided to write a similar story to the one written in the New York Times about the sinking of
the Titanic. Your best friend is writing another story about the famous oceanographer and filmmaker, Jacques Cousteau. Your story will use 2 & 2/3 pages. Your friends story will use 1 & 5/7 pages.
How many pages will these two stories use of your school’s newspaper?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
Want More Tutorials?
Discover the transformative power of learning with TeachersDungeon today. Dive into a world where education meets adventure, empowering students from grades 3 to 6 with personalized math instruction
that adapts to their needs. Whether you’re an educator looking to enrich classroom learning or a parent seeking to support your child’s academic journey, The Teacher’s Dungeon offers interactive
gameplay, instant help with video tutorials, and comprehensive progress tracking through its Stats Page. Visit The Teacher’s Dungeon’s website now to explore how our innovative approach can elevate
your child’s math education. Embark on this exciting educational journey with us and watch your students thrive!
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Electricity Costs for 10 Key Household Products | Arcadia
Electricity Costs for 10 Key Household Products
Electricity powers many of your key household products and appliances, but how much is actually needed to run them, and how much does this electricity cost?
Opower recently did a study on how much it costs to charge an iPhone 6. They calculated how long it took to get the iPhone's battery from 0% to 100%, and found that it only took 10.5 watt-hours (Wh)
of electricity to fully charge. Surprisingly, when crunching the numbers, they realized it only costs you a mere $0.47 to fully charge your iPhone everyday for a year.
We were inspired by this data, so we collected our own. We looked at 10 household products and calculated how many kilowatt-hours (kWh) it takes for them to run if you use them everyday for a year.
We then used 12.29 cents as the average price per kWh to come up with typical electricity costs. Here's a look at which of your household products are using the most electricity and are likely
busting your energy budget each month.
Hair Dryer Electricity Costs
The estimated time you use a hair dryer when you get ready is 30 minutes. Since it takes 1200 watts for a hair dryer to run for a full hour, that means it takes 600 watts, or 600 Wh, or 0.6 kWh to
run for 30 minutes. When multiplying this usage by days in a year, at a rate of 12.19 cents per kWh, we find that you pay $26.92 per year to dry your hair everyday.
• Average run: 30 minutes per day
• Wh per use: 1200 watts per hour (per ½ hour use) = 600 Wh
• kWh per use: 600 Wh/1000 = 0.6 kWh
• Cost: 0.6 kWh x $0.1229 x 365 = $26.92 per year
This single appliance is costing you almost $30 a year, and even if you don't use it often, the amount of wattage it requires for a single use is what may be increasing your energy bill. To reduce
the cost of this appliance, try using it less, if possible. One suggestion is to take fewer showers or let your hair air dry. You'll use less water and, and therefore, less electricity that would
normally be used to dry your hair. Make sure to unplug the appliance, as well.
Refrigerator Electricity Costs
A refrigerator must run 24 hours a day in order to keep your food fresh. This means that the 180 watts it produces per hour must be multiplied by its 24 hours of use. So, a refrigerator is actually
using 4320 Wh, or 4.32 kWh, everyday of the year. Multiplying this by 365 days at the average price per kWh, we find that your fridge is costing you $193.70 per year.
• Average run: 24 hours per day
• Wh per use: 180 watts (per 24 hour use) = 4,320 Wh
• kWh per use: 4320 Wh/1000 = 4.32 kWh
• Cost: 4.32 kWh x $0.1229 x 365 = $193.70 per year
You're spending almost $200 each year to keep your food fresh. A refrigerator is a necessity, but its high electricity costs are not. To reduce the cost of this appliance, purchase one that uses less
electricity. Energy Star appliances, and ones alike, can save you hundreds of dollars simply because of how they are made. These appliances are built more efficiently and effectively to save you
money. You can also stock your fridge with cold items, so it will require less work to keep the food cool. Also, let hot items, such as soup and pasta, cool down before you put them in the fridge.
Laptop Electricity Costs
An average 14-15 inch laptop uses 60 watts when it charges for one hour, the recommended charging time. Therefore, it takes 60 Wh, or 0.06kWh, to fully charge. When multiplying this by days in a year
at the average cost per kWh, we find that it costs $2.69 a year to fully charge your laptop everyday.
• Average run: 1 hour per day (recommended charge)
• Wh per use: 60 watts (per 1 hour charge) = 60 Wh
• kWh per use: 60 Wh/1000 = 0.06 kWh
• Cost: 0.06 kWh x $0.1229 x 365 = $2.69 per year
Alike to the iPhone, this is an extremely minimal cost given how frequently a laptop is used. If you believe this electricity cost is too high, you can simply use the laptop until it's completely out
of battery then set it aside to recharge. When a laptop is used while charging, it takes it longer and requires more energy to get to a full 100% battery.
Light Bulb Electricity Costs
An average incandescent bulb uses 60 watts an hour, whereas an average CFL bulb uses only 14 watts on average. The average home keeps their lights on for about 3 hours each day. This means that one
incandescent bulb needs 180 Wh, or 0.18 kWh, to run for just three hours, and a CFL needs 42 Wh, or 0.042 kWh. So, one incandescent bulb costs $8.07 per year and one CFL bulb only costs $1.88 per
year, a whole $6 less.
Both of these bulb costs might seem low, but given that an average household has over 40 bulbs, your lighting, especially incandescent, can easily hike up your electricity bills.
Incandescent Light Bulb
• Average run: 3 hours per day
• Wh per use: 60 watts (per 3 hour use) = 180 Wh
• kWh per use: 180 Wh/1000 = 0.18 kWh
• Cost per bulb: 0.18 kWh x $0.1229 x 365 = $8.07 per year
• Total Cost: $8.07 x 40 bulbs = $322.80 per year
CFL Light Bulb
• Average run: 3 hour per day
• Wh per use: 14 watts (per 3 hour use) = 42 Wh
• kWh per use: 42 Wh/1000 = 0.042 kWh per use
• Cost per bulb: 0.042 kWh x $0.1229 x 365 = $1.88 per year
• Total Cost: $1.88 x 40 bulbs = $75.20 per year
The best way to save electricity and reduce the your lighting costs is to switch over to CFL or LED light bulbs. If you switch just one lightbulb, you will save $6 per year. Imagine the savings you
would get from switching over 10, 20, or even all 40 or so bulbs in your home.
Dishwasher Electricity Costs
Many of us think a dishwasher uses more water and electricity than hand washing dishes, and they may be right, at least about the electricity. It takes about 1800 watts for a dishwasher to run for
one hour, and the average washer runs for more than 2 hours. This means it uses 3600 Wh, or 3.6 kWh, per average use, costing you over $161.50 if used everyday for a year.
To lower the cost of running your dishwasher, use it less frequently. If you use it only once per week, for example, your costs will drop from $161 per year to just $23.
• Average run: 2 hours per day
• Wh per use: 1800 watts (per 2 hour wash) = 3,600 Wh
• kWh per use: 3600 Wh/1000 = 3.6 kWh per use
• Cost to run daily: 3.6 kWh x $0.1229 x 365 = $161.50
• Cost to run weekly: 3.6 kWh x $0.1229 x 52 = $23
Coffee Maker Electricity Costs
Your average coffee maker is used for 10 minutes to brew 4 cups of coffee. It takes about 800 watts per hour for a coffee maker to run, which is about 133.33 Wh, or 0.133 kWh. This means
you're spending $5.9o each year if you brew coffee everyday. The electricity costs from your coffee maker are almost as high as the incandescent light bulbs, but luckily, there is usually only one
coffee maker per household, not 40.
• Average run: 10 minutes
• Wh per use: 800 watts (per 10 minute brew) = 133.33 Wh
• kWh per use: 133.33 Wh/1000 = 0.133 kWh per use
• Cost: 0.133 kWh x $0.1229 x 365 = $5.90 per year
To reduce any extra costs from your coffee maker, make sure to unplug it after every use. A coffee maker will use energy simply by being turned on or plugged, even if it isn't being used. For
example, its other functions, like its clock or cleaning mechanism, automatically run throughout the day and drain electricity.
Washer & Dryer Electricity Costs
Washers and dryers are frequently used and well known for their large energy use. What's surprising, however, is that a washing machine requires way less electricity than a dryer.
An average cycle for a washing machine is 30 minutes. This appliance, which is a widely used Energy Star model, needs 500 watts per hour to run, which means it requires 250 Wh, or 2.25 kWh, to run
for 30 minutes. If used everyday for a year, a washing machine's electricity costs are only $11.21. If run only once per week, it would only cost $1.60 per year.
• Average run: 30 minutes
• Wh per use: 500 watts (per 30 minute cycle) = 250 Wh
• kWh per use: 250 Wh/1000 = 0.25 kWh per use
• Cost to run daily: 0.25 kWh x $0.1229 x 365 = $11.21
• Cost to run weekly: 0.25 kWh x $0.1229 x 52 = $1.60
Your dryer, however, requires 3000 watts per hour use, and runs for an average of 45 minutes or more, depending on the load. One dryer cycle requires 2250 Wh, or 2.25 kWh. That means you pay $100.93
for electricity if you run it everyday for an entire year. However, if you run it only once a week, your cost goes down to $14.38.
• Average run: 45 minutes
• Wh per use: 3000 watts (per 45 minute cycle) = 2250 Wh
• kWh per use: 2250 Wh/1000 = 2.25 kWh per use
• Cost to run daily: 2.25 kWh x $0.1229 x 365 = $100.93
• Cost to run weekly: 2.25 kWh x $0.1229 x 52 = $14.38
A simple way to reduce your laundry's electricity costs is to air dry your clothes or only use the appliances once a week. Washing your clothes less and choosing to air dry not only saves energy, but
it saves the quality of your clothing as well. You could also replace your appliances with Energy Star models. They are becoming more popular in the market, and, as shown, can save you a lot of money
on your electricity bills.
Microwave, Oven & Stove Electricity Costs
We estimated that a microwave is used, on average, about 15-30 minutes per day. It takes about 1200 watts per hour for an average microwave to run. Therefore, it requires 300 Wh, or 0.3 kWh, for 15
minutes of use, and costs about $13.46 to use your everyday for a year.
• Average run: 15 minutes
• Wh per use: 1200 watts (per 15 minute use) = 300 Wh
• kWh per use: 300 Wh/1000 = 0.3 kWh per use
• Cost: 0.3 kWh x $0.1229 x 365 = $13.46 per year
An oven, however, takes longer and requires much more energy to get to a high temperature. An oven on medium to high heat uses 2400 watts per hour and a stovetop uses 1500 watts per hour on medium to
high heat. So even though the cost of a microwave seems high, it's a quicker and more efficient way to cook if you’re looking to save energy and money.
If you want to reduce the electricity costs coming from your microwave, make sure you set the appropriate time and cooking level for your food so it's not running any longer than it needs to.
The cheapest household product on our list, other than the iPhone, was the washing machine used on a weekly basis. An Energy Star washer's electricity costs came out to just $1.60 per year. The most
expensive appliance was the refrigerator, which runs for 24 hours a day, and costs a whopping $193.70 each year.
Overall, unplug, shut down, and refrain from using your energy-intensive appliances as much as you can. Your iPhone, luckily, is the least of your worries.
Reduce your footprint with Arcadia by matching your home's electricity usage with clean energy from wind farms. Track your usage, impact, and get home efficiency tips in your personal energy
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Trapezoid Area Calculator with Fractions - GEGCalculators
Trapezoid Area Calculator with Fractions
Trapezoid Area Calculator
1. How to find the area of a trapezoid without the height calculator?
Finding the area of a trapezoid without the height can be challenging. You need either the height or additional information, such as the lengths of both bases and an angle or diagonal, to determine
the height.
2. What is the calculation for trapezoid area?
The formula for calculating the area of a trapezoid is:
Area = (1/2) * (base1 + base2) * height
where ‘base1’ and ‘base2’ are the lengths of the two parallel sides (bases), and ‘height’ is the perpendicular distance between the bases.
3. How do you find the area of a trapezoid with 4 points?
To find the area of a trapezoid with 4 points (vertices), you first need to identify the lengths of the two parallel sides (bases) and the height. You can do this by measuring the distances between
the points or by using geometric properties of the trapezoid.
4. How do you find the fourth side of a trapezoid?
A trapezoid has four sides, but the lengths of the sides can vary. To find the length of the fourth side, you need additional information, such as the lengths of the two bases, the height, or other
angles and side lengths.
5. How do you find the unknown height of a trapezoid?
To find the unknown height of a trapezoid, you typically need to use the formula for the area of a trapezoid:
Area = (1/2) * (base1 + base2) * height
Rearrange the formula to solve for the height:
height = (2 * Area) / (base1 + base2)
You’ll need the values of both bases and the area to calculate the height.
6. How do you find area when height is not given?
When the height is not given, you can’t directly find the area of a trapezoid. You need either the height or additional information, such as the lengths of both bases and an angle or diagonal, to
determine the height.
7. How do you find the missing length of a trapezoid with the area?
If you have the area of the trapezoid and the lengths of both bases, you can rearrange the area formula to find the missing length:
Area = (1/2) * (base1 + base2) * height
base2 = (2 * Area) / height – base1
8. How do you find the height of a trapezoid with only the area and base?
You can find the height of a trapezoid with only the area and one of the bases using the formula:
height = (2 * Area) / (base1 + base2)
where ‘Area’ is the given area, ‘base1’ is the given base, and ‘base2’ is the other base.
9. What is the base and height of a trapezoid?
In a trapezoid, the bases are the two parallel sides of the shape, and the height is the perpendicular distance between the bases.
10. How do you find the area of 4 unequal sides?
If you have 4 unequal sides, you can’t directly calculate the area of a trapezoid. You need either the height or additional information, such as the lengths of both bases and an angle or diagonal, to
determine the height and find the area.
11. How do you find the area of a trapezoid 6th grade?
In 6th grade, to find the area of a trapezoid, you’ll need the lengths of both bases and the height. Once you have this information, you can use the formula:
Area = (1/2) * (base1 + base2) * height
12. How do you work out the area of a 4-sided?
If you have a 4-sided polygon, like a trapezoid, you can calculate its area using the trapezoid area formula:
Area = (1/2) * (base1 + base2) * height
13. What is the formula for trapezoid method?
The formula for calculating the area of a trapezoid is:
Area = (1/2) * (base1 + base2) * height
14. What is the formula to solve the problem trapezoid?
The formula to find the area of a trapezoid is:
Area = (1/2) * (base1 + base2) * height
15. Does a trapezoid have 4 unequal sides?
A trapezoid has four sides, but not all four sides are necessarily equal in length. In a trapezoid, there are two sides that are parallel (the bases), and the other two sides are typically
non-parallel, which can have different lengths.
16. How to calculate the area of the trapezoid in the figure in square feet?
To calculate the area of a trapezoid in square feet, you need the lengths of both bases and the height, all measured in feet. Then use the formula:
Area = (1/2) * (base1 + base2) * height
17. How do you find the area with only the base and height?
If you have the length of one base and the height of the trapezoid, you can find the area using the formula:
Area = (1/2) * (base1 + base2) * height
You’ll need to know the length of the other base or have more information to find the area.
18. How do you find area with height?
To find the area of a trapezoid with the height, you need the lengths of both bases. Once you have the lengths of both bases and the height, you can use the formula:
Area = (1/2) * (base1 + base2) * height
19. What is the formula for the area rule?
The formula for calculating the area of a trapezoid is:
Area = (1/2) * (base1 + base2) * height
20. How do you find the area of a trapezium without the height and area?
Finding the area of a trapezium (trapezoid) without the height or area is not possible. You need either the height or the area, along with the lengths of the bases, to calculate the area.
21. Can the formula used to find the area of a trapezoid be used to find areas of parallelograms?
No, the formula for the area of a trapezoid is different from the formula for the area of a parallelogram. The formula for a parallelogram is:
Area = base * height
where ‘base’ is one side length, and ‘height’ is the perpendicular distance between the base and its opposite side.
22. Is it possible to find a formula for the area of a trapezoid in terms of the lengths of its sides?
No, it is not possible to find a simple formula for the area of a trapezoid solely based on the lengths of its sides. The area formula involves the height, which is typically not determined solely by
the side lengths.
23. How to find the height of a trapezoid using Pythagorean Theorem?
To find the height of a trapezoid using the Pythagorean Theorem, you need to have a right triangle formed by the height, one of the bases, and a segment perpendicular to the bases. The square of the
height can be calculated using the Pythagorean Theorem.
24. How to find the height of a rectangle with only the area and base?
To find the height of a rectangle with only the area and base, you can rearrange the formula for the area of a rectangle:
Area = base * height
height = Area / base
25. What is the area of a trapezoid with height 5 m and bases 8 m and 1 m?
To find the area of the trapezoid with a height of 5 meters and bases of 8 meters and 1 meter, use the formula:
Area = (1/2) * (base1 + base2) * height
Area = (1/2) * (8 + 1) * 5 = 22.5 square meters
26. Is the altitude of a trapezoid the same as the height?
Yes, in the context of a trapezoid, the term “altitude” is used interchangeably with “height.” Both terms refer to the perpendicular distance between the bases of the trapezoid.
27. Which height is used to calculate the area of a trapezoid – the height of the figure is needed?
To calculate the area of a trapezoid, you need the height (altitude) of the trapezoid. The height is the perpendicular distance between the bases.
28. What is the formula for the shorter base of a trapezoid?
There is no specific formula for the shorter base of a trapezoid. The trapezoid has two bases, and the length of each base must be given or determined using other information.
29. How do you find area with all sides different?
If all the sides of a quadrilateral are different, it does not necessarily mean that it is a trapezoid. To find the area of such a shape, you will need additional information such as the height or
other angles and side lengths.
30. How do you find the area of each irregular figure in each figure?
To find the area of each irregular figure, you need to break the figure down into simpler shapes (e.g., triangles, rectangles, trapezoids) whose areas can be calculated using standard formulas. Then,
calculate the area of each component and add up the areas to get the total area of the irregular figure.
31. How do you find the area of a shape with different side lengths?
To find the area of a shape with different side lengths, you need to identify the shape’s properties and break it down into smaller, familiar shapes (e.g., triangles, rectangles, trapezoids) whose
areas can be calculated using standard formulas. Then, calculate the area of each component and add up the areas to get the total area of the shape.
32. How do you find the area of an irregular shape?
To find the area of an irregular shape, you need to break it down into simpler, familiar shapes (e.g., triangles, rectangles, trapezoids) whose areas can be calculated using standard formulas. Then,
calculate the area of each component and add up the areas to get the total area of the irregular shape.
33. What is an irregular shape with 4 sides?
An irregular shape with 4 sides is a quadrilateral. A quadrilateral is any polygon with four sides and four vertices.
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MHD Ekman layer on a porous plate
An exact solution of the steady three-dimensional Navier-Stokes equations is obtained for the case of flow past a porous plate at zero incidence in a rotating frame of reference by using similarity
analysis. The behavior of the MHD Ekman layer on a flat plate, subjected to suction and blowing, is studied. It is shown that the Ekman-layer thickness is inversely proportional to suction and
directly proportional to blowing for a given Taylor number and magnetic parameter. The Ekman-layer thickness is found to be inversely proportional to both the Taylor number and the magnetic parameter
either under suction or blowing.
Nuovo Cimento B Serie
Pub Date:
October 1975
□ Ekman Layer;
□ Magnetohydrodynamic Flow;
□ Porous Boundary Layer Control;
□ Porous Plates;
□ Rotating Fluids;
□ Blowing;
□ Earth Planetary Structure;
□ Magnetic Effects;
□ Navier-Stokes Equation;
□ Suction;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1975NCimB..29..296M/abstract","timestamp":"2024-11-07T22:09:38Z","content_type":"text/html","content_length":"36430","record_id":"<urn:uuid:6826d76f-8652-4c6b-a04b-65fd5dcff620>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00353.warc.gz"} |
PhysicsLAB: Circular Orbits
NextTime Question
Circular Orbits
Printer Friendly Version
A cannonball is fired horizontally from a tall mountain to the ground below. Because of gravity, it strikes the ground with increased speed. A second cannonball is fired fast enough to go
into circular orbit ... but gravity does not increase its speed, Why?
View Correct Answer
Related Documents
Resource Lesson:
A cannonball is fired horizontally from a tall mountain to the ground below. Because of gravity, it strikes the ground with increased speed. A second cannonball is fired fast enough to go into
circular orbit ... but gravity does not increase its speed, Why?
View Correct Answer
A cannonball is fired horizontally from a tall mountain to the ground below. Because of gravity, it strikes the ground with increased speed. A second cannonball is fired fast enough to go into
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Problem with Formula when inserting a row
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Problem with Formula when inserting a row
I'm not sure how to handle the following situation... my data looks as follows:
The formula in Status2 (and filled down) reads as follows:
=IF([Act Finish Date]2 = "", "Open", IF(AND([Act Finish Date]3 = "", [Act Finish Date]2 <> ""), "Closed, Last Event", "Closed"))
In other words, the formula is checking not only if a row is Open or Closed but also determining which was the last closed event.
All of that works great until I insert a row between rows 5 and 6. Two things go wrong...
1. 1. The formula in Status5 no longer looks at the next row down. Instead it keeps looking at the row it was looking at before I inserted the new row (which is now row 7... it skips the newly
inserted Row 6).
3. 2. The newly inserted row doesn't automatically fill down (presumably because of problem #1)
Any thoughts on how to get the newly inserted row to automatically get the formula and also have Row 5 always look at the "next" row even if I insert a new row below it?
• This is a really convoluted way to do this, which makes me think there must be a better way. If someone finds it, definitely let me know! Here's the best I could come up with, taking advantage of
the new INDEX function.
First, I'm going to assume your Row # column is an actual column. If that's the case, you could do something like:
=IF(INDEX([Col 1]:[Col 1], [Row #]1 + 1, 1) = "", "Next Row Blank", "OK")
So rather than referencing the cell directly, we're referencing "the cell one row down from me." Since we're no longer referencing a cell outside the current row, inserting a new row won't affect
this formula.
I hope that makes sense! Let me know if that doesn't solve your issue.
• I have updated my original post... I tried to simplify my situation too much. Now it refelcts my real situation.
• Thanks Greg... I understand what you mean. I'll give it a try later today.
• Greag - I tried what you suggested but the Row # part doesn't work right. When I add a new row, the Row # columns is blank. If I try to add the system row # data type, the new row # is not in
order so the calculation is off (it takes the next available row # for the entire sheet). :-(
• Could you set your Row # column to be the "Auto-Number" column type (or add this column in if you don't have it already)? You could even hide this column once you add it in if you don't want to
see it.
• That's exactly what I did (set the Row# to AutoNumber). But when you add a new row in the middle it doesn't put the next # in order... it puts the next # from the entire sheet. So, my numbering
looks like this:
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Understanding Mathematical Functions: Is An Absolute Value Function On
Mathematical functions play a crucial role in various fields, from engineering to economics and even in daily life. These functions help us understand and represent relationships between different
quantities or variables. One important aspect of functions is whether they are one-to-one or not. A one-to-one function is a function where each element in the domain maps to exactly one element in
the range, and no two elements in the domain map to the same element in the range. Today, we'll delve into the concept of absolute value functions and explore whether they are one-to-one.
Key Takeaways
• Mathematical functions are crucial in various fields and help represent relationships between quantities or variables.
• A one-to-one function maps each element in the domain to exactly one element in the range, with no two elements in the domain mapping to the same element in the range.
• Absolute value functions are explored to determine if they are one-to-one, involving graphical representation and algebraic methods.
• Understanding one-to-one functions in absolute value functions has implications in mathematical analysis and real-life applications.
• The one-to-one property affects the behavior of the absolute value function and is important to understand in mathematics.
Understanding Absolute Value Functions
An absolute value function is a mathematical function that returns the absolute value of a number, which is its distance from zero on the number line. Absolute value functions are represented using
the notation |x|.
When dealing with real numbers, the absolute value of a number is always non-negative. For example, the absolute value of -5 is 5, and the absolute value of 3 is also 3.
Definition of absolute value function
• Absolute value function definition: The absolute value of a number x, denoted as |x|, is defined as follows:
□ If x is greater than or equal to 0, then |x| = x.
□ If x is less than 0, then |x| = -x.
Graphical representation of absolute value function
• Graph of the absolute value function: The graph of the absolute value function is a V-shaped graph, with its vertex at the origin (0,0). It has a slope of 1 for x > 0 and a slope of -1 for x < 0.
• Key characteristics of the graph: The graph of |x| reflects the distance of x from 0, without considering the direction. This results in a symmetrical graph about the y-axis.
Characteristics of absolute value function
• Domain and Range: The domain of the absolute value function is all real numbers. The range is also all real numbers, but the output is always non-negative.
• One-to-One Function: An absolute value function is not a one-to-one function because it fails the horizontal line test. A horizontal line intersects the graph of an absolute value function at two
points, indicating that it is not one-to-one.
Understanding Mathematical Functions: Is an absolute value function one-to-one
Mathematical functions are essential in understanding relationships between variables and their outputs. One important aspect of functions is determining if they are one-to-one, which plays a crucial
role in various mathematical concepts and applications.
A. Definition of one-to-one function
A one-to-one function, also known as an injective function, is a function in which each element in the domain maps to a unique element in the range. In other words, no two distinct elements in the
domain map to the same element in the range.
B. Criteria for determining if a function is one-to-one
• Horizontal Line Test: One way to determine if a function is one-to-one is by using the horizontal line test. If any horizontal line intersects the graph of the function at most once, then the
function is one-to-one.
• Algebraic Approach: Another method is to use algebraic techniques to analyze the function. For a function f(x) to be one-to-one, if two different inputs x1 and x2 lead to the same output f(x1) =
f(x2), then the function is not one-to-one.
C. Importance of one-to-one functions in mathematics
One-to-one functions are important in various mathematical concepts such as inverse functions, logarithms, and solving equations. Inverse functions, for example, rely on the property of one-to-one
functions to ensure that each input in the range corresponds to a unique output in the domain. Logarithms, on the other hand, are based on the inverse relationship of exponential functions, which are
Furthermore, one-to-one functions are essential in solving equations, especially when it comes to finding unique solutions for different variables. They help in ensuring that each input has only one
corresponding output, making it easier to analyze and solve mathematical problems.
Understanding Mathematical Functions: Is an absolute value function one-to-one
In the realm of mathematical functions, one important property to consider is whether a function is one-to-one, also known as injective. In this post, we will delve into the absolute value function
and analyze whether it possesses this property.
Testing the absolute value function for one-to-one property
Before we dive into the analysis, it is crucial to understand the concept of a one-to-one function. A function f is said to be one-to-one if no two different inputs produce the same output, in other
words, for any two distinct inputs x1 and x2, f(x1) does not equal f(x2).
Using algebraic methods to analyze the absolute value function
One way to test whether the absolute value function is one-to-one is by using algebraic methods. We can examine the equation f(x) = |x| and evaluate its behavior for different input values. By
testing various pairs of input values and observing the corresponding outputs, we can determine whether the function satisfies the one-to-one property.
Graphical representation to determine if the absolute value function is one-to-one
Another approach to analyzing the one-to-one property of the absolute value function is by examining its graphical representation. By plotting the function on a coordinate plane, we can visually
inspect whether the function passes the horizontal line test. If every horizontal line intersects the graph at most once, then the function is one-to-one.
Understanding Mathematical Functions: Is an Absolute Value Function One-to-One?
In mathematics, functions are a fundamental concept that describes the relationship between input and output values. One important type of function is the absolute value function, which is denoted as
|x| and returns the magnitude of a real number without considering its sign.
A. Explanation of the properties of the absolute value function
The absolute value function is defined as follows:
• |x| = x if x is greater than or equal to 0
• |x| = -x if x is less than 0
This means that the absolute value of a non-negative number is the number itself, while the absolute value of a negative number is its positive counterpart.
B. Determining if the absolute value function satisfies the criteria for being one-to-one
A function is considered one-to-one if each element of the domain maps to a unique element in the range. In other words, no two different inputs can produce the same output.
1. Using the horizontal line test
To determine if the absolute value function is one-to-one, we can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function is
not one-to-one. In the case of the absolute value function, it fails the horizontal line test because a horizontal line at y = 0 intersects the graph at two points, indicating that multiple inputs
map to the same output.
2. Analyzing the slope of the function
Another way to determine if a function is one-to-one is to analyze its slope. For the absolute value function, the slope changes abruptly at x = 0, as the function transitions from a slope of 1 to a
slope of -1. This sudden change in slope indicates that the function is not one-to-one, as different inputs produce the same output.
Implications of One-to-One Property in Absolute Value Functions
The one-to-one property in absolute value functions has significant implications in mathematical analysis, real-life applications, and the behavior of the function.
A. Advantages of one-to-one property in mathematical analysis
• Uniqueness: One-to-one functions ensure that each input corresponds to a unique output, allowing for straightforward analysis and interpretation of the function.
• Solvability: In mathematical equations involving absolute value functions, the one-to-one property helps in finding unique solutions, reducing ambiguity and simplifying the process of solving
• Consistency: One-to-one property ensures that the function preserves the order and relationships between input and output values, leading to consistent and predictable behavior.
B. Real-life applications of understanding one-to-one functions in absolute value functions
• Distance and direction: In real-world scenarios such as navigation and physics, absolute value functions represent distance and direction, where understanding the one-to-one property is crucial
for accurate measurements and calculations.
• Optimization problems: Applications in economics, engineering, and optimization rely on one-to-one functions to identify optimal solutions and make informed decisions based on unique
relationships between variables.
• Biomedical analysis: In medical research and analysis, absolute value functions with one-to-one property are used to model relationships between variables, leading to insights and advancements in
healthcare and pharmaceuticals.
C. How the one-to-one property affects the behavior of the absolute value function
The one-to-one property influences the behavior of the absolute value function in several ways:
• Injective nature: The one-to-one property makes the absolute value function an injective function, ensuring that distinct inputs correspond to distinct outputs, leading to a consistent and
predictable mapping.
• Reflection symmetry: Understanding the one-to-one property helps in visualizing the reflection symmetry of the absolute value function, where the function's graph reflects across the y-axis due
to the unique mapping of inputs and outputs.
• Strict monotonicity: The one-to-one property ensures that the absolute value function exhibits strict monotonicity, where the function's values either consistently increase or decrease,
reflecting the unique relationships between inputs and outputs.
Understanding one-to-one functions in mathematics is crucial for analyzing relationships between inputs and outputs. It helps us determine whether a function has a unique inverse and provides
valuable insight into the behavior of mathematical expressions.
Final thoughts on the one-to-one property of the absolute value function:
• The absolute value function is not one-to-one because it fails the horizontal line test, meaning that there are multiple inputs that result in the same output.
• Despite not being one-to-one, the absolute value function still plays a significant role in many mathematical applications and is valuable for solving equations and inequalities.
Overall, a deep understanding of mathematical functions, including whether they are one-to-one, enhances our ability to analyze and interpret mathematical models, ultimately strengthening our
problem-solving skills.
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Lesson 24
Quadratic Situations
• Let’s work with situations and quadratic equations.
24.1: Growing Plants
Plant A’s height over time is represented by \(y=\frac{1}{2}x+4\). Plant B’s height is \(y=\frac{1}{3}x+3\) for which \(x\) represents the number of weeks since the plants were found, and \(y\)
represents the height in inches.
1. Which graph goes with which equation? How do you know?
2. What is a pair of values that works for Plant A but not B? What does it represent?
3. What is a pair of values that works for Plant B but not A? What does it represent?
4. What is a pair of values that works for both plants? What does it represent?
24.2: Diego’s Plant
1. The height, in centimeters, of Diego’s plant is represented by the equation \(p(t) = \text{-}0.5(t-10)^2+58\) where \(t\) represents the number of weeks since Diego has started nurturing the
plant. Determine if each statement is true or false. Explain your reasoning.
□ Diego’s plant shrinks each week.
□ Diego’s plant is 8 cm tall when he starts to nurture it.
□ Diego’s plant grows to be 58 cm tall.
□ The plant shrinks 4 weeks after Diego begins to nurture it.
2. Write your own true statement about Diego’s plant.
24.3: Making the Grades
Jada’s quiz grade after \(h\) hours of studying is given by the equation \(Q(h) = 10h + 70\). Her test grade after \(h\) hours of studying is given by the equation \(T(h) = 6h + 76\).
Here’s a graph of both functions:
1. Which graph represents Jada’s quiz grade after \(h\) hours of studying?
2. What do the \(y\)-intercepts of the lines mean in this situation?
3. Find the coordinates of the \(y\)-intercepts.
4. The 2 lines intersect at a point. What does that point represent in this situation?
5. Find the coordinates of the intersection point. Explain or show your reasoning. | {"url":"https://curriculum.illustrativemathematics.org/HS/students/4/7/24/index.html","timestamp":"2024-11-04T10:56:12Z","content_type":"text/html","content_length":"79687","record_id":"<urn:uuid:b8ba94c8-993a-45c7-89e2-55aa91ea3235>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00464.warc.gz"} |
Revision history
The method you are looking for is called lift:
sage: g = f.lift()
sage: g
7*x^9 + 9*x^8 + 9*x^7 + 7*x^6 + 8*x^5 + 8*x^4 + 8*x + 8
sage: g.parent()
Univariate Polynomial Ring in x over Ring of integers modulo 14
sage: g.discriminant()
As a side comment: I created Zn = Zmod(14). If you want to work in a finite field with p elements, you should better use K = GF(p) than Zn = Zmod(p) since you are telling Sage that K is a field.
The method you are looking for is called lift:
sage: g = f.lift()
sage: g
7*x^9 + 9*x^8 + 9*x^7 + 7*x^6 + 8*x^5 + 8*x^4 + 8*x + 8
sage: g.parent()
Univariate Polynomial Ring in x over Ring of integers modulo 14
sage: g.discriminant()
As a side comment: I created Zn = Zmod(14). If you want to work in a finite field with p elements, you should better use K = GF(p) than Zn = Zmod(p) since you are telling Sage that K is a field. | {"url":"https://ask.sagemath.org/answers/33880/revisions/","timestamp":"2024-11-14T13:23:16Z","content_type":"application/xhtml+xml","content_length":"16857","record_id":"<urn:uuid:8d49b293-14a9-4345-b321-872a76d5b008>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00120.warc.gz"} |
Types of heat like latent heat
Understand the Problem
The question is asking about different types of heat, specifically mentioning latent heat as one type. It suggests an interest in understanding various forms of heat energy and their characteristics,
such as sensible heat and others.
Latent heat of fusion and latent heat of vaporization.
The two main types of latent heat are the latent heat of fusion and the latent heat of vaporization.
Answer for screen readers
The two main types of latent heat are the latent heat of fusion and the latent heat of vaporization.
More Information
Latent heat of fusion is involved in melting or freezing, while latent heat of vaporization is involved in boiling or condensation.
Sometimes people confuse latent heat with sensible heat. Remember that latent heat relates to phase changes and does not alter the temperature even though energy is absorbed or released. | {"url":"https://quizgecko.com/q/types-of-heat-like-latent-heat-xogjk","timestamp":"2024-11-07T12:19:01Z","content_type":"text/html","content_length":"172024","record_id":"<urn:uuid:61a472e5-7e7d-44f8-8416-11d3bd2753a8>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00600.warc.gz"} |
Mathematics 9 (MYP 4) (3rd Edition)
About the Book
Mathematics 9 MYP 4 third edition has been designed and written for the International Baccalaureate Middle Years Programme (IB MYP) Mathematics framework, providing complete coverage of the content
and expectations outlined.
This book is suitable for students completing the Standard or Extended content of the MYP Mathematics framework, with Extended only content labeled blue within the text.
Discussions, Activities, Investigations, and Research exercises are used throughout the chapters to develop conceptual understanding. Material is presented in a clear, easy-to-follow style to aid
comprehension and retention, especially for English Language Learners. Each chapter ends with extensive review sets and an online multiple-choice quiz.
The associated digital Snowflake subscriptions supports the textbook content with interactive and engaging resources for students and educators.
The Global Context projects highlight the use of mathematics in understanding history, culture, science, society, and the environment. We have aimed to provide a diverse range of topics and styles to
create interest for all students and illustrate the real-world application of mathematics.
We have developed this book in consultation with experienced teachers of IB Mathematics internationally, but independent of the International Baccalaureate Organisation (IBO). It is not endorsed by
the IBO.
We have endeavored to publish a stimulating and thorough textbook and digital resource to develop and encourage student understanding and nurture an appreciation of mathematics.
Year Published: 2022
Page Count: 560
ISBN: 978-1-922416-34-6 (9781922416346)
Online ISBN: 978-1-922416-35-3 (9781922416353) | {"url":"https://www.haesemathematics.com/books/mathematics-9-myp-4-3rd-edition","timestamp":"2024-11-10T02:39:13Z","content_type":"text/html","content_length":"160007","record_id":"<urn:uuid:25cbd07e-f763-4171-9f7f-cc7a37757161>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00659.warc.gz"} |
Roulette Algorithm Probability Loop
Commented: Gaven Henry on 22 Jun 2021
I am making a program that will run a roulette style program 1000000 times. The program should give me an output with my average winnings per trip to the casino. I am starting with 315 dollars and
the conditions for betting are: if I lose i will double my bet and play again. If I win, I will start my bid back at my first initial bid amount (5 dollars) I have created a random number generator
and an equation to determine my probability, I have having trouble with the loop. while numbers are generating I am trying to have the program do the bidding for me and my results are less than
successful. Any ideas? I realize now that one of my equations in the loop is incorrect, I am just concerned with getting the loop to work at the moment. Thanks!
A = 315 %starting amount of money
initial = 5 %Initial bet amount
WIN = 335 %Condition for a win
LOSE = 0 %Condition for loss
profit = 0
g = 0 %counter for random numbers when multiplied by %probability
N = ((log(A+5)-log(5))/log(2)) %Number of bets
Pwinner = (1 - (1/2^N))^4 %Probability of a win
a = initial * (2^N -1) %bid condition for a loss
Gwinner = (1 - (10/19)^N)^4 %probability of win with green slots
%Allow Green Slots?
x = input('Do you want to allow for the green slots (0=No, 1=Yes)? ');
if x == 0;
r = randi([1,36],[1000000 ,1]);
r = randi([1,38],[1000000 ,1]);
fprintf('Please wait while computer simulates game 1000000 times');
%Solution while A < WIN && A > LOSE
g = Pwinner .* r;
if g > 18;
A = A - initial
profit = (1/2^N) * A
fprintf('this isnt working')
3 Comments
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Roulette Algorithm Probability Loop
Are you interested in learning something about proper formatting of code in the forum? As you can see in many other threads, it is possible to format the code much nicer and without inserting blank
lines after each line of code. Could you be so kind and explain, where these lines are coming from? Did you insert the code by copy&paste, and if so, under which OS and browser? Did you see any other
information about proper formatting in the forum? Where could they be placed more prominently, such that beginners will see and consider them? Please help us to improve the forum's layout.
Jan on 5 Apr 2013
Thanks for answering my questions, Mark. Does the opriginal code contain an empty line after each line of code? You can follow the "? Help" link to learn, how to improve the readability of your code.
And making the reading as easy as possible is a good strategy in a forum.
Accepted Answer
More Answers (2)
To summarize the strategie: You allways lose all your money. This is for 37 slots, only one zero slot.
bet0 = 5; % initial bet amount
bank0 = 315; % initial bankroll
display = 1;
bet = bet0;
bank = bank0;
n = 0;
while 1
n = n+1;
r = randi([0 36]); % Roulette Wheel
win = 18 < r;
if win
bank(n+1) = bank(n)+bet(n);
bet(n+1) = bet0;
else % lost
bank(n+1) = bank(n)-bet(n);
bet(n+1) = 2*bet(n);
if bet(n+1) > bank(n+1)
if display, disp_state(win,bank(n:n+1),bet(n:n+1)); end
if display, disp_state(win,bank(n:n+1),bet(n:n+1)); end
plot(bet,'k'), hold on, plot(bank,'r')
function disp_state(win,bank,bet)
wonstr = {'lost','won'};
fprintf('--- %4s\n',wonstr{win+1})
fprintf('Bank [$] %6d -> %4d\n',bank)
fprintf('Bet [$] %6d -> %4d\n',bet)
Guys, if you really want to know how the algorithms for roulette in online casinos are correctly composed, you need to study this process in more detail.
1 Comment
Gaven Henry on 22 Jun 2021
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Displacement MDCAT MCQs with Answers - Youth For Pakistan
Welcome to the Displacement MDCAT MCQs with Answers. In this post, we have shared Displacement Multiple Choice Questions and Answers for PMC MDCAT 2024. Each question in MDCAT Physics offers a chance
to enhance your knowledge regarding Displacement MCQs in this MDCAT Online Test.
Displacement MDCAT MCQs Test Preparations
Displacement is defined as:
a) The total distance traveled by an object
b) The change in position of an object
c) The speed of an object
d) The average velocity of an object
If a person walks 3 km east and then 4 km north, the displacement is:
a) 5 km
b) 7 km
c) 1 km
d) 12 km
Displacement is a:
a) Scalar quantity
b) Vector quantity
c) Rate of change
d) Force
The magnitude of displacement can be found using:
a) Pythagorean theorem
b) Newton’s second law
c) Einstein’s theory
d) Conservation of energy
If an object moves from point A to point B and then returns to point A, its displacement is:
a) Zero
b) Equal to the total distance traveled
c) Half of the total distance traveled
d) The same as the distance between A and B
An object moves 5 meters north and then 5 meters south. The displacement of the object is:
a) 5 meters
b) 10 meters
c) 0 meters
d) 25 meters
Displacement is:
a) Always equal to distance
b) Independent of direction
c) A measure of how far out of the way an object has gone
d) The shortest path between two points
If a car travels 50 km east and then 30 km west, the displacement of the car is:
a) 20 km east
b) 20 km west
c) 80 km
d) 30 km
The displacement of an object in a circular path is:
a) The radius of the circle
b) The circumference of the circle
c) Zero
d) The diameter of the circle
If a person walks 10 meters north, 5 meters east, and 10 meters south, the displacement is:
a) 5 meters east
b) 15 meters
c) 10 meters
d) 10 meters south
Displacement is always:
a) Less than or equal to distance
b) Equal to distance
c) Greater than distance
d) Zero
If a ball is thrown straight up and comes back down to the starting point, its displacement is:
a) Equal to the height it reached
b) Zero
c) The total distance traveled
d) The same as the distance it fell
A person moves 2 km north and then 2 km east. The displacement of the person is:
a) 2√2 km
b) 4 km
c) 2 km
d) 4√2 km
The displacement of an object moving in a straight line at a constant speed is:
a) Zero
b) Equal to the distance traveled
c) Less than the distance traveled
d) Dependent on acceleration
If a car travels 100 km east and then 100 km back to the starting point, the displacement is:
a) 200 km
b) 100 km
c) 0 km
d) 50 km
The displacement of a particle moving in a circular path is:
a) The radius of the circle
b) The circumference of the circle
c) Zero
d) The diameter of the circle
An object moves 3 km west and then 4 km north. The resultant displacement is:
a) 5 km
b) 7 km
c) 1 km
d) 12 km
The shortest distance between two points is represented by:
a) Displacement
b) Distance
c) Path length
d) Velocity
If a person walks 6 km south and then 8 km east, their displacement is:
a) 10 km
b) 14 km
c) 8 km
d) 6 km
The magnitude of displacement is always:
a) Equal to or less than the distance traveled
b) Equal to or greater than the distance traveled
c) The same as the distance traveled
d) Zero if the distance is zero
If an object moves in a closed loop and returns to the starting point, its displacement is:
a) The length of the path
b) The radius of the path
c) Zero
d) The area enclosed by the path
The displacement vector from point A to point B is always:
a) The same as the distance
b) The shortest path from A to B
c) The total distance traveled
d) The longest path from A to B
The displacement of an object moving along a straight path with constant velocity is:
a) Zero
b) Directly proportional to time
c) Inversely proportional to time
d) Constant
A person walks 10 meters east, 5 meters north, and then 5 meters west. The displacement of the person is:
a) 10 meters east
b) 5 meters north
c) 5 meters east
d) 10 meters north
If a car moves from A to B and then from B to C in a straight line, the displacement from A to C is:
a) The sum of the distances from A to B and B to C
b) The distance from A to B plus the distance from B to C
c) The straight-line distance from A to C
d) The average of the distances from A to B and B to C
An object moves 8 meters south, then 6 meters east, and then 8 meters north. The total displacement is:
a) 6 meters east
b) 14 meters
c) 8 meters north
d) 8 meters south
Displacement can be zero if an object:
a) Moves back to its starting point
b) Is moving at constant speed
c) Changes direction constantly
d) Moves in a circular path
If a ball rolls 4 meters to the right and then 3 meters to the left, the displacement is:
a) 1 meter to the right
b) 7 meters
c) 1 meter to the left
d) 4 meters
The direction of displacement is:
a) Always the same as the direction of motion
b) The shortest path between start and end points
c) The average direction of travel
d) Perpendicular to the direction of travel
A particle moves 4 meters north and then 3 meters west. The displacement can be found using:
a) The Law of Sines
b) Pythagorean theorem
c) Newton’s laws
d) Conservation of momentum
A cyclist travels 30 km west and then 40 km east. The displacement of the cyclist is:
a) 10 km east
b) 70 km east
c) 30 km west
d) 40 km east
A person walks 20 meters north and then 10 meters south. The displacement is:
a) 10 meters north
b) 20 meters
c) 30 meters
d) 10 meters south
If a car moves from position A to position B, the displacement is:
a) The distance traveled by the car
b) The direction of travel
c) The shortest distance between A and B
d) The total distance traveled plus the direction
Displacement is affected by:
a) Only the initial and final positions
b) The total distance traveled
c) The path taken
d) The time taken to travel
An object moves 10 meters in one direction and then 10 meters in the opposite direction. The displacement is:
a) 10 meters
b) 20 meters
c) 0 meters
d) 5 meters
If a person moves 5 meters east and then 5 meters north, the displacement is:
a) 10 meters
b) 5√2 meters
c) 5 meters
d) 10√2 meters
The displacement of an object in uniform circular motion is:
a) The circumference of the circle
b) Zero
c) The radius of the circle
d) The diameter of the circle
A person walks 4 meters south, 3 meters west, and then 4 meters north. The displacement of the person is:
a) 3 meters west
b) 4 meters north
c) 4 meters south
d) 7 meters west
An object is displaced from (2,3) to (5,7). The displacement vector is:
a) (3,4)
b) (7,5)
c) (3,-4)
d) (5,7)
Displacement is measured in:
a) Meters
b) Meters per second
c) Kilograms
d) Newtons
If you are interested to enhance your knowledge regarding Physics, Chemistry, Computer, and Biology please click on the link of each category, you will be redirected to dedicated website for each
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SLANSB - Linux Manuals (3)
SLANSB (3) - Linux Manuals
slansb.f -
REAL function slansb (NORM, UPLO, N, K, AB, LDAB, WORK)
SLANSB returns the value of the 1-norm, or the Frobenius norm, or the infinity norm, or the element of largest absolute value of a symmetric band matrix.
Function/Subroutine Documentation
REAL function slansb (characterNORM, characterUPLO, integerN, integerK, real, dimension( ldab, * )AB, integerLDAB, real, dimension( * )WORK)
SLANSB returns the value of the 1-norm, or the Frobenius norm, or the infinity norm, or the element of largest absolute value of a symmetric band matrix.
SLANSB returns the value of the one norm, or the Frobenius norm, or
the infinity norm, or the element of largest absolute value of an
n by n symmetric band matrix A, with k super-diagonals.
SLANSB = ( max(abs(A(i,j))), NORM = 'M' or 'm'
( norm1(A), NORM = '1', 'O' or 'o'
( normI(A), NORM = 'I' or 'i'
( normF(A), NORM = 'F', 'f', 'E' or 'e'
where norm1 denotes the one norm of a matrix (maximum column sum),
normI denotes the infinity norm of a matrix (maximum row sum) and
normF denotes the Frobenius norm of a matrix (square root of sum of
squares). Note that max(abs(A(i,j))) is not a consistent matrix norm.
NORM is CHARACTER*1
Specifies the value to be returned in SLANSB as described
UPLO is CHARACTER*1
Specifies whether the upper or lower triangular part of the
band matrix A is supplied.
= 'U': Upper triangular part is supplied
= 'L': Lower triangular part is supplied
N is INTEGER
The order of the matrix A. N >= 0. When N = 0, SLANSB is
set to zero.
K is INTEGER
The number of super-diagonals or sub-diagonals of the
band matrix A. K >= 0.
AB is REAL array, dimension (LDAB,N)
The upper or lower triangle of the symmetric band matrix A,
stored in the first K+1 rows of AB. The j-th column of A is
stored in the j-th column of the array AB as follows:
if UPLO = 'U', AB(k+1+i-j,j) = A(i,j) for max(1,j-k)<=i<=j;
if UPLO = 'L', AB(1+i-j,j) = A(i,j) for j<=i<=min(n,j+k).
LDAB is INTEGER
The leading dimension of the array AB. LDAB >= K+1.
WORK is REAL array, dimension (MAX(1,LWORK)),
where LWORK >= N when NORM = 'I' or '1' or 'O'; otherwise,
WORK is not referenced.
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
September 2012
Definition at line 129 of file slansb.f.
Generated automatically by Doxygen for LAPACK from the source code. | {"url":"https://www.systutorials.com/docs/linux/man/3-SLANSB/","timestamp":"2024-11-07T13:10:21Z","content_type":"text/html","content_length":"9781","record_id":"<urn:uuid:64fba814-b771-4cfd-8983-92d790c40f3d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00649.warc.gz"} |
6.5 Z Scores
Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Models with a Quantitative Explanatory Variable
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
list High School / Statistics and Data Science I (AB)
6.5 Z Scores
We have looked at the mean as a model; and we have learned some ways to quantify total error around the mean, as well as some good reasons for doing so. But there is another reason to look at both
mean and error together. Sometimes, by putting the two ideas together it can give us a better way of understanding where a particular score falls in a distribution.
A student (let’s call her Zelda) has a thumb length of 65.1 mm. What does this mean? Is that a particularly long thumb? How can we know? By now you may be getting the idea that just knowing the
length of one thumb doesn’t tell you very much.
To interpret the meaning of a single score, it helps to know something about the distribution the score came from. Specifically, we need to know something about its shape, center and spread.
We know that this student’s thumb is about 5 mm longer than the average. But because we have no idea about the spread of the distribution, we still don’t have a very clear idea of how to judge 65.1
mm thumb length. Is a 5 mm distance still pretty close to the mean, or is it far away? It’s hard to tell without knowing what the range of thumb lengths looks like.
Although SS will be really useful later, for this purpose it stinks. 65.1 and 11,880 don’t seem like they belong in the same universe! Variance will also be useful, but its units are still somewhat
hard to interpret. It’s hard to use squared millimeters as a unit when trying to make sense of unsquared millimeters.
Standard deviation, on the other hand, is really useful. We know that Zelda’s thumb is about 5 mm longer than the average thumb. But now we also know that, on average, thumbs are 8.7 mm away from the
mean, both above and below. Although Zelda’s thumb is above average in length, it is definitely not one of the longest thumbs in the distribution. Check out the histogram below to see if this
interpretation is supported.
The mean of thumb length is shown in blue, and Zelda’s 65.1 mm thumb is shown in red.
Combining Mean and Standard Deviation
In the Thumb situation, we find it valuable to coordinate both mean and standard deviation in order to interpret the meaning of an individual score. Now, let’s introduce a measure that will combine
these two pieces of information into a single score: the z score.
Let’s say you know that the mean score across all players of the game is 35,000. How would that help you? Clearly it would help. You would know that the score of 37,000 is above the average by 2,000
points. But even though it helps you interpret the meaning of the 37,000, it’s not enough. What it doesn’t tell you is how far above the average 37,000 points is in relation to the whole distribution
Let’s say the distribution of scores on Kargle is represented by one of these histograms. Both distributions have an average score of 35,000. But in the distribution on the top (#1), the standard
deviation is 1,000 points, while on the bottom (#2) the standard deviation is 5,000 points. The blue line depicts the mean, and the red line depicts our friend’s score of 37,000.
Clearly your friend would be an outstanding player if Distribution 1 were true. But if Distribution 2 were true, they would be just slightly above average.
We can see this visually just by looking at the two histograms. But is there a way to quantify this intuition? One way to do this is by transforming the score we are trying to interpret into a z
score using this formula:
Let’s apply this formula to our video game score of 37,000 based on each of the two hypothetical distributions (#1 and #2) above.
We show you the R code for calculating the z score for a score of 37,000 if Distribution 1 is true. Write similar code to calculate z score if Distribution 2 is true.
require(coursekata) # z-score if distribution 1 were true (37000 - 35000)/1000 # z-score if distribution 2 were true (37000 - 35000)/1000 (37000 - 35000)/5000 ex() %>% { check_output_expr(., "(37000
- 35000)/1000") check_output_expr(., "(37000 - 35000)/5000", missing_msg="Did you divide by the sd of the second distribution?") }
CK Code: ch6-8
[1] 2
[1] 0.4
In both cases, the numerator is the same: 37,000 (the individual score) minus the mean of the distribution, which equals 2,000. The denominators for the two z scores are different, though, because
the distributions have different standard deviations. In distribution #1, the standard deviation is 1,000. So, the z score is 2,000 divided by 1,000, or 2. For the other distribution, the standard
deviation is 5,000. So, the z score is 2,000 divided by 5,000, or .40.
If we did this calculation without parentheses, the calculation would be 37,000 - (35,000 / 5,000) because order of operations, our cultural conventions for how we do arithmetic, says that division
is done before subtraction.
A z score represents the number of standard deviations a score is above (if positive) or below (if negative) the mean. So, the units are standard deviations. A z score of 2 is two standard deviations
above the mean. A z score of 0.4 is 0.4 standard deviations above the mean.
A z score of 2 is more impressive—it’s two standard deviations above the mean. It should be harder to score two standard deviations above the mean than to score 0.4 (or less than one half) a standard
deviation above the mean.
Standard deviation (SD) is roughly the average deviation of all scores from the mean. It can be seen as an indicator of the spread of the distribution. A z score uses SD as a sort of ruler for
measuring how far an individual score is above or below the mean.
A z score tells you how many standard deviations a score is from the mean of its distribution, but doesn’t tell you what the standard deviation is (or what the mean is). Another way to think about it
is that a z score is a way of comparing a deviation of a score (the numerator) to the standard deviation of the distribution (the denominator).
Let’s use z scores to help us make sense of our Thumb data. Calculate the z score for a 65.1 mm thumb.
require(coursekata) # Thumb_stats will have mean and sd in it Thumb_stats <- favstats(~ Thumb, data = Fingers) # write code to calculate the z score for a 65.1 mm Thumb Thumb_stats <- favstats
(~Thumb, data = Fingers) (65.1 - Thumb_stats$mean) / Thumb_stats$sd ex() %>% { check_object(., "Thumb_stats") %>% check_equal() check_output_expr(., "(65.1 - Thumb_stats$mean) / Thumb_stats$sd") }
CK Code: ch6-9
[1] 0.5725349
A single z score tells us how many standard deviations away this particular 65.1 mm thumb is from the mean. Because the standard deviation is roughly the average distance of all scores from the mean,
it is likely that most scores are clustered between one standard deviation above and one standard deviation below the mean. It is less likely to find scores that are two or three standard deviations
away from the mean. Z scores give us a way to characterize scores in a bit finer way than just bigger or smaller than the mean.
Using Z Scores to Compare Scores From Different Distributions
One more use for the z score is to compare scores that come from different distributions, even if the variables are measured on different scales.
Here’s the distribution of scores for all players of the video game Kargle again. We know that the distribution is roughly normal, the mean score is 35,000, and the standard deviation is 5,000.
Her z score is +2. Wow, two standard deviations from the mean! Not a lot of scores are way up there.
Now let’s say you have another friend who doesn’t play Kargle at all. She plays a similar game, though—Spargle! Spargle may be similar, but it has a completely different scoring system. Although the
scores on Spargle are roughly normally distributed, their mean is 50, and the standard deviation is 5. This other friend has a high score of 65 on Spargle.
Now: what if we want to know which friend, in general, is a better gamer? The one who plays Kargle, or the one who plays Spargle? This is a hard question, and there are lots of ways to answer it. The
z score provides one way.
We’ve summarized the z scores for your two friends in the table below.
Player Player Score Game Mean Game SD Player Z Score
Kargle Player 45,000 35,000 5,000 +2.0
Spargle Player 65 50 5 +3.0
Looking at the z scores helps us to compare the abilities of these two players, even though they play games with different scoring systems. Based on the z scores, we could say that the Spargle player
is a better gamer, because she scored three standard deviations above the mean, compared with only two standard deviations above the mean for the Kargle player.
Of course, nothing is really definite with such comparisons. Someone might argue that Spargle is a much easier game, and so the people who play it tend to be novices. Maybe the Kargle player is
better, because even though her z score is lower, she is being compared to a more awesome group of gamers! | {"url":"https://staging.coursekata.org/preview/book/7846adcd-3aea-416d-abb6-f499aa45584e/lesson/9/4","timestamp":"2024-11-14T20:24:49Z","content_type":"text/html","content_length":"86284","record_id":"<urn:uuid:f1cf611e-dfe2-4b92-9aa3-3bc42d082071>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00529.warc.gz"} |
Trend - Overview
Goal: Overview of trend, from data modelling in excel to python.
I have been craving to learn statistic, but I don’t know where to start. The lights comes up when I get a challenge, to get a curve fitting from a series data. From this I can step up to regression
and correlation.
I never skip any statistic class in my college day. But the thing is, they didn’t teach me that far. Wht I got is only basic statistic. While in te same time, I understand other college teach them
ANOVA. Now I have to learn on my own, without class or any course.
Actually I started by reading a statistics book in 2017, this help me get started. There are a lot of books and video that comes up with concept. There are also ready to use online calculator. But
there are not many excel sheet examples online, for daily practical use. But then, how do I suppose to know if my calculation right, if I can’t calculate manually. I need to know how the math works
Even with calculator online, How do I suppose to understand how it works, if it doesn’t comes with the math? Of course there are books for this. But even when I know the concept and the math, how do
I automate my job with scripting?
The knowledge related to statistics has evolve. So I decide to choose another approach:
1. The math behind
2. The manual calculation with excel
3. The implementation with python
Habit Change
How do you start solving math problem?
In college days, I used to take a piece of paper, and I can just right the equation down, and derive stuff over the paper.
Things have changed, there is no much paper anymore. I do not bring pen in my bag, just like the oldschool. So the write down things has also changed.
What tools do we used these days?
We have spreadsheet all around since about 25 years ago. My intuition say that I can just use excel right away, for data modelling. Then we can just take a screenshot, or copy paste the result into
whatsapp. And also attach the spreadsheet file in that very conversation.
How do you communicate the equation?
Sure Excel can produce equation perfectly. The thing is we need to also share, the equation source, in human readable or even machine readable. This can be done perfectly in LaTex.
What is the final form?
Depends on the requirement, and preference. For the polynomial case,I need a visual chart, that I can compare both source data and the XY lines, that the result can be send via any social media.
Although this chart can be easily produce with matplotlib. You may use any programming language.
How do Start
We start from the polynomial curve fitting.
Then we are going to step up to least square along with regression and correlation. Strating from equation cheatsheet for linear regression for samples.
I also make complete worksheet helper for beginner, to solve linear regression along with statistics properties, using manual calculation. After manual calculation, we are going to continue to
utilize built-in formula in Excel/Calc.
We are also going to cover python, from manual calculation, to methods to from numpy and also statistic related library. And then we visualize the interpretation, of the statistics properties using
matplotlib. For example the standard error with level of confidence in shaded region below:
Since we also need to analyze additional statistic properties, and compared with the data distribution, we need to learn the basic of plotting the distribution curve.
So we can put additional statistics in the same chart with the histogram.
Then we are going to enhance matplotlib visualization view with Seaborn. All also provided with Jupyter Lab counterpart in github.
We also need to verify our manual calculation with PSPPire.
Then you can imagine how this can be implemented in different language, such as R, Julia, Typescript, and Golang.
What Comes Next 🤔?
Our journey begins with the built in linset formula in spreadsheet, and built-in polyfit method in python.
Consider diving into the next step by exploring [ Trend - Built-in Method ]. | {"url":"https://epsi.bitbucket.io/statistics/2020/03/01/trend-overview/","timestamp":"2024-11-04T23:08:44Z","content_type":"text/html","content_length":"43560","record_id":"<urn:uuid:1e94dc3f-5d51-4c2a-a1a8-d0e796d6672c>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00436.warc.gz"} |
Maxim Kontsevich, Feb 23 2022
Maxim KontsevichIHES, France
Introduction to wall-crossing
Wall-crossing structures appeared several years ago in several mathematical contexts, including cluster algebras and theory of generalized Donaldson-Thomas invariants. In my lecture I will describe
the general formalism based on a graded Lie algebra and an additive map from the grading lattice to an oriented plane ("central charge").
A geometric example of a wall-crossing structure comes from theory of translation surfaces. The number of saddle connections in a given homology class is an integer-valued function on the parameter
space (moduli space of abelian or quadratic differentials), which jumps along certain walls. The whole theory can be made totally explicit in this case. Also, I'll talk about another closely related
example, which can be dubbed a "holomorphic Morse-Novikov theory". | {"url":"https://vinberg.combgeo.org/maxim-kontsevich/","timestamp":"2024-11-03T12:53:18Z","content_type":"text/html","content_length":"56436","record_id":"<urn:uuid:acb1f037-e5f6-4d9e-ba57-dbd0d5b3031b>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00418.warc.gz"} |
Can't Covert to Column Formula
I am trying to convert some formulas to column formulas, but the feature is greyed out and will not let me. Is this because of the way it is written, or the type of formula being used?
Two specific scenarios are as follows:
1. Target Duration - formula is: =IF([Project Type]@row = "Quote", 10, IF([Project Type]@row = "Profile", 5, IF([Project Type]@row = "Schedule", 5, IF([Project Type]@row = "Special Project", 15,
2. Actual Duration: =IF(ISBLANK([Actual End Date]@row), "TBD", (NETDAYS([Start Date]@row, [Actual End Date]@row)))
…The other formulas I was able to make column formulas as indicated by the icon.
• Create a new column and try it. If that does not work, create a new sheet and try it. I can create column formulas with your formulas.
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/121613/cant-covert-to-column-formula","timestamp":"2024-11-09T04:40:45Z","content_type":"text/html","content_length":"393052","record_id":"<urn:uuid:59807d53-bf2c-41de-ba40-476c41388a66>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00353.warc.gz"} |
Find the range of \
Here we need to find the range of the given function. Range of a function is defined as the set of values which are given out from the function by substituting the value of the domain of the
function. Here, the given function is the square root of some function. So the expression inside the square root can’t be less than zero but can be equal or greater than zero. We will use this
condition to find the value of the variable and hence the range of the given function.
Complete step by step solution:
The given function is \[f\left( x \right) = \sqrt {9 - {x^2}} \].
Here, we need to find the range of this function.
We know that the terms inside the square root can’t be zero but can be equal or greater than zero.
Now, we will write the condition mathematically.
\[ \Rightarrow 9 - {x^2} \ge 0\]
Now, subtracting 9 on both sides, we get
\[\begin{array}{l} \Rightarrow 9 - {x^2} - 9 \ge - 9\\ \Rightarrow - {x^2} \ge - 9\end{array}\]
Multiplying \[ - 1\] on both sides, we get
\[ \Rightarrow {x^2} \le 9\]
Taking square roots on both sides, we get
\[ \Rightarrow \left| x \right| \le 3\]
We can write this inequality as
\[ \Rightarrow - 3 \le x \le 3\]
Therefore, we can say that
\[x \in \left[ { - 3,3} \right]\]
Now substituting \[x = 0\] in \[f\left( x \right) = \sqrt {9 - {x^2}} \], we get
\[ \Rightarrow f\left( x \right) = \sqrt {9 - {0^2}} = \sqrt 9 = 3\]
Now substituting \[x = 3\] in \[f\left( x \right) = \sqrt {9 - {x^2}} \], we get
\[ \Rightarrow f\left( x \right) = \sqrt {9 - {3^2}} = \sqrt {9 - 9} = 0\]
Hence, the range of the function is \[\left[ {0,3} \right]\]
Given range of the function is \[\left[ {a,b} \right]\].
On comparing these ranges, we get
\[\begin{array}{l}a = 0\\b = 3\end{array}\]
Here, I need to calculate \[b - a\].
On substituting the obtained values, we get
\[ \Rightarrow b - a = 3 - 0 = 3\]
1) We have solved the given inequality to find the value of the variable \[x\]. If the expression on the left-hand side and right-hand sides are unequal, that would be called an inequality equation.
We need to know some basic and important properties of inequality. Some important properties of inequality are as follows:-
2) If we add or subtract a term from both sides of an inequality, then inequality remains the same.
3) If we multiply or divide a positive number on both sides of an inequality, then inequality remains the same.
4) But if we multiply or divide a negative number on both sides of an inequality, then inequality does not remain the same. | {"url":"https://www.vedantu.com/question-answer/find-the-range-of-fleft-x-right-sqrt-9-x2-range-class-11-maths-cbse-60106f1d03e4507026fb22e5","timestamp":"2024-11-14T04:53:28Z","content_type":"text/html","content_length":"165226","record_id":"<urn:uuid:a9051d9b-f853-420d-b859-9e098ba0914f>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00880.warc.gz"} |
How Do You Calculate Probability With Examples
Probability is an important concept in mathematics and it can be used in many different fields. It can also be a great way to help you decide which career path is right for you.
To Female Reality Calculator probability, you first need to identify the event you want to determine and then identify the number of outcomes that could occur. This can be as simple as rolling a die
or as complex as choosing an investment strategy.
Probability as a percentage
Probability is a very important concept in many aspects of life. From the odds of your baby having a genetic disorder to the chance of your vehicle breaking down, probability is used in virtually
every field that involves making decisions about something.
When you need to calculate the probability of an event, there are a few different ways to do it. One way is to use percentages.
Using probability as a percentage is easy to do, and it can help you make informed decisions. For example, if you’re planning to launch a new product, probability can be a very useful tool in
predicting its success.
The probability of an event can be calculated as a number between 0 and 1. This number is determined by dividing the favorable number of outcomes by the total number of outcomes.
Probability as a fraction
Probabilities are often expressed in fraction form. They have values between zero and one, with probabilities closer to one indicating that an event is more likely to occur.
In some cases, probabilities are expressed as percentages. They have values between zero and one hundred percent. These percentages are more accurate than fractions and have a greater range of
To calculate probability as a fraction, you need to know the number of ways an outcome can happen (the numerator) and the total number of ways all possible outcomes can occur (the denominator). The
numerator is the same as the probability and the denominator is the same as the ratio.
A probability can be expressed as a fraction if it involves two events that are independent, which means that one event does not affect the other. In other cases, it can be expressed as a probability
where the two events depend on each other.
Probability as an integer
There are several ways to calculate probability. The first step is to identify the event you are trying to determine probability for. This can be an event like winning the lottery or rolling a
certain number with a die.
The second step is to find the number of outcomes that can occur from that event. For example, if you have a jar with 4 blue marbles and 5 red marbles, there are 20 total possible outcomes from
drawing one of the marbles.
Once you have identified the events and their corresponding outcomes, you can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. For
example, if you roll a “3” on a die and it rolls six other numbers, the probability is 1/6.
Probability as an exponent
Probability is a key element of many jobs, from statisticians to meteorologists. It allows you to estimate the likelihood of something happening and make decisions based on it.
When calculating probability, you’ll want to use an exponent. This is because probability is a random variable whose logarithm is normally distributed, and therefore has more small values than larger
To calculate probability as an exponent, you must first determine the mean and variance of the distribution. The mean is a number that represents the average rate of occurrence for independent
The mean of an exponential distribution can be determined by integrating by parts. The distribution’s variance is a number that indicates how much the probability of a specific event changes as you
increase or decrease the scale parameter.
The exponential distribution is often used to model the lifetime of an electrical or mechanical device, like a car battery. The distribution has a key property known as memorylessness, which means
that knowledge of what has happened in the past doesn’t affect the future probability of an event occurring. | {"url":"https://businessgracy.com/female-reality-calculator/","timestamp":"2024-11-04T21:29:20Z","content_type":"text/html","content_length":"56176","record_id":"<urn:uuid:5598c8ad-cc6a-477f-8674-ea41fd13012d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00723.warc.gz"} |
Linear Multilayer Independent Component Analysis for Large Natural Scenes
Part of Advances in Neural Information Processing Systems 17 (NIPS 2004)
Yoshitatsu Matsuda, Kazunori Yamaguchi
In this paper, linear multilayer ICA (LMICA) is proposed for extracting independent components from quite high-dimensional observed signals such as large-size natural scenes. There are two phases in
each layer of LMICA. One is the mapping phase, where a one-dimensional mapping is formed by a stochastic gradient algorithm which makes more highly- correlated (non-independent) signals be nearer
incrementally. Another is the local-ICA phase, where each neighbor (namely, highly-correlated) pair of signals in the mapping is separated by the MaxKurt algorithm. Because LMICA separates only the
highly-correlated pairs instead of all ones, it can extract independent components quite efficiently from ap- propriate observed signals. In addition, it is proved that LMICA always converges. Some
numerical experiments verify that LMICA is quite ef- ficient and effective in large-size natural image processing.
1 Introduction
Independent component analysis (ICA) is a recently-developed method in the fields of signal processing and artificial neural networks, and has been shown to be quite useful for the blind separation
problem [1][2][3] [4]. The linear ICA is formalized as follows. Let s and A are N -dimensional source signals and N N mixing matrix. Then, the observed signals x are defined as x = As. (1)
The purpose is to find out A (or the inverse W ) when the observed (mixed) signals only are given. In other words, ICA blindly extracts the source signals from M samples of the observed signals as
follows: ^ S = W X, (2)
where X is an N M matrix of the observed signals and ^ S is the estimate of the source signals. This is a typical ill-conditioned problem, but ICA can solve it by assuming that the source signals are
generated according to independent and non-gaussian probability dis- tributions. In general, the ICA algorithms find out W by maximizing a criterion (called the contrast function) such as the
higher-order statistics (e.g. the kurtosis) of every com- ponent of ^ S. That is, the ICA algorithms can be regarded as an optimization method of such criteria. Some efficient algorithms for this
optimization problem have been proposed, for example, the fast ICA algorithm [5][6], the relative gradient algorithm [4], and JADE [7][8].
Now, suppose that quite high-dimensional observed signals (namely, N is quite large) are given such as large-size natural scenes. In this case, even the efficient algorithms are not much useful
because they have to find out all the N 2 components of W . Recently, we pro- posed a new algorithm for this problem, which can find out global independent components by integrating the local ICA
modules. Developing this approach in this paper, we propose a new efficient ICA algorithm named " the linear multilayer ICA algorithm (LMICA)." It will be shown in this paper that LMICA is quite
efficient than other standard ICA algo- rithms in the processing of natural scenes. This paper is an extension of our previous works [9][10].
This paper is organized as follows. In Section 2, the algorithm is described. In Section 3, numerical experiments will verify that LMICA is quite efficient in image processing and can extract some
interesting edge detectors from large natural scenes. Lastly, this paper is concluded in Section 4. | {"url":"https://proceedings.nips.cc/paper_files/paper/2004/hash/dbd22ba3bd0df8f385bdac3e9f8be207-Abstract.html","timestamp":"2024-11-10T04:40:02Z","content_type":"text/html","content_length":"11571","record_id":"<urn:uuid:6e9ee386-4274-4d79-83cc-16e0640f8a2a>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00515.warc.gz"} |
How Much Are The Exams Worth In This Course?
How much is a 20% final exam worth?
Aug 06, 2015 · Exit Full Screen. End-of-course tests could count for up to 30 percent of a high school student’s final grade, beginning this year. The …
How much does it cost to take an exams?
Your overall grade depends on how low your lowest test grades are. If your final replaces your lowest test grade, then tell the calculator that your lowest 1 test is dropped and your final also
counts as 1 test. Your current grade is %. You want (at least) a % in the class. Tests are worth % of your grade. Your have taken tests already.
How much is my final grade worth?
Final exam grade = (Target Grade - Current Grade x (100% - Weight of Final (%))) / Weight of Final (%) To use the calculator, you just need to know your current grade, the weight given to the final
exam grade in the overall scoring of the course or class, and to set a goal for your overall grade. If you do not know your current weighted average grade, then use our grade calculator instead.
Is it worth it to take the same classes in college?
End-of-Course Exam (2 Hours)—45% of AP Seminar Score. Component. Scoring Method. Weight. Understanding and analyzing an argument (3 short-answer questions); suggested time: 30 minutes. College Board
scored. 30% of 45%. Evidence-Based argument essay (1 long essay); suggested time: 90 minutes. College Board scored.
How much of your course grade are the exams worth?
Calculating the Grade
For example, if you got a score of 90 percent on the test and the test is worth 20 percent of your overall grade, you would multiply 90 by 0.2 for a value of 18 points out of the possible 20 points.
Mar 13, 2018
How much are final exams usually worth?
And most of the finals are worth 20% of your grade which is what it was in high school. And finals are not super hard, designed to make you fail. So then what is the difference between college finals
and high school finals since that is the title of this article?Dec 17, 2014
How much points are exams worth?
So, 100% on the final exam gives the student 30 points. If you need to calculate a "worth" score, choose the value above.
Sample Test Score Chart: 25 Point Test (Worth 30 Points)
Points Off Score Worth
-0 100% 30
-1 96% 28.8
-2 92% 27.6
-3 88% 26.4
23 more rows
How much are exams worth in high school?
End-of-course tests could count for up to 30 percent of a high school student's final grade, beginning this year. The exams — in core subjects such as math and English — now count for 15 percent of
the final grade in Lafayette Parish.Aug 6, 2015
How much does 20% affect your grade?
If the Final Exam is worth 20%, then every point above your current average will raise your grade . 2 points, and every point below your current average will lower your grade .May 30, 2017
How much is a final exam worth in college?
The final exam is worth 22 percent of your final grade.
19 reading assignments 38 percent
1 final exam 23 percent
Total Points 100 percent
2 more rows
How much will a 76 affect my grade?
Use this calculator to find out the grade needed on the final exam in order to get a desired grade in a course. It accepts letter grades, percentage grades, and other numerical inputs.
Your final is worth:
Letter Grade GPA Percentage
C+ 2.3 77-79%
C 2 73-76%
C- 1.7 70-72%
D+ 1.3 67-69%
9 more rows
What is my grade out of 200 points?
The total answers count 200 - it's 100%, so we to get a 1% value, divide 200 by 100 to get 2.00. Next, calculate the percentage of 200: divide 200 by 1% value (2.00), and you get 100.00% - it's your
percentage grade.
How much will a 70 affect my grade?
80% to 89% mark: Good job. Ranges from 70% to 79%: Satisfactory results.
Grade Calculator – Frequently Asked Questions.
Letter Grade Percentage 4.0 Scale
C+ 77-79 2.3
C 73-76 2.0
C- 70-72 1.7
D+ 67-69 1.3
8 more rows
Is a 76 percent good?
C - this is a grade that rests right in the middle. C is anywhere between 70% and 79% D - this is still a passing grade, and it's between 59% and 69% F - this is a failing grade.Jan 10, 2022
How much will a 0 affect my grade if I have a 88?
88 average get a zero on final exam and final exam counts 20% of grade.May 10, 2016
How to calculate minimum grade required for final exam?
Therefore, you can calculate the minimum grade you need to score on the final exam using the formula: Required = (Goal − Current × (100% − Final Weight)) / Final Weight.
How to calculate final grade?
Therefore, your final grade can be calculated using the formula: Grade = Exam Worth × Exam Score + (1 – Exam Worth) × Current Grade.
How much do I need to get on my final exam?
When the end of the year, semester, or course approaches, students often want to find out what grade is needed on the final exam in order to get a desired grade in a course or class. The final grade
calculator can tell you what final grade is needed, at minimum, in order to achieve the target overall grade.
How will my final affect my grade?
The effect of the final exam grade on the overall grade for a school class or college course depends both on the score you obtain and on its weight towards the overall grade.
How to convert my letter grade to percentage?
In case you only know your current weighted grade as a letter grade and not a percentage, you will need to convert it to a percentage before you can calculate the final grade you need to acquire.
When can students try out the test day experience?
Starting April 8, students can try out the test-day experience by answering example questions in the digital testing application. See the Digital Practice page for general information about practice
Can you use videos as homework?
Videos can be assigned as homework to encourage students to watch on their own, so you can use class time to focus discussions where students need more help, whether teaching online, in person, or
both. Sign in to AP Classroom to access the AP Seminar question bank and AP Daily videos.
Why do students use annotations?
Students can use the digital annotation tool to highlight key elements within texts, organize their thoughts, and create brief notes. While the annotations that students construct will not be scored,
annotating is an opportunity for students to analyze texts to help them as they write their responses.
Final Grade Calculator
Use this calculator to find out the grade needed on the final exam in order to get a desired grade in a course. It accepts letter grades, percentage grades, and other numerical inputs.
Brief history of different grading systems
In 1785, students at Yale were ranked based on "optimi" being the highest rank, followed by second optimi, inferiore (lower), and pejores (worse). At William and Mary, students were ranked as either
No. 1, or No. 2, where No. 1 represented students that were first in their class, while No.
An alternative to the letter grading system
Letter grades provide an easy means to generalize a student's performance.
Is AP class worth it?
Learning the reputation of AP classes at your high school can help you decide if taking certain AP classes is worth your time. There is a huge amount of variation in how AP classes are taught at
different high schools. You'll struggle if an AP class is poorly taught or overly difficult. On the flipside, a well-taught class might not only help you pass the AP test, but also help you discover
a new academic interest.
What happens if you don't pass the AP exam?
If you didn't pass, the AP class loses a lot of its admissions benefit. Even (or especially!) if you get an A in the class, that would just mean your class was too easy compared with those in the
rest of the country. In short, how good an AP class will be for you hinges on whether you pass the exam or not.
What happens if you pass the AP test?
If you pass the AP test at the end of the school year (meaning you get a score of 3 or higher ), you will be eligible to receive college credit for that test.
Can AP classes hurt your college chances?
For example, if you neglect studying for the SAT/ACT because you're so busy with AP classes, this can really hurt your college admission chances.
Why do we need AP classes?
AP classes exist to expose high school students to college-level courses. Even though you're taking the class at your high school, AP classes tend to have harder, more detailed curriculums than your
typical high school classes do.
Final Exam Grade Calculation
Final Class Grade Calculation
• Sometimes you'll get your final exam grade but the instructor hasn't yet posted final class grades. You can use this calculator to find your final class grade once you know your final exam score.
Use the formula: Where: 1. G = Grade you'll receive for the class 2. F = Final exam grade 3. w = Weight of the final exam, divided by 100 (put weight in decimal form vs. percentage form) 4. C = …
See more on calculatorsoup.com
Example Final Exam Grade Calculation
• My grade in Statistics class is 85%. I want to get at least an A- or 90% in the class for the term. What score do I need on the final exam if it is worth 40% of my grade? Using the Final Exam
Grade formula above, I want a 90 in the class and I currently have an 85. The final is worth 40% of the term grade. First, convert the weight of the final exam from percent to decimal: 40 ÷ 100 =
0.40 …
See more on calculatorsoup.com
Example Final Course Grade Calculation
• Going into finals my grade in Economics was 91%. My final exam score was 88.6% and it was worth 15% of my grade for the course. What is my final grade in the course? Using the Final Class Grade
formula above, F = 88.6, w = 15, and C = 91. First, convert the weight of the final exam from percent to decimal: 15 ÷ 100 = 0.15 So I went into the final w...
See more on calculatorsoup.com | {"url":"https://course-faq.com/how-much-are-the-exams-worth-in-this-course","timestamp":"2024-11-12T20:35:42Z","content_type":"text/html","content_length":"52192","record_id":"<urn:uuid:e06b9331-4e6d-4b36-b18a-98fb551535ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00314.warc.gz"} |
Interview Questions - Algorithm, Data Structure | Youngho Chaa cha cha
Interview Questions - Algorithm, Data Structure
— algorithm, interview — 1 min read
These are my revision notes for my job interviews. As a contractor, I have job interviews more often than a permanent employee. Writing down the revision notes does not imply that I am incapable of
answering these questions naturally. During interviews, I naturally get nervous, and due to my introversion, I often struggle to clearly articulate what I already know and understand. I simply want
to avoid those situations and prepare myself to present myself in the best possible way.
What is big O?
• Mathematical notation
• Describe the time complexity or space complexity of an algorithm.
• Time complexity of O(n) will have its execution time grow linearly as the input size increases.
• Time complexity is O(1), the algorithm executes in constant time, regardless of input size.
• O(n²) means the time taken will be proportional to the square of the size of input, and so on.
What is recurrence?
In the context of computer science and algorithm analysis, recurrence is a way of defining a problem in terms of one or more smaller instances of the same problem.
A good example of this is the Fibonacci sequence, where each number is the sum of the two preceding ones. In terms of a recurrence relation, it could be defined as:
with base cases: | {"url":"https://andrewchaa.me.uk/interview-questions-algorithm-data-structure/","timestamp":"2024-11-02T10:52:49Z","content_type":"text/html","content_length":"117422","record_id":"<urn:uuid:0b6358c7-42b7-41a4-acb8-b62487be3f63>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00246.warc.gz"} |
RSI with Dynamic levels
I am posting after a gap. Got hold of some original Wyckoff stuff and has been spending some time on it. More about it in a later post. Today I am presenting a Regular Classical Indicator RSI with a
variation, a RSI with dynamic levels. No I am not talking about the RSI with Bollinger bands around it.
In the Stocks & Commodities V15:7 Leo Zamansky and David Stendahl talked about dynamic zones. They said that oscillator driven systems lack the ability to evolve with the market because they use
fixed buy and sell zones. Typically the set of buy and sell zones for a bull market will be substantially different zones for a bear market. We need to have a system automatically define its own buy
and sell zones and thereby profitably trade in any market — bull or bear. Dynamic zones offer a solution to the problem of fixed buy and sell zones for any oscillator-driven system. The idea of their
system was to create a distribution of the signals in the given look back period. Then we have to find the value which is equal to the desired probability.
First of all the assumption is the distribution is normal distribution is a little far fetched actually. On that assumption we need to do the laborious calculation of the making the distribution and
calculating the probability. But I would like to keep things simple. Trading is art and not complicated science. Precise calculation will not immensely improve your trading system. In order to
simplify the matters we can just assume uniform distribution and calculate the probability accordingly. I knew people will find it difficult to accept this. However even this simplification will
provide a adequately dynamic zones. Here I am presenting the conventional RSI with Dynamic Levels. Also in order not to confuse with widely available Dynamic Zone indicator with Bollinger bands I
will call this indicator RSI with Dynamic Levels.
6 comments:
1. hello sir, i am using ur MAMA positioning system since a year and it is working fine. Sir i want the logic used in it in simple English so i can use it in another charting software. please guide
me thru it will help a lot. the code is: _SECTION_BEGIN("MAMA"); SetBarsRequired( 10000, 10000 ); SetChartOptions(0,chartShowArrows|chartShowDates); prc = ( High + Low ) / 2; fastlimit = 0.5;
slowlimit = 0.05; pi=4*atan(1); RTD=180/pi; DTR=pi/180; Cyclepart=Optimize("Alpha",Param ("Alpha",0.2,0.1,1,0.1),0.1,1,0.1); Smooth[0] = Period = Detrender[0] = I1[0] = Q1[0] = 0; phase[0]=
deltaphase[0]=MAMA[0]=FAMA[0]=0; for ( i = 6; i < BarCount; i++ ) { Smooth[i] = ( 4 * prc[i] + 3 * prc[i-1] + 2 * prc[i-2] + prc[i-3] ) / 10; AmpCorr[i] = 0.075 * Period[i-1] + 0.54; Detrender[i]
= ( 0.0962 * Smooth[i] + 0.5769 * Smooth[i-2] - 0.5769 * Smooth[i-4] - 0.0962 * Smooth[i-6] ) * AmpCorr[i]; Q1[i] = ( 0.0962 * Detrender[i] + 0.5769 * Detrender[i-2] - 0.5769 * Detrender[i-4] -
0.0962 * Detrender[i-6] ) * AmpCorr[i]; I1[i] = Detrender[i-3]; if (I1[i] != 0) phase[i] = DTR*360/atan(q1[i]/I1[i]); deltaphase[i]=phase[i-1]-phase[i]; if (deltaphase[i] <1) deltaphase[i]=1;
alpha[i]=fastlimit[i]/deltaphase[i]; if (alpha[i] < slowlimit[i]) alpha[i]=slowlimit[i]; MAMA[i]=alpha[i] * prc [i] +(1-alpha[i])*MAMA[i-1]; FAMA[i]=Cyclepart*alpha[i] * prc [i] +
(1-Cyclepart*alpha[i])*FAMA[i-1]; } //Plot( MAMA, "MAMA", colorTurquoise, styleLine|styleThick ); //Plot( FAMA, "FAMA", colorYellow, styleLine|styleThick ); PlotOHLC(O,H,L,C,"MAMA",IIf(MAma>
fama,colorLime,colorRed),styleCandle|styleThick); _SECTION_END(); _SECTION_BEGIN("TWI SYSTEM"); BuySetupValue=ValueWhen(Cross(MAMA,FAMA),H,1); shortsetupValue=ValueWhen(Cross(FAMA,MAMA),L,1);
Buysetup =Cross(MAMA,FAMA) ; shortsetup = Cross(FAMA,MAMA); Longa = Flip(Buysetup,shortsetup); shrta = Flip(shortsetup,Buysetup); Buy=Longa AND Cross(C,BuySetupValue); Short=shrta AND Cross
(shortsetupValue,C); Buy = ExRem(Buy,Short); Short = ExRem(Short,Buy); Sell = Short; Cover = Buy; t1= Flip(Buy,Short); t2= Flip(Short,Buy); BPrice=ValueWhen(t1 AND Ref(t1,-1)==0,C,1); SPrice=
ValueWhen(t2 AND Ref(t2,-1)==0,C,1); GraphXSpace = 20; dist = 4.5*ATR(20); for( i = 0; i < BarCount; i++ ) { if( Buy[i] ) PlotText( "" + C[ i ], i, L[ i ]-dist[i], colorLime); if( Short[i] )
PlotText( "" + C[ i ], i, H[ i ]+dist[i], colorYellow ); } _SECTION_END();
2. Hello sir,....How do i get exploration AFL for this indicator.Kindly help
3. Can you code this for Tradestation?
4. Great afl sir,
do you have "stochastic with dynamic level" for amibroker ?
5. how do I use this thing????
6. Hi Karthik, I came across your RSI Indicator, liking it a lot. I was hoping that you could help me to understand how to interpret the info its giving. I will really appreciate that | {"url":"http://karthikmarar.blogspot.com/2013/08/rsi-with-dynamic-levels.html","timestamp":"2024-11-02T21:55:42Z","content_type":"text/html","content_length":"87355","record_id":"<urn:uuid:fa6b77ed-4bbd-4fb4-a6e4-2b7d94dd8837>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00626.warc.gz"} |
Kim Group
As a condensed matter theory group, we find ideas and motivations from experiments. Naturally, our projects evolve with experimental developments. Condensed matter physics is nestled in the
intersection where material science, nanotechnology, chemistry, physics and mathematics meet. Rapid experimental developments fueled by new materials and probing techniques offer many opportunities
to work on new problems and impact experimental research. We enjoy bringing real world materials together with mathematical and beautiful theoretical concepts and calculations.
We tend to focus on the physics of strongly correlated systems. In weakly correlated systems, such as semiconductors and most metals, the electron-electron interaction energy is far smaller than
individual electron's kinetic energy. Hence the system can be sucessfully thought in terms of the motion of independent electrons: electrons behave like molecules in gas. On the other hand,
interaction energy is either more dominant or on par with kinetic energy in strongly correlated systems. As a result, the system as a whole display properties that cannot be boiled down to the
properties of individual electrons: electrons behave like molecules in liquid, well aware of each other. Examples of such systems include electrons in a two-dimensional(2D) plane under strong
magnetic field (quantum Hall regime), high temperature superconductors, and low-dimensional (1D or 2D) systems in the clean limit. Strongly interaction effect open up many new possibilities for the
material properties. For instance, they may lead to exotic ground states and excitations that act like a fraction of an electron. We may need to use sophisticated mathematics to describe such
excitations. Also, analogies to seemingly unrelated systems can be very useful in understanding new phenomena. For example, the concept of electronic analogue of liquid crystals is turning out to be
quite useful in describing a number of systems.
We take "holistic approach" in our research. On one hand, we search for the common thread within experimental observations and use it as the foundation and check for our theoretical research. At the
same time, we combine various analytical methods such as field theory and renormalization group, with computational techniques such as quantum Monte Carlo through collaborations.
Research Areas
The diversity of emergent phenomena and opportunities to connect theoretical ideas and calculations to experimetanlly observed phenomena are major attractions for condensed matter theory. While much
effort of our group go into the study of strongly correlated systems such as high temperature superconductors and quantum Hall effect, we work on a diverse set of topics as you can see from our
publication list. For instance, we have worked on fractional vortices in chiral superconductors, curvature effects on graphene, edge states of topological insulator, topological quantum phase
transition, charge density waves in rare-earth-Tellurides, nodal nematic quantum phase transition, and intra-unit-cell nematicity in high Tc superconductors. See below for topics of current research. | {"url":"http://eunahkim.ccmr.cornell.edu/research","timestamp":"2024-11-09T12:08:17Z","content_type":"text/html","content_length":"10610","record_id":"<urn:uuid:dd9eff59-fdd7-4561-b5de-138c9029996b>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00419.warc.gz"} |
Interpolate digital signal and translate it from baseband to IF band
The dsp.DigitalUpConverter System object™ interpolates a digital signal, and translates it from baseband to intermediate frequency (IF) band.
To digitally upconvert the input signal:
1. Create the dsp.DigitalUpConverter object and set its properties.
2. Call the object with arguments, as if it were a function.
To learn more about how System objects work, see What Are System Objects?
This object supports C/C++ code generation and SIMD code generation under certain conditions. For more information, see Code Generation.
upConv = dsp.DigitalUpConverter returns a digital up-converter (DUC) System object, upConv.
upConv = dsp.DigitalUpConverter(Name=Value) returns a DUC System object with the specified property Name set to the specified value Value. You can specify one or more name-value pair arguments in any
order as (Name1=Value1,...,NameN=ValueN). For example, create an object that upsamples the input signal by a factor of 20, using a filter with the specified qualities.
upConv = dsp.DigitalUpConverter(InterpolationFactor=20,...
Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the release function unlocks them.
If a property is tunable, you can change its value at any time.
For more information on changing property values, see System Design in MATLAB Using System Objects.
InterpolationFactor — Interpolation factor
100 (default) | positive integer | vector of positive integers
Interpolation factor, specified as a positive integer, or a 1-by-2 or 1-by-3 vector of positive integers.
When you set this property to a scalar the object automatically chooses the interpolation factors for each of the three filtering stages.
When you set this property to a 1-by-2 vector, the object bypasses the first filter stage and sets the interpolation factor of the second and third filtering stages to the values in the first and
second vector elements, respectively. Both elements of this InterpolationFactor vector must be greater than 1.
When you set this property to a 1-by-3 vector, the ith element of the vector specifies the interpolation factor for the ith filtering stage. The second and third elements of this InterpolationFactor
vector must be greater than 1 and the first element must equal 1 or 2.
Data Types: double
MinimumOrderDesign — Minimum order filter design
true (default) | false
Minimum order filter design, specified as true or false.
When you set this property to true, the object designs filters with the minimum order that meets the passband ripple, stopband attenuation, passband frequency, and stopband frequency specifications
that you set using the PassbandRipple, StopbandAttenuation, Bandwidth, StopbandFrequencySource, and StopbandFrequency properties.
When you set this property to false, the object designs filters with orders that you specify in the FirstFilterOrder, SecondFilterOrder, and NumCICSections properties. The filter designs meet the
passband and stopband frequency specifications that you set using the Bandwidth, StopbandFrequencySource, and StopbandFrequency properties.
Data Types: logical
SecondFilterOrder — Order of CIC compensation filter stage
12 (default) | positive integer
Order of CIC compensation filter stage, specified as a positive integer.
To enable this property, set the MinimumOrderDesign property to false.
Data Types: double
FirstFilterOrder — Order of first filter stage
10 (default) | positive even integer
Order of first filter stage, specified as a positive even integer.
To enable this property, set the MinimumOrderDesign property to false. When you set the InterpolationFactor property to a 1-by-2 vector, the object ignores the FirstFilterOrder property, because the
first filter stage is bypassed.
Data Types: double
NumCICSections — Number of sections of CIC interpolator
3 (default) | positive integer
Number of sections of CIC interpolator, specified as a positive integer.
To enable this property, set the MinimumOrderDesign property to false.
Data Types: double
Bandwidth — Two-sided bandwidth of input signal in Hz
200000 (default) | positive integer
Two-sided bandwidth BW of the input signal, specified as a positive integer in Hz or in normalized frequency units (since R2024b). The object sets the passband frequency of the cascade of filters to
half of the value that you specify in this Bandwidth property.
Data Types: double
StopbandFrequencySource — Source of stopband frequency
Auto (default) | Property
Source of the stopband frequency, specified as Auto or Property.
When you set this property to Auto and set:
• NormalizedFrequency to false –– The object places the cutoff frequency of the cascade filter response at approximately F[c] = Fs/2 Hz and computes the stopband frequency as F[stop] = F[c] + TW/2.
TW is the transition bandwidth of the cascade response, computed as 2×(F[c]–F[p]). F[p] is the passband frequency computed by BW/2, where BW is the two-sided bandwidth of the input signal.
• NormalizedFrequency to true –– The object places the cutoff frequency of the cascade filter response at approximately F[c] = 1/L, where L is the total interpolation factor specified in the
InterpolationFactor property, and computes the stopband frequency as F[stop] = F[c] + TW/2. TW is the transition bandwidth of the cascade response, computed as 2×(F[c]–F[p]), and the passband
frequency F[p] equals BW/2, where BW is the two-sided bandwidth of the input signal.
When you set this property to Property, you can specify the stopband frequency value using the StopbandFrequency property.
StopbandFrequency — Stopband frequency
150000 (default) | positive scalar
Stopband frequency F[stop], specified as a positive scalar in Hz or in normalized frequency units (since R2024b).
To enable this property, set the StopbandFrequencySource property to Property.
Data Types: double
PassbandRipple — Passband ripple of cascade response in dB
0.1 (default) | positive scalar
Passband ripple of cascade response in dB, specified as a positive scalar. When you set the MinimumOrderDesign property to true, the object designs the filters so that the cascade response meets the
passband ripple that you specify in this PassbandRipple property.
To enable this property, set the MinimumOrderDesign property to true.
Data Types: double
StopbandAttenuation — Stopband attenuation of cascade response in dB
60 (default) | positive scalar
Stopband attenuation of cascade response in dB, specified as a positive scalar. When you set the MinimumOrderDesign property to true, the object designs the filters so that the cascade response meets
the stopband attenuation that you specify in this StopbandAttenuation property.
To enable this property, set the MinimumOrderDesign property to true.
Data Types: double
Oscillator — Type of oscillator
Sine wave (default) | NCO
Type of oscillator, specified as one of these:
• Sine wave –– The object frequency-upconverts the output of the interpolation filter cascade by using a complex exponential signal obtained from samples of a sinusoidal trigonometric function.
• NCO –– The object frequency-upconverts the output by using a complex exponential obtained from a numerically controlled oscillator (NCO).
CenterFrequency — Center frequency of output signal in Hz
14000000 (default) | positive scalar
Center frequency of the output signal Fc, specified as a positive scalar in Hz or in normalized frequency units (since R2024b). The value of this property must be less than or equal to half the
product of the SampleRate property and the total interpolation factor. The object upconverts the input signal so that the output spectrum centers at the frequency you specify in the CenterFrequency
Data Types: double
NormalizedFrequency — Option to set frequencies in normalized units
false (default) | true
Since R2024b
Option to set frequencies in normalized units, specified as one of these values:
• true –– The center frequency, stopband frequency, and bandwidth must be in the normalized frequency units (0 to 1).
When you set the NormalizedFrequency property to true while creating the object and you do not set the frequency specifications, the object automatically sets the default values to normalized
frequency units. The object computes the frequencies in normalized units by normalizing the absolute frequency values in Hz with respect to the output sample rate, Fs×L, where L is the
interpolation factor.
duc = dsp.DigitalUpConverter(NormalizedFrequency=true)
duc =
dsp.DigitalUpConverter with properties:
InterpolationFactor: 100
MinimumOrderDesign: true
Bandwidth: 0.0133
StopbandFrequencySource: 'Auto'
PassbandRipple: 0.1000
StopbandAttenuation: 60
Oscillator: 'Sine wave'
CenterFrequency: 0.9333
NormalizedFrequency: true
When you set the NormalizedFrequency property to true after you create the object, you must specify the center frequency, stopband frequency, and bandwidth in normalized units before you run the
object algorithm.
duc = dsp.DigitalUpConverter
duc =
dsp.DigitalUpConverter with properties:
InterpolationFactor: 100
MinimumOrderDesign: true
Bandwidth: 200000
StopbandFrequencySource: 'Auto'
PassbandRipple: 0.1000
StopbandAttenuation: 60
Oscillator: 'Sine wave'
CenterFrequency: 14000000
NormalizedFrequency: false
SampleRate: 300000
To specify the normalized frequency values, set NormalizedFrequency to true and manually convert the frequency values in Hz to the normalized values using the output sample rate in Hz, Fs×L. The
bandwidth value in normalized units is BW[Hz]/(Fs×L/2), the center frequency in normalized units is F[c][Hz]/(Fs×L/2), and the stopband frequency in normalized units is F[stop][Hz]/(Fs×L/2).
duc = dsp.DigitalUpConverter;
duc.NormalizedFrequency = true;
duc.Bandwidth = 200000/(300e3*100/2);
duc.CenterFrequency = 14e6/(300e3*100/2)
duc =
dsp.DigitalUpConverter with properties:
InterpolationFactor: 100
MinimumOrderDesign: true
Bandwidth: 0.0133
StopbandFrequencySource: 'Auto'
PassbandRipple: 0.1000
StopbandAttenuation: 60
Oscillator: 'Sine wave'
CenterFrequency: 0.9333
NormalizedFrequency: true
• false –– The bandwidth, stopband frequency, and center frequency values are in Hz. You can specify the input sample rate through the SampleRate property.
Data Types: logical
SampleRate — Sample rate of input signal
300000 (default) | positive scalar
Sample rate of the input signal Fs, specified as a positive scalar value. The value of this property multiplied by the total interpolation factor must be greater than or equal to twice the value of
the CenterFrequency property.
To enable this property, set NormalizedFrequency to false. (since R2024b)
Data Types: single | double
NCO Properties
NumAccumulatorBits — Number of NCO accumulator bits
16 (default) | integer in the range [1, 128]
Number of NCO accumulator bits, specified as an integer in the range [1,128]. For more details, see the dsp.NCO System object.
To enable this property, set the Oscillator property to NCO.
Data Types: double
NumQuantizedAccumulatorBits — Number of NCO accumulator bits
12 (default) | integer in the range [1, 128]
Number of NCO accumulator bits, specified as an integer in the range [1,128]. The value you specify for this property must be less than the value you specify in the NumAccumulatorBits property. For
more details, see the dsp.NCO System object.
To enable this property, set the Oscillator property to NCO.
Data Types: double
Dither — Dither control for NCO
true (default) | false
Dither control for NCO, specified as true or false. When you set this property to true, the object uses the number of dither bits specified in the NumDitherBits property when applying dither to the
NCO signal. When this property is false, the NCO does not apply dither to the signal. For more details, see the dsp.NCO System object.
To enable this property, set the Oscillator property to NCO.
Data Types: logical
NumDitherBits — Number of NCO dither bits
4 (default) | positive integer
Number of NCO dither bits, specified as a positive integer scalar smaller than the number of accumulator bits that you specify in the NumAccumulatorBits property. For more details, see the dsp.NCO
System object.
To enable this property, set the Oscillator property to NCO and the Dither property to true.
Data Types: double
Fixed-Point Properties
FiltersOutputDataType — Data type at output of each filter stage
Same as input (default) | Custom
Data type at the output of the first (if it has not been bypassed), second, and third filter stages, specified as Same as input or Custom. The object casts the data at the output of each filter stage
according to the value you set in this property. For the CIC stage, the casting is done after the signal is scaled by the normalization factor.
CustomFiltersOutputDataType — Fixed-point data type at output of each filter stage
numerictype([],16,15) (default) | numerictype object
Fixed-point data type at output of each filter stage, specified as a scaled numerictype (Fixed-Point Designer) object with the Signedness property set to Auto.
To enable this property, set the FiltersOutputDataType property to Custom.
OutputDataType — Data type of output
Same as input (default) | Custom
Data type of output, specified as Same as input or Custom.
CustomOutputDataType — Fixed-point data type of output
numerictype([],16,15) (default) | numerictype object
Fixed-point data type of output, specified as a scaled numerictype object the Signedness property set to Auto.
To enable this property, set the OutputDataType property to Custom.
y = upConv(x)returns an upsampled and frequency-upconverted signal y, for a real or complex input column vector x.
Input Arguments
x — Input signal
column vector
Input signal, specified as a column vector of real or complex values.
When the data type of x is double or single, the data type of y is the same as that of x. When the data type of x is of a fixed-point type, the data type of y is defined by the OutputDataType
Data Types: single | double | int8 | int16 | int32 | int64 | fi
Complex Number Support: Yes
Output Arguments
y — Upconverted and upsampled signal
column vector
Upconverted and upsampled signal, returned as a column vector. The length of y is equal to the length of x multiplied by the value in the InterpolationFactor property. When the data type of x is
double or single, the data type of y is the same as that of x. When the data type of x is of a fixed-point type, the data type of y is defined by the OutputDataType property.
Data Types: single | double | int8 | int16 | int32 | int64 | fi
Object Functions
To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named obj, use this syntax:
Specific to dsp.DigitalUpConverter
getInterpolationFactors Get interpolation factors of each filter stage of digital upconverter
getFilterOrders Get orders of digital down converter or digital up converter filter cascade
getFilters Get handles to digital down converter or digital up converter filter cascade objects
groupDelay Group delay of digital down converter or digital up converter filter cascade
visualize Display response of digital down converter or digital up converter filter cascade
generatehdl Generate HDL code for quantized DSP filter (requires Filter Design HDL Coder)
Common to All System Objects
step Run System object algorithm
release Release resources and allow changes to System object property values and input characteristics
reset Reset internal states of System object
Upconvert Sine Wave Signal
Create a DUC System object™ that upsamples a 1-kHz sinusoidal signal by a factor of 20 and upconverts it to 50 kHz.
Create a sine wave generator to obtain the 1-kHz sinusoidal signal with a sample rate of 6 kHz.
Fs = 6e3; % Sample rate
sine = dsp.SineWave(Frequency=1000,...
x = sine(); % generate signal
Create a DUC System object. Use minimum order filter designs and set the passband ripple to 0.2 dB and stopband attenuation to 55 dB. Set the double-sided signal bandwidth to 2 kHz.
upConv = dsp.DigitalUpConverter(...
Create a spectrum estimator to visualize the signal spectrum before and after upconverting.
window = hamming(floor(length(x)/10));
figure; pwelch(x,window,[],[],Fs,'centered')
title('Spectrum of baseband signal x')
Upconvert the signal and visualize the spectrum.
xUp = upConv(x);
window = hamming(floor(length(xUp)/10));
title('Spectrum of upconverted signal xUp')
Visualize the response of the interpolation filters.
More About
Fixed Point
The block diagram represents the DUC arithmetic with signed fixed-point inputs.
• WL is the word length of the input, and FL is the fraction length of the input.
• The output of each filter is cast to the filter output data type. In the dsp.DigitalUpConverter object, you can specify the filter output data type through the FiltersOutputDataType and
CustomFiltersOutputDataType properties. In the Digital Up-Converter block, you can specify the filter output data type through the Stage output parameter. The casting of the CIC output occurs
after the scaling factor is applied.
• The oscillator output is cast to a word length equal to the filter output data type word length plus one. The fraction length is equal to the filter output data type word length minus one.
• The scaling at the output of the CIC interpolator consists of coarse-gain and fine-gain adjustments. The coarse gain is achieved using the reinterpretcast (Fixed-Point Designer) function on the
CIC interpolator output. The fine gain is achieved using full-precision multiplication.
The figure shows the coarse-gain and fine-gain operations.
If the normalization gain is G (where 0<G≦1), then:
• WL[cic] is the word length of the CIC interpolator output, and FL[cic] is the fraction length of the CIC interpolator output.
• F1 = abs(nextpow2(G)), indicating the part of G achieved by using bit shifts (coarse gain).
• F2 is the fraction length specified by the filter output data type.
• fg = fi((2^F1)*G,true,16), which indicates that the remaining gain cannot be achieved with a bit shift (fine gain).
The digital up converter upsamples the input signal using a cascade of three interpolation filters. This algorithm frequency-upconverts the upsampled signal by multiplying it with a complex
exponential that has the specified center frequency. In this case, the filter cascade consists of an FIR interpolation stage, a second stage for CIC compensation, and a CIC interpolator. The block
diagram shows the architecture of the digital up converter.
The scaling section normalizes the CIC gain and the oscillator power. It can also contain a correction factor to achieve the desired ripple specification. Depending on how you set the interpolation
factor, the block bypasses the first filter stage. When the input data type is floating point, the algorithm implements an N-section CIC interpolation filter as a FIR filter with a response that
corresponds to a cascade of N boxcar filters. The algorithm emulates a CIC filter with an FIR filter so that you can run simulations with floating-point data. When the input data type is a
fixed-point type, the algorithm implements a true CIC filter with actual comb and integrator sections.
This block diagram represents the DUC arithmetic with floating-point inputs.
For details about fixed-point operation, see Fixed Point.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
See System Objects in MATLAB Code Generation (MATLAB Coder).
This object also supports SIMD code generation using Intel^® AVX2 code replacement library when the input signal has a data type of single or double.
The SIMD technology significantly improves the performance of the generated code. For more information, see SIMD Code Generation. To generate SIMD code from this object, see Use Intel AVX2 Code
Replacement Library to Generate SIMD Code from MATLAB Algorithms.
HDL Code Generation
Generate VHDL, Verilog and SystemVerilog code for FPGA and ASIC designs using HDL Coder™.
This object supports HDL code generation with the Filter Design HDL Coder™ product. For workflows and limitations, see Generate HDL Code for Filter System Objects (Filter Design HDL Coder).
Version History
Introduced in R2012a
R2024b: visualizeFilterStages has been renamed to visualize
The visualizeFilterStages function has been renamed to visualize. Existing instances of this function continue to run. For new instances, use visualize.
R2024b: visualize launches MATLAB figure
The visualize function now launches a MATLAB^® figure to display the magnitude response of the digital up converter filter cascade.
R2024b: Support for normalized frequencies
When you set the NormalizedFrequency property to true, you must specify the bandwidth, stopband frequency, and the center frequency in normalized frequency units (0 to 1). For more information, see
the NormalizedFrequency property description.
See Also | {"url":"https://uk.mathworks.com/help/dsp/ref/dsp.digitalupconverter-system-object.html","timestamp":"2024-11-05T06:48:03Z","content_type":"text/html","content_length":"169486","record_id":"<urn:uuid:2f0faebb-4df0-4739-b892-6bb599f4aa2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00461.warc.gz"} |
Understanding the Cost of Capital.
Understanding Cost of Capital
The cost of capital is a financial metric that serves as a benchmark for an array of investment decisions. It represents the rate of return that a company must earn on its investments to satisfy its
investors and maintain its capital structure. Essentially, it’s the price the company pays for using capital to fund its operations and growth initiatives.
Investors also use the cost of capital as a gauge for the risk associated with investing in a firm. An analyst may measure it as a discount rate, which is applied to forecast future cash flows to
their present value. The discount rate reflects both the time value of money and the risk of the investment.
Components of Cost of Capital:
• Cost of Equity: Return expected by shareholders for investing equity into the company.
• Cost of Debt: Also factored into the overall cost of capital, this is the effective interest rate a company pays on its borrowed funds.
Component Description
Equity Return demanded by equity investors (dividends/price growth).
Debt Interest rates on bonds, loans, and other forms of debt.
When they assess an investment opportunity, investors look at the expected return to ensure it exceeds the cost of capital. If it does, the investment can be deemed profitable in terms of risk-reward
ratio. The market risk premium plays a role here, which represents the additional return over the risk-free rate that investors require for choosing a risky investment.
It should be noted that the cost of capital can vary with market conditions and the specific risk profile of the company. Calculating the accurate cost of capital requires considering both the cost
of debt and equity, adjusting for taxes, and incorporating the impact of financial policies.
Components of Capital
Understanding the components of capital is essential for businesses and investors as they consider the mix of debt and equity in a company’s capital structure and the costs associated with each
Cost of Debt
The Cost of Debt refers to the effective rate that a company pays on its borrowed funds. These funds can come from sources like loans or bonds. The interest rate on debt is often lower than the cost
of equity due to its tax-deductible nature, which can reduce the after-tax cost. To calculate the after-tax cost of debt, one subtracts the corporate tax rate from the interest rate. For instance, if
the interest rate on a bond is 5% and the corporate tax rate is 30%, the after-tax cost of debt is 3.5% after considering tax savings.
Cost of Equity
The Cost of Equity represents the compensation that the market demands in exchange for owning the asset and bearing the risk of ownership. Two common models used to estimate the cost of equity are
the Dividend Capitalization Model and the Capital Asset Pricing Model (CAPM). CAPM calculates the cost of equity using the formula: Risk-Free Rate + Beta*(Market Return – Risk-Free Rate). The beta
reflects the investment’s volatility relative to the market, a key determinant of equity risk and, therefore, its cost. The risk-free rate is usually based on the yield of government bonds,
signifying a return with minimal risk.
Weighted Average Cost of Capital
The Weighted Average Cost of Capital (WACC) integrates the cost of debt and cost of equity into a single figure that reflects the overall cost of capital for a company. It is calculated by weighting
the cost of each capital component by its proportion in the company’s capital structure. WACC is crucial as it represents the minimum return a company must earn to satisfy its debtors, investors, and
shareholders. The formula for WACC is:
(WACC) = (E/V) * (Cost of Equity) + (D/V) * (Cost of Debt) * (1 – Corporate Tax Rate)
where E is the market value of equity, D is the market value of debt, V is E + D, E/V is the proportion of equity in the company’s capital structure, and D/V is the proportion of debt. WACC is a
crucial determinant in investment decisions and corporate finance strategies. It serves as a benchmark for evaluating investment opportunities, as investments should aim to deliver returns above this
rate to be considered viable.
Calculating Cost of Capital
In corporate finance, determining the cost of capital is essential as it influences investment decisions, corporate strategies, and acquisition analysis. It’s a pivotal factor in assessing the
relative attractiveness of various opportunities by identifying the minimum return necessary to undertake a specific investment.
Formulas and Calculation Methods
Calculating the cost of capital involves several key components, each with its own formula. The Weighted Average Cost of Capital (WACC) is widely used and encompasses the costs of both equity and
debt financing. The formula for WACC is expressed as:
WACC = (E / V) × Re + (D / V) × Rd × (1 – Tc)
E represents the market value of equity,
D is the market value of debt,
V equals E + D (total value),
Re is the cost of equity,
Rd the cost of debt, and
Tc the corporate tax rate.
The Cost of Equity can be calculated using the Capital Asset Pricing Model (CAPM), which relates the expected return on an investment to its risk:
Cost of Equity = Risk-Free Rate + Beta × Market Risk Premium
Beta (β) measures a stock’s volatility relative to the overall market, and Market Risk Premium is the difference between the expected market return and the risk-free rate.
For stable companies with consistent dividends, the Dividend Capitalization Model might be more suitable:
Cost of Equity = (Dividends per Share in Next Year / Current Market Value of Stock) + Growth Rate of Dividends
Practical Applications in Valuation
Cost of capital acts as a hurdle rate or discount rate when evaluating future cash flows of investment opportunities. It is utilized to determine the present value of expected cash flows for projects
or enterprises. For instance, during valuation, it is imperative to discount future cash flows back to present value using WACC to determine if the Net Present Value (NPV) or Internal Rate of Return
(IRR) justifies the investment.
Factors Influencing Cost
Various elements affecting the cost of capital include credit spread, default risk, and the industry‘s inherent financial risk. A company’s balance sheet, funding structure, and even the weight of
free cash flows can modify the overall cost due to changes in perceived market value of the securities involved. As businesses operate, these components must be analyzed continually to ensure optimal
capital structure and to adhere to the principle of value maximization.
Capital Allocation Strategies
In the realm of financial management, crafting astute capital allocation strategies is paramount for organizations looking to optimize their balance sheets and ensure the most profitable use of
resources. These strategies encompass discerning investment decisions, evaluating opportunities for return on investment, and determining how the composition of capital structure influences cost.
Investment Decisions and Project Evaluation
When allocating capital, companies meticulously assess various investment opportunities to gauge potential return on investment. This involves a comparison of the expected profitability against the
opportunity cost of alternative projects. Often financial managers use tools like net present value (NPV) and internal rate of return (IRR) to evaluate projects, considering factors such as future
cash flows, cost of capital, and the time value of money.
• Evaluation Criteria:
□ Net Present Value (NPV)
□ Internal Rate of Return (IRR)
□ Payback Period
Strategic project selection aligns with long-term corporate goals and drives sustained growth. It is one of the most consequential aspects of capital allocation and necessitates robust analysis to
avoid suboptimal fund deployment.
Influence of Capital Structure on Cost
A company’s capital structure—its mix of debt and equity—profoundly affects its cost of capital. The delicate balance on a company’s balance sheet between these sources of finance must account for
the risk and cost associated with each. Debt financing can be less expensive due to tax deductibility of interest, but it elevates financial risk. Conversely, equity is costlier but does not require
periodic interest payments, influencing the company’s financial leverage.
• Capital Composition:
□ Debt: Lower cost, higher risk.
□ Equity: Higher cost, no repayment obligation.
Understanding the interplay between debt and equity empowers financial leaders to optimize their capital structure, thereby potentially reducing the overall cost of capital and reinforcing the
company’s financial health.
Advanced Considerations for Analysts
In their efforts to accurately assess a company’s cost of capital, analysts integrate complex market dynamics and rigorous financial methodologies. This process not only informs investment decisions
but also influences strategic financial management within the firms.
Market Conditions and Industry Benchmarks
Analysts assess market conditions to gauge the cost of capital, underpinned by industry benchmarks and the prevailing market rate of return. By drawing comparisons with peers and sector standards,
they estimate the premium investors require. These comparisons are essential in ensuring that estimates reflect the economic landscape and industry-specific risks.
Leveraging Financial Metrics and Tools
Employing financial metrics and tools like Excel is commonplace for an analyst working in financial management or an accounting department. Metrics such as the Weighted Average Cost of Capital (WACC)
provide quantitative backing for estimates. Analysts also factor in variables like interest rates to figure the cost of different capital components, pivotal for strategic decision-making.
The Role of Assumptions in Cost Estimation
The foundation of cost estimation lies in the assumptions made by the analyst. Estimates heavily rely on these assumptions, ranging from macroeconomic conditions to firm-specific factors. Accurate
and well-founded assumptions are vital; they determine the realism and reliability of the cost of capital figures as determined by the analysts. | {"url":"https://richriddles.com/terms/c/cost-of-capital/","timestamp":"2024-11-14T22:00:09Z","content_type":"text/html","content_length":"51584","record_id":"<urn:uuid:95845cb8-51a1-41d0-b3eb-8a71d2b626e7>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00166.warc.gz"} |
Best Feet to Inches Conversion Calculator (ft to in) with a fraction - convertmastertoolFeet to Inches Calculator with Fraction - Convert Feet to Inches Online
Feet to Inches with Fraction
Welcome to the "Feet to Inches Converter with fraction"
How to Convert Feet to Inches
To convert a measurement from feet to inches, you can utilize a straightforward formula. Multiply the length in feet by the conversion ratio of 12 inches per foot. This is because there are 12 inches
in one foot.
The conversion formula can be expressed as follows:
inches = feet × 12
By applying this formula, you can determine the length in inches by multiplying the length in feet by 12.
The foot is a unit of length used for measurement, and it has various characteristics and uses. Here is some information about the foot:
• The foot is equivalent to 12 inches or 1/3 of a yard. This means that there are 12 inches in one foot and three feet in one yard.
• In terms of the metric system, since the international yard is defined as exactly 0.9144 meters, one foot is equal to 0.3048 meters.
• The foot is primarily used as a unit of length in the United States customary and imperial systems of measurement.
• The abbreviation for feet is “ft.” For example, you can represent one foot as 1 ft.
• An alternative way to denote feet is by using the prime symbol (′), but it is common to use a single-quote (‘) instead for simplicity. Thus, 1 ft can be written as 1’ as well.
• When measuring in feet, a standard 12-inch ruler or a tape measure is typically used, although there are other measuring devices available for this purpose.
• The term “linear feet” is sometimes used to refer to measurements expressed in feet. It simply indicates a measurement of length in feet.
• If you need to perform calculations involving feet and other units like inches, centimeters, or meters, you may find our “inches -to-feet with fraction” calculator useful.
• An inch is equal to 1/12 of a foot or 1/36 of a yard. In terms of the metric system, with the international yard defined as precisely 0.9144 meters, one inch is equivalent to 2.54 centimeters.
• The inch is predominantly used as a unit of length in the United States customary and imperial systems of measurement.
• The abbreviation for inches is “in.” For instance, you can represent one inch as 1 in.
• Alternatively, inches can be denoted using the double-prime symbol (″). However, it is common to substitute it with a double-quote (“) for simplicity. Thus, 1 inch can be expressed as 1″.
• A quarter has a diameter of approximately .955 inches, just slightly smaller than 1 inch.
• The standard ruler is 12 inches long and serves as a widely utilized tool for measuring length in inches. Additionally, a tape measure, ranging from 6 feet to 35 feet in length, is frequently
employed for inch measurements. Other devices employed for measuring in inches include scales, calipers, measuring wheels, micrometers, yardsticks, and even lasers.
• If you wish to delve deeper into the concept of inches and its usage in measuring length, you can explore further information.
To measure length accurately in inches, it is recommended to utilize a ruler or tape measure, which can be obtained from local retailers or home centers. Ensure that you select the appropriate type
of measurement device—imperial, metric, or a combination—to suit your specific requirements.
An inch is a unit of length commonly used for measurement, and it possesses several characteristics and applications. Here is a breakdown of information about the inch:
Link this Tool to Own Website
Link this Tool to Your Website
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First-Order Algorithms for Convex Smooth Optimization Problems With Homogeneous Linear Constraints
Document Type
Degree Name
Doctor of Philosophy (PhD)
Mathematical Sciences
Committee Chair/Advisor
Yuyuan Ouyang
Committee Member
Matthew Saltzman
Committee Member
Boshi Yang
The purpose of this dissertation is to explore the first-order methods that can be used to solve an approximate solution for convex smooth problems with homogeneous linear constraints. It consists of
three interconnected research projects.
In the first project, we study the problem of computing the projection of a given vector onto the kernel of a symmetric positive semi-definite matrix. The complexity of an algorithm for computing a
numerical solution is evaluated by the total number of matrix-vector multiplications required for computing an approximate solution. Such problems arise commonly in consensus optimization, in which
the total number of matrix-vector multiplications is the same as the total number of communications needed to reach a consensus. We show that an accelerated gradient descent algorithm can compute an
$\varepsilon$-approximate solution with at most $\cO(\log(1/\varepsilon))$ matrix-vector multiplications.
In the second project, we study convex smooth optimization with general homogeneous linear constraints. This optimization problem includes the decentralized multi-agent optimization problem as a
special case. Motivated by \cite{lihuan2020decentralized}, we show the classical accelerated gradient method can be adapted to solve this problem. The adaptation we make is to allow the accelerated
gradient method to solve the subproblem approximately. In order to solve the subproblem, we call the algorithm designed in the first chapter. Our proposed algorithm is called the accelerated penalty
method (APM). We show that to compute an $\varepsilon$-approximate solution when the objective function is strongly convex, the total number of gradient evaluations of the objective function and
matrix-vector multiplications involving the constraint matrix are bounded by $\cO(\log(1/\varepsilon))$ and $\cO(\log^2(1/\varepsilon))$ respectively. When the objective function is non-strongly
convex, the corresponding complexity bounds are $\cO(1/\sqrt{\varepsilon})$ and $\cO((1/\sqrt{\varepsilon})\log(1/\varepsilon))$ respectively.
In the third project, we show that one other variant of the accelerated gradient method could be adapted to develop an APM analogy. It should be noted that the adaptation is non-trivial since APM is
not a straightforward implementation of the accelerated gradient method but requires a penalty strategy and inexact solutions to the subproblem. We show that by a slightly different penalty strategy,
we can develop a second version of the accelerated penalty method (APM2). Our proposed APM2 has the same order of complexity as that of APM.
Recommended Citation
Guo, Yidan, "First-Order Algorithms for Convex Smooth Optimization Problems With Homogeneous Linear Constraints" (2024). All Dissertations. 3759.
Available for download on Sunday, August 31, 2025
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Equivalence | Bridging Practices Among Connecticut Mathematics Educators
This task is designed for sixth grade students developing skills with geometric formulas for area while also working on algebraic interpretations of expressions and equations. Students are given the
formula for the area of a triangle, written in two different ways. Students must use logic and algebraic knowledge to determine if the two expressions are equivalent, and why. This task highlights
the relationship between multiplying by a fraction, and dividing. Students are critiquing a student interpretation.
Microsoft Word version: 6_EE_ExpressionsEquivalence_Problem_Critique
PDF version: 6_EE_ExpressionsEquivalence_Problem_Critique
Gr 5_NBT_DecimalsEquivalence_Problem_Critique
This task is designed for fifth graders and uses a visual picture of a 100% block. It asks students to critique two student responses that state how much of the block is shaded. Each student takes a
different approach, but essentially states equivalent answers. One student looks at all of the pieces individually, as 23 ones, and the other student looks at the pieces like place values, as 2 tens
and 3 ones. Students must recognize that these are two different ways to arrive at the same answer. Students are asked to draw pictures and use words to explain how the two students came about the
answer in different ways.This task is an excellent example of how students may use different interpretations and methods to solve the same problem.
Microsoft Word version: 5_NBT_DecimalsEquivalence_Problem_Critique
PDF version: 5_NBT_DecimalsEquivalence_Problem_Critique
Gr 4_NF_FractionsEquivalence_Problem_Critique_AgreeingOrDisagreeingWithSally
Agreeing or Disagreeing with Sally is a task designed for fourth graders working on fraction fluency. Students are asked to compare two fractions with different numerators and different denominators.
Students must critique an argument that states that two fractions are equivalent by agreeing or disagreeing and explaining the decision. The task will also allow students to demonstrate ability to
create equivalent fractions.
Microsoft Word version: 4_NF_FractionsEquivalence_Problem_Critique_AgreeingOrDisagreeingWithSally
PDF version: 4_NF_FractionsEquivalence_Problem_Critique_AgreeingOrDisagreeingWithSally
Gr 3_NF_FractionsEquivalence_ThinkPairShare_Critique
I used this task in my 3rd grade classroom as an introduction to equivalent fractions. Students were given a visual, as well as a written description of 2 different ways that students split a pizza.
The idea was for students to shade in fraction pieces and see that the 2 fractions were equivalent, however students were not told how to go about solving the problem. The task gives students a place
to write their claim and argument, as well as a place to write what their partner thinks.
Microsoft Word version: 3_NF_FractionsEquivalence_ThinkPairShare_Critique
PDF version: 3_NF_FractionsEquivalence_ThinkPairShare_Critique
Gr 3_NF_FractionsEquivalence_Problem_Critique_WhatDoYouThink
What Do You Think is for third graders to understand the concept of fraction equivalence. Students are given a square divided into sections, with some of them shaded, and two responses as to the
amount of squares shaded. Argumentation language is used when asking students to critique the given responses and explain their own thinking.
Microsoft Word version: 3_NF_FractionsEquivalence_Problem_Critique_WhatDoYouThink
PDF version: 3_NF_FractionsEquivalence_Problem_Critique_WhatDoYouThink | {"url":"https://bridges.education.uconn.edu/tag/equivalence/","timestamp":"2024-11-11T17:50:34Z","content_type":"text/html","content_length":"64804","record_id":"<urn:uuid:ff0dd6a4-a3dc-4f29-ac0d-280a66f96ca9>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00824.warc.gz"} |
Combination of lenses.. can you solve it? | HIX Tutor
Combination of lenses.. can you solve it?
Answer 1
Let us solve it using the following lens maker formula
$\textcolor{b l u e}{\frac{1}{f} = \left(\mu - 1\right) \left(\frac{1}{R} _ 1 - \frac{1}{R} _ 2\right) \ldots \ldots . . \left[1\right]}$
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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The Stacks project
Remark 101.19.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ be a point. To indicate the equivalent conditions of Lemma 101.19.4 are satisfied
for $f$ and $x$ in the literature the terminology $f$ is stabilizer preserving at $x$ or $f$ is fixed-point reflecting at $x$ is used. We prefer to say $f$ induces an isomorphism between automorphism
groups at $x$ and $f(x)$.
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Mathematics Is More Than Just A Language. It Is Language Plus Logic.
Mathematics Is More Than Just A Language. It Is Language Plus Logic.
Isaac Newton’s Principia
There is no model of the theory of gravitation today, other than the mathematical form.
It this were the only law of this character it would be interesting and rather annoying. But what turns out to be true is that the more we investigate, the more laws we find, and the deeper we
penetrate nature, the more this disease persists. Every one of our laws is a purely mathematical statement in rather complex and abstruse mathematics. Newton’s statement of the law of gravitation is
relatively simple mathematics. It gets more and more abstruse and more and more difficult as we go on. Why? I have not the slightest idea. It is only my purpose here to tell you about this fact. The
burden of the lecture is just to emphasize the fact that it is impossible to explain honestly the beauties of the laws of nature in a way that people can feel, without their having some deep
understanding of mathematics. I am sorry, but this seems to be the case.
You might say, “All right, then if there is no explanation of the law, at least tell me what the law is. Why not tell me in words instead of symbols? Mathematics is just a language, and I want to be
able to translate the language.” In fact I can, with patience, and I think I partly did. I could go a little further and explain in more detail that the equation means that if the distance is twice
as far the force is one fourth as much, and so on. I could convert all the symbols into words. In other words I could be kind to the layman as they all sit hopefully waiting for me to explain
something. Different people get different reputations for their skill at explaining to the layman in layman’s language these difficult and abstruse subjects. The layman then searches for book after
book in the hope that he will avoid the complexities which eventually set in, even with the best expositor of this type. He finds as he reads a generally increasing confusion, one complicated
statement after another, one difficult-to-understand thing after another, all apparently disconnected from one another. It becomes obscure, and he hopes that maybe in some other book there is some
explanation… The author almost made it- maybe another fellow will make it right.
But I do not think it is possible, because mathematics is not just another language. Mathematics is a language plus reasoning; it is like a language plus logic. Mathematics is a tool for reasoning.
It is in fact a big collection of the results of some person’s careful thought and reasoning. By mathematics it is possible to connect one statement to another. For instance, I can say that the force
is directed towards the sun. I can also tell you, as I did, that the planet moves so that if I draw a line from the sun to the planet, and draw another line at some definite period, like three weeks,
later, then the area that is swung out by the planet is exactly the same as it will be in the next three weeks, and the next three weeks, and so on as it goes around the sun. I can explain both of
those statements carefully, but I cannot explain why they are both the same. The apparent enormous complexities of nature, with all its funny laws and rules, each of which has been carefully
explained to you, are really very closely interwoven. However, if you do not appreciate the mathematics, you cannot see, among the great variety of facts, that logic permits you to go from one to
It may be unbelievable that I can demonstrate that equal areas will be swept out in equal times if the forces are directed towards the sun. So if I may, I will do one demonstration to show you that
those two things really are equivalent, so that you can appreciate more than the mere statement of the two laws. I will show that the two laws are connected so that reasoning alone will bring you
from one to the other, and that mathematics is just organized reasoning. Then you will appreciate the beauty of the relationship of the statements. I am going to prove the relationship that if the
forces are directed towards the sun then equal areas are swept out in equal times.
Figure 1
We start with a sun and a planet (Figure 1), and we imagine that at a certain time the planet is at position 1. It is moving in such a way that, say, one second later it has moved to position 2. If
the sun did not exert a force on the planet, then, by Galileo’s principle of inertia, it would keep right on going in a straight line. So after the same interval of time, the next second, it would
have moved exactly the same distance in the same straight line, to the position 3. First we are going to show that if there is no force, then equal areas are swept out in equal times. I remind you
that the area of a triangle is half the base times the altitude, and that the altitude is the vertical distance to the base. If the triangle is obtuse (Figure 2), then the altitude is the vertical
height AD and the base is BC.
Figure 2
Now let us compare the areas which would be swept out if the sun exerted no force whatsoever (Figure 1). The two distances 1-2 and 2-3 are equal, remember. The question is, are the two areas equal?
Consider the triangle made from the sun and the two points 1 and 2. What is its area? It is the base 1-2, multiplied by half the perpendicular height from the baseline to S. What about the other
triangle, the triangle in the motion from 2 to 3? Its area is the base 2-3, times half the perpendicular height to S. The two triangles have the same altitude, and, as I indicated, the same base, and
therefore they have the same area. So far so good. If there were no force from the sun, equal areas would be swept out in equal times. But there is a force from the sun. During the interval 1-2-3 the
sun is pulling and changing the motion in various directions towards itself. To get a good approximation we will take the central position, or average position, at 2, and say that the whole effect
during the interval 1-3 was to change the motion by some amount in the direction of the line 2-S. (Figure 3).
Figure 3
This means that though the particles were moving on the line 1-2, and would, were there no force, have continued to move on the same line in the next second, because of the influence of the sun the
motion is altered by an amount that is poking in a direction parallel to the line 2-S. The next motion is therefore a compound of what the planet wanted to do and the change that has been induced by
the action of the sun. So the planet does not really end up at position 3, but rather at position 4. Now we would like to compare the areas of the triangles 23S and 24S, and I will show you that
those are equal. They have the same base, S-2. Do they have the same altitude? Sure, because they are included between parallel lines. The distance from 4 to the line S-2 is equal to the distance
from 3 to line S-2 (extended). Thus the area of the triangle S24 is the same as S23. I proved earlier that S12 and S23 were equal in area, so we now know S12 = S24. So, in the actual orbital motion
of the planet the areas swept out in the first second and the second second are equal. Therefore, by reasoning, we can see a connection between the fact that the force is towards the sun, and the
fact that the areas are equal. Isn’t that ingenious? I borrowed it straight from Newton. It comes right out of the Principia, diagram and all. Only the letters are different, because he wrote in
Latin and these are Arabic numerals…
Mathematics, then, is a way of going from one set of statements to another. It is evidently useful in physics, because we have these different ways in which we can speak of things, and mathematics
permits us to develop consequences, to analyze the situations, and to change the laws in different ways to connect the various statements. In fact the total amount that a physicist knows is very
little. He has only to remember the rules to get him from one place to another and he is all right, because all the various statements about equal times, the force being in the direction of the
radius, and so on, are all interconnected by reasoning.
Similar Posts: | {"url":"https://www.collectedthoughts.com/2017/11/05/mathematics-is-more-than-just-a-language-it-is-language-plus-logic/","timestamp":"2024-11-03T07:15:52Z","content_type":"text/html","content_length":"105862","record_id":"<urn:uuid:ae96fb32-0965-4b2c-aed2-f14efd886efc>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00050.warc.gz"} |
Average True Range - Traders Log
Average True Range
A volatility measurement indicator introduced by J. Welles Wilder in his book: New Concepts in Technical Trading Systems. Wilder originally developed the ATR for commodities but the indicator is also
be used for stocks and indices. The ATR measures a commodity or security’s volatility. High ATR values reflects high volatility while low ATR readings indicate range bound movement. True Range is
defined as the largest difference of:
-current high minus the current low.
-the absolute value of the current high minus the previous close.
-the absolute value of the current low minus the previous close.
Average True Range (ATR) is a moving average of true range calculated over a series of days.
Chart courtesy of Prophet Financial Systems | {"url":"https://www.traderslog.com/average-true-range","timestamp":"2024-11-03T14:04:44Z","content_type":"text/html","content_length":"239492","record_id":"<urn:uuid:3c44b3f5-571f-4977-b98f-d4da35777c90>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00200.warc.gz"} |
Cinema Problem
A cinema has 100 seats. How can ticket sales make £100 for these different combinations of ticket prices?
Cinema Problem printable sheet
Alison's cinema has 100 seats.
One day, Alison notices that her cinema is full,
and she has taken exactly £100.
The prices were as follows:
Adults £3.50
Pensioners £1.00
Children £0.85
She knows that not everyone in the audience was a pensioner!
Is it possible to work out how many adults, pensioners and children were present?
You may want to start by trying different ways of filling all 100 seats.
e.g. 5 adults, 20 pensioners and 75 children
Does this earn you £100?
Too much? Too little?
Can you tweak the numbers to get closer to £100?
You may find this spreadsheet useful.
What other interesting mathematical questions can you think of to explore next?
We have thought of some possibilities:
Is there only one possible combination of adults, pensioners and children that add to 100 with takings of exactly £100?
Can there be 100 people and takings of exactly £100 if the prices are:
Adults £4.00 Adults £5.00
Pensioners £1.00 or Pensioners £2.50
Children £0.50 Children £0.50
Can you find alternative sets of prices that also offer many solutions? What about exactly one solution?
If I can find one solution, can I use it to help me find other solutions?
If a children's film has an audience of 3 children for every adult (no pensioners), how could the prices be set to take exactly £100 when all the seats are sold?
What about a family film where adults, children and pensioners come along in the ratio 2:2:1?
This problem is based on Cinema Problem from SIGMA 1, by David Kent and Keith Hedger
Getting Started
What happens to the amount of money taken if an adult is replaced by a pensioner? Or a pensioner with a child? Or...?
Student Solutions
We had loads of really great solutions to this problem, so many that unfortunately we can't mention you all by name, but thanks to anyone who submitted a solution.
Thank you to Jaclyn from Brighton College for her solution to this problem.
3 adults, 47 pensioners, 50 children.
1. Children can only go up in 10s.
2. I tried 10 children but it didn't add up to 100.
2. I tried 20 children but it didn't work.
2. I tried 30 children with 3 adults and it didn't work.
2. I tried 40 children and ... it didn't' work.
3. Finally, I tried 50 children and it worked perfectly with 3 adults and 47 pensioners!
We have a couple of questions about this solution. Firstly, why does the number of children have to be a multiple of 10? Secondly, what do you think Jaclyn meant by 'it didn't work'?
Jaclyn also gave these solutions for the later ticket prices in the problem.
Solution for adult price A=£4, pensioner price P=£1 and child price C=£0.50.
Solution for adult price A=£5, pensioner price P=£2.50 and child price C=£0.50.
These give us one solution for each set of ticket prices, but are there any more solutions for these?
Here's a solution from Ellie from Hardwick Middle School - she used trial and error, but made in a very logical way.
My solution was that there were $3$ adults, $47$ pensioners and $50$ children. I mainly used trial and error, which is where you pick numbers from an educated guess and see if it works. If it doesn't
you then change your answers. I knew that there would be very few adults because their prices were too high at £3.50 and if $28$ adults visited that would take the price to £98 meaning there would
not be enough money remaining to have $100$ people in the cinema. So I started with the adults at two because that made a whole number( £6). I then had $50$ pensioners and $48$ children. That
equalled $100$ people but £97.80 which was too little. To take the price up I changed $2$ adults to $3$ and $50$ pensioners to $49$. This equalled £100.30 though, which was too much. I changed the
$49$ pensioners to $48$ pensioners and $48$ children to $49$. It equalled £100.15 which was £0.15 too much but I had realized that the difference between pensioners and children was £0.15. Therefore
I changed $48$ pensioners to $47$ and $49$ children to $50$ which saved £0.15 and then the number of people in the cinema was 100 and ticket sales equalled £100 meaning I had solved the problem.
Abhishek from William Law CofE sent us this solution, which uses the idea of lowest common multiples:
From the question we know that everyone in the cinema hall is not a pensioner. If everyone is a child, the total cost will be too low because $100 \times 0.85 = 85$. If everyone is an adult, the
total cost will be too high because $100 \times 3.50 = 350$. On an average, everybody's ticket costs £1 so if there are only pensioners and children, the total cost will be too low. If there are only
pensioners and adults, the total cost will be too high. If there are only adults and children, we can find a solution but the answer is not a whole number. Therefore, there are adults, children and
pensioners in the cinema hall. For every adult, the price is £2.50 more than the average price. For every child, the price is 15p less than the average price. The LCM of $250$p and $15$p is $750$p,
i.e. £7.50. £2.50 $\times3$ is £7.50 and $15$p $\times 50$ is £7.50. Therefore, for every $50$ children, there are $3$ adults.
We can conclude that there are $3$ adults, $47$ pensioners and $50$ children.
9MAT1A from Diocesan Girls School in New Zealand sent us this solution, which takes a similar approach using simultaneous equations:
Initially we have two pairs of simultaneous equations and $3$ variables. If $A$
is the number of adults, $C$ the number of children and $P$ the number of
pensioners then we find:
$$3.5A + 0.85C + P = 100$$
$$A + C + P = 100$$
We realised that the number of pensioners would be equal to the amount the
pensioners paid so really there are only two variables:
$$3.5A +0.85C = 100-p = A + C$$
so $$2.5A = 0.15C$$
We decided to make the coefficients whole numbers, since $A, C$ and $P$ are all
discrete variables.
$$250A = 15C$$
so $$50A = 3C$$
We can see then that one solution is $A=3$ and $C=50$. This will force $P = 47$.
All other whole number solutions will be in the ratio $3:50$ adults to children and this means other solutions have $C$ greater than $100$ which cannot be true as there are only $100$ seats in the
Niharika from Rugby School also investigated other combinations of ticket prices:
$l$ is number of adults
$m$ is number of pensioners
$n$ is number of children
Using prices in pence, if adults pay $400$p, pensioners pay $100$p, children pay $50$p
(they pay £100 i.e. 10 000p in total)
Multiplying by $400$,
$$400l+400m+400n=40 000$$
Subtract this from the first equation:
$$300m+350n=30 000$$
so $$m=\frac{600-7n}{6}$$
so $$m=100-\frac{7n}{6}$$
So $n$ is a multiple of $6, 0 \le n \le 100$. Values of $l$ and $m$ can be found for any of these values of $n$.
If adults pay $500$p, pensioners pay $250$p, children pay $50$p
$$500l+250m+50n=10 000$$
so $$10l+5m+n=200$$
and $$l+m+n=100$$
Multiply the last equation by $10$ and subtract the first equation:
so $$m=40-\frac{9n}{5}$$
So there are solutions if $n$ is a multiple of $5$ and $0\le n\le 100$.
If we want to find a set of prices so that $25$ adults and $75$ children pay £100 in total
An adult pays $a$p
A child pays $c$p
so $$25a+75c=10 000$$
so $$a+3c=400$$
If we assume that $a \ge c$ we get that $a \ge 100$ and $c\le 100$.
If we want to find a set of prices so that there are $40$ adults, $40$ children and $20$ pensioners,
A pensioner pays $p$ p
then $$40a+40c+20p=10 000$$
so $$2a+2c+p=500$$
Some examples of when there are many solutions or just one solution:
Going back to earlier situations where children paid $85$p, pensioners paid $100$p and adults paid $350$p, we had
$n$ must be a multiple of $50$. If $n>50$, $\frac{53n}{50}>100$ so $m<0$, which is impossible. So we only have one possible value of $n$, and one possible solution.
If children paid $50$p, pensioners paid $100$p and adults paid $400$p, we had
Then we have solutions as long as $n$ is a multiple of $6$, so there are many solutions.
Some more examples:
If adults pay £3.00, pensioners pay £2.00 and children pay £1.50 there are many solutions.
If adults pay £3.50, pensioners pay £1.00 and children pay £0.65 there is only one solution.
Well done!
Teachers' Resources
Why do this problem?
This problem requires flexibility of thought and can be solved in many different ways. Once students have solved the initial problem, there are alternative pricings so they can adapt their solution
method to other situations. For each part of the problem, the question "How many solutions are there? How do you know you have found them all" can be considered.
Possible approach
Display the initial problem to fill the cinema with 100 people for £100 with the prices:
Adults £3.50
Pensioners £1.00
Children £0.85
The problem is available on slides: CinemaProblem
Give students some time to try to come up with a solution in pairs, and then encourage them to work in small groups to share strategies.
If strategies are not forthcoming, the following questions might help:
What's the maximum number of adults I could include?
What must be true about the number of children?
If I swap an adult for a child, how does the total change?
If I swap an adult for a pensioner, how does the total change?
If I swap a pensioner for a child, how does the total change?
Once students have had a chance to tackle the original problem, the second part of the task invites them to consider varying the prices:
Can there be 100 people and takings of exactly £100 if the prices are:
Adults £4.00
Pensioners £1.00
Children £0.50
What if the prices are:
Adults £5.00
Pensioners £2.50
Children £0.50
Students could create a poster or presentation showing all the possible solutions and how they know they have found them all.
Possible support
Students could start by exploring the different possible totals if the cinema contains just adults and pensioners, or just pensioners and children.
Possible extension
Invite students to come up with their own pricing schemes where there is exactly one solution, exactly two solutions, exactly three solutions... | {"url":"https://nrich.maths.org/problems/cinema-problem","timestamp":"2024-11-14T22:10:10Z","content_type":"text/html","content_length":"51506","record_id":"<urn:uuid:b3143679-6043-4133-a00f-355265811909>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00592.warc.gz"} |
YTF 11
Nejc Ceplak (Queen Mary, University of London)
The aim of the talk is to present the construction of a new family of smooth horizonless solutions of supergravity that have the same charges as the supersymmetric D1-D5-P black hole. We will begin
with a brief review of the Fuzzball proposal for black holes, which states that at the length scale of the horizon a new, fuzzy, phase takes over, allowing outside observers to distinguish between
different microstates of the black hole. We will then focus on the three charge supersymmetric D1-D5-P black hole and review some of its microstate geometries. We then present a method of obtaining a
new family of solutions using supersymmetry generators. The motivation behind this construction is coming from the dual CFT multiplet structure, where these fermionic generators are used to create
new linearly independent states in the theory. On the gravity side the geometries dual to these new states are generated by the Killing spinors of AdS$_3 \times S^3 \times T^4$. Hence we present the
explicit form of these spinors and use them to construct new solutions to the supergravity equations. Finally we present these new solutions and show that they are simpler than the ones previously
known with having a fewer number of excited fields. | {"url":"https://conference.ippp.dur.ac.uk/event/748/contributions/4309/","timestamp":"2024-11-02T08:44:07Z","content_type":"text/html","content_length":"103753","record_id":"<urn:uuid:dc60bf2c-e6b2-4853-b44c-c52f5765bbca>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00779.warc.gz"} |
Question Writing Oddities
Purpose of this document
This document describes some of the common pitfalls and oddities of the IMathAS question language. For detailed question language reference, please refer to the help file.
IMathAS Question Writing Oddities
Fractional Exponent Display
Fractional exponents do not seem to display well with MathML. For example, x^(2/3) will display as `x^(2/3)`. The best approach is to try x^(2//3), which renders as `x^(2//3)`. If you want to raise
up the exponent higher, a silly trick to try is x^({::}^(2/3)). The {::} creates an invisible item. This renders as `x^({::}^(2/3))`.
Curly Braces
Beware of using curly braces {}. While curly braces can be used for display or for grouping, like in the TeX-style \frac{3}{5}, strange things can happen if you place variables inside the curly
braces. This is because PHP, the back-end interpreter, uses curly braces to isolate variables from surrounding text.
For example, if you wanted to display `3x` rather than `3*x`, then you need to enter 3x rather than 3*x. With a variable coefficient, writing $ax doesn't work, since the interpreter thinks that "$ax"
is the variable. Curly braces can avoid this, allowing you to write {$a}x to achieve the desired result. Alternatively, writing $a x works as well. In rendered math (inside backticks), extra spaces
are removed.
As a side effect, writing \frac{$a}{$b} causes problems, since the interpreter essentially removes the curly braces during variable interpolation, leaving \frac34 (if $a=3,$b=4). A simple way to
avoid this is to add spaces: enter \frac{ $a }{ $b } instead, and the interpreter will leave the curly braces alone, leaving \frac{ 3 }{ 4 }, which will correctly display as the desired `3/4`.
Dollar sign
Because dollar signs are used for variables, entering a dollar sign in question text requires caution. If $a=5, entering $$a will display correctly as $5, but entering ${$a} will not (it's something
called a "variable variable" in PHP). To be extra safe, entering $ $a is recommended, or \$$a (the backslash says "don't try to interpret the next symbol").
Array Variables
You can define array variables, like $a = rands(1,5,3). $a is now an array of three numbers; the elements can be accessed as $a[0], $a[1], and $a[2] (note that arrays are zero-indexed). If you use
this approach, enclose the variable reference in parenthesis in calculations, like $new = ($a[0])^2, and in curly brackets inside strings, like $string = "there were {$a[0]} people".
Variables with numbers in the name
Variables like $a1 are fine to use, but like array variables, should be enclosed in parentheses to prevent misinterpretation. For example, use ($a1)^($a2) instead of $a1^$a2
Function type $variables that share letters with functions
When defining variables for Function type answer, beware that if the variable shares a letter with a function being used, you have to be a bit careful. For example, if $variables="r", and you typed
$answer = "rsqrt(2)", the system will get confused. This can be solved by putting an explicit multiplication between the r and the square root: $answer = "r*sqrt(2)". Students in their entry will
also need to either put an explicit multiplication sign, or at least leave a space between the variable and the function name
If you define:
$a,$b,$c = rand(-5,5,3)
$eqn = "$a x^2 + $b x + $c"
then there is potential your $eqn would display as `4x^2+-3x+2` (that's 4x^2+-3x+2). To clean up the double sign issue, use the makepretty function:
$eqn = makepretty("$a x^2 + $b x + $c")
Makepretty is automatically run on $answer for Function type problems
Less than and Greater than signs
Because HTML uses angle brackets to denote HTML tags, and since IMathAS allows HTML tags for formatting purposes, the use of < and > in problem text can sometimes be problematic. The system attempts
to differential between HTML tags and inequalities, but does not always do so successfully.
Generally, same direction inequalities are handled okay, such as 3 < x < 5. But mixed inequalities, such as "x < 3 and x > 1" are sometimes mishandled. To avoid this, it is recommended that you use
the HTML < and > in place of < and >. Inside backticks (rendered as math), lt and gt are sufficient to denote < and >. You can also use le and ge or leq and geq inside backticks for `le` and
© 2006 David Lippman
This guide was written with development grant support from the WA State Distance Learning Council | {"url":"https://aitmath.com/docs/questionoddities.html","timestamp":"2024-11-14T05:02:46Z","content_type":"text/html","content_length":"5548","record_id":"<urn:uuid:ead1e6b8-edcc-47e8-ba2e-6619493301ad>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00310.warc.gz"} |
R(a|ela)tional Design | 8th Light
In my last post, I talked about Expressive HTML and the importance of writing your markup from the content out. This idea—that you start at a granular level and work your way outward, building
context and meaning as you go—holds equally true as a methodology for web design.
Ever since attending last year’s Build Conference I’ve been using what I’ll call “The Modular Scale Method,” which is a technique I’ve found invaluable for building my designs from the content out.
We’ll talk about this method more in-depth in just a little while as we look briefly at a simple Cookbook application. But given that the modular scale concerns itself with typography—the most
granular level of our content—it makes sense to start today with a review of a few articles I’ve written on the basic rules of good web typography, which you can find here:
Modular Scales
I first encountered the idea of the “modular scale,” at least as it applies to typography, during my first read-through of Robert Bringhurst’s The Elements of Typographic Style.
I’ll freely admit that at the time my eyes simply glazed over this section of his book, most likely due to the fact that I was intimidated by the 2+ pages of complicated looking ratios presented
Pages of complicated-looking ratios in Bringhurst’s “The Elements of Typographics Style.”
And though, according to Bringhurst, “The mathematics are not here to impose drudgery upon anyone,” to me, at the time, it just all seemed so complicated. And I suppose, at least to some degree, I
still find it to be so.
Nonetheless, my appreciation of modular scales (and the benefits that a thoughtful application of them can provide) began about a year ago now, after a wonderful talk I saw Tim Brown give, titled
More Perfect Typography.
If you haven’t heard this talk, and you are interested in the subject of web typography, I highly encourage you to take the time to watch it. It’s fascinating stuff.
I think one of the things that was so compelling about Tim’s talk was the way he reduced a decidedly complex set of ideas (a la Bringhurst) down into a simple methodology for typesetting on the web.
It is this methodology that I would like to explore today.
What’s a Modular Scale?
Well, it’s a pretty basic idea. We can see this fact evidenced by the axioms that are central to the examinations (of this approach to typesetting) made by both Bringhurst and Brown:
A modular scale, like a musical scale, is a prearranged set of harmonious proportions. Robert Bringhurst, The Elements of Typographic Style
A modular scale is a sequence of numbers that relate to one another in a meaningful way. Tim Brown, More Meaningful Typography
Simple, right? And, at its core, the idea is pretty basic. But, oddly enough, it’s quite easy to both under-estimate its implications and over-estimate its complexity (at least once you start to dig
into it a bit). It's my hope to usher you all swiftly past this seeming paradox.
As you may know, the Golden Ratio is an irrational mathematical constant that is equal to approx: 1.61803398874989. Technically speaking, two amounts or quantities are said to “be in the Golden
Ratio” if:
The ratio of the sum of the quantities to the larger quantity is equal to the ratio of the larger quantity to the smaller one. —Wikipedia
Now, the reason I mention the Golden Ratio is because, throughout history, artists, architects, printers, typographers, musicians, furniture makers, anatomists, etc. have used such numerical
relationships as a tool to develop harmonious proportions for their various works.
Interestingly, the Golden Ratio seems to have an almost universal human appeal. The Parthenon, to cite just one famous example, is known to exhibit proportions that approximate the Golden Ratio.
The Parthenon is known to exhibit proportions that approximate the Golden Ratio. ( source )
There are, of course, other famous ratios, many of which come form the realm of music theory, in which they are more commonly known as “intervals”—the measure of the ratio between the frequencies of
two different notes.
Many famous ratios come from the realm of music theory, where they are commonly known as “musical intervals”.
Some of the better known musical intervals (and applicable to our cause here) are the:
• Perfect Fourth (4:3)
• Perfect Fifth (3:2)
• Perfect Octave (2:1)
• Major Third (5:4)
• Major Sixth (5:3)
What has any of this got to do with Web Design?
Just as musicians have employed the inherent magic of certain numerical relationships to lend consonance and/or dissonance to their musical scores, and just as architects, scribes, typographers, and
craftsmen have used these ratios to lend humane or aesthetically pleasing proportions to their buildings, books, passages, and furniture so too can we [web professionals] use these numbers to satisfy
the ends of our designs.
Recognizing type as the atomic element in web design affords us the opportunity to make better design decisions that resonate upward and outward into the experience. Tim Brown, More Meaningful
In the same way that certain objects—a computer mouse for example, or a desk chair, even a spoon—are consistently produced at sizes that are comfortable to the human body/hand, certain proportions
reappear in the creative output of humans throughout the centuries (and in cultures around the world) because they are similarly pleasing to our eyes, our sensibilities and our expectations for how
things should look.
Moreover, many of these special numerical relationships occur naturally in “the structures of molecules, mineral crystals, soap bubbles, flowers” and the human body.
Using strands of numbers derived from such consonant ratios as The Perfect Fourth (4:3), The Perfect Fifth (3:2), and The Golden Ratio (1:1.618) (among others) we can generate predefined measurements
that will allow us to establish hierarchy and lend pleasing harmonies to our compositions—harmonies that might otherwise be difficult, if not impossible, to find in layouts patched together through
arbitrary guesswork alone.
Sizing and spacing type, like composing and performing music or applying paint to canvas, is largely concerned with intervals and differences. As the texture builds, precise relationships and
very small discrepancies are easily perceived. Robert Bringhurst, The Elements of Typographic Style
What’s more, such a rational, systematic approach provides a reliable and resuable tool on which to base a myriad of difficult design decisions. In short, using modular scales to drive your designs
can make the terror of the blank page (every designer’s sworn enemy) a little less stultifying.
How do we fly this Spaceship?
The two most important steps in implementing the “Modular Scale Method” are to:
1. Choose a scale.
2. Choose an important number.
Now hopefully some of you are wondering: how exactly does one go about choosing these values?
Using a modular scale on the web means choosing numbers from the scale for type sizes, line height, line length, margins, column widths, and more. Tim Brown, More Meaningful Typography
In establishing a reusable system based on meaningful relationships between numbers, we are conscientiously moving our decision making process away from the realm of the arbitrary, firmly supplanting
it into the realm of the mathematical.
So it follows, that these values should be anything but arbitrary. But, is it immediately obvious how you might meaningfully choose one of these values?
In my personal experience, I most often struggle with choosing a scale, of which there are many to choose from.
Tips for choosing a scale
1. Consider the content you are designing for. What’s it about? Does it have any sort of cultural or historical reference? How about the typeface you've chosen?
Scales, like the Golden Ratio or Perfect Fourth, have historical significance and cultural relevance. For instance, as mentioned previously, the Golden Section was a favorite mechanism for
classical Greek and Renaissance mathematicians and architects, and scribes.
In the 20th century, during the tail end of the Common Practice Period, harmony “built explicitly on fourths became important…and quartal harmony became an important means of expression by
Debussy, Ravel and others.”
By leveraging the historical or cultural context of your content or your chosen typeface, you can narrow down your choices considerably and insure that you are making a choice of scale that is
conceptually aligned with your project.
2. Consider how much variation in type size you think you are going to need. Some scales feature larger increases between values—for instance an Octave (1:2). While others, like The Major Second
(1:1.25), feature smaller increments.
In the event that your project requires a good deal of typographic variety (maybe you’re using all 6 header levels) you’ll likely want to choose a scale that allows for more subtlety between
3. Be a synesthete. Many of the ratios we’ve examined thus far find their roots in musical intervals. Consider playing sound clips of them. Do they evoke different feelings/moods in you? Are these
feelings or moods some how in-line with the mood or feeling you are trying to convey with your design?
Consider listening to musical intervals to determine their mood or emotional implications.
In short, try to assign feelings to your scales based on what they sound like and make your selections accordingly.
It sounds a little crazy, I know, but it is sure to present some interesting results:
Choosing an important number
The value of your “Important Number”—what is effectively the prime mover of your modular scale—cannot be overlooked. Every number that is generated by the scale you choose will be based upon it.
This fact presents us with an amazing opportunity to design from the content out.
I’ve found that a variety of things can serve as an important number. The size at which caption text looks best, for instance, or a large number like the width of a piece of media—if the project
at hand mandates ads or embedded videos, for instance—ensures that something about those elements resonates with the layout as a whole. Tim Brown, More Meaningful Typography
In the world of typesetting there exists a divide between “text” and “display” faces. For setting longer passages of type, you’ll want to use text faces: those fonts whose proportions and design
details are suited to extended reading, on a screen, at relatively small sizes.
Unfortunately, it is a sad fact that, as a medium, the screen has a few shortcomings that can hamper designers, not the least of which is a question of resolution. Without diving too deeply into the
technical limitations of low resolution media, suffice it to say that type size can have a tremendous impact on how well a given typeface will look.
The fine details of typefaces get lost at smaller sizes (10px Didot maginified 600 times).
Due to the process of rasterization and the limited resolution that most monitors possess, there are a very limited number of pixels that are available to represent letters at sufficiently small type
sizes. As a general rule, bigger is better (there are more pixels with which to accurately render the outlines of your typeface).
And chances are, if you’ve spent the time to make an informed typeface selection, there will be a range of sizes (generally below 18-20px) at which your chosen typeface will just sing.
For example, let’s examine a type specimen for Verdana. Here we are presented with an overview of how our typeface renders at a myriad of different sizes. Immediately, we can begin to see the drastic
difference between the way this typeface renders at say 9 pixels, and the way it renders at 16 pixels.
We can see the drastic difference between the way this typeface renders 9 pixels, and the way it renders at 16 pixels.
At a type size of just 9 points, Verdana begins to break down—there just aren’t enough pixels available to accurately approximate the vector outlines of the original letterforms. If you look closely
it looks jagged, unkempt and worst of all, unreadable.
Conversely, at a size between 16-18px, we notice that the face looks considerably heavier, fuller in body and weight. The letterforms themselves are considerably less jagged and their shape is more
For the purposes of our Cookbook design exercise, I've made a few assumptions based on personal experience. I envision that this application will be used in-situ (ever use a cookbook while cooking?)
and decided that Verdana at a size of 16px (though perhaps at the very upper limit for this typeface, as it begins to drift apart at this size) would be best.
Given that this is an application that is likely to be read on a screen at a considerable distance, I felt that a large, open typeface, with clean forms, would lend itself well to the cause and
contrast nicely with frantic mess I often find the kitchen to be.
Do the math
Here is what our simple cookbook application looks like currently, featuring a smattering of arbitrarily chosen type sizes and line-lengths:
Here is what our simple cookbook application looks like currently, featuring a smattering of arbitrarily chosen type sizes.
In order to start using a “Modular Scale Method,” we must generate the resonate numbers of our scale. And to do this, all that is required is a little simple math:
Our generated modular scale, using the Major Sixth (5:3).
We can now use our scale to implement some preliminary design changes such as:
1. Choosing a measure.
2. Choosing a line-height.
3. Choosing heading sizes.
It’s important to keep in mind that a modular scale is not a guarantee. It’s a design tool. And, like any tool, it’s just an aid.
A successful design will still rely, to large extent, on a designer’s eye and on his or her aesthetic sensibility and sensitivity to the needs of the project.
What a “Modular Scale Method” does present is a rational guideline (a set of meaningful constraints) that can help inform your decision making. As Tim Brown says:
Consider the scale’s numbers educated suggestions. Round them if you like (22.162 becomes 22). Combine them (3.56 + 16 = 19.56). Tim Brown, More Meaningful Typography
If you feel there is a need to do so, break from the scale entirely. Improvisation can lend (or reintroduce) the organic back into your design and in so doing allow you to create tension, asymmetry
and originality.
For those of you who “need to see it to believe it,” Tim Brown has generously provided a couple finished examples which you can find here:
Also, this blog, and the 8th Light website itself, were designed using this method.
And with that I say: Go forth and Modularize! | {"url":"https://8thlight.com/insights/ra-elational-design","timestamp":"2024-11-06T01:26:14Z","content_type":"text/html","content_length":"132704","record_id":"<urn:uuid:3a271191-bb4f-49be-b936-d6e9521248c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00026.warc.gz"} |
: Numerical compute difficulty
It depends on what result you want. Without quotes, perl will do the computations before passing the result to Compute(), which will then turn the result into a MathObject Real (unless the perl
computation already involves MathObjects and results in some other MathObject type). It is perfectly fine to do that, and it is somewhat more efficient since it does not involve the expensive step of
parsing the expression by the MathObject parser. For a Real value, that is probably a reasonable thing to do, as long as you remember things like the difference (in perl) between ^ and **.
When you pass a string to Compute(), it will first parse it using the MathObject parser, as controlled by the current Context object. That means that the operations and functions and so on that are
defined there will be the ones that control the result, no the built-in perl expression parser. That makes the result work exactly like student answers, which is good for consistency, and also makes
it easier to produce results that are not easy to produce in perl (like Intervals, and Vectors). Finally, Compute() saves the original string as the correct answer string, so that if you used
$ans = Compute("sqrt(2)");
to produce a MathObject that was used for an answer checker via
then the correct answer would show as
rather than
as it would have been had you used
without the quotes.
I don't think there is a hard-and-fast rule for when to use quotes and when not to. You have to think about what result you want to have in terms of the correct answer as well as the computed result.
Hope that clears things up for you. | {"url":"https://webwork.maa.org/moodle/mod/forum/discuss.php?d=452&parent=1844","timestamp":"2024-11-07T02:31:16Z","content_type":"text/html","content_length":"69458","record_id":"<urn:uuid:f0259c06-92f3-4f8e-9fdd-9aa67154332d>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00242.warc.gz"} |
How can I convert an Elixir tuple into a bitstring?
In Elixir, you can convert a tuple to a bitstring using the :erlang.term_to_binary/1 function, which converts any Erlang term (including Elixir tuples) to its binary representation. Here's an
1 tuple = {:hello, "world", 123}
2 binary = :erlang.term_to_binary(tuple)
In this example, the tuple variable contains a tuple with three elements. The :erlang.term_to_binary/1 function is then used to convert the tuple to a binary representation, which is stored in the
binary variable.
You can then convert the binary to a bitstring using the <<>> binary syntax:
1 bitstring = <<binary::bitstring>>
In this example, the <<>> syntax is used to convert the binary variable to a bitstring, which is stored in the bitstring variable.
Note that when using the <<>> syntax, you need to specify the binary as binary::bitstring, which tells Elixir to interpret the binary as a bitstring. If you don't include the ::bitstring type
specifier, Elixir will interpret the binary as a sequence of bytes, which may not be what you want. | {"url":"https://devhubby.com/thread/how-can-i-convert-an-elixir-tuple-into-a-bitstring","timestamp":"2024-11-14T20:54:14Z","content_type":"text/html","content_length":"121075","record_id":"<urn:uuid:5f59fba9-8f5a-48ef-a355-6aa8e0a882ea>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00436.warc.gz"} |
How to Convert an Integer to Binary in C# - CSharp Academy
How to Convert an Integer to Binary in C#
In C#, converting an integer to its binary representation can be useful in many situations, such as low-level programming, bit manipulation, or debugging. The binary system uses only two digits, 0
and 1, making it a fundamental aspect of computing. In this article, we’ll explore multiple ways to convert an integer to binary in C#.
Why Convert an Integer to Binary?
There are several reasons why you might need to convert an integer to binary in C#:
• Bitwise operations: Many low-level operations require binary representation.
• Networking: IP addressing and subnet masking involve binary numbers.
• Cryptography: Certain encryption algorithms rely on binary conversions.
• Data Compression: Binary is used in encoding schemes like Huffman coding.
Methods for Converting an Integer to Binary in C#
There are a few different ways you can convert an integer to binary in C#. We’ll explore the following approaches:
1. Using Convert.ToString method
2. Using custom bitwise operations
3. Using recursion
Method 1: Using Convert.ToString
The simplest and most straightforward way to convert an integer to binary in C# is to use the built-in Convert.ToString method. This method allows you to specify a base (binary in this case) as a
second argument.
int number = 42;
string binaryRepresentation = Convert.ToString(number, 2);
Console.WriteLine(binaryRepresentation); // Output: 101010
In this example, the number 42 is converted into its binary representation: 101010. The second argument in the Convert.ToString method specifies the base, with 2 representing binary.
Method 2: Using Bitwise Operations
Another approach is to use bitwise operations. This method is more manual and requires a loop to continuously shift the bits to the right while examining the least significant bit (LSB).
int number = 42;
string binaryRepresentation = "";
while (number > 0)
binaryRepresentation = (number % 2) + binaryRepresentation;
number = number / 2;
Console.WriteLine(binaryRepresentation); // Output: 101010
How It Works:
• In each iteration, the remainder when dividing the number by 2 is appended to the binary string.
• The number is divided by 2 in each step, effectively shifting its bits to the right.
Method 3: Using Recursion
If you prefer a more functional programming style, you can use recursion to convert an integer to binary. This approach repeatedly divides the number by 2 and builds the binary string from the bottom
string ConvertToBinary(int number)
if (number == 0)
return "";
return ConvertToBinary(number / 2) + (number % 2).ToString();
int number = 42;
string binaryRepresentation = ConvertToBinary(number);
Console.WriteLine(binaryRepresentation); // Output: 101010
How It Works:
• The function calls itself, dividing the number by 2 in each recursive call.
• It appends the remainder (0 or 1) to the result when unwinding from the recursive stack.
• The recursion stops when the number reaches 0.
Edge Case: Converting Zero
When converting the number 0, the binary representation is simply "0". Here’s how you can handle this special case:
int number = 0;
string binaryRepresentation = Convert.ToString(number, 2);
Console.WriteLine(binaryRepresentation); // Output: 0
Alternatively, in the bitwise and recursive approaches, you can add a simple condition to return "0" when the input is zero.
Edge Case: Handling Negative Numbers
By default, negative numbers are represented using two’s complement in binary. If you need to convert a negative integer to its two’s complement binary representation, you can still use
Convert.ToString, but it will include the sign bit.
int number = -42;
string binaryRepresentation = Convert.ToString(number, 2);
Console.WriteLine(binaryRepresentation); // Output: 11111111111111111111111111010110
If you only want to display the binary representation without the sign bit, you can work with the absolute value:
int number = -42;
string binaryRepresentation = Convert.ToString(Math.Abs(number), 2);
Console.WriteLine(binaryRepresentation); // Output: 101010
In C#, converting an integer to binary can be done in multiple ways, from using the built-in Convert.ToString method to more manual techniques like bitwise operations and recursion. The
Convert.ToString method is the easiest and most efficient option for most cases, but understanding bitwise manipulation and recursion is valuable for deeper control and learning.
By mastering these methods, you can confidently handle binary conversions in your C# applications, whether you’re working on bitwise operations, networking, cryptography, or other low-level
programming tasks.
Need Help with Your C# Projects?
We offer expert support and development services for projects of any size. Contact us for a free consultation and see how we can help you succeed. | {"url":"https://csharp.academy/how-to-convert-an-integer-to-binary-in-c/","timestamp":"2024-11-13T09:01:59Z","content_type":"text/html","content_length":"155173","record_id":"<urn:uuid:81031f76-1eae-4e2d-8afd-a7132c189f00>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00792.warc.gz"} |
As part of deep learning, we mostly would like to optimize the value of a function that either minimizes or maximizes f(x) with respect to x. A few examples of optimization problems are
least-squares, logistic regression, and support vector machines. Many of these techniques will get examined in detail in later chapters. | {"url":"https://subscription.packtpub.com/book/data/9781788390392/1/ch01lvl1sec04/optimization","timestamp":"2024-11-04T18:55:22Z","content_type":"text/html","content_length":"111759","record_id":"<urn:uuid:d027ddc1-ab93-4e27-8541-7033f1e03818>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00455.warc.gz"} |
VISP/MA Student Post Graduation
Spring 24:
27 students graduated, 15 of them have been admitted to PhD programs.
University of Wisconsin-Madison, Math(5 students)
Northwestern University, Math
Purdue University, Math
Notre Dame, Applied and Computational Mathematics and Statistics
Texas A&M University, Math(2 students)
University of Connecticut, Math
Georgia Tech University, Math
Stevens Institute of Technology, Financial Engineering
University of California-Irvine, Math
Chinese University of Hong Kong, Math
Spring 23:
30 students graduated, 21 of them have been admitted to Ph.D. programs.
University of Connecticut, Math (2 students)
University of Wisconsin-Madison, Math, (7 students)
University of Wisconsin-Madison, Engineering,
University of Wisconsin-Madison, Biomedical
University of Wisconsin-Madison, Electrical and Computer Engineering
University of Missouri, Math
Duke University, Computer Science
University of Huston , Industrial Engineering
University of Illinois- Chicago, Math
University of Utah, Math
North Carolina State University, Applied Math
Northwestern University, Math
Hong Kong University of Science and Technology, Math
International Graduate School of Tsinghua University, Math
Spring 22
29 students graduated, 14 of them were admitted to Ph.D. programs
University of Wisconsin-Madison, Math, (6 students)
Vanderbilt University, Biomedical
University of Georgia, Statistics
Indiana University-Purdue University Indianapolis, Math
University of Wisconsin-Madison, Mechanical Engineering
University of Wisconsin-Madison, Industrial and Systems Engineering
Boston College, Math
Texas A&M University, Math
University of Arizona, Math
Spring 21
28 students graduated, 10 of them were admitted into Ph.D. programs
University of Wisconsin-Madison, Math, (5 students)
Purdue University, Computer Information and Technology
University of Minnesota, JD
University of Science and Technology of China, Information and Computational Science
University of Toronto, Statistics
Shanghai Jiaotong University, Math | {"url":"https://math.wisc.edu/visp-ma-student-post-graduation/","timestamp":"2024-11-03T07:13:37Z","content_type":"text/html","content_length":"61411","record_id":"<urn:uuid:cd331fcf-752a-4a95-b99e-9182d1dd7aae>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00629.warc.gz"} |
Optimal Design of Lower Limb Rehabilitation System Based on Parallel and Serial Mechanisms
Research Center Robotics and Control Systems, Belgorod State Technological University Named after V.G. Shukhov, 308012 Belgorod, Russia
LARM2—Laboratory of Robot Mechatronics, Department of Industrial Engineering, University of Rome Tor Vergata, 00133 Roma, Italy
Author to whom correspondence should be addressed.
Submission received: 2 January 2024 / Revised: 25 January 2024 / Accepted: 30 January 2024 / Published: 1 February 2024
This paper presents the structure and model of a hybrid modular structure of a robotic system for lower limb rehabilitation. It is made of two modules identical in structure, including an active
3-PRRR manipulator for moving the patient’s foot and a passive orthosis based on the RRR mechanism for supporting the lower limb. A mathematical model has been developed to describe the positions for
the links of the active and passive mechanisms of two modules, as a function of the angles in the joints of the passive orthosis, considering constraints for attaching the active manipulators to the
moving platform and their configurations. A method has been formulated for a parametric synthesis of the hybrid robotic system proposed with modular structure, taking into account the generated
levels of parametric constraints depending on the ergonomic and manufacturability features. The proposed design is based on a criterion in the form of a convolution, including two components, one of
which is based on minimizing unattainable points of the trajectory, considering the characteristics of anthropometric data, and the other is based on the compactness of the design. The results of the
mathematical modeling are discussed as well as the analysis results towards a prototype validation.
1. Introduction
One of the most pressing and complex problems of medicine and neurology is the rehabilitation of patients. The number of people who need rehabilitation is growing every day. According to
investigations [
], 419,000 people diagnosed with stroke were registered in the Russian Federation in 2014. According to the Ministry of Health of the Russian Federation, in 2017, 427,963 people were registered for
whom acute cerebrovascular disorders were first detected. Patients who have experienced acute attacks of this disease type either cannot move without help or are completely deprived of the
opportunity to move independently. Currently, over one million stroke survivors live in Russia. A third of this number are people of working age. At the same time, only every fourth of working people
return to work after a stroke [
]. Violations of the limbs must undergo a rehabilitation process in order to restore the normal functioning of the limbs. Lower limb rehabilitation or treatment has been a hot topic in recent years,
since robotized systems promised effective results and led to significant improvements in the recovery of patients using robotic physiotherapy, as pointed out in [
It Is widely recognized that a person’s locomotion depends both on basic patterns generated at the level of the spine and on the prognosis and reflex-dependent precise control of these patterns at
different levels [
]. These physiological movements recorded in healthy people are applied as exercises in patients with disorders of the lower limbs. These data sequences for the joints of the lower limbs are called
gait data (walking pattern) [
]. In addition to gait data, the lower limbs perform movements such as hip flexion and extension, knee flexion and extension, ankle flexion, and back flexion. Gait training restores synchronization
of muscle action in lower limbs and is processed separately to strengthen each motor joint, to then strengthen each joint of the leg.
Currently, there are many devices available for lower limb rehabilitation. The Gait Trainer [
] is a wheeled device that helps a person who cannot walk independently. The BIT LExoChair [
] mobility-assisted wheelchair has a modular design for user’s locomotion mobility with the aim of assisting autonomy, exercising, and rehabilitation. The design is based on a traditional wheelchair
design that uses a force sensor for controlled operation with different equipment as a function of the environment and aim of usage. The rehabilitation system WalkTrainer [
] has a mobile frame design, which includes a system for unloading weight. Thus, it is possible to regulate the dynamic load on the patient’s lower limbs. There are also rehabilitation devices that
are based on treadmills. It is worth noting the Lokomat rehabilitation system [
] and the CDLR cable robot for lower limb rehabilitation [
]. The functionality of those devices does not allow for the rehabilitation of the lower limbs of patients in the early stages of rehabilitation; therefore, these devices are not suitable for
patients who are only able to be in a sitting and lying position, impaired with lower limbs.
Among rehabilitation systems, which are available in the early stages of rehabilitation, one can point out the LokoHelp movement therapy station [
], which is an electromechanical gait trainer with a weight-unloading system. However, this system is not applicable to patients who cannot be suspended. A gait rehabilitation device based on a 3DOF
parallel arm [
] generates the required gait pattern by moving the patient’s foot while the body weight is supported by a seat belt system. The KARR rehabilitation system [
] allows the rehabilitation of patients in a sitting position. Also worth noting is the Lambda robot rehabilitation system [
], which can be applied to the rehabilitation of bedridden patients by mobilizing the ankle joint. An upright table with an integrated orthopedic device and synchronized functional electrical
stimulation, Erigo Pro [
], allows for intensive cyclic movement therapy in the form of passive dynamic movements of the lower limbs of recumbent patients. However, the above-mentioned devices for physiotherapeutic movements
provide flexion–/extension in the knee joint, while a movement such as adduction–abduction in the hip joint is impossible, with a significant drawback. The CUBE cable rehabilitation device [
] allows for the rehabilitation of both the upper and lower limbs. However, with this design, active robot-assisted therapy for flexion/extension in the ankle joint is rather difficult.
In general, in rehabilitation procedures, the treatment of patients with disabilities of impaired motor functions of the lower limbs occurs in a critical sitting or lying position. These conditions
are because patients at this stage cannot control the movement of their limbs. In this regard, the treatment of patients using only the BWS (body weight support) system is difficult, since this
requires a certain level of physical fitness. CPM (continuous passive motion) is one of the conventional therapies at the initial stage of treatment when patients have weak or even uncontrolled
Analysis of the literature sources has revealed that most rehabilitation systems have an active orthosis within their mechanical design. This orthosis has a simple structure, but its dynamic
characteristics are low due to the presence of drives in the orthosis system. Other types of rehabilitation systems have linear drives and high dynamic performance. However, these types of
rehabilitation systems have a limited workspace with limited movement of linear drives, or do not include supporting frames or orthopedic systems. Many manipulators are based on CPM and are intended
for continuous or repeated treatments. In addition, these manipulators cannot provide the required quality when performing continuous passive movements due to their complex design, high dynamic
loads, and bulkiness.
In order to overcome the above limitations, in this work, the popular Cartesian parallel manipulator, Tripteron [
], is taken as a basis for a novel solution. It consists of three legs, each having a prismatic–revolute–revolute–revolute joint arrangement with an active prismatic joint and each leg being mounted
on the base platform [
]. The major difficulty in off-planar movement is the movement of the actuator assembly, as they are mounted in single plane. Thus, the proposed system combines a parallel manipulator based on
Tripteron, with a serial passive orthosis which can give four degrees of freedom to the lower limb. The dynamics of the separate right and left modules of the system were considered in [
]. The Euler–Lagrangian approach is used to formulate the dynamics of the manipulator for a simulated performance analysis. The papers [
] also present an augmented proportional-derivative (PD) controller for gravity compensation in motion control. This control method transforms the closed-loop dynamics of the manipulator into a
decoupled behavior for a more convenient analysis of motion performance. The numerical simulations showed that the augmented PD controller enables us to obtain a reliable control strategy. In the
works [
], a method for optimizing the geometric parameters of a robotic system based on one Tripteron module for the rehabilitation of one limb is discussed. Nevertheless, it has a number of limitations for
the case of application to the proposed two-module system, which include as follows:
• The location of the drive guides of the active manipulator, because of the intersection of their axes at one point in the upper right side of a patient user;
• Not a convenient shape for the moving platform, as in the form of an equilateral triangle;
• Limited location of fastenings of mechanisms to the platform center;
• The used optimization algorithms for the generation of random parameters and search by climbing to the top;
• Consideration of intersections of only active manipulator links;
• The use of the average size of a human limb with no customizing capability.
Based on the aforementioned flaws, the main contributions of this paper can be summarized as follows:
• A novel hybrid modular structure of a robotic system for the rehabilitation of the lower limbs is designed as based on two modules identical in structure, including an active 3-PRRR manipulator
for moving the patient’s foot and a passive orthosis based on the RRR mechanism for supporting the lower limbs;
• A method for parametric synthesis of a hybrid robotic system with a modular structure is formulated, taking into account the generated levels of parametric constraints depending on the ergonomics
and manufacturability of the proposed design.
The paper is organized as follows:
Section 2
provides motion requirements from the lower limb biomechanics. The dimensions of the lower limbs of people of different nationalities are considered and the necessary trajectory is formulated.
Section 3
provides a mathematical model of the proposed robotic system. Design parameters for the robotic system are considered.
Section 4
presents an optimization problem for the design of the proposed solution.
Section 5
is devoted to numerical simulation.
Section 6
reviews experimental studies for validation and characterization purposes.
2. Motion Requirements from Lower Limb Biomechanics
In early rehabilitation after injuries, surgical interventions on the musculoskeletal system and neurological treatments such as the so-called CPM therapy (Continuous Passive Motion) have shown high
efficiency [
]. The last is a rehabilitation technique that is associated with continuous passive exercise of human joints. This technique is based on implementing long-term repetitive movements of the joints
using a specialized robot simulator without the participation of the patient’s own muscle strength.
Biomechanical parameters for a motion therapy activity may include volume, direction of movement, degree of applied force, velocity, and accuracy of task reproduction. An assisting device should also
be used for safe exercise with no functional overstrain, cognitive discomfort, and psycho-emotional over excitation.
In general, the main movements of the lower limb are carried out in the sagittal plane. However, the possibility of additional rotational and frontal movements due to the hip joint has a synergistic
effect on the restoration of motion also in the main plane. Therefore, a statically balanced passive orthosis model, where the limb can be placed, should allow for synchronization of movements in the
hip, knee, and ankle joints. In this case, the main requirement for the correct operation of an assisting device is the precise positioning of the hip joint and of the device, relative to the
patient’s hip joint.
The hip joint can be modeled as a ball-shaped joint of a limited type (cup-shaped joint), and therefore allows movement, although not as extensive as in a free ball joint (for example, in the
shoulder), around three main axes: frontal, sagittal and vertical. Lateral and medial rotation occurs relative to the vertical axis, extension and flexion occurs relative to the frontal axis, and
abduction and adduction occurs relative to the sagittal axis, as indicated in
Figure 1
Referring to
Figure 1
], flexion and extension of the lower limb occurs around the frontal axis. Flexion has the greatest range due to the absence of tension in the fibrous capsule of the joint, which has no attachment to
the femoral neck from behind. When the knee joint is bent, it can real (118–130°), so that the lower limb, with its maximum flexion, can be pressed against the stomach. With an extended knee joint,
the range of motion is less as in the round (84–90°), since it is inhibited by the tension of the muscles on the back of the thigh. Extension from the vertical position, taken as 0°, is limited by
the tension of the joint ligaments and in most cases is limited to 10° (but can reach 20°).
Around the sagittal axis, the limb is abducted and adducted. The counting is carried out from the center line (0°). Abduction is possible at up to 50°, and reduction at up to 30°.
During the rehabilitation process, it is required to activated required angles in the hip, knee, and ankle joint. In particular, the angle of abduction in the hip joint is 30 degrees, as well as
angles corresponding to the imitation of human gait; flexion in the hip joint is from −20 to 10 degrees, as well as flexion in the knee joint being from −60 to 0 degrees.
The trajectory of the limb movement, which will be used further during optimization, is built on the basis of repetitive movements corresponding to the human gait and abduction of the limb at the hip
By imitation of gait, we mean the movement of each limb in accordance with the following law:
$α i t = 0.5 α m a x + α m i n + sin t + 180 ° ( i − 1 ) α m a x − α m i n$
$β i t = 0.5 β m a x + β m i n − cos t + 180 ° ( i − 1 ) β m a x − β m i n ,$
is the limb index (1—left, 2—right),
$α m i n$
$α m a x$
are minimum and maximum angle of flexion of the hip joint, and
$β m i n$
$β m a x$
are the knee joint. To increase the working range of hip joint flexion training,
$α m i n$
was taken equal to
$α m i n = − 20 °$
. The remaining constraints on the angles in the joints were adopted in accordance with the clinical information above, that is,
$α m a x = 20 ° ,$$β m i n = − 60 °$
$β m a x = 0 °$
Thus, a sequence of movements has been planned with the following steps:
• The limbs are not abducted (leg abduction angle $γ i = 0$) while performing one cycle ($t ∈ 0 ; 360$) of gait simulation;
• Abduction of the left leg to $γ 1 = − 30 °$;
• Performing one cycle of gait simulation;
• Simultaneous adduction of the left leg to the angle $γ 1 = 0 °$ and abduction of the right leg to $γ 2 = 30 °$;
• Performing one cycle of gait simulation;
• Abduction of the right leg $γ 2 = 0 °$.
Figure 2
shows the planned sequence of movements as a reference for joint trajectories.
Figure 2
a shows the sequence for flexion of the hip joint,
Figure 2
b shows the knee, and
Figure 2
c shows abduction of the hip joint.
The proposed system should provide the ability to rehabilitate patients with different anthropometric data.
Table 1
presents the main anthropometric measurements of the lower limbs based on statistical data from national populations, in accordance with [
]. Each size is presented for P95 (95th percentile, that is, the maximum statistical size without taking into account the five percent of people who have unusually small limb sizes) based on
individual samples for adults in different countries, taking into account gender. As a result of the analysis of the statistical data, a data set was generated as in
Table 2
, which is used as a reference for designing the geometric parameters of the robotic system. The set of sizes, without taking into account the orthosis, was obtained based on the choice of the
maximum value of each limb size among the values in
Table 1
. Since the limb dimensions are used according to P95 (95th percentile, that is, the maximum statistical size without taking into account the five percent of people who have unusually small limb
sizes) this will ensure the possibility of lower limb rehabilitation for more than 95 percent of people. The choice for maximum dimensions is justified by the presence, in this case, of the maximum
area of movement of the human foot during the development of the trajectory, as well as the most complicated task of eliminating the collision of the links of the robotic system with the limbs.
Considering the dimensions of the orthosis, which includes the frame and structural elements used to protect the limb, the dimensions have been adjusted. Based on the thickness of the human leg
orthosis, equal to 15 mm, the thigh width is taken to be 30 mm larger (15 mm∙2), and the thigh circumference is 95 mm larger (15 mm
$· 2 · π$
= 94.25 mm). Considering the increased thickness of the calf orthosis, equal to 30 mm, taking into account the possibility of adding additional safety elements, the circumference of the calf muscle
is increased by 188 mm, the length of the foot by 30 mm, and the width of the foot by 60 mm. Also, taking into account the possibility of adding safety-structural elements, the length of the lower
leg has been increased by 200 mm. Taking into account the dimensions of the reference data, the trajectory of the reference movement is designed as in
Figure 2
, as the assumed trajectory of assisted joints by the rehabilitation system.
Figure 3
shows the change in the position of the center of the ankle joint, taking into account the planned sequence of movements and the selected dimensions of the orthosis.
3. Mathematical Model
The proposed hybrid robotic system consists of two modules identical in structure, each of which is based on an active 3-PRRR mechanism of a parallel structure, which ensures the movement of the
patient’s fixed foot, and a planar RRRR mechanism of a serial structure as a passive cut to support the lower limb, as sketched in
Figure 4
. The modular structure allows for a change to the parameters of the systems depending on the method of rehabilitation and, depending on the anthropometry and characteristics of the disease, the use
of one and two modules simultaneously. The mutual movement of the end-effector of the two manipulators makes it possible to simulate the gait of a healthy person, while safe movement algorithms are
implemented taking into account possible intersections (collisions) of the links.
The design in
Figure 4
is better explained by the model in
Figure 5
Figure 5
a shows the kinematic design with labels for all of the joints of the active manipulators, as well as the joints and angles of the passive orthoses. In
Figure 5
a, where I is for the left module for the rehabilitation of the left limb, and II is for the rehabilitation of the right limb. The 3-PRRR mechanism, which provides movement of the patient’s limb
through active actuators, consists of three kinematic chains, a fixed base, and a moving platform. The 3-PRRR mechanism (Tripteron) provides the necessary degrees of freedom (translational movements
along three axes) and the absence of singularity, since the Jacobian matrix of the mechanism is a unit matrix. Each chain contains one drive prismatic joint (P) and three rotation joints (R). Linear
actuators are connected to active linear translational pairs, which in turn are connected to guides and passive RRR chains connected to the moving platform. The configurations of kinematic chains,
$A i j B i j C i j D i j$
(hereinafter in the formulas i is the module index: 1—left module, 2—right module, j—index of kinematic chains of active manipulators), are variable, which means the possibility of bending chains in
different directions is present. Each of the chains have two possible configurations that are designated as
$l i j$
Figure 5
b. Also, the presented structure, in comparison with a single-module structure, assumes the use of the platform shape
$D i 1 D i 2 D i 3$
, not in the form of a regular triangle, but in the form of a right triangle with variability in the options for attaching kinematic chains to the platform (
Figure 6
). This assumes that the joints
$D i j$
can be located on different sides, relative to the center of the platform (
$D i 1$
—can be located in both the negative and positive directions along the Y–axis,
$D i 2$
—along the X–axis,
$D i 3$
—along the Z–axis). The active manipulator is connected to the passive orthosis through the link GP, connecting the G joint from the ankle joint of the passive orthosis and the center P from the
movable platform of the active manipulator. Each of the passive sections includes four rotational joints, two of which correspond to the hip joint (
$E i$
with angles
$α i$
flexion/extension and
$γ i$
abduction/adduction of the joint), one knee (
$F i$
with angle
$β i$
flexion/extension of the joint) and one ankle (
$G i$
with the angle
$θ i$
of joint flexion/extension). The point corresponding to the toe of the human foot is designated as
$H i$
. The relative position of the active manipulator and the passive orthosis, as well as the relative position of the two active manipulators, are determined by the value of two constant coordinates of
each of the active manipulator’s guides, relative to the base coordinate system located in the center of the patient’s pelvis.
Figure 5
a shows the effect of changing the coordinates of the guides on the system configuration, using the example of the second guide of the left module (
$A 12$
), which has a lower Z coordinate value compared to the first guide (
$A 11$
) of the left leg module and the first and second guides of the right leg module (
$A 21$
$A 22$
To check the reachability of the robotic system position, taking into account the intersections of the links, it is necessary to determine the position of all links of the active manipulator and
passive orthosis. The input data are the dimensions of the links and the angles in the patient’s joints:
$α i$
of the hip joint flexion,
$γ i$
of the hip joint abduction,
$β i$
of the knee joint flexion, and
$θ i$
of the ankle joint flexion. In this case referring to X, Y, Z in
Figure 5
, we define the coordinates of the joint centers
$E i$
as follows:
$E 1 = − L O E 0 0 , E 2 = L O E 0 0$
Taking into account the angles
$α i$
$γ i$
, the coordinates of the joint centers
$F i$
are given as follows:
$F i = E i + L E F cos α i sin γ i L E F cos α i cos γ i L E F sin α i ,$
Let’s substitute (3) into (4) to achieve the following:
$F 1 = L E F cos α 1 sin γ 1 − L O E L E F cos α 1 cos γ 1 L E F sin α 1 , F 2 = L E F cos α 2 sin γ 2 + L O E L E F cos α 2 cos γ 2 L E F sin α 2$
Taking into account the angle
$β i$
, the coordinates of the centers of the joints
$G i$
are given as follows:
$G i = F i + L F G cos α i + β i sin γ i L F G cos α i + β i cos γ i L F G sin α i + β i ,$
Let’s substitute (7) into (6) to achieve the following:
$G 1 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) L F G sin α 1 + β 1 + L E F sin α 1 ,$
$G 2 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) L F G sin α 2 + β 2 + L E F sin α 2$
Taking into account the angle
$θ i$
, we define the coordinates of the extreme point of the link
$G i H i$
corresponding to the human foot as follows:
$H i = G i + L G H cos α i + β i + θ i sin γ i L G H cos α i + β i + θ i cos γ i L G H sin α i + β i + θ i ,$
Let’s substitute (7), (8) into (9) to achieve the following:
$H 1 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 + L GH cos α 1 + β 1 + θ 1 ) − L O E cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 + L G H cos α 1 + β 1 + θ 1 ) L F G sin α
1 + β 1 + L E F sin α 1 + L G H sin α 1 + β 1 + θ 1 ,$
$H 2 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 + L GH cos α 2 + β 2 + θ 2 ) + L O E cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 + L G H cos α 2 + β 2 + θ 2 ) L F G sin α
2 + β 2 + L E F sin α 2 + L G H sin α 2 + β 2 + θ 2$
We define the coordinates of the centers
$P i$
of the moving platforms of active manipulators as follows:
$P i = G i + 0 0 − L G P$
Let’s substitute (7), (8) into (12) to achieve the following::
$P 1 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) L F G sin α 1 + β 1 + L E F sin α 1 − L G P ,$
$P 2 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) L F G sin α 2 + β 2 + L E F sin α 2 − L G P$
The center $P i$ of the moving platforms is understood as a point that is located along the X– and Y–axes in the center of a circle inscribed in the triangle of the platform with legs $d x$ and $d y$
, and along the Z–axis in the middle of the platform, having a thickness of $d z$.
To determine the coordinates of the joint centers
$D i j$
, we will consider options for attaching the kinematic platforms (
Figure 6
Let us denote the options for fastening the joint
$D i j$
by the variable
$p i j$
, which can take the value 1 (in the case of fastening in a negative direction along the corresponding coordinate axis relative to a right angle) or 2 (in a positive direction). To use values 1 and 2
in joint coordinate formulas, introduce the following function, which in the case of argument 1 returns the value −1, and in the case of argument 2 returns the value 1, as follows:
Taking into account (15), we define the coordinates of the centers of the joints as the centers
$D i j$
as follows:
$D i 1 = P i + − r λ p i 2 d y − r λ p i 1 0 , D i 2 = P i + d x − r λ p i 2 − r λ p i 1 0 ,$
$D i 3 = P i + − r λ p i 2 − r λ p i 1 d z λ p i 3 2 ,$
is the index of kinematic chains of active manipulators: 1—a chain with linear movement of the guide along the Y–axis; 2—along the X–axis; 3—along the Z–axis
—the radius of the inscribed circle of the platform, which is determined by the following formula:
$r i = d x i + d y i − d x i 2 + d y i 2 2$
Let’s substitute (13), (14) into (16), (17) and achieve the following:
$D 11 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E − r λ p 12 cos γ 1 L F G cos α 1 + β 1 + L E F cos α 1 + d y − r λ p 11 L F G sin α 1 + β 1 + L E F sin α 1 − L G P$
$D 12 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E + d x − r λ p 12 cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) − r λ p 11 L F G sin α 1 + β 1 + L E F sin α 1 − L
G P$
$D 13 = sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E − r λ p 12 cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) − r λ p 11 L F G sin α 1 + β 1 + L E F sin α 1 − L G P +
d z λ p 13 2$
$D 21 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E − r λ p 22 cos γ 2 L F G cos α 2 + β 2 + L E F cos α 2 + d y − r λ p 21 L F G sin α 2 + β 2 + L E F sin α 2 − L G P$
$D 22 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E + d x − r λ p 22 cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) − r λ p 21 L F G sin α 2 + β 2 + L E F sin α 2 − L
G P$
$D 23 = sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E − r λ p 22 cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) − r λ p 21 L F G sin α 2 + β 2 + L E F sin α 2 − L G P +
d z λ p 23 2$
The coordinates of the joint centers
$B i j$
in two dimensions depend on the position of the guides (coordinates that do not depend on the position of the platform) and in the third, they correspond to the coordinate of the joint center
$D i j$
, that is as follows:
$B i 1 = x B i 1 y D i 1 z B i 1 , B i 2 = x D i 2 y B i 2 z B i 2 , B i 3 = x B i 3 y B i 3 z D i 3$
Let’s substitute (18)–(23) into (24) to achieve the following:
$B 11 = x B 11 y cos γ 1 L F G cos α 1 + β 1 + L E F cos α 1 + d y − r λ p 11 z B 11 ,$
$B 12 = x sin γ 1 ( L F G cos α 1 + β 1 + L EF cos α 1 ) − L O E + d x − r λ p 12 y B 12 z B 12 ,$
$B 13 = x B 13 y B 13 z L F G sin α 1 + β 1 + L E F sin α 1 − L G P + d z λ p 13 2 ,$
$B 21 = x B 21 y cos γ 2 L F G cos α 2 + β 2 + L E F cos α 2 + d y − r λ p 21 z B 21 ,$
$B 22 = x sin γ 2 ( L F G cos α 2 + β 2 + L EF cos α 2 ) + L O E + d x − r λ p 22 y B 22 z B 22$
$B 23 = x B 23 y B 23 z L F G sin α 2 + β 2 + L E F sin α 2 − L G P + d z λ p 23 2$
Let us denote the variants of configurations of kinematic chains as
$l i j$
; they take values 1 and 2 in accordance with
Figure 5
a. Considering that the center of the joint
$C i j$
is located at the intersection of a circle with center
$B i j$
and radius
$L B C i j$
and a circle with center
$D i j$
and radius
$L C D i j$
, the coordinates
$C i j$
can be determined as follows:
$C i 1 = x B i 1 + s i 1 x D i 1 − x B i 1 − λ l i 1 g i 1 ( z D i 1 − z B i 1 ) L B D i 1 y D i 1 z B i 1 + s i 1 z D i 1 − z B i 1 + λ l i 1 g i 1 ( x D i 1 − x B i 1 ) L B D i 1 ,$
$C i 2 = x D i 2 y B i 2 + s i 2 y D i 2 − y B i 2 − λ l i 2 g i 2 ( z D i 2 − z B i 2 ) L B D i 2 z B i 2 + s i 2 z D i 2 − z B i 2 + λ l i 2 g i 2 ( y D i 2 − y B i 2 ) L B D i 2 ,$
$C i 3 = x B i 3 + s i 3 x D i 3 − x B i 3 − λ l i 3 g i 3 ( y D i 3 − y B i 3 ) L B D i 3 y B i 3 + s i 3 y D i 3 − y B i 3 + λ l i 3 g i 3 ( x D i 3 − x B i 3 ) L B D i 3 z D i 3 ,$
$s i j = L B C i j 2 − L C D i j 2 + L B D i j 2 2 L B D i j$
$g i j = L B C i j 2 − s i j 2$
$L B D i j = D i j − B i j$
$x D i j , y D i j , z D i j$
are coordinates of the joint centers
, which in accordance with (18)–(23) are defined as follows:
$x D 11 = sin γ 1 ( L FG cos α 1 + β 1 + L EF cos α 1 ) − L O E − r λ p 12 ,$
$z D 13 = L F G sin α 1 + β 1 + L E F sin α 1 − L G P ,$
$y D 12 = cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) − r λ p 11 ,$
$z D 12 = L F G sin α 1 + β 1 + L E F sin α 1 − L G P ,$
$x D 13 = sin γ 1 ( L FG cos α 1 + β 1 + L EF cos α 1 ) − L O E − r λ p 12 ,$
$y D 13 = cos γ 1 ( L F G cos α 1 + β 1 + L E F cos α 1 ) − r λ p 11 ,$
$x D 21 = sin γ 2 ( L FG cos α 2 + β 2 + L EF cos α 2 ) + L O E − r λ p 22 ,$
$z D 23 = L F G sin α 2 + β 2 + L E F sin α 2 − L G P ,$
$y D 22 = cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) − r λ p 21 ,$
$z D 22 = L F G sin α 2 + β 2 + L E F sin α 2 − L G P ,$
$x D 23 = sin γ 2 ( L FG cos α 2 + β 2 + L EF cos α 2 ) + L O E − r λ p 22 ,$
$y D 23 = cos γ 2 ( L F G cos α 2 + β 2 + L E F cos α 2 ) − r λ p 21 .$
To ensure the operability of the robotic system and the accessibility of the positions of the working platforms of two modules in the space required for the rehabilitation process, while excluding
their possible collisions during operation and mutual intersections in the presence of a large number of links of two modules, it is necessary to provide a condition that defines these criteria. To
do this, using the expressions obtained for the coordinates of the connection centers (3)–(33), which can be used for the design and operation of the system, we check the orientation of the
mechanisms for possible intersections of their links. For links connected by the joint, intersection checking can be performed by calculating the angle between the links and comparing it to the
minimum acceptable value. For links that are not connected to each other, intersections can be checked using the geometric approach discussed in [
]. The method is based on determining the minimum distance between the segments drawn between the center of the joints of each link; it is as follows. Let us imagine the links in the form of
spherocylinders (capsules). Let
$A 1 A 2$
$A 3 A 4$
be the segments connecting the centers of the joints of the links (
Figure 7
In this case, the condition of no intersections is written as follows:
$r l i n k 1 + r l i n k 2 < x ′ 2 + y ′ 2 + z ′ 2$
$r l i n k 1 , r l i n k 2$
are the radii of the links and
$x ′$
$y ′$
$z ′$
are the distance between the nearest points of the segments along each of the axes, defined as follows:
$x ′ = min i ∈ 1,2 x A i − max j ∈ 3,4 x A j i f min i ∈ 1,2 x A i > max i ∈ 1,2 x A i , min i ∈ 3,4 x A j − max i ∈ 1,2 x A i i f min i ∈ 3,4 x A j > max i ∈ 1,2 x A i , 0 i f min i
∈ 1,2 x A i ; max i ∈ 1,2 x A i ∩ min i ∈ 3,4 x A j ; max j ∈ 3,4 x A j .$
The values of $y ′$ and $z ′$ are determined similarly. If condition (35) is not met, the following check is carried out for the absence of intersections, when the minimum distance between the
segments $A 1 A 2$ and $A 3 A 4$ is greater than the sum of the radii of the links $r l i n k 1 + r l i n k 2$.
Let’s consider two cases of a mutual arrangement of links.
In case 1, the links are parallel. Let’s determine the minimum distance between the segments by rotating the segments relative to point
$A 1$
so that they become perpendicular to the YOZ plane. Let’s designate points
$A 2$
$A 3$
$A 4$
after the rotation as
$A 12$
$A 13$
$A 14$
, respectively. The distance between the segments is defined as follows:
$u 1$
is the distance between the segments
$A 1 A 12$
$A 13 A 14$
in the projection onto the YOZ plane, and
$u 2$
is the distance between the nearest points of the segments along the X–axis.
Figure 7
b shows an example of checking for the first case of intersection of links, when
$u 1 > 0$
$u 2 > 0$
, with
$u > r l i n k 1 + r l i n k 2$
. In this case there is no intersection.
In case 2, the links are not parallel. Let’s construct an auxiliary plane in which the segment $A 3 A 4$ will lie, with the segment $A 1 A 2$ parallel to this plane. In this case, the distance
between the segments is determined by Formula (36), where $u 1$ is the distance between the segment.
$A 1 A 2$ and the auxiliary plane, $u 2$, is the distance between the nearest points of the segments $A 3 A 4$ and the projection $A 11 A 12$ of the segment $A 1 A 2$ onto the auxiliary plane.
To determine $u 1$, we calculate the normal vector $N = N x N y N z T$.
Let’s determine the distance
$u 1$
using the following:
$u 1 = N x k 2 + N y k 2 + N z k 2 ,$
$k = N x x A 1 − x A 3 + N y y A 1 − y A 3 + N z z A 1 − z A 3 x A 1 N x + y A 1 N y + z A 1 N z$
To determine $u 2$, we rotate the auxiliary plane around point $A 11$, which is the projection of point $A 1$ onto the auxiliary plane, so that the auxiliary plane becomes parallel to the YOZ plane.
Let’s denote points $A 3$,$A 4$, and the projection of point $A 2$ onto the auxiliary plane after rotation as $A 13$,$A 14$ and $A 12$, respectively. As a result, determining $u 2$ is reduced to the
problem of calculating the distance between the nearest points of the segments $A 11 A 12$ and $A 13 A 14$ on a two-dimensional plane.
Figure 7
c shows an example of checking for the second case of intersection of links,
$u 2 > 0$
, that is, the segments do not intersect in the projection, but for
$u < r l i n k 1 + r l i n k 2$
, respectively, the segments intersect.
Let us write down the condition for the unattainability of the position of the moving platform, expressed by the lengths of the active manipulator links, which has the following form:
$L B D i j > L B C i j + L C D i j ,$
$L B D i j$
$D i j − B i j$
We will further use the resulting inequality to solve problems of parametric synthesis and select the optimal robotic system configurations.
4. Optimization Design Problem
The optimization problem consists in the selection of parameters to ensure the compactness of the mechanism when performing the trajectory of rehabilitation of the lower limbs. Various methods can be
used for optimization including multi-objective robust design optimization method for a mechatronic system [
] and mechanical structure optimization followed by FEM analysis to verify the changes made [
]. By optimization of the task placement, a reduction in energy costs in parallel manipulators is achieved [
], and nonlinear optimization problems can be solved using the reconfiguration of a parallel manipulator [
]. The problem under consideration is non-trivial. Let us formulate the optimization problem with the following steps:
• Selection of optimization parameters, which includes both continuous and discrete.
We use the link lengths as continuous optimization parameters
$L B C i j$
$L C D i j$
, guide positions
$x B i 1$
$z B i 1$
$y B i 1$
$z B i 1$
$x B i 3$
$y B i 3$
Figure 5
), and horizontal dimensions of the platforms.
We use the options as discrete parameters
$p i j$
for fastening kinematic chains to moving platforms using joints
$D i j$
Figure 6
) and options
$l i j$
for configurations of kinematic chains (
Figure 5
• Selection of optimization criterion. Due to the fact that as a result of optimization it is necessary to determine the geometric parameters at which the compactness of the structure is ensured,
we write the criterion function in the following form:
$F = ∑ i = 1 2 ∑ j = 1 3 L B C i j + L C D i j → m i n$
• The optimization constraint is the reachability of all points of the trajectory described earlier in
Section 2
and the absence of intersections for each of these points, that is as follows:
$N −$
is the number of trajectory points in discrete form, taking into account the given accuracy
$Δ t$
, which are unattainable.
Due to the significant reduction in the range of permissible parameter values and the optimization limitation, as well as the possibility of using the limitation as part of the criterion function in
the form
$N − → m i n$
, we will exclude the optimization limitation, but we will take it into account in the criterion function (39) as follows:
$F ′ = ϑ ∑ i = 1 2 ∑ j = 1 3 L B C i j + L C D i j + N − + ρ 1 − ϑ → m i n$
is a given penalty coefficient,
is the Heaviside function in the following:
$ϑ = 1 , i f N − = 0 0 − o v e r w i s e$
As an optimization algorithm, we use a parallel modification of the PSO algorithm [
], which has successfully proven itself for solving a wide range of optimization problems.
The requirements of ergonomics and manufacturability of a robotic system may impose additional constraints on the optimized parameters and thereby limit the range of feasible solutions. For a
balanced selection of geometric parameters from a wide range of possible ones but taking into account the ergonomics and manufacturability of the robot design, modeling was performed for four levels
of constraints as in
Table 3
5. Numerical Simulation
Let us assign the initial platform data and algorithm optimization parameters. Based on the perpendicular position of the foot relative to the shin, we take the angle at the ankle joints
$θ i = 90 °$
. The dimensions of the links of passive orthoses (
Table 4
) are set in accordance with the anthropometric data given in
Table 2
Let us take the diameters of the active manipulator links to be equal to
$d l i n k = 80$
mm. To ensure that
$C i 3 D i 3$
can move under the platform without colliding with the links
$C i 1 D i 1$
$C i 2 D i 2$
the condition
$d Z > 2 d l i n k$
must be met, therefore we will take the size
$d Z = 170$
mm. To avoid collisions between the orthosis links and the links
$C i 3 D i 3$
, we calculate the size
$L G P$
as follows:
$L G P = d F G + 1,1 d l i n k + 0.5 d Z = 270 mm$
Optimization parameter ranges can be considered as follows:
• Continuous:
□ Link sizes: $L B C i j ∈ 200 ; 900$, $L C D i j ∈ 200 ; 900$;
□ Coordinates of the guides:
$x B 11 ∈ − 2000 ; − 50 , z B 11 ∈ − 1500 ; 1500 , y B 12 ∈ 0 ; 2000 ,$
$z B 12 ∈ − 1500 ; 1500 , x B 13 ∈ − 50 ; 2000 , y B 13 ∈ 0 ; 2000 ,$
$x B 21 ∈ 50 ; 2000 , z B 21 ∈ − 1500 ; 1500 , y B 22 ∈ 0 ; 2000 ,$
$z B 22 ∈ − 1500 ; 1500 , x B 23 ∈ 50 ; 2000 , y B 23 ∈ 0 ; 2000 ,$
□ Platform sizes: $d x ∈ 100 ; 300$, $d y ∈ 100 ; 300$.
• Discrete:
□ Options for attaching kinematic chains to moving platforms $p i j ∈ 1,2$;
□ Variants of configurations of kinematic chains $l i j ∈ 1 , 2 .$
The time step when checking trajectory points is $Δ t = 5$, and penalty coefficient is $ρ = 100,000$. The PSO algorithm parameters are the number of individuals in the initial population $H = 10,000$
, number of generations $W = 4$, number of groups $G = 2$, values of free parameters $α P S O = 0.7 , β P S O = 1.4 , γ P S O = 1.4$. To increase efficiency, each iteration of searching for optimal
configurations is performed in two stages. At the first stage, the optimal solution is searched for with the original parameter ranges; at the second stage, the range of each parameter is reduced
five times; and the center of the new ranges coincides with the best solution found based on the result of the first stage.
To perform optimization of the parameters, a software package has been developed, including an optimization module in the C++ programming language, using parallel computing for simultaneous
calculation of the criterion function of various individuals of the PSO algorithm population. In addition, visualization modules are formulated in Python, which allow us to construct graphs of
changes in the position of joint centers and to visualize the movement of the robotic system with link interference to check optimal configurations. An example of visualization using the developed
software package is shown in
Figure 8
. In the case of intersection, the links are depicted in blue.
Numerical simulation is performed not only to obtain structural parameters, but also to verify the correctness of the selected structure.
5.1. Constraint Level 1
Due to the non-monotonic nature of the criterion function (40) and the large dimension of the search space during the iterative search for optimal configurations, as a result of executing the
algorithm, various configuration options corresponding to local extrema can be obtained. Iterative optimization was performed. Iterative computations are worked out to obtain optimal solutions using
the PSO (Particle Swarm Optimization) algorithm [
]. Each iteration is performed independently of each other to obtain results at different local minima. In each iteration, an optimal set of parameters is obtained. The distribution of each of the
continuous parameters is shown in the box plot (
Figure 9
). Local minima were obtained with a value of the criterion function within the range from 7040.7 (the minimum value) to 8700.3, whereas the arithmetic mean value of the criterion function in local
minima was computed as 7686.3. The corresponding design parameter values for the minimum value of the criterion function are shown in
Table 5
Table 6
Table 7
in the first row for Level 1.
Let us introduce additional constraints related to ergonomics and manufacturability of the design, corresponding to the second level.
5.2. Constraint Level 2
At the second level, a constraint has been added on the equality of the
coordinates of all guides, that is,
$y B 12 = y B 13 = y B 22 = y B 23$
, as well as the equality of the
coordinates of the guides for the left and right legs, that is,
$x B 11 = x B 13$
$x B 21 = x B 23$
. This makes it possible to reduce the dimension of the parameter space and provide a more ergonomic and technological design of the robotic system.
Figure 10
shows a comparison example of the relative position of the drive guides for the first and second level of constraints. As can be seen from the figure, the constraint allows us to obtain more ordered
robotic system configurations.
Iterative optimization was performed. In each iteration, an optimal set of parameters is obtained. The distribution of each of the continuous parameters is shown in the box plot (
Figure 11
Local minima were obtained with value of the criterion function within the range from 7265.1 (the minimum value) to 7898.2, whereas the arithmetic mean value of the criterion function in local minima
was computed as 7552.4. The corresponding design parameters values for the minimum value of the criterion function are shown in
Table 5
Table 6
Table 7
, in the second row for Level 2. The addition of the constraint affected the minimum value of the criterion function, which is 3.19% more than the value of the criterion function for level 1, which
is 7040.7.
5.3. Constraint Level 3
Within the third level, in addition to the constraints of the second level, constraints have been added based on the ranges of variable coordinates of the guides, suggesting the location of the
guides for each of the chains within 300 mm above or below the range of changes in the variable coordinates of the guides (
Figure 12
). In addition to the second level limitation, it also increases the ergonomics and manufacturability of the robotic system design.
This constraint can be written as follows:
$z B i j ∈ z B i 3 _ − 300 ; z B i 3 _ ∨ z B i j ∈ z B i 3 ¯ ; z B i 3 ¯ + 300 , j ∈ 1,2 , x B 1 j ∈ x B 12 _ − 300 ; x B 12 _ , x B 2 j ∈ x B 22 ¯ ; x B 22 ¯ + 300 , j ∈ 1,3 , y B i j ∈ y B i 1 ¯ ;
y B i 1 ¯ + 300 , j ∈ 2,3 ,$
$x B i 2 _$
$x B i 2 ¯$
are the limits of the range of movement
$x B i 2$
of the guides,
$z B i 3 _$
$z B i 3 ¯$
are the range of
$z B i 3$
$y B i 1 ¯$
is the upper limit of the range of movement y_Bi1. Based on the kinematics of the robotics system, which assumes the equalities
$x B 12 = x P 1$
$x B 22 = x P 2$
$y B i 1 = y P i$
$z B i 3 = z P i ± d z / 2$
, it follows that the boundaries of the ranges of movement of the guides are determined based on the ranges of movement of the center of the platform. The graph of changes in the coordinates of the
centers of the platforms
$P i$
during the trajectory development process is shown in
Figure 13
Due to the ambiguity in determining the position of the platform, according to the equality
$z B i 3 = z P i ± d z / 2$
, we accept the assumption
$z B i 3 ≈ z P i$
to calculate the parametric constraints. Taking into account the range of changes in the coordinates of the center of the platforms
$P i$
, the ranges of optimization parameters corresponding to the coordinates of the guides for constraint level 3 have the following values:
$x B 11 = x B 13 ∈ − 1155.5 ; − 855.5 , x B 21 = x B 23 ∈ 855.5 ; 1155.5 ,$
$y B 12 = y B 13 = y B 12 = y B 13 ∈ 1441 ; 1741 ,$
$z B i j ∈ − 25 ; 275 ∨ z B i j ∈ − 1132,96 ; − 1432,96$
The distribution of each of the continuous parameters is shown in the box plot (
Figure 14
Local minima were obtained with value of the criterion function within the range from 7739.7 (the minimum value) to 9320.9, whereas the arithmetic mean value of the criterion function in local minima
was computed as 8332.9. The corresponding design parameter values for the minimum value of the criterion function are shown in
Table 5
Table 6
Table 7
, in the third row for Level 3. The consequence of adding additional constraints of the third level is an increase in the value of the criterion function (7739.7) by 6.5%, compared to the second
level. However, this made it possible to significantly improve the ergonomics and manufacturability of the design, which can be clearly assessed when comparing configurations of level 2 (
Figure 10
b) and level 3 (
Figure 12
5.4. Constraint Level 4
For level 4, in addition to the constraints of the 3rd level, we will add a constraint on the equality of the lengths of module links
$L B C 1 j = L B C 2 j$
$L C D 1 j = L C D 2 j$
. In this case, the number of parameters is reduced by six to twenty-seven. The distribution of each of the continuous parameters is shown in the box plot (
Figure 15
). Local minima were obtained with the value of the criterion function within the range from 7784.1 (the minimum value) to 8735.5, whereas the arithmetic mean value of the criterion function in local
minima was computed as 8439.7. The corresponding design parameters values for the minimum value of the criterion function are shown in
Table 5
Table 6
Table 7
, in the fourth row for Level 4.
Platform Connection Chain Configuration
Level Target Function Left (I) Module Chains Right (II) Module Chains Left (I) Module Chains Right (II) Module Chains
1 7040.7 1 2 2 1 2 2 2 2 1 1 2 1
2 7265.1 1 1 1 2 1 1 1 1 1 1 1 1
3 7739.7 2 2 2 2 1 2 1 1 1 1 1 1
4 7784.1 1 1 1 2 2 2 1 2 1 1 1 1
Guide Positions (mm) Platforms
Level Left (I) Module Right (II) Module d_x (mm) d_y
x_B11 z_B11 y_B12 z_B12 x_B13 y_B13 x_B21 z_B21 y_B22 z_B22 x_B23 y_B23 (mm)
1 −1028.0 −629 2000 −957 −1114 1227 419 −1162 1844 −355 682 2000
2 −1042.6 −1031.0 1588.2 −1092.6 −1042.6 1588.2 1165.1 −861.4 1588.2 −1096.6 1165.1 1588.2 140.8 272.7
3 −1066.5 −1239.2 1567.9 −1208.0 −1066.5 1567.9 1043.0 −1280.4 1567.9 −1283.0 1043.0 1567.9 152.5 300.0
4 −884.6 247.1 1677.1 −1300.6 −884.6 1677.1 891.6 148.7 1677.1 −25.0 891.6 1677.1 148.9 281.7
Link Sizes (mm)
Level Left (I) Module Right (II) Module
L_ L_ L_ L_ L_ L_ L_ L_ L_ L_ L_ L_
BC11 CD11 BC12 CD12 BC13 CD13 BC21 CD21 BC22 CD22 BC23 CD23
1 519.3 573.8 900.0 443.4 408.5 569.0 656.7 631.8 685.9 544.6 285.2 822.5
2 603.1 777.0 467.4 726.6 674.2 397.4 675.7 661.9 450.7 704.3 382.7 744.1
3 780.6 725.0 568.8 783.5 559.8 465.5 725.0 814.9 628.0 682.1 494.2 512.3
4 790.9 799.0 599.1 757.0 393.8 552.2 790.9 799.0 599.1 757.0 393.8 552.2
The increase in the minimum value of the criterion function for the fourth level in comparison with the third level was 0.57%. The increase in the value of the criterion function in comparison with
the first level was 10.56%. Configuration 9, with the minimum value of the criterion function significantly exceeds the other nine found configurations of the fourth level and provides the
opportunity to unify the links in comparison with level 3. To verify configuration 9, a visualization of the movement along the generated trajectory was visualized, successfully completed without
link interference (
Figure 16
Based on the insignificant difference in the criterion function, with a significant increase in manufacturability and ergonomics, we will choose the level 9 configuration as the final one for the
production of the prototype.
6. Experimental Investigations
A prototype rehabilitation robot was built with the optimal design solution. The prototype was made according to drawings obtained as a result of the design algorithm implemented to NX CAD/CAE system
to make the system assembly. The prototype consists of two modules (I and II
$a c c o r d i n g t o t h e$Figure 5
) identical in structure, which can be controlled independently of each other (
Figure 17
). Each module according to the proposed design, it includes 3-PRRR manipulator (with three kinematic chains
$A i j B i j C i j D i j$
according to the
Figure 5
) and passive orthosis (
$E i F i G i H i$
A controller has been selected that provides precise control of the system and maximum safety for the patient. To control the prototype, an industrial logic controller OWEN PLC 210 (LLC "Production
Association OWEN", Moscow, Russia) was used. The controller uses the development environment (IDE) CODESYS V3.5. CODESYS supports IEC 61131-3 programming languages and the additional CFC language,
which allows to develop a human–machine interface and configure data exchange with devices. To control the robot system, a control interface for the operator was created and visualized. Data exchange
with servo motor drivers are carried out via the RS-485 interface.
When the prototype runs, calibration is performed using limit sensors, and moving the slider
$A i j$
Figure 5
) along the guide during operation is determined from data received from an encoder mounted on the servomotor. The control system, as shown in
Figure 18
, provides for standard motion control of six engines, implementation of control blocking for all engines, as well as an emergency stop button. To test and verify the operating modes of the
prototype, control programs have been created that implement the developed motion trajectories for conducting rehabilitation exercises.
To ensure the seating of patients with different anthropometric data, a chair is used that can be adjusted to three positions: tilting the back of the chair, as well as longitudinal and transverse
movement of the chair in the plane of the base of the rehabilitation system. For longitudinal movement of the chair, a stepper motor with a torque of 20 kg/cm is used, and for transverse movement, a
stepper motor is used, the torque of which is 34 kg/cm. To control these motors, programmable stepper motor controllers SMSD-4.2 are used. The chair is moved using ball screws along two linear
To prevent injury to the patient when the limb moves into a position not intended by physiology, the orthosis is connected to the end-effector (platform
$D i 1 D i 2 D i 3$
with center
$P i$
according to the
Figure 5
) of the manipulator by a suspended safety device with elastic elements that provide shock absorption when a dangerous load occurs. Gas lift GL105 was used as elastic elements, providing the required
forces to meet the safety requirements of rehabilitation exercises.
Experimental tests of the developed prototype were carried out, which include as follows:
• Operation of the mechanical safety device to ensure patient safety.
• Practicing the movement of limbs in the sagittal plane.
The safety device is the link between the active 3-PRRR mechanism and the passive orthosis. In the initial position, there are no external forces acting on the orthosis, all elastic elements of the
safety device are in a relaxed state, their length is maximum (
Figure 19
When the orthosis is exposed to a force exceeding the rigidity of the elastic elements, directed along the Y–axis, the upper pair of elastic elements is compressed, and the orthosis is displaced in
the direction opposite to the direction of the force.
Similarly, when the orthosis is exposed to a force exceeding the rigidity of the elastic elements, directed along the X–axis, the lower pair of elastic elements is compressed, and the orthosis is
displaced in the opposite direction (
Figure 20
Based on the results of experimental tests, it can be noted that the safety device functions correctly, operates under the required load, and allows for safety when performing rehabilitation
movements of the lower limbs.
The prototype was used to test the trajectory of movement during the rehabilitation process. The acquired trajectory of movement of the end-effector point
of the active manipulator is shown in
Figure 21
. The trajectory is obtained using Equations (7)–(14), taking into account the following dimensions of a real patient’s orthosis:
$L E F = 483$
$L F G = 630$
$L G P = 90$
mm. The obtained trajectory clearly shows the movement of the end-effector during the experimental investigations with suitable characteristics in reproducing a typical human trajectory.
A comparison was made of the simulated and experimentally obtained values of the angles in the hip and knee joints when moving along the rehabilitation trajectory. The discrepancy was measured to
assess how accurately the experimental sample allows for a change in angles in the patient’s joints (
Figure 22
Figure 23
graphically shows the difference between the experimental and theoretical values of the patient’s joint angles.
The maximum discrepancy between the theoretical and experimental angle values was 2.82 deg, and the average discrepancy was 1.2 deg. This discrepancy is due to the design features of the passive
orthosis, which are difficult to take into account when designing and calculating the mechanism. Experimental tests have shown that the final version of the prototype of a robotic system for the
rehabilitation of the lower extremities allows you to perform the required therapeutic movements with a given accuracy, and a safety device ensures the safety of performing exercises. Discrepancies
between the simulation and practical angles in the patient’s joints do not have a critical effect within the framework of therapeutic movements of rehabilitation of the lower limbs, but they can be
compensated by correcting errors in the control system of the active mechanism.
7. Conclusions
Based on the results of a numerical experiment, the best configuration for design was obtained. It has been established that an increase in the level of parametric constraints reduces the design
compactness in the range from 0.57 to 10.56%. But at the same time, a significant improvement in the ergonomics and manufacturability of the design is achieved. An experimental sample of a two-module
hybrid robotic system was designed and manufactured, which successfully passed experimental tests. To ensure safety, the use of a suspended safety device has been proposed and experimentally
confirmed to compensate for the excess load acting on the patient’s limb from the active manipulator, which allows, due to elastic elements, compensation for the movements of the active manipulator
that are unacceptable by the patient’s physiology.
Author Contributions
Conceptualization, D.M. and M.C.; methodology, D.M. and V.P.; software, D.M.; validation, M.C. and V.P.; formal analysis, V.P.; investigation, D.M. and V.P.; resources, V.P.; data curation, V.P.;
writing—original draft preparation, D.M. and V.P.; writing—review and editing, M.C.; visualization, D.M.; supervision, M.C.; project administration, V.P.; funding acquisition, D.M. and V.P. All
authors have read and agreed to the published version of the manuscript.
This work was supported by the state assignment of Ministry of Science and Higher Education of the Russian Federation under Grant FZWN-2020-0017.
Data Availability Statement
Data are contained within the article.
The work realized using equipment of High Technology Center at BSTU named after V.G. Shukhov.
Conflicts of Interest
The authors declare no conflicts of interest.
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Figure 1.
Axes of rotations of the hip joint [
Figure 2. The planned sequence of movements as reference joint trajectories: (a) Hip joint flexion $α i$; (b) Knee joint flexion $β i$; (c) Hip joint abduction $γ i$.
Figure 3. Reference movement of the center of the ankle joint: (a) three-dimensional view; (b) in projection on the XY plane; (c) in projection on the XZ plane; (d) in projection on the YZ plane.
Figure 5. Kinematic scheme of rehabilitation system: (a) the kinematic design (I is the left module for the left limb; II is the right module for the right limb), (b) an alternative design for the
kinematic design.
Figure 7. Checking the links intersection: (a) model for links; (b) first case type; (c) second type case.
Figure 9. Distribution of continuous parameters for constraint level 1: (a) guide positions; (b) link sizes.
Figure 10. An example of the relative position of guides at different levels of constraints: (a) Level 1; (b) Level 2.
Figure 11. Distribution of continuous parameters for constraint level 2: (a) guide positions; (b) link sizes.
Figure 13. Changing the coordinates of the center $P i$ of the platforms during the trajectory development process.
Figure 14. Distribution of continuous parameters for constraint level 3: (a) guide positions; (b) link sizes.
Figure 15. Distribution of continuous parameters for constraint level 4: (a) guide positions; (b) link sizes.
Figure 17.
The built prototype of the designed solution from
Figure 4
Figure 5
Figure 18.
Control system block diagram for the prototype in
Figure 17
Figure 19. Test position of the safety device: (a) without applying load, (b) compression of the upper pair of elastic elements and movement of the orthosis when the load directed along the Y–axis is
Figure 20. Test compression of elastic elements with simultaneous excess load along: (a) X–axis, simultaneously along the X and Y, (b) when a movement causes a simultaneous excess of load on both
axes, compression of all pairs of elastic elements occurs.
Figure 21. Acquired trajectory of test movement of the end-effector of the active manipulator during the rehabilitation process.
Figure 23. Discrepancy between experimental and simulated values of angles when practicing rehabilitation movements.
Table 1.
Anthropometric measurements of the lower limbs, [
Measurement Country with Max Value for Women Value, mm Country with Max Value for Men Value, mm
Thigh circumference Kenya 720 Thailand 660
Calf muscle circumference Kenya 416 Japan 422
Length buttock-knee The Netherlands 664 The Netherlands 703
Foot length Kenya 270 The Netherlands 296
Foot width The Netherlands 107 The Netherlands 116
Calf length The Netherlands 483 The Netherlands 538
Thigh width in sitting position USA 501 The Netherlands 438
Without Orthosis Taking into Account the Orthosis
Thigh circumference 720 815
Calf muscle circumference 422 610
Length buttock-knee 703 703
Foot length 296 326
Foot width 116 176
Calf length 538 738
Thigh width in sitting position 501 531
№ Level Constraints Level Number of Optimization Parameters
1 No assumptions 38
2 Constraint on equality of $Y$ coordinates of all guides, equality of $X$ coordinates of guides for the left and right legs 33
3 Level 2 + constraint on the coordinate ranges of guides based on the ranges of their variable coordinates 33
4 Level 3 + constraint on equal lengths of module links $L B C 1 j = L B C 2 j$, $L C D 1 j = L C D 2 j$ 27
$θ i °$ $L O E ,$ $L E F ,$ $L F G ,$ $L G H ,$ $d E F ,$ $d F G ,$ $d G H ,$ $d Z ,$ $d l i n k ,$
$m m$ $m m$ $m m$ $m m$ $m m$ $m m$ $m m$ $m m$ $m m$
90 135.5 703 738 326 259 194 176 170 80
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Malyshev, D.; Perevuznik, V.; Ceccarelli, M. Optimal Design of Lower Limb Rehabilitation System Based on Parallel and Serial Mechanisms. Machines 2024, 12, 104. https://doi.org/10.3390/
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Malyshev D, Perevuznik V, Ceccarelli M. Optimal Design of Lower Limb Rehabilitation System Based on Parallel and Serial Mechanisms. Machines. 2024; 12(2):104. https://doi.org/10.3390/machines12020104
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Malyshev, Dmitry, Victoria Perevuznik, and Marco Ceccarelli. 2024. "Optimal Design of Lower Limb Rehabilitation System Based on Parallel and Serial Mechanisms" Machines 12, no. 2: 104. https://
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Anderson localization in optical lattices with correlated disorder
arXiv:1510.05121v1 [cond-mat.quant-gas] 17 Oct 2015
E. Fratini and S. Pilati1 1
The Abdus Salam International Centre for Theoretical Physics, 34151 Trieste, Italy We study the Anderson localization of atomic gases exposed to simple-cubic optical lattices with a superimposed
disordered speckle pattern. The two mobility edges in the first band and the corresponding critical filling factors are determined as a function of the disorder strength, ranging from vanishing
disorder up to the critical disorder intensity where the two mobility edges merge and the whole band becomes localized. Our theoretical analysis is based both on continuous-space models which take
into account the details of the spatial correlation of the speckle pattern, and also on a simplified tight-binding model with an uncorrelated distribution of the on-site energies. The mobility edges
are computed via the analysis of the energy-level statistics, and we determine the universal value of the ratio between consecutive level spacings at the mobility edge. We analyze the role of the
spatial correlation of the disorder, and we also discuss a qualitative comparison with available experimental data for interacting atomic Fermi gases measured in the moderate interaction regime.
PACS numbers: 03.75.-b, 67.85.-d,05.60.Gg
Anderson localization, namely the complete suppres-sion of wave diffusuppres-sion due to sufficiently strong disor-der [1], is one of the most important and intriguing phe-nomena studied in condensed
matter physics [2, 3]. Mak-ing reliable predictions for the critical disorder strength required to induce complete localization is a major the-oretical challenge. In the theory of solid-state
systems, studies that aim at a quantitative comparison between theory and experiments, and thus employ realistic mod-els taking into account the details of a specific material, have appeared only
recently [4].
Following the first experimental observations of Ander-son localization in quantum matter waves [5–9], ultra-cold atomic gases have emerged as the ideal setup to investigate the effects due to
disorder in quantum sys-tems [10, 11]. Feshbach resonances provide experimen-talists with a knob to turn off the interatomic scattering, allowing them to disentangle the effects due to disorder from
those due to interactions. Furthermore, using the optical speckle fields produced by shining coherent light through a diffusive plate, they can introduce disorder in a controlled manner, and even
manipulate the structure of its spatial correlations [12]; this kind of control is not possible in solid-state devices. Techniques to accurately measure the mobility edge, namely the energy threshold
which separates the localized states from the extended states, have also been implemented [13].
Several previous theoretical studies on Anderson local-ization have disclosed the fundamental role played by the disorder correlations. In low dimensional systems, the characteristics of the
correlations determine the pres-ence or abspres-ence of an effective mobility edge [14–19]. In three dimensions, varying the correlation structure dras-tically changes the localization length and the
transport properties [20, 21]. In two recent studies, the mobility edge of ultracold atoms in the presence of isotropic and anisotropic optical speckle patterns has been precisely
FIG. 1: (color online) Cross section of the three-dimensional intensity profiles of a simple-cubic optical lattice with a super-imposed blue-detuned isotropic optical speckle pattern. The optical
lattice intensity is V0 = 4Er, the disorder strength is
Vdis= 1.3Er. The speckles patterns in the two panels have
different correlation lengths: σ = d/π in panel (a) and σ = d in panel (b). The color scale represents the potential intensity in units of recoil energy Er.
determined [22, 23], highlighting again the importance of taking into account the details of the disorder correla-tions. However, the experimental configuration which re-sembles more closely the
behavior of electrons in solids is the one where the atoms are exposed to the deep periodic potential due to an optical lattice with, additionally, the disorder due to a superimposed optical speckle
pattern (see intensity profiles in Fig. 1). This configuration with both an optical lattice and a speckle field has been imple-mented in experiments performed with Bose and Fermi gases [24–26], so
far considering interacting atoms.
In this Article, we investigate the Anderson localiza-tion of noninteracting atomic gases in a simple-cubic op-tical lattice plus an isotropic blue-detuned opop-tical speckle field. The first two
mobility edges and the corresponding critical filling factors are determined as a function of the
disorder strength (see Fig. 2). Our computational proce-dure is based on the analysis of the energy-level statistics familiar from quantum-chaos theory [27] and on the de-termination of the universal
critical adjacent-gap ratio.
We employ both continuous-space models which de-scribe the spatial correlation of an isotropic speckle pat-tern, and also an uncorrelated discrete-lattice model de-rived within a tight-binding
scheme. This allows us to measure the important effect of changing the disorder correlations length, and to shed light on the inadequacy of the simple tight-binding approximation in the strong
disorder regime. Our (unbiased) results are important as a guide for future experiments performed with noninter-acting atoms in disordered optical lattices, and also as a stringent benchmark for
(inevitably approximate) theo-retical calculations of the properties of disordered inter-acting fermions based on realistic models of disorder. The rest of the Article is organized as follows: in
Sec-tion II we define our model Hamiltonians, describing the details of the optical speckle patterns; we explain our theoretical formalism and analyze the universality of the critical adjacent-gap
ratio; furthermore, we pro-vide benchmarks of our predictions with previous results for tight-binding models with box and with exponential disorder-intensity distributions. In Section III our
pre-dictions for the mobility edges and the critical filling fac-tors are reported, with an analysis on the role played by the correlation length and on the validity of the tight-binding
approximation. We also discuss the compari-son with a recent transport experiment performed with atomic Fermi gases in the regime of moderate interaction strength [26]. Section IV summarizes the main
findings of this Article and reports our conclusions.
We consider noninteracting atoms exposed to a simple-cubic optical lattice with a superimposed optical speckle pattern. The single-particle Hamiltonian which describes the system is:
ˆ H = −~
2m∆ + V (r), (1)
where ~ is the reduced Planck’s constant, m is the atomic mass, and the external potential V (r) = VL(r) + VS(r)
is the sum of the simple-cubic optical lattice VL(r =
(x, y, z)) = V0P[ι]sin2(πι/d) (here, ι = x, y, z, d is the
lattice periodicity, and V0is the optical lattice intensity),
and the disordered potential VS(r) which represents the
isotropic optical speckle pattern. This intensity-based sum corresponds to the incoherent superposition of the optical-lattice and optical-speckle fields. In the follow-ing, it will be convenient to
express V0 in units of the
recoil energy Er = ~2/(2md2). The size L of the
three-dimensional box is chosen to be a multiple of d, consis-tently with the use of periodic boundary conditions.
0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 0 5 10 15 20 Ns / Ω V[dis]/E[r] V[dis]d /t σ = d/π σ = d TB[ELSS] 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.5 1 1.5 2 0 5 10 15 0 5 10 15 20 (E c -E 0 )/E r (E c -E 0 )/t V
[dis]/E[r] V[dis]d /t TB[ELSS] TB[TMM] σ = d/π σ = d
FIG. 2: (color online) Phase diagrams of an atomic gas ex-posed to three-dimensional simple-cubic optical lattices with a superimposed blue-detuned isotropic disordered speckle pat-tern: (a) First
two mobility edges Ec as a function of the
disorder strength Vdis/Er (or Vdisd/t for the discrete-lattice
model, in the top axis). Empty symbols correspond to the first mobility edge Ec1, full symbols to the second mobility
edge Ec2(see text). The energies are measured with respect
to the bottom of the first band of the clean system E0. The
red rhombi and the green circles correspond to the continuos-space Hamiltonian (1) with correlation lengths σ = d/π and σ = d, respectively. The blue squares correspond to our re-sults for the
tight-binding model with exponential on-site en-ergy distribution, obtained via analysis of the enen-ergy-level spacings statistic (TBELSS). The results obtained in Ref. [28]
using the transfer-matrix method (TBTMM) are represented
by pink crosses. The optical lattice intensity is V0 = 4Er,
and the corresponding hopping energy t ∼= 0.0855Er is used
to compare the continuous-space data (bottom-left axes) with the discrete-lattice data (top-right axes). (b) Critical filling factors Ns/Ω as a function of the disorder strength,
corre-sponding to the mobility edges represented in panel (a). Ns
is the number of states below the mobility edge, Ω the adi-mensional volume.
Disordered speckle patterns are realized in cold-atom experiments by shining lasers through diffusive plates, and then focusing the diffused light onto the atomic
0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 5 5.5 6 6.5 7 <r> E/E[r] L/d=9 L/d=7 L/d=5 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.4 0.5 0.6 0.7 0.8 0.9 1 <r> E/E[σ] L=11πσ L=15πσ L=21.25πσ
FIG. 3: (color online) Ensemble-average adjacent-gap ratio hri as a function of the energy E for the continuous-space Hamil-tonian (1). Left panel: simple-cubic optical lattice with intensity V0 =
4Er plus an isotropic optical speckle pattern with
correlation length σ = d/π and intensity Vdis= Er. Right panel: optical speckle field with intensity Vdis= Eσ (without an
optical lattice, namely V0= 0). The three datasets correspond to different system sizes. The horizontal cyan solid line indicates
the value for the Wigner-Dyson distribution hriWD, the dashed magenta line the one for the Poisson distribution hriP. The
dash-dot black line indicates the universal critical adjacent-gap ratio hriC, and the light-gray bar represents its error bar. The
energy units are the recoil energy Er and the correlation energy Eσ (see text).
cloud [10, 11]. Fully developed speckle fields are char-acterized by an exponential distribution of the local in-tensities [29]. In the case of a blue-detuned optical field, the atoms experience a
repulsive potential with the local-intensity distribution: Pbd(V ) = exp (−V /Vdis) /Vdis, if
the local intensity is V > 0, and Pbd(V ) = 0 otherwise.
The (global) intensity parameter Vdisfixes both the
spa-tial average of the disordered potenspa-tial Vdis= hVS(r)i and
also its standard deviation: V2
dis =VS(r) 2[ − hV]
For sufficiently large systems, spatial averages coincide with averages over disorder realizations.
The spatial correlations of the speckle pattern depend on the details of the illumination on the plate and of the optical setup used for focusing. We consider the ide-alized case of isotropic spatial
correlations described by the following two-point correlation function [22]: Γ(r = |r|) = hVS(r′+ r)VS(r′)i /Vdis2 − 1 = [sin(r/σ)/(r/σ)]
(averaging over the position of the first point r′ [is ]
as-sumed). The parameter σ determines the length scale of the spatial correlations and, therefore, the typical size of the speckle grains. The full width at half maximum of the correlation function Γ
(r) (defined by the condition Γ(ℓc/2) = Γ(0)/2) is ℓc ∼= 0.89πσ, while the first zero is
at rz = πσ. To generate this isotropic speckle pattern
we employ the numerical recipe described in Ref. [23]. For further details on speckle pattern generation, see Refs. [22, 30, 31].
We determine the positions of the mobility edges by analyzing the statistical distribution of the spacings between consecutive energy levels. The spectrum is obtained via exact diagonalization of the
matrix represented in momentum space, using the PLASMA library [32] for large-scale linear algebra computations on multi-core architectures. Special care is taken in analyzing the convergence of the
results with the basis-set size. Further details on the numerical procedure can be found in Ref. [23].
The mobility edges can be identified as the energy thresholds where the level-spacing distribution trans-forms from the Wigner-Dyson distribution characteristic of chaotic systems in the ergodic
phase, to the Poisson distribution characteristic of localized systems, or vice versa [27]. To distinguish the Wigner-Dyson and the Poisson distributions, it is convenient to consider the parameter r
= min {δn, δn−1} / max {δn, δn−1}, where
δn = En+1− En is the spacing between the (n + 1)th
and the nth energy levels, ordered for ascending energy values [33]. Its average over disorder realizations (later on referred to as adjacent-gap ratio) is known to be hriWD ≃ 0.5307 for the
Wigner-Dyson distribution and
hri[P]≃ 0.38629 for the Poisson distribution [34].
While in an infinite system hri would change abruptly at the mobility edge Ec, in finite systems one observes a
smooth crossover from hriPto hriWD, or vice versa. The
critical point can be determined from the crossing of the curves representing hri versus energy E corresponding to different system sizes L. We fit the data using the scaling function hri = g(E − Ec)
L1/ν (universal up to
a rescaling of the argument) [3] , where ν is the critical exponent of the correlation length. We Taylor expand the function g[x] up to second order and obtain Ec from
the best-fit analysis.
0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 -7.8 -7.6 -7.4 -7.2 <r> E/t L=24 L=28 L=32 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 -3 -2.8 -2.6 -2.4 -2.2 -2 -1.8 <r> E/t L=24 L=28 L=32
FIG. 4: (color online) Ensemble-average adjacent-gap ratio hri as a function of the energy E for the tight-binding Hamilto-nian (2). Left panel: three dimensional Anderson model with box disorder
distribution with intensity Vd
dis= 5t. Right panel:
three dimensional Anderson model with the exponential distribution with intensity Vd
dis= 7t. The three datasets correspond to
different system sizes. The horizontal cyan solid line indicates the value for the Wigner-Dyson distribution hriWD, the dashed
magenta line the one for the Poisson distribution hriP. The dash-dot black line indicates the universal critical adjacent-gap
ratio hriC, and the light-gray bar represents its error bar. The energy unit is the hopping energy t.
was previously employed in Ref. [23] for speckle pat-terns without optical lattices, requires several datasets corresponding to different system sizes with very small statistical error bars. A less
computationally expensive procedure is obtained by exploiting the universal properties of the critical point. Indeed, the level-spacing distribution at the critical point differs both from the
Wigner-Dyson and from the Poisson distributions [35, 36]; it is expected to be system-size independent and universal, meaning that it does not depend on the details of the disorder. This implies a
universal value of the critical adjacent-gap ratio, which we denote as hriC, different from hriWD and from hriP.
We verified this universality by performing the finite-size scaling analysis for various models, determining hri[C] as the value of the scaling function at vanishing argument g[0]. In Fig. 3 we
report the finite-size scaling analysis for a simple-cubic optical lattice with a superimposed speckle pattern, and also for a speckle pattern without the optical lattice (data from Ref. [23]). The
critical adjacent-gap ratios hri[C] of the two models (for the dis-ordered optical lattice we consider the first two mobility edges) agree within statistical error bar. Furthermore, we verified that
hri[C] does not depend on the disorder strength Vdis, and that a compatible value of hri[C] is
obtained also for red-detuned optical speckle fields, which have the same spatial correlations of blue-detuned speckle fields Γ(r) defined above, but the opposite local-intensity distribution Prd(V )
= Pbd(−V ).
A further verification of the universality of the critical adjacent-gap ratio hri[C] can be obtained by considering single-band models in a tight-binding scheme. The
corre-sponding discrete-lattice Hamiltonian can be written in Dirac notation as:
ˆ Hd= −t X hi,ji |ii hj| +X i Vi|ii hi| , (2)
where the indices i, j = 1, . . . , L label the sites of the cubic discrete lattice of adimensional volume Ω = L3[,]
t is the hopping energy, and the brackets hi, ji indi-cate nearest neighbor sites. The on-site energies Vi
are chosen according to a random probability distri-bution. The most commonly adopted choice in the theory of Anderson localization is the box distribution Pb(Vi) = θ(
Vi− Vdisd
)/(2V[dis]d). The parameter V[dis]d determines the disorder strength. We also consider the exponential distribution Pe(Vi) = exp Vi/Vdisd /V
d dis,
analogous to the exponential distribution Pbd(V )
de-scribed above for blue-detuned speckle patterns in the continuous-space Hamiltonian. This discrete-lattice model with the exponential on-site energy distribution is relevant to describe deep
optical lattices with super-imposed weak and uncorrelated speckle patterns, as explained more in detail in the Section III.
The finite-size scaling analyses for these two lattice models (box and exponential distributions) are shown in Fig. 4. The spectrum is obtained via exact diago-nalization of the matrix representing
the Hamiltonian
Hd, defined on the three dimensional lattice. The
universality of the critical adjacent-gap ratio is, again, confirmed within statistical uncertainty.
The average of the critical adjacent-gap ratios of the various models described above, including both the continuous-space models with correlated speckle patterns
and the uncorrelated tight-binding models, is hri[C] = 0.513 ± 0.05; the error bar represents the standard devia-tion of the populadevia-tion. This predicdevia-tion provides us with a computationally
convenient criterion to locate the transi-tion, consisting in identifying the mobility edge Ecas the
energy threshold at which the adjacent-gap ratio crosses the critical value hri[C]; the standard deviation of hri[C]will be used to define the error bar on Ec. By applying this
criterion to the isotropic speckle pattern (without optical lattice) analyzed in Fig. 3, we obtain Ec = 0.562(10)Eσ
(Eσ = ~2/mσ2 is the correlation energy), in agreement
with the transfer matrix theory of Ref. [22], which pre-dicts Ec = 0.570(7)Eσ. We further confirm the validity of
this criterion by reproducing the complete phase diagram of the discrete-lattice model with box disorder distribu-tion (typically refereed to as Anderson model), making comparison with older results
obtained using transfer-matrix theory [37] and multi-fractal analysis [38], as well as with the recent data from Ref. [39] obtained using the typical medium dynamical cluster approximation; see Fig.
5. Furthermore, in the case of the exponential disor-der distribution, our results perfectly agree with the very recent transfer-matrix theory from Ref. [28] (see Fig. 2). It is worth specifying that
our prediction for the uni-versal critical adjacent-gap ratio hri[C] applies to a cu-bic box with periodic boundary conditions. In fact, it has been predicted that the critical energy-level
distri-bution, and so possibly the corresponding value of hri[C], depends on the box shape [40] and on the boundary con-ditions [41, 42].
The continuous-space Hamiltonian (1) accurately describes atomic gases exposed to optical lattices with superimposed optical speckle patterns, for any optical lattice intensity V0 and disorder
strength Vdis. In
particular, it takes into account the spatial correlations of the optical speckle pattern. In order to make com-parison with recent experimental data, we consider the intermediate optical lattice
intensity V0 = 4Er, and we
determine the lowest two mobility edges as a function of the disorder strength Vdis, up to the critical value
where the two mobility edges merge and the whole band becomes localized.
We consider two isotropic speckle patterns with corre-lation lengths σ = d/π and σ = d. We recall that the first zero of the spatial correlation function Γ(r) (see def-inition in Section II) is at rz
= σπ. Beyond this distance
the speckle-field intensities are almost uncorrelated. The intensity profiles of the total potential V (r) correspond-ing to these two correlation lengths are shown in Fig. (1). The deformation of
the regular structure of the simple-cubic optical lattice due to the speckle pattern is evi-dent in both cases. In the first case the intensity values in nearest-neighbor wells of the optical lattice
are only
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Ec /t V[dis]d /t TMM MFA TMDCA ELSS
FIG. 5: (color online) Mobility edge Ec as a function of
the disorder strength Vd
dis for the three dimensional
Ander-son model with box disorder distribution. Our data com-puted via the analysis of the energy-level spacings statistics (ELSS, red diamonds) are compared with previous results ob-tained via
transfer-matrix method (TMM, green circles, from Ref. [37]), via multi-fractal analysis (MFA, blue squares, from Ref. [38]), and via typical medium dynamical cluster approx-imation (TMDCA, black
crosses, from Ref. [39]). The energy unit is the hopping energy t.
weakly correlated, while in the second case the correla-tions extend to a few lattice sites.
The phase diagram obtained using the methods presented in Section II, namely the analysis of the energy-level spac-ing statistics and the universality of the critical adjacent-gap ratio, is presented
in Fig. 2. The empty symbols in-dicate the lowest mobility edge Ec1, where the orbitals
transform from localized (for energies E < Ec1) to
ex-tended (for E > Ec1), while the solid symbols indicate
the second mobility edge Ec2 > Ec1, where the orbitals
transform from extended (for E < Ec2) to localized (for
E > Ec2). Other mobility edges are located at
signifi-cantly higher energies, outside the energy range investi-gated in this Article.
The data reported in Fig. 2 shed light on the fun-damental role played by the spatial correlations of the disorder pattern. The critical disorder strength beyond which the first band is fully
localized strongly depends on the correlation length. Indeed, for the short correlation length σ = d/π, full localization occurs already at the disorder strength Vdis ≃ 1.32Er, while for the longer
correlation length σ = d full localization occurs only at the much stronger disorder intensity Vdis ≃ 1.95Er.
This indicates that the disorder is more effective in inhibiting particle diffusion if the correlation length is short compared to the lattice spacing; also, it implies that, in order to
quantitatively describe experiments performed with noninteracting atomic gases exposed to disordered optical lattices, it is necessary to take into
0 50 100 0 0.5 1 1.5 2 2.5 3 3.5 4 dos (E-E[0])/E[r] σ = d/π, V[dis]=1.3E[r] σ = d,V[dis]=1.3E[r] TB, V[dis]=1.3E[r] σ = d/π, V[dis]=0.2E[r]
FIG. 6: (color online) Density of states dos (in arbitrary units) as a function of the energy E measured from the bottom of first band of the clean system E0. The energy unit is the
recoil energy Er. The continuous red, dashed green, and the
double-dash black curves correspond to the continuos-space model (1) with different correlation lengths σ and disorder intensities Vdis. The dotted blue curve corresponds to the
tight-binding (TB) model.
account the details of the optical speckle pattern. In Fig. 2, we report the two critical filling factors (defined as the number of eigenstates Nsper adimensional volume
Ω = (L/d)3 [with energy E < E]
c1 and E < Ec2) as a
function of the disorder strength. The role of the spatial correlations is again manifest. Both the mobility-edge data and the critical filling-factors data display a strong asymmetry around the band
center; this originates from the asymmetry of the exponential intensity distribution of the optical speckle pattern Pbd(V ).
Most theoretical studies of atomic gases exposed to clean optical lattices are based on single-band tight-binding Hamiltonians analogous to the one defined in Eq. (2). The conventional procedure to
map optical lattice systems to tight-binding models is based on the computation of the maximally localized Wannier function from the band-structure analysis of the periodic system. For sufficiently
deep optical lattices V0 ≫ Er,
the effect of higher Bloch bands and of hopping pro-cesses between non-adjacent Wannier orbitals can be ignored, leading to single-band tight-binding models in the discrete-lattice form defined by
Eq. (2). At the optical lattice intensity addressed in this Article, namely V0 = 4Er, the deep-lattice condition is marginally
fulfilled, with a next-nearest neighbor hopping energy |t2| ∼= 6.1 · 10−3Er, which is only one order of magnitude
smaller than the nearest neighbor hopping energy t ∼= 0.0855Er.
In the presence of additional disordered optical fields, the conventional mapping procedure [43, 44] based on band-structure calculation cannot be applied. A more
generic approach, valid also in the presence of weak optical speckle patterns with intensity Vdis ≪ V0, has
been developed in Ref. [45]; this method allows one to construct an orthonormal basis of localized Wannier-like orbitals which describes the correct low-energy properties of weakly disordered
optical-lattice systems. In the corresponding discrete-lattice Hamiltonian, the on-site energies Vi have, with good approximation,
the exponential distribution Pe(Vi), with a disorder
intensity Vd
dis ≃ Vdis, essentially coinciding with the
intensity of the optical speckle field Vdis. The on-site
energies on nearby lattice sites have significative cor-relations which depends on the details of the optical speckle pattern. Also, the nearest-neighbor hopping energies have an (asymmetric) random
distribution, characterized by strong correlations with the difference between the on-site energies of the corresponding lattice sites. In first approximation, one might neglect the hopping-energy
fluctuations and the on-site energy correlations, and retain only the exponential on-site energy distribution. This approximate model of optical lattices with superimposed speckle patterns - which
leads (in the noninteracting case) to the tight-binding Hamiltonian (2) with the on-site energy distribution Pe(Vi) - has been adopted in Ref. [46] to describe a
recent transport experiment performed with interacting ultracold atoms [46]. In this experiment, a drifting force was applied by introducing a magnetic-field gradient for a short interval of time;
after this impulse, the confining potential was switched off, and the velocity of the center of mass of the atomic cloud was measured by absorption imaging and band mapping after a time of flight;
the measurement was repeated with different intensities of the optical speckle field . Also, various optical lattice intensities were considered, ranging from V0 = 4Er to
V0 = 7Er. The authors of Ref. [46] considered mainly
the case of the deep optical lattice V0 ≃ 7Er, where
the Hubbard interaction energy of two opposite-spin fermions on the same lattice site is large: U ≃ 9t. They argue that in this strongly interacting regime the details of the correlations of the
hopping and of the on-site energies are not relevant, since transport is dominated by effective quasi-particles (not the original particles which are obviously relevant in the noninteracting case),
which experience correlated hopping and interaction processes even in the simplified model. They indeed found satisfactory agreement between the computed center-of-mass velocities and the
experimental data. Our findings indicate that in the absence of interac-tions the details of the speckle pattern are, instead, important. The mobility edges of the uncorrelated tight-binding model
(2) (with the exponential on-site energy distribution) are shown in Fig. 2, together with the results for the continuous-space model (1). To make comparison between the two models, the energies in
the lattice model have to be converted using the hopping energy t ∼= 0.0855Ercorresponding to the optical lattice
that certain qualitative features of the phase diagram are captured also by the tight-binding model. However, while at very weak disorder Vdis≈ 0.2Erthe
continuous-space and the discrete-lattice models quantitatively agree, important discrepancies appear at strong disor-der. In particular, the critical disorder strength where the whole band is
localized in the discrete-lattice model, namely Vd
dis ≃ 12t (corresponding to Vdis ≃ 0.95Er),
significantly underestimates the results obtained with the more accurate correlated continuous-space models. In principle, the details of the speckle pattern could be included also in a
discrete-lattice Hamiltonian, following the numerical procedure of Ref. [45]. This approach has been adopted in Ref. [47] to investigate an interacting Anderson-Hubbard model with correlated speckle
fields. However, the dynamical mean-field theory employed in Ref. [47] does not correctly describe the Anderson localization in the noninteracting limit, probably due to the assumed Bethe-lattice
structure. More recently, the dynamical mean-field theory has been improved using the typical medium dynamical cluster approximation [39], allowing researchers to give more accurate predictions for
the localization transition in the (uncorrelated) Anderson model with box distribution; the data from Ref. [39] are reported in Fig. 5.
Nevertheless, it should be emphasized that the numerical technique of Ref. [45] converges only as long as there is a well defined gap between the first and the second band. As shown in Fig. 6, in our
optical lattice the gap is well defined only for very weak disorder, while it is substantially filled when the intensity of the optical speckle field approaches the strength required to localize the
whole band, making that numerical technique inapplicable.
Experimental data for noninteracting atomic gases in disordered optical lattices are not available. However, in the experiment of Ref. [26] (described above), which was performed with interacting
atoms, the optical-lattice in-tensity was tuned down to V0= 4Er, corresponding to a
relatively small Hubbard interaction parameter, namely U ≃ 2.3t. It is then reasonable to discuss the comparison of these latter results with our theoretical predictions. It should be taken into
account that the optical speckle pattern employed in the experiment is anisotropic, with an axial correlation length approximately 5 times larger than the radial correlation length, and that its
spatial correlations decays as a Gaussian function. However, the propagation axis of the optical speckle field is dis-aligned with respect to the optical lattice axes; this is expected to strongly
reduce the role of the correlation anisotropy. If we consider the geometrically-averaged correlation length, we obtain a Gaussian correlation func-tion with similar full width at half maximum as our
speckle pattern with σ = d (within ≈ 15%). Further-more, in the experiment the density is inhomogeneous due to the confinement (with approximately 0.3 − 0.7 particles per lattice well in the trap
center, per spin
com-ponent) and the energy distribution is not precisely char-acterized. In Fig. 7 we plot the center-of-mass velocities vc.m. measured in the experiment, as a function of the
disorder strength. The critical point where vc.m.
van-ishes has been interpreted in Ref. [26] as the average disorder strength required to localize the whole band, since all extended states are expected contribute to trans-port. We indeed observe
that vc.m. reaches negligible
values (compatible with the experimental resolution) in the regime where we predict full localization to occur, depending on the details of the optical speckle pattern. Clearly, a quantitative
comparison with the experimen-tal data would require a precise characterization of the experimental atomic density and of the energy distribu-tion. This would also allow us to clarify the potential
role played by states in higher-energy bands. Never-theless, this qualitative agreement between experimen-tal data and theoretical predictions is encouraging, and should stimulate further
experimental efforts aiming at observing Anderson localization in noninteracting atomic gases in disordered optical lattices. All details of the op-tical speckle pattern could be included in our
theoreop-tical formalism. 0.01 0.1 1 0 0.5 1 1.5 2 0 5 10 15 20 TB
=d vc.m. (mm/sec) V
d /t
FIG. 7: (color online) Experimental data from Ref. [26]: cen-ter of mass velocity vc.m.of the atomic cloud (black squares)
as a function of the disorder strength Vdis/Er (or V d dis/t for
the tight-binding model, in the top axis). The vertical lines represent our predictions for the critical disorder strength where the whole band becomes localized in the continuos-space Hamiltonian
(1) (dashed red and dotted green lines) and in the uncorrelated tight-binding model (2) with expo-nential disorder distribution (dot-dash blue line). The gray band represents the experimental
resolution in detecting a vanishing velocity.
In summary, we have investigated the Anderson localization of noninteracting atomic gases in disordered
optical lattices. We considered both continuous-space models which describe the effect of a simple-cubic opti-cal lattice with a superimposed isotropic blue-detuned optical speckle field, taking into
account the spatial correlations of the disorder, and also an uncorrelated discrete-lattice Hamiltonian in a tight-binding scheme. Our predictions for the mobility edges and for the critical filling
factors indicate that the details of the speckle pattern play an important role; the critical disorder intensity where the whole band becomes localized strongly depends on the disorder correlation
length. The tight-binding model with an uncorrelated (exponential) disorder distribution significantly underestimates this critical disorder strength.
Our theoretical formalism is based on the analysis of the energy-level statistics familiar from random-matrix and quantum-chaos theories and on the determination of the universal critical
adjacent-gap ratio. The prediction for this universal value will be useful also in future studies of Anderson localization in different models belonging to the same universality class.
We have shown that the findings of a recent trans-port experiment performed with an atomic gas in the moderate interaction regime [26] are qualitatively consistent with our predictions; this
encouraging com-parison should stimulate further experimental efforts to accurately measure the critical point of the Anderson transition in noninteracting atomic gases exposed to controlled and well
characterized disordered fields. Such experiments would allow us to quantitatively benchmark sophisticated theories for Anderson localization based on realistic models which take into account all
details of the disorder. This would be beneficial for the field of ul-tracold atoms, and likely beyond, possibly including the research on disordered materials, on randomized optical fibers [48], and
on disordered photonic crystals [49].
We acknowledge fruitful discussions with Brian De-Marco, Vito Scarola, Estelle Maeva Inack and Giuliano Orso. Brian DeMarco is also acknowledge for providing us the data from Ref. [26]
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K. W. Koch, and A. Mafi, Optics letters 37, 2304 (2012). [49] M. Segev, Y. Silberberg, and D. N. Christodoulides, | {"url":"https://123dok.org/document/6qm3l44y-anderson-localization-in-optical-lattices-with-correlated-disorder.html","timestamp":"2024-11-09T22:03:34Z","content_type":"text/html","content_length":"191831","record_id":"<urn:uuid:384241e5-bf65-4b7d-a319-5afa32c2fc9e>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00202.warc.gz"} |
Mat. stat. seminarium 9 december 2002
Tid: 9 december 2002 kl 1515-1700
Plats : Seminarierummet 3733, Institutionen för matematik, KTH, Lindstedts väg 25, plan 7. Karta!
Föredragshållare: Ola Hössjer, Matematisk statistisk, Stockholms Universitet.
Titel: Computing genomwise significance levels in linkage analysis.
Sammanfattning: Statistical linkage analysis is a method for localizing genes causing or contributing to an inheritable disease. The idea is to test, at each chromosomal position, whether inheritance
of the disease is correlated (linked) to inheritance of DNA at that position. Because of the large number of tests being performed, it is important to adjust significance levels for multiple testing.
Standard Bonferroni corrections are too crude since tests at nearby positions are very correlated. In this talk, I describe a method based on extreme value theory for Gaussian processes, which
generalizes previous work by e.g. Kruglyak and Lander (1995). The score function (test statistic) is viewed as a stochastic process, with chromosomal position as `time index'. Two important steps of
the methods are 1) to calculate the slope of the covariance function at zero and 2) to correct for non-Gaussianity by means of a transformation. | {"url":"https://www.math.kth.se/matstat/seminarier/021209.htm","timestamp":"2024-11-10T01:47:12Z","content_type":"text/html","content_length":"2520","record_id":"<urn:uuid:5f4f401a-8fdd-444c-b42f-a71332d5c878>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00139.warc.gz"} |
L-4.1: Introduction to Greedy Techniques With Example | What is Greedy Techniques - RD2SUCCESS
L-4.1: Introduction to Greedy Techniques With Example | What is Greedy Techniques
Hello friends, welcome to Gate Smashers In this video we are going to discuss
introduction to greedy algorithms or what you can call greedy techniques which is a very important topic in algorithm So guys, like the video quickly, subscribe
the channel if you haven't done it yet and please press the bell button so that
you get all the latest notifications So let's start what is actually a greedy
technique or what are greedy algorithms Algorithms which follow local optimal choice at each stage with intent of finding the global optimum What does it mean to say? We go towards the best choice at
every stage locally Like we see with a simple example Let's say this is my source I have multiple paths from this source And these paths take me to some destination Let's say this is my destination D
This is my source and that is my destination Now if we talk here, as soon as I started from the source I am at this stage It is saying that it follows the local optimal choice What does local optimal
choice mean? At this stage, let's say the cost to go here is 10 The cost to go here is 20 The cost to go here is let's say 5 So what does it want to do in the greedy algorithm? Greedy algorithm at
this stage, the minimum cost Cost wise means if we talk The cost to go here is 5, 20 and 10 So what does it want to do? First of all, it wants to follow this choice What is the reason for that? It
follows the best or the optimal locally first And on the basis of that, it says With intent of finding the global optimum Means it wants on the basis that I can find global optimum result Or what I
can do with global maximum result And here if we talk In this we talk about solution space,
feasible solutions or optimal solution I actually want to tell the greedy algorithm from a real life example Like if we talk As a student we have multiple career options Like we have an option that
let's say if I have chosen non-medical So I have non-medical option that I can go to engineering line Or if I have chosen medical option then I can go to medical line Or I have banking sector option
SSE etc.
We can go for government jobs Or we can choose our business entrepreneurship Means I have multiple career options You can call this a solution space Means all these options are a solution space But
out of this solution space We first find out feasible solution What is the meaning of feasible solution? Based on some selection criteria Means we on the basis of selection criteria Out of all the
feasible, out of all the solution space I will find out the feasible condition I will find out the feasible solution To find out feasible solution means Let's say if I have done arts in 11th and 12th
Then obviously I can't go to engineering field Means out of all the solution space One field will come out from there So what is my selection criteria? Based on some selection criteria You have to
find out feasible solution So let's say if I have arts then on my selection criteria Engineering came out from here Medical came out from here Now what are the feasible solutions I have? I can go to
banking line I can go to banking line after taking IBPS exam I can go to SSE I can go to government teacher or
government assistant professor jobs I can do business I can open coaching centers These are the career choices I have now Out of all the feasible solutions Now what we have to find out from that?
Optimal solution Now what does optimal solution mean? Because greedy algorithms are totally based on what? Optimal solution And what we have to do in optimal solution? We focus here on minimum cost
What does minimum cost mean? Let's say if we talk from this example I have multiple feasible solutions One is 10, one is 20 and one is 5 But which is the minimum cost among these? The minimum cost is
5 So what I will do is choose the path with minimum cost Because this is my optimal path The minimum cost of such real life example that we were discussing Means where my course fees are less I will
try to choose that path Where my coaching is a little cheaper I will try to choose that path And the second thing we can do here is Maximize the profit What does maximize the profit mean? That
obviously I will try to choose that career path Where I have maximum profit Means where I get the highest pay scale Or where the location can be You have different criteria That on which criteria you
want to find maximum value So here I will try to maximize my profit Because what is the name of the algorithm? Greedy So what is greedy? That you are greedy regarding profit So whatever solution you
give Whatever problem you have In that problem you will go towards the same solution Which will give you maximum profit Like if there is time for offers So what do we do during the time of offers? We
will try to go to the same shop or mall We will go to the same shop Where I am getting maximum offers Or the highest offer I am getting Regardless of how the cloth is Means whatever we are buying
Let's say if we buy clothes then how is the cloth? How is the stuff? It may be that in the future I may not get the best solution But locally at that stage It is giving me the best solution This is
the main point of Greedy That you try to find the best solution locally So if you are finding the cost Then you will go towards the minimum cost If you are finding the profit Then you will try to
maximize the profit If you are talking about risk Then you will try to go towards the minimum risk Means you will try to choose that part Where the risk is less Means like if we talk about investment
So we try to invest there Where my risk is less So this is the main point of Greedy That at that stage At the stage where you are standing At that stage it tries to find the best result So if your
problem is related to cost Then we will find the minimum cost It is related to profit So we will find the maximum profit It is related to risk So we will try to find the minimum risk Regardless of
what my find What is my global solution here It may be that there are many problems Which I have written here Like there is a problem with Napsack Job sequencing Minimum cost spanning tree Optimal
merge pattern Huffman coding Digestra algorithm That is called a singer-sourcer test path So all these problems Are in the Greedy solution They come in the Greedy algorithm So basically we are doing
this in them Either I am finding the minimum cost Or I am finding the maximum profit Or I am trying to minimize the risk But we focus on this At the stage where we are standing At that stage we find
the best result Globally your result is coming best or not This does not give any guarantee So this is a basic introduction About Greedy technique So in the next videos We will discuss these problems
one by one Thank You. | {"url":"https://www.rd2success.co.uk/2023/06/21/l-4-1-introduction-to-greedy-techniques-with-example-what-is-greedy-techniques/","timestamp":"2024-11-12T05:56:06Z","content_type":"text/html","content_length":"175519","record_id":"<urn:uuid:9873fc69-cdae-4bc5-b2b3-012f56f163c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00877.warc.gz"} |
Ronald Fisher
From New World Encyclopedia
(Redirected from R. A. Fisher)
Ronald Fisher
Sir Ronald Aylmer Fisher
17 February 1890
Born East Finchley, London
29 July 1962
Adelaide, Australia
Residence UK, Australia
Nationality UK
Field Statistics, Genetics
Rothamsted Experimental Station
Institutions University College London
Cambridge University
Alma mater Cambridge University
Academic advisor Sir James Jeans
F.J.M. Stratton
Notable students C. R. Rao
Maximum likelihood
Known for Fisher information
Analysis of variance
Notable prizes Royal Medal (1938)
Copley Medal (1955)
Religious stance Church of England
Sir Ronald Aylmer Fisher, Fellow of the Royal Society (FRS) (February 17, 1890 â July 29, 1962) was a British statistician, evolutionary biologist, and geneticist. He was described by Anders Hald
(1998) as "a genius who almost single-handedly created the foundations for modern statistical science" and Richard Dawkins (1995) described him as "the greatest of Darwin's successors."
Contrary to the popular conception of an either-or dichotomy between evolution and belief in Godâ either evolutionary theory is correct or belief in God is correctâ Ronald Fisher successfully
juxtaposed the two viewpoints (Orr 1999). Fisher was a deeply devout Anglican and a leader in evolutionary theory. Orr (1999) finds it surprising that so few evolutionists seem to know that many of
the brightest stars of evolutionary biology, such as Fisher and Theodosius Dobzhansky, were ardent believers in Godâ almost as if an "unconscious censorship" is going on because the facts are "a bit
too embarrassing."
Early life
Fisher was born in East Finchley, London to George and Katie Fisher. His father was a successful fine arts dealer. He had three older sisters and an older brother. His mother died when Fisher was 14.
His father lost his business in several ill-considered transactions only 18 months later (Box 1978).
Although Fisher had very poor eyesight, he was a precocious student, winning the Neeld Medal (a competitive essay in mathematics) at Harrow School at the age of 16. Because of his poor eyesight, he
was tutored in mathematics without the aid of paper and pen, which developed his ability to visualize problems in geometrical terms, as opposed to using algebraic manipulations. He was legendary in
being able to produce mathematical results without setting down the intermediate steps. Fisher also developed a strong interest in biology and, especially, evolution.
In 1909, Fisher won a scholarship to Gonville and Caius College, Cambridge. There he formed many friendships and became enthralled with the heady intellectual atmosphere. At Cambridge, Fisher learned
of the newly rediscovered theory of Mendelian genetics; he saw biometryâ and its growing corpus of statistical methodsâ as a potential way to reconcile the discontinuous nature of Mendelian
inheritance with continuous variation and gradual evolution.
However, Fisher's foremost concern was eugenics, which he saw as a pressing social as well as scientific issue that encompassed both genetics and statistics. In 1911, he was involved in forming the
Cambridge University Eugenics Society with such luminaries as John Maynard Keynes, R. C. Punnett, and Horace Darwin (Charles Darwin's son). The group was active and held monthly meetings, often
featuring addresses by leaders of mainstream eugenics organizations, such as the Eugenics Education Society of London, founded by Francis Galton in 1909 (Box 1978).
After graduating in 1913, Fisher was eager to join the army in anticipation of Great Britain's entry into World War I; however, he failed the medical examinations (repeatedly) because of his
eyesight. Over the next six years, he worked as a statistician for the City of London. For his war work, he took up teaching physics and mathematics at a series of public schools, including Bradfield
College in Berkshire, as well as aboard H.M. Training Ship Worcester. Major Leonard Darwin (another of Charles Darwin's sons) and an unconventional and vivacious friend he called Gudruna were almost
his only contacts with his Cambridge circle. They sustained him through this difficult period.
A bright spot in his life was that Gudruna matched him to her younger sister Ruth Eileen Gratton Guinness. The father of Ruth Eileen and Gudruna, Dr. Henry Gratton Guinness, had died when they were
young and Ruth Eileen, only 16 years of age, knew that her mother would not approve of her marrying so young. As a result, Fisher married Ruth Eileen at a secret wedding ceremony without her mother's
knowledge, on April 26, 1917, only days after Ruth Eileen's 17th birthday. They set up a subsistence farming operation on the Bradfield estate, where they had a large garden and raised animals,
learning to make do on very little. They lived through the war without ever using their food coupons (Box 1978). Fisher and Rush Eileen were to have two sons and seven daughters, one of whom died in
infancy. His daughter Joan married George E. P. Box and wrote a well-received biography of her father.
During this period of the war, Fisher started writing book reviews for the Eugenic Review and gradually increased his interest in genetics and statistical work. He volunteered to undertake all such
reviews for the journal, and was hired to a part-time position by Major Darwin. He published several articles on biometry during this period, including the ground-breaking "The Correlation between
Relatives on the Supposition of Mendelian Inheritance," written in 1916 and published in 1918. This paper laid the foundation for what came to be known as biometrical genetics, and introduced the
very important methodology of the analysis of variance, which was a considerable advance over the correlation methods used previously. The paper showed very convincingly that the inheritance of
traits measurable by real values, the values of continuous variables, is consistent with Mendelian principles (Box 1978).
At the end of the war, Fisher went looking for a new job and was offered one at the famed Galton Laboratory by Karl Pearson. Because he saw the developing rivalry with Pearson as a professional
obstacle, however, he accepted instead a temporary job as a statistician with a small agricultural station in the country in 1919, the Rothamsted Experimental Station.
Early professional years
The Rothamsted Experimental Station is now one of the oldest agricultural research institutions in the world. In 1919, Fisher started work at this station, which was (and is) located at Harpenden in
Hertfordshire, England. Here he started a major study of the extensive collections of data recorded over many years. This resulted in a series of reports under the general title Studies in Crop
Fisher was in his prime and he began a period of amazing productivity. Over the next seven years, he pioneered the principles of the design of experiments and elaborated his studies of "analysis of
variance." He furthered his studies of the statistics of small samples. Perhaps even more important, he began his systematic approach of the analysis of real data as the springboard for the
development of new statistical methods. He began to pay particular attention to the labor involved in the necessary computations, and developed ingenious methods that were as practical as they were
founded in rigor. In 1925, this work culminated in the publication of his first book, Statistical Methods for Research Workers (Box 1978). This went into many editions and translations in later
years, and became a standard reference work for scientists in many disciplines. In 1935, this was followed by The Design of Experiments, which also became a standard.
In addition to "analysis of variance," Fisher invented the technique of maximum likelihood and originated the concepts of sufficiency, ancillarity, Fisher's linear discriminator, and Fisher
information. His 1924 article "On a distribution yielding the error functions of several well known statistics" presented Karl Pearson's chi-squared and Student's t in the same framework as the
Gaussian distribution, and his own "analysis of variance" distribution z (more commonly used today in the form of the F distribution). These contributions made him a major figure in twentieth-century
In defending the use of the z distribution when the data were not Gaussian, Fisher developed the "randomization test." According to biographers Yates and Mather (1963), "Fisher introduced the
randomization test, comparing the value of t or z actually obtained with the distribution of the t or z values when all possible random arrangements were imposed on the experimental data.â However,
Fisher wrote that randomization tests were "in no sense put forward to supersede the common and expeditious tests based on the Gaussian theory of errors." Fisher thus effectively began the field of
non-parametric statistics, even though he did not believe it was a necessary move.
His work on the theory of population genetics also made him one of the three great figures of that field, together with Sewall Wright and J. B. S. Haldane, and as such was one of the founders of the
modern evolutionary synthesis (neo-Darwinism).
In addition to founding modern quantitative genetics with his 1918 paper, Fisher was the first to use diffusion equations to attempt to calculate the distribution of gene frequencies among
populations. He pioneered the estimation of genetic linkage and gene frequencies by maximum likelihood methods, and wrote early papers on the wave of advance of advantageous genes and on clines of
gene frequency. His 1950 paper on gene frequency clines is notable as first application of computers to biology.
Fisher introduced the concept of Fisher information in 1925, some years before Claude E. Shannon's notions of information and entropy. Fisher information has been the subject of renewed interest in
the last few years, both due to the growth of Bayesian inference in artificial intelligence, and due to B. Roy Frieden's book Physics from Fisher Information, which attempts to derive the laws of
physics from a Fisherian starting point.
Genetical Theory of Natural Selection
An ardent promoter of eugenics, this subject stimulated and guided much of Fisher's work in human genetics. His book The Genetical Theory of Natural Selection was started in 1928 and published in
1930. It contained a summary of what was already known in the literature. Fisher developed ideas on sexual selection, mimicry, and the evolution of dominance. He famously showed that the probability
of a mutation increasing the fitness of an organism decreases proportionately with the magnitude of the mutation. He also proved that larger populations carry more variation so that they have a
larger chance of survival. He set forth the foundations of what was to become known as population genetics.
About a third of the book concerned the applications of these ideas to humans and summarized the data available at the time. Fisher presented a theory that attributed the decline and fall of
civilizations to the arrival of a state where the fertility of the upper classes is forced down. Using the census data of 1911 for England, he showed that there was an inverse relationship between
fertility and social class. This was partly due, he believed, to the rise in social status of families who were not capable of producing many children but who rose because of the financial advantage
of having a small number of children. Therefore, he proposed the abolishment of the economic advantage of small families by instituting subsidies (he called them allowances) to families with larger
numbers of children, with the allowances proportional to the earnings of the father. He himself had two sons and six daughters. According to Yates and Mather (1963), "His large family, in particular,
reared in conditions of great financial stringency, was a personal expression of his genetic and evolutionary convictions."
The book was reviewed, among others, by physicist Charles Galton Darwin, a grandson of Charles Darwin, and following publication of his review, C. G. Darwin sent Fisher his copy of the book, with
notes in the margin. The marginal notes became the food for a correspondence running at least three years (Fisher 1999).
Between 1929 and 1934, the Eugenics Society also campaigned hard for a law permitting sterilization on eugenic grounds. They believed that it should be entirely voluntary and a right, rather than
compulsory or a punishment. They published a draft of a proposed bill, and it was submitted to Parliament. Although it was defeated by a 2:1 ratio, this was viewed as progress, and the campaign
continued. Fisher played a major role in this movement, and served in several official committees to promote it.
In 1934, Fisher moved to increase the power of scientists within the Eugenics Society, but was ultimately thwarted by members with an environmentalist point of view, and he, along with many other
scientists, resigned.
Method and personality
As an adult, Fisher was noted for his loyalty to his friends. Once he had formed a favorable opinion of any man, he was loyal to a fault. A similar sense of loyalty bound him to his culture. He was a
patriot, a member of the Church of England, politically conservative, and a scientific rationalist. Much sought after as a brilliant conversationalist and dinner companion, he very early on developed
a reputation for carelessness in his dress and, sometimes, his manners. In later years, he was the archetype of the absent-minded professor.
Fisher knew the biblical scriptures well and was deeply devout. Orr (1999) describes him as "deeply devout Anglican who, between founding modern statistics and population genetics, penned articles
for church magazines." But he was not dogmatic in his religious beliefs. In a 1955 broadcast on Science and Christianity, he said (Yates and Mather 1963):
The custom of making abstract dogmatic assertions is not, certainly, derived from the teaching of Jesus, but has been a widespread weakness among religious teachers in subsequent centuries. I do
not think that the word for the Christian virtue of faith should be prostituted to mean the credulous acceptance of all such piously intended assertions. Much self-deception in the young believer
is needed to convince himself that he knows that of which in reality he knows himself to be ignorant. That surely is hypocrisy, against which we have been most conspicuously warned.
Later years
It was Fisher who referred to the growth rate r (used in equations such as the logistic function) as the Malthusian parameter, as a criticism of the writings of Thomas Robert Malthus. Fisher referred
to "â Śa relic of creationist philosophyâ Ś" in observing the fecundity of nature and deducing (as Darwin did) that this therefore drove natural selection.
He received the recognition of his peers in 1929 when he was inducted into the Royal Society. His fame grew and he began to travel more and lecture to wider circles. In 1931, he spent six weeks at
the Statistical Laboratory at Iowa State College in Ames, Iowa. He gave three lectures a week on his work, and met many of the active American statisticians, including George W. Snedecor. He returned
again for another visit in 1936.
In 1933, Fisher left Rothamsted to become a professor of eugenics at University College London. In 1937, he visited the Indian Statistical Institute (in Calcutta), which at the time consisted of one
part-time employee, Professor P. C. Mahalanobis. He revisited there often in later years, encouraging its development. He was the guest of honor at its 25th anniversary in 1957, when it had grown to
2,000 employees.
In 1939, when World War II broke out, University College London tried to dissolve the eugenics department, and ordered all of the animals destroyed. Fisher fought back, but he was then exiled back to
Rothamsted with a much-reduced staff and resources. He was unable to find any suitable war work, and though he kept very busy with various small projects, he became discouraged of any real progress.
His marriage disintegrated. His oldest son, a pilot, was killed in the war.
In 1943, Fisher was offered the Balfour Chair of Genetics at Cambridge University, his alma mater. During the war, this department was also pretty much destroyed, but the university promised him that
he would be charged with rebuilding it after the war. He accepted the offer, but the promises were largely unfilled, and the department grew very slowly. A notable exception was the recruitment in
1948 of the Italian researcher Cavalli-Sforza, who established a one-man unit of bacterial genetics. Fisher continued his work on mouse chromosome mapping and other projects. They culminated in the
publication in 1949 of The Theory of Inbreeding.
In 1947, Fisher co-founded with Cyril Darlington the journal Heredity: An International Journal of Genetics.
Fisher eventually received many awards for his work and was dubbed a Knight Bachelor by Queen Elizabeth II in 1952.
Fisher was opposed to the conclusions of Richard Doll that smoking caused lung cancer. Yates and Mather (1963) conclude: "It has been suggested that the fact that Fisher was employed as consultant by
the tobacco firms in this controversy casts doubt on the value of his arguments. This is to misjudge the man. He was not above accepting financial reward for his labours, but the reason for his
interest was undoubtedly his dislike and mistrust of puritanical tendencies of all kinds; and perhaps also the personal solace he had always found in tobacco."
After retiring from Cambridge University in 1957, Fisher spent some time as a senior research fellow at the CSIRO in Adelaide, Australia. He died of colon cancer there in 1962.
Fisher's important contributions to both genetics and statistics are emphasized by the remark of L. J. Savage, "I occasionally meet geneticists who ask me whether it is true that the great geneticist
R. A. Fisher was also an important statistician" (Aldrich 2007).
A selection from Fisher's 395 articles
These are available on the University of Adelaide website (Retrieved November 15, 2007):
• Fisher, R. A. 1915. Frequency distribution of the values of the correlation coefficient in samples from an indefinitely large population. Biometrika 10: 507â 521.
• Fisher, R. A. 1918. The correlation between relatives on the supposition of Mendelian inheritance. Trans. Roy. Soc. Edinb. 52: 399â 433. It was in this paper that the word variance was first
introduced into probability theory and statistics.
• Fisher, R. A. 1922. On the mathematical foundations of theoretical statistics. Philosophical Transactions of the Royal Society, A 222: 309â 368.
• Fisher, R. A. 1922. On the dominance ratio. Proc. Roy. Soc. Edinb. 42: 321â 341.
• Fisher, R. A. 1924. On a distribution yielding the error functions of several well known statistics. Proc. Int. Cong. Math. 2: 805â 813.
• Fisher, R. A. 1925. Theory of statistical estimation. Proceedings of the Cambridge Philosophical Society 22: 700â 725.
• Fisher, R. A. 1925. Applications of Student's distribution. Metron 5: 90â 104.
• Fisher, R. A. 1926. The arrangement of field experiments. J. Min. Agric. G. Br. 33: 503â 513.
• Fisher, R. A. 1928. The general sampling distribution of the multiple correlation coefficient. Proceedings of Royal Society, A 121: 654â 673.
• Fisher, R. A. 1934. Two new properties of mathematical likelihood. Proceedings of Royal Society, A 144: 285â 307.
Books by Fisher
Full publication details are available on the University of Adelaide website (Retrieved November 15, 2007):
• Fisher, R. A. 1925. Statistical Methods for Research Workers. Edinburgh: Oliver and Boyd. ISBN 0050021702.
• Fisher, R. A. 1930. The Genetical Theory of Natural Selection. Oxford: Clarendon Press. ISBN 0198504403.
• Fisher, R. A. 1935. The Design of Experiments. Edinburgh; London: Oliver and Boyd.
• Fisher, R. A. 1949. The Theory of Inbreeding. New York: Academic Press.
• Fisher, R. A. 1950. Contributions to Mathematical Statistics. John Wiley.
• Fisher, R. A. 1956. Statistical Methods and Statistical Inference. New York: Hafner Press. ISBN 0028447409.
• Fisher, R. A., with F. Yates. 1938. Statistical Tables for Biological, Agricultural and Medical Research. London: Oliver and Boyd.
ISBN links support NWE through referral fees
• Aldrich, J. 1997. R. A. Fisher and the making of maximum likelihood 1912â 1922. Statistical Science 12(3): 162â 176. Retrieved May 17, 2007.
• Aldrich, J. 2007. A guide to R. A. Fisher. University of Southampton. Retrieved May 17, 2007.
• Box, J. F. 1978. R. A. Fisher: The Life of a Scientist. New York: Wiley. ISBN 0471093009.
• Dawkins, R. 1995. River out of Eden: A Darwinian View of Life. New York: Basic Books. ISBN 0465016065.
• Fisher, R. A. [1930] 1999. The Genetical Theory of Natural Selection. Oxford University Press. ISBN 0198504403.
• Hald, A. 1998. A History of Mathematical Statistics from 1750 to 1930. New York: Wiley. ISBN 0471179124.
• Howie, D. 2002. Interpreting Probability: Controversies and Developments in the Early Twentieth Century. Cambridge University Press. ISBN 0521812518.
• Orr, H. A. 1999. Gould on God: Can religion and science be happily reconciled? Boston Review October/November. Retrieved May 17, 2007.
• Salsburg, D. 2002. The Lady Tasting Tea: How Statistics Revolutionized Science in the Twentieth Century. New York: W.H. Freeman. ISBN 0805071342.
• Yates, F., and K. Mather. 1963. Ronald Aylmer Fisher. Biographical Memoirs of Fellows of the Royal Society of London 9: 91â 120.
 Topics in population genetics
Key concepts: Hardy-Weinberg law | genetic linkage | linkage disequilibrium | Fisher's fundamental theorem | neutral theory
Selection: natural | sexual | artificial | ecological
Effects of selection on genomic variation: genetic hitchhiking | background selection
Genetic drift: small population size | population bottleneck | founder effect | coalescence
Founders: R.A. Fisher | J. B. S. Haldane | Sewall Wright
Related topics: evolution | microevolution | evolutionary game theory | fitness landscape | genetic genealogy
List of evolutionary biology topics
│ Preceded by: │ Presidents of the Royal Statistical Society │ Succeeded by: │
│ Austin Bradford Hill │ 1952â 1954 │ Lord Piercy of Burford │
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Ultimate Models for the Universe: A New Kind of Science | Online by Stephen Wolfram [Page 466]
behavior. And in fact, this is precisely why it is conceivable that a simple program could reproduce all the complexity we see in physics.
Given a particular underlying program, it is always in principle possible to work out what it will do just by running it. But for the whole universe, doing this kind of explicit simulation is almost
by definition out of the question. So how then can one even expect to tell whether a particular program is a correct model for the universe? Small-scale simulation will certainly be possible. And I
expect that by combining this with a certain amount of perhaps fairly sophisticated mathematical and logical deduction, it will be possible to get at least as far as reproducing the known laws of
physics—and thus of determining whether a particular model has the potential to be correct.
So if there is indeed a definite ultimate model for the universe, how might one set about finding it? For those familiar with existing science, there is at first a tremendous tendency to try to work
backwards from the known laws of physics, and in essence to try to "engineer" a universe that will have particular features that we observe.
But if there is in fact an ultimate model that is quite simple, then from what we have seen in this book, I strongly believe that such an approach will never realistically be successful. For human
thinking—even supplemented by the most sophisticated ideas of current mathematics and logic—is far from being able to do what is needed.
Imagine for example trying to work backwards from a knowledge of the overall features of the picture on the facing page to construct a rule that would reproduce it. With great effort one might
perhaps come up with some immensely complex rule that would work in most cases. But there is no serious possibility that starting from overall features one would ever arrive at the extremely simple
rule that was actually used.
It is already difficult enough to work out from an underlying rule what behavior it will produce. But to invert this in any systematic way is probably even in principle beyond what any realistic
computation can do.
So how then could one ever expect to find the underlying rule in such a case? Almost always, it seems that the best strategy is a simple one: to come up with an appropriate general class of rules,
and then just | {"url":"https://www.wolframscience.com/nks/p466--ultimate-models-for-the-universe/","timestamp":"2024-11-15T04:19:33Z","content_type":"text/html","content_length":"87447","record_id":"<urn:uuid:833214b0-b56d-4e89-93cf-39884cb3f809>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00777.warc.gz"} |
How does C++ do math?
How does C++ do math?
You can give that same pleasure to your computer. C++ uses operators to do arithmetic. It provides operators for five basic arithmetic calculations: addition, subtraction, multiplication, division,
and taking the modulus. Each of these operators uses two values (called operands) to calculate a final answer.
What is left to right associativity in C?
Operators Associativity is used when two operators of same precedence appear in an expression. Associativity can be either Left to Right or Right to Left. For example: ‘*’ and ‘/’ have same
precedence and their associativity is Left to Right, so the expression “100 / 10 * 10” is treated as “(100 / 10) * 10”.
What is unary minus?
The – (unary minus) operator negates the value of the operand. For example, if quality has the value 100 , -quality has the value -100 . The result has the same type as the operand after integral
promotion. Note Any minus sign in front of a constant is not part of the constant.
How do you convert mathematical expressions to C equivalent?
How to convert mathematical expressions into C statement?
1. 1 / (x^2 + y^2)
2. square root of (b^2 – 4ac)
Is a unary operator?
A unary operator is one that takes a single operand/argument and performs an operation. A unary operation is an operation with only one operand. This operand comes either before or after the
operator. Additionally, unary operators can not be overridden, therefore their functionality is guaranteed.
How do you write mathematical expressions in C++?
The following mathematical function may be translated in to C++ in several different ways. Valid translations of the above formula include the following: P = F * r / (pow(1 + r, n) – 1) / (1 + r); P
= F * r / (pow(1 + r, n) – 1) * 1 / (1 + r); P = F * (r / (pow(1 + r, n) – 1)) * (1 / (1 + r));
What is correct order of precedence in C?
Operators Precedence in C
Category Operator Associativity
Additive + – Left to right
Shift << >> Left to right
Relational < <= > >= Left to right
Equality == != Left to right
What is unary operator in C?
Unary operator is operators that act upon a single operand to produce a new value. The result of the unary plus operator (+) is the value of its operand. The operand to the unary plus operator must
be of an arithmetic type. Unary negation operator (-) The – (unary minus) operator negates the value of the operand.
Which operator is evaluated first?
An operator’s precedence is meaningful only if other operators with higher or lower precedence are present. Expressions with higher-precedence operators are evaluated first. Precedence can also be
described by the word “binding.” Operators with a higher precedence are said to have tighter binding.
Which has higher precedence * or?
Operator precedence is the evaluating order which is followed by multiple operators in a same mathematical expression. In the above expression, * has a higher precedence over +.
What is associativity C?
Associativity: It defines the order in which operators of the same precedence are evaluated in an expression. Associativity can be either from left to right or right to left. In C, each operator has
a fixed priority or precedence in relation to other operators.
Which operator has the lowest priority?
LOWEST PRECEDENCE The compound logical operators, &&, ||, -a, and -o have low precedence. The order of evaluation of equal-precedence operators is usually left-to-right.
What is unary plus and minus?
A unary operator works on one operand. The unary operators in JavaScript are: Unary plus ( + ) – convert an operand into a number. Unary minus ( – ) – convert an operand into a number and negate the
value after that. prefix / postfix decrements ( — ) – subtract one from its operand.
Which operator has highest precedence C?
operator precedence
Precedence Operator Associativity
++ —
1 () Left-to-right
How do you remember the order of precedence?
An expression like p++->x is parsed as (p++)->x ; both postfix ++ and -> operators have the same precedence, so they’re parsed from left to right. This is about as short as shortcuts get; when in
doubt, use parentheses. There is a shortcut to remember C operator Precedence. PUMA IS REBL ( spell “REBL” as if “REBEL”).
Is size of a unary operator?
List of Unary Operators in C programming language
SrNo Operators Symbols
6 Size of Operator sizeof()
7 Dereferencing Operator *
8 Logical NOT !
9 Bitwise NOT/ Bitwise Negation/ One’s Compliment ~
How does unary operator work?
The unary operators require only one operand; they perform various operations such as incrementing/decrementing a value by one, negating an expression, or inverting the value of a boolean. The
increment/decrement operators can be applied before (prefix) or after (postfix) the operand.
Is a pair of parenthesis used as a C++ multiplication symbol?
Parentheses are used in C++ expressions in the same manner as in algebraic expressions. For example, to multiply a times the quantity b + c we write a * ( b + c ).
Does C++ do order of operations?
Math in C++ is very simple. Keep in mind that C++ mathematical operations follow a particular order much the same as high school math. For example, multiplication and division take precedence over
addition and subtraction. The order in which these operations are evaluated can be changed using parentheses.
Which type of language is C?
C (/siː/, as in the letter c) is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, with a static type system. By
design, C provides constructs that map efficiently to typical machine instructions.
How do you write square root in C++?
Syntax of sqrt() function: sqrt(x); Parameter(s): x – a number whose square root to be calculated. Return value: double – it returns double value that is the square root of the given number x.
Is coding considered math?
Coding is math. Coding, at the bottom line, is math. In order to write a line of code that works well, and that is completely bug-free, coders need to strengthen their algorithmic thinking and
computational thinking.
How do you write an expression in C++?
An expression can consist of one or more operands, zero or more operators to compute a value. Every expression produces some value which is assigned to the variable with the help of an assignment
operator….Examples of C++ expression:
1. (a+b) – c.
2. (x/y) -z.
3. 4a2 – 5b +c.
4. (a+b) * (x+y)
Which is a unary operation *?
In mathematics, a unary operation is an operation with only one operand, i.e. a single input. This is in contrast to binary operations, which use two operands. An example is the function f : A → A,
where A is a set. The function f is a unary operation on A.
Does C++ follow Bedmas?
One is that C++ does follow standard mathematical precedence rules, which you refer to as BODMAS. However, if any of the expressions involved in the operation have side effects, then C++ is not
guaranteed to evaluate them in what one might consider to be standard mathematical order.
How many unary operators are there?
6.4. C has two unary operators for incrementing and decrementing scalar objects. The increment operator ++ adds 1 to its operand; the decrement operator – subtracts 1. Both ++ and – can be used
either as prefix operators (before the variable: ++n ) or postfix operators (after the variable: n++ ).
What is the only ternary operator in C?
In computer programming, ?: is a ternary operator that is part of the syntax for basic conditional expressions in several programming languages. It is commonly referred to as the conditional
operator, inline if (iif), or ternary if. An expression a ? b : c evaluates to b if the value of a is true, and otherwise to c . | {"url":"https://somme2016.org/blog/how-does-c-do-math/","timestamp":"2024-11-06T01:47:07Z","content_type":"text/html","content_length":"49076","record_id":"<urn:uuid:ebc01439-e610-44c4-864f-104187b2b15d>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00574.warc.gz"} |
CFA Vs. Conventional Theory | James Sawle (MD0MDI)
CFA vs. Conventional Theory
Undoubtedly there is more for everyone to learn about CFAs—including ourselves, the inventors. However, let me outline a few concepts of CFAs in relation to “conventional” theory.
The D plate of the CFA is a parallel capacitor, which with an AC signal creates a surrounding H field. This is well known and is even used to prove continuity of the fourth Maxwell equation i.e. curl
H = J + D’. Feynam’s lectures on physics even evaluates how to calculate the H field surrounding a capacitor. If anyone is not convinced, they can simply measure the existence of the H field around a
capacitor and prove that it is 90 degrees phase advanced from the applied voltage and also the E field lines between the plates.
The E plate produces E-lines since it is also a capacitor. There can be no question of this either.
If both the D and E plate structures were fed together with voltage signals which were in phase, then, I’m afraid we would have E and H fields which are 90 degrees out of phase close to the D and E
plates. This is similar in some ways to a conventional antenna. There should be little radiation and the behaviour would mimic an extremely short antenna. Radiation would be produced from
conventional theory in the very far field and have little power. Indeed some publications have presented this feature—ignoring the phase details of the feeders and also the phase properties of the
field distributions surrounding the antenna which actually produce the radiation.
Now phase shift the D plate voltage by 90 degrees. Then we have E and H in time phase close to the structure. However, a point of note which has unfortunately not been communicated by us very well in
the past. If the phasing is 90 degrees one way, the Poynting vector S = E x H is inward. No power radiation is possible as the radiation resistance has a high impedance. I believe many people have
dismissed the CFA because they have attained the 90 degrees but no or extremely small levels of radiation. When the correct 90 degrees is achieved, i.e. phase shifted the other way, S = E x H is
outward and power radiation occurs.
This is the ultimate test of the CFA. Where does the power radiation start? If it is conventional theory, then close to the antenna will be inductive reactive field, which has a well-established
relationship with distance from the antenna. If it is power radiation and most of the fields participate in this radiation, then the distribution with distance will be different showing the CFA is a
different type of antenna system. In addition, because the fields close to the CFA are “strong” in comparison to fields in the far field which produce radiation in conventional theory, then we may
expect strong radiation.
I have not suggested anything which is unusual or breaks any electromagnetic laws. Simply, E and H in time phase produce radiated power—the Poynting theorem. The D and E structures produce E and H
fields, easily experimentally proved, and these are simply made to be in time phase.
A further proof is also the voltage levels on the antenna structures. The CFA appears to have about 1/6 the voltage level of conventional antennas. In conventional theory this should result in much
less radiated power than is measured, even for the “worst” CFAs!!
In all of this, one must also get the ratio of E/H to match space impedance for maximum power transfer, so you can appreciate that 90 phase and little power is actually a space impedance matching
problem. The voltages on the E and D plates must be altered to change the field strengths for maximum radiation. This has been how Dr. Kabbary has maximised radiation on the Egyptian CFAs.
Conventional Theory Does Not Apply
Now, in terms of conventional theory, well the CFA is fundamentally different and standard antenna formulae and techniques cannot be invoked when E and H exist close to the structure in time phase.
Conventional theory cannot accommodate this fact and thus should not be applied. I suppose it is obvious from another viewpoint. Conventional monopoles have one signal feed to the antenna structure
and an earth. The Groundplane CFA has two signal feeds to two structures and an earth. There is no way conventional theory of monopole structures can be transferred directly without alteration to
another type of antenna.
A further thought relates to phasing circuits. One remarkable feature is that the phasing circuit to provide 90 phase shift appears to compensate frequency movements when radiating. The E and D
plates go “off frequency” together as they are both capacitive thus introducing “wider” band radiation as E and H are still then in time phase.
There has been some question about the CFA in regard to the FCC 73.186 et al. Let me mention a few details. The purpose of the National Association of Broadcasters paper scheduled for the April 1999
convention in Las Vegas is to present broadcasting CFAs to the US. For that reason, and because no one has actually built a broadcast CFA in the US, we have not been motivated to satisfy every demand
of the FCC yet. Give us time! However, in our paper relative normalised radiation field strength patterns of a broadcast CFA (at reduced power) taken at about 600m (not the 1km for the FCC 73.186)
are presented. We have the measurement values and will bring these to the NAB conference. We also hope that prior to NAB, some more information regarding FCC criteria will be available as further
measurements are taken in this respect.
One need only take a field strength meter and measure the shear signal strength levels of a CFA. Having done this for the CFAs in Egypt, there is no comparison. The levels are so strong.
In this respect, last year I asked the BBC to make reception checks on a 30kW 1161kHz CFA in Cyprus—a distance of about 500km from the CFA. The BBC reported fair-strong signal strengths during
daytime and evening with no fading characteristics. No other distant broadcast programmes using even higher power levels compare. This is reported in the NAB conference.
I realise the CFA will still face opposition and still come in for criticism, some perhaps justified, a lot of it not. However, what concerns me is that those who have been most vocal in dismissing
it have never actually built one and properly phased it! If I truly felt that they didn’t work, I’d pack it in—since I’d be wasting my time.
Originally posted on the AntennaX Online Magazine by Dr. Brian Stewart, MM1DVD
Last Updated : 3rd May 2024 | {"url":"https://www.md0mdi.im/cfa-vs-conventional-theory/","timestamp":"2024-11-07T05:41:06Z","content_type":"text/html","content_length":"433744","record_id":"<urn:uuid:d5b30bbc-ec16-40c8-b279-507d989cf960>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00033.warc.gz"} |
Shortest Distance In A Plane LeetCode Solution - Leetcode Solution - TO THE INNOVATION
Shortest Distance in a Plane LeetCode Solution
/ Leetcode Solution / LeetCode Solution, Medium, MySql, SQL
Last updated on October 9th, 2024 at 10:11 pm
This Leetcode problem Shortest Distance in a Plane LeetCode Solution is done in SQL.
Level of Question
Shortest Distance in a Plane LeetCode Solution
Table of Contents
Problem Statement
Table point_2d holds the coordinates (x,y) of some unique points (more than two) in a plane.
Write a query to find the shortest distance between these points rounded to 2 decimals.
The shortest distance is 1.00 from point (-1,-1) to (-1,2). So the output should be:
1. Shortest Distance in a Plane LeetCode Solution MySQL
) as shortest
a.x = b.x
and a.y = b.y,
power(a.x - b.x, 2) + power(a.y - b.y, 2)
) as dist
point_2d as a,
point_2d as b
) as d;
Shortest Unsorted Continuous Subarray LeetCode Solution | {"url":"https://totheinnovation.com/shortest-distance-in-a-plane-leetcode-solution/","timestamp":"2024-11-02T11:24:18Z","content_type":"text/html","content_length":"194779","record_id":"<urn:uuid:53bfa6dc-b52c-4fe5-a094-d008c5f135e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00449.warc.gz"} |
21 October 2018 Archives
Here’s a thought: what’s the minimum number of votes your party would need to attract in order to be able to secure a majority of seats in the House of Commons and form a government? Let’s try to
work it out.
The 2017 general election reportedly enjoyed a 68.8% turnout. If we assume for simplicity’s sake that each constituency had the same turnout and that votes for candidates other than yours are
equally-divided amongst your opposition, that means that the number of votes you need to attract in a given constituency is:
68.8% × the size of its electorate ÷ the number of candidates (rounded up)
For example, if there was a constituency of 1,000 people, 688 (68.8%) would have voted. If there were 3 candidates in that constituency you’d need 688 ÷ 3 = 229⅓, which rounds up to 230 (because you
need the plurality of the ballots) to vote for your candidate in order to secure the seat. If there are only 2, you need 345 of them.
It would later turn out that Barry and Linda Johnson of 14 West Street had both indented to vote for the other candidate but got confused and voted for your candidate instead. In response, 89% of the
nation blame the pair of them for throwing the election.
The minimum number of votes you’d need would therefore be this number for each of the smallest 326 constituencies (326 is the fewest number of seats you can hold in the 650-seat House of Commons and
guarantee a strict majority; in reality, a minority government can sometimes form a government but let’s not get into that right now). Constituencies vary significantly in size, from only 21,769
registered voters in Na h-Eileanan an Iar (the Western Isles of Scotland, an SNP/Labour marginal) to 110,697 in the Isle of Wight (which flip-flops between the Conservatives and the Liberals), but
each is awarded exactly one seat, so if we’re talking about the minimum number of votes you need we can take the smallest 326.
Win these constituencies and no others and you control the Commons, even though they’ve tiny populations. In other news, I think this is how we end up with a SNP/Plaid coalition government.
By my calculation, with a voter turnout of 68.8% and assuming two parties field candidates, one can win a general election with only 7,375,016 votes; that’s 15.76% of the electorate (or 11.23% of the
total population). That’s right: you could win a general election with the support of a little over 1 in 10 of the population, so long as it’s the right 1 in 10.
I used a spreadsheet and everything; that’s how you know you can trust me. And you can download it, below, and try it for yourself.
I’ll leave you to decide how you feel about that. In the meantime, here’s my working (and you can tweak the turnout and number-of-parties fields to see how that affects things). My data comes from
the following Wikipedia/Wikidata sources: [1], [2], [3], [4], [5] mostly because the Office of National Statistics’ search engine is terrible.
JACOB PRETENDS TO BE STRAIGHT… | The Trueman Show! Ep 13
This is a repost promoting content originally published elsewhere. See more things Dan's reposted.
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Previously, a brief introduction was made on linear regression by the traditional method in order to understand the mathematics behind it. However, Python has built-in machine learning libraries to
make coding easier and shorter. In this second part of linear regression, you will learn how to use this powerful library.
Previously, a brief introduction was made on linear regression by the traditional method in order to understand the mathematics behind it. However, Python has built-in machine learning libraries to
make coding easier and shorter. In this second part of linear regression, you will learn how to use this powerful library.
In the previous tutorial, you have learned how to build a linear regression using matrix multiplication (please go to Python Machine Learning: Linear Regression (I)). Now, in this tutorial, a Machine
Learning Python library called scikit-learn will be used for this purpose.
Once we have imported the data from the text file, let's set our x- and y-values.
#Importing library
import numpy as np
#Importing text file
data = np.loadtxt('points.txt', skiprows=(2), dtype=float)
#Setting x values
x = data[:,0]
#Setting y values
y = data[:,1]
From the figure above (an extract of the whole data), we can notice that $x$ and $y$ are 1D arrays. If we want to work with the scikit-learn Machine Learning Python library, it is necessary to
convert our 1D arrays into 2D. For this, the function reshape(-1,1).
#Reshaping the array into a vector-column
x2 = data[:,0].reshape(-1,1)
#Reshaping the array into a vector-column
y2 = data[:,1].reshape(-1,1)
Now, we are able to build our linear regression model using the LinearRegression module from the scikit-learn library. Do not forget to import the library.
#Importing library
from sklearn.linear_model import LinearRegression
#Building the linear regression model
linear_regression = LinearRegression()
linear_model = linear_regression.fit(x2,y2)
As explained in the previous tutorial, the linear relationship can be as \[y = c_{0} + c_{1}*x\], where $c_0$ is the intercept with the y-axis, and $c_1$ is the slope of the line. These two
coefficients can be found easier and faster thanks to the function LinearRegression().fit(). In order to get both coefficients, the functions intercept_ and coef_ are needed.
#Getting the intercept with y-axis
intercept_yaxis = linear_model.intercept_
#Getting the coefficient
slope = linear_model.coef_
In contrast to the matrix multiplication approach where the coefficient matrix is an array of two elements, both elements are now got in two different arrays of one element each. If comparing both
approaches, both intercept and slope should be exactly the same. The coefficient matrix from the previous tutorial was the following:
As seen from both pictures, we notice that both coefficients (intercept and slope) are exactly the same. This means we did a great job making the linear regression! Finally, let's establish the
linear relationship and plot it.
#Importing library
import matplotlib.pyplot as plt
#Establishing the linear relationship
y_lineal2 = slope*x2 + intercept_yaxis
#Initially given x- and y-points
#Linear regression points
plt.plot(x2, y_lineal2, color='red')
#Naming the graph, x- and y-axis
plt.title('scikit-learn library')
The plot we got in the previous tutorial was the following:
As seen from both graphics, we can say they are exactly the same! The final Python code will look like the following:
#Importing libraries
import numpy as np
from sklearn.linear_model import LinearRegression
import matplotlib.pyplot as plt
#Importing text file
data = np.loadtxt('points.txt', skiprows=(2), dtype=float)
#Setting x values
x = data[:,0]
#Setting y values
y = data[:,1]
#Reshaping the array into a vector-column
x2 = data[:,0].reshape(-1,1)
#Reshaping the array into a vector-column
y2 = data[:,1].reshape(-1,1)
#Building the linear regression model
linear_regression = LinearRegression()
linear_model = linear_regression.fit(x2,y2)
#Getting the intercept with y-axis
intercept_yaxis = linear_model.intercept_
#Getting the coefficient
slope = linear_model.coef_
#Establishing the linear relationship
y_lineal2 = slope*x2 + intercept_yaxis
#Initially given x- and y-points
#Linear regression points
plt.plot(x2, y_lineal2, color='red')
#Naming the graph, x- and y-axis
plt.title('scikit-learn library')
Congratulations! You just made your first Machine Learning regression. In the next tutorial, polynomial regression will be explained. To download the complete code and the text file containing the
data used in this tutorial, please click here.
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extended functorial field theory
added this pointer under relation to condensed matter:
• Davide Gaiotto, Theo Johnson-Freyd, Condensations in higher categories (arXiv:1905.09566)
diff, v43, current
Added Ruth J. Lawrence’s paper.
diff, v42, current
Typo fixed,
diff, v41, current
added pointer to:
• Lukas Müller, Extended Functorial Field Theories and Anomalies in Quantum Field Theories (arXiv:2003.08217)
diff, v38, current
Some discussion of that paper at MO.
also pointer to:
• Daniel Freed, Michael Hopkins, Reflection positivity and invertible topological phases, Geometry & Topology (arXiv:1604.06527)
(I can’t find any publication record of this, but the last comment on the arXiv page suggests it has been accepted with G&T.)
diff, v43, current
added pointer to today’s:
• Daniel S. Freed, Gregory W. Moore, Constantin Teleman, Topological symmetry in quantum field theory [arXiv:2209.07471]
diff, v44, current
Renamed the article to indicate that extended field theories need not be topological.
diff, v46, current
Made adjustments for the nontopological case.
diff, v46, current
added pointer to:
• Shawn X. Cui, Higher Categories and Topological Quantum Field Theories, Quantum Topology 10 4 (2019) 593-676 [[arXiv:1610.07628] (https://arxiv.org/abs/1610.07628), doi:10.4171/QT/128&
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Milliliters (mL) in an Ounce (oz) for the NAPLEX Exam
Pharmacists play a crucial role in patient care by ensuring accurate medication dispensing. A key skill in pharmacy practice is […]
Pharmacists play a crucial role in patient care by ensuring accurate medication dispensing. A key skill in pharmacy practice is mastering dosage calculations, especially when converting between units
like milliliters (mL) and ounces (oz). For students preparing for the North American Pharmacist Licensure Examination (NAPLEX), understanding such conversions is essential.
In this article, we’ll explore the mL-to-oz conversion, why it matters in pharmacy, and how to confidently tackle these calculations. By the end, you’ll have a solid understanding of how to handle
these conversions for NAPLEX success.
Why Are Conversions Important in Pharmacy?
Pharmacy is a field that demands precision. Small errors in dosage can lead to ineffective treatment or dangerous outcomes for patients. This is why pharmacists must be adept at converting
measurements, especially when working with different medication forms (liquids, tablets, capsules, etc.).
Liquid medications, in particular, often require conversion between milliliters and ounces. Whether dispensing oral solutions, injectable drugs, or IV fluids, understanding these conversions is
For the NAPLEX, these conversions are likely to appear in calculation-based questions, making it crucial to master the basic math behind them.
The Basics: What is an Ounce?
An ounce (oz) is a unit of volume commonly used in the United States, especially for liquid measurements. When discussing fluid ounces in the pharmacy context, we typically refer to the US fluid
ounce, which is different from the weight-based ounce used for solid items. One fluid ounce is equivalent to approximately 29.5735 milliliters (mL).
This may seem like an odd number, but it’s important to memorize this conversion for pharmacy calculations.
How Many mL in an Ounce?
The conversion between ounces and milliliters is a fundamental pharmacy calculation:
1 ounce (oz) = 29.5735 milliliters (mL)
This value is key when converting from ounces to milliliters or vice versa. Since medications are often dispensed in milliliters but may be prescribed in ounces, understanding how to convert between
the two is critical.
Here’s a simple way to remember this:
1 oz ≈ 30 mL (rounded for convenience)
This rounded value is often sufficient for everyday use in pharmacies unless extreme precision is required. However, for the NAPLEX exam, you should know the exact conversion and be prepared to use
it when calculating precise doses.
Practical Application of mL to Ounce Conversions in Pharmacy
To illustrate the importance of this conversion in real-life pharmacy scenarios, let’s look at some examples that could appear on the NAPLEX.
Example 1: Converting from Ounces to Milliliters
Imagine a patient has been prescribed a medication with a dosage of 2 fluid ounces. The medication is available as a liquid suspension, and you need to dispense the correct amount in milliliters.
Using the conversion formula:
\text{Volume in mL} = 2 \, \text{oz} \times 29.5735 \, \text{mL/oz}
\text{Volume in mL} = 59.147 \, \text{mL}
Therefore, you would dispense approximately 59.15 mL of the medication.
Example 2: Converting from Milliliters to Ounces
Suppose a patient has been prescribed 120 mL of a liquid medication. How many ounces is this?
Using the inverse of the conversion formula:
\text{Volume in oz} = \frac{120 \, \text{mL}}{29.5735 \, \text{mL/oz}}
\text{Volume in oz} = 4.06 \, \text{oz}
So, the patient would receive roughly 4.06 ounces of the medication.
Handling Complex NAPLEX Questions
The NAPLEX may present these conversions in a more complex format, often integrating other variables like concentration, weight-based dosing, or IV flow rates. Let’s look at a more advanced example.
Example 3: Adjusting for Concentration
A liquid medication has a concentration of 10 mg/mL, and the doctor prescribes a 2 oz dose. How many milligrams of the drug will the patient receive?
First, convert 2 oz to milliliters:
2 \, \text{oz} \times 29.5735 \, \text{mL/oz} = 59.147 \, \text{mL}
Next, calculate the amount of drug in 59.15 mL:
59.15 \, \text{mL} \times 10 \, \text{mg/mL} = 591.5 \, \text{mg}
Thus, the patient will receive 591.5 mg of the medication.
This type of multi-step question is common on the NAPLEX, so it’s important to be comfortable with performing these calculations quickly and accurately.
Tips for Mastering mL and Ounce Conversions for the NAPLEX
1. Memorize Key Conversions: While approximations like 1 oz ≈ 30 mL can be useful in daily practice, for the NAPLEX, it’s important to know the exact value (1 oz = 29.5735 mL).
2. Practice, Practice, Practice: The more you practice conversions, the more second nature they will become. Use practice questions and past NAPLEX exams to test yourself under time constraints.
3. Watch for Units: One common mistake is confusing different measurement units. Always double-check whether you are dealing with fluid ounces, milliliters, or another unit.
4. Break Down Complex Problems: Many NAPLEX questions involve multiple steps. Break down each problem into manageable parts, and focus on one conversion or calculation at a time.
5. Use Dimensional Analysis: Dimensional analysis is a helpful tool for converting between units. It allows you to set up the problem so that the units cancel out, ensuring your calculations are
Converting between milliliters and ounces is a fundamental skill for pharmacists, especially when preparing for the NAPLEX. By understanding and mastering these conversions, you’ll be well-prepared
to tackle dosage calculations and provide accurate medication doses for your patients. Remember to practice regularly and become comfortable with applying these conversions in both simple and complex
scenarios. The ability to confidently and accurately perform these calculations will serve you well throughout your pharmacy career.
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Regression model outputting probability density distribution
For a classification problem (let say output is one of the labels R, G, B), how do we predict ?
There are two formats that we can report our prediction
1. Output a single value which is most probable outcome. e.g. output "B" if P(B) > P(R) and P(B) > P(G)
2. Output the probability estimation of each label. (e.g. R=0.2, G=0.3, B=0.4)
But if we look at regression problem (lets say we output a numeric value v), most regression model only output a single value (that minimize the RMSE). In this article, we will look at some use cases
where outputting a probability density function is much preferred.
Predict the event occurrence time
As an illustrative example, we want to predict when would a student finish her work given she has already spent some time s. In other words, we want to estimate E[t | t > s] where t is a random
variable representing the total duration and s is the elapse time so far.
Estimating time t is generally hard if the model only output an expectation. Notice that the model has the same set of features, expect that the elapse time has changed in a continuous manner as time
Lets look at how we can train a prediction model that can output a density distribution. Lets say our raw data schema: [feature, duration]
• f1, 13.30
• f2, 14.15
• f3, 15.35
• f4, 15.42
Take a look at the range (ie. min and max) of the output value. We transform into the training data of the following schema:
[feature, dur<13, dur<14, dur<15, dur<16]
• f1, 0, 1, 1, 1
• f2, 0, 0, 1, 1
• f3, 0, 0, 0, 1
• f4, 0, 0, 0, 1
After that, we train 4 classification model.
• feature, dur<13
• feature, dur<14
• feature, dur<15
• feature, dur<16
Now, given a new observation with corresponding feature, we can invoke these 4 model to output the probability of binary classification (cumulative probability). If we want the probability density,
simply take the difference (ie: differentiation of cumulative probability).
At this moment, we can output a probability distribution given its input feature.
Now, we can easily estimate the remaining time from the expected time in the shade region. As time passed, we just need to slide the red line continuously and recalculate the expected time, we don't
need to execute the prediction model unless the input features has changed.
Predict cancellation before commitment
As an illustrative example, lets say a customer of restaurant has reserved a table at 8:00pm. Time now is 7:55pm and the customer still hasn't arrive, what is the chance of no-show ?
Now, given a person (with feature x), and current time is S - t (still hasn't bought the ticket yet), predict the probability of this person watching the movie.
Lets say our raw data schema: [feature, arrival]
• f1, -15.42
• f2, -15.35
• f3, -14.15
• f4, -13.30
• f5, infinity
• f6, infinity
We transform into the training data of the following schema:
[feature, arr<-16, arr<-15, arr<-14, arr<-13]
• f1, 0, 1, 1, 1
• f2, 0, 1, 1, 1
• f3, 0, 0, 1, 1
• f4, 0, 0, 0, 1
• f5, 0, 0, 0, 0
• f6, 0, 0, 0, 0
After that, we train 4 classification models.
• feature, arr<-16
• feature, arr<-15
• feature, arr<-14
• feature, arr<-13
Notice that P(arr<0) can be smaller than 1 because the customer can be no show.
In this post, we discuss some use cases where we need the regression model to output not just its value prediction but also the probability density distribution. And we also illustrate how we can
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Methods of Data Manipulation Using Pandas
Pandas is a tool which is powerful , flexible and is easy to use built on top of python programming to help in data analysis.We will 5 major methods how this tools is used for data manipulation i.e
Applying Function , Pivot Table,Imputing Missing Values, Multi Indexing and Plotting.
To start with panda we need a very basic code to import the pandas library into our notebook which is
import numpy as np
import pandas as pd
Applying Function
Whenever we need a new column in data frame according to the user requirement we use this . We put value of the position at which the column needs to inserted.Here, name_col represents name of column
and new_col which act as and array for the field.
Sample This method is utilise to select the sample values as random to check the data over a distribution.
"WHERE” is used as a condition to get the specific value which is totally dependent . Here, new_val where values are greater than 1 will replaced the value with 0.
Pivot Table
Pivot table is a table which very flexible and used as summarisation of the data analysis.It works as an excel hence do not contain any missing values. The basic things in order to create a pivot
table is data and indexing . Indexing is like feature on which the whole data representation will revolve around.
You can always select more than one index in the for multi index pivot table.
table = pd.pivot_table(df,index=[’Gen,’’City’])
Imputing Missing Values
The most interesting and challenging part of data analysis is to deal with missing values .There are three major methods to deal with it .
Null Keep values Null means do nothing to the data and use as it is this will maintain the integrity of the data and also keep the analysis raw but real.
Put mean or median values of the column in all the null areas of the missing value columns .This will make the data analysis more challenging and useful as the algorithms will give more accurate
results.But the data integrity is lost hence should only be used with small numerical datasets.
Use the values zero and 1 in pivot table and make a categorical features which helps you to analyse the data in more than one way and makes manipulation easy.It also works with algorithms and
numerical data sets, ye or no datasets and trees.
Multi Indexing
It is one of the advanced ways to use the data analysis. We use multi- indexing as it helps us to analyse things by reshaping ,selecting and grouping of the hirerarichal index data in more than one
dimension. Syntax:
pandas.MultiIndex(levels=None, codes=None, sortorder=None, names=None, dtype=None, copy=False, name=None, verify_integrity=True)
# importing pandas as pd
import pandas as pd
# Create the MultiIndex
midx = pd.MultiIndex.from_tuples([(10, 'Ten'), (10, 'Twenty'),(20, 'Ten'), (20, 'Twenty')], names =['Num', 'Char'])
# Print the MultiIndex
One of my favourite things about pandas is plotting helps to summarise the data and represent in visual format .There are many plotting library such matplotlib , seaborn and many more.Types of plot
The basic plot which is representation of the line is called line plot.This connect all the points represented by the data over a graph with the help of line chart.
df = pd.DataFrame(np.random.randn(500), columns=["B"]).cumsum()
df["A"] = pd.Series(list(range(len(df))))
df.plot(x="A", y="B"); # df.plot.line(x="A", y="B")
As the name suggest this is representation of the area underneath the line plot this can contain multi-values such as A,B,C,D used as different variables of representation.
df = pd.DataFrame(np.random.rand(20, 4),
columns =['A', 'B', 'C', 'D'])
The bar plot allows to represent the values represented by categorical value which co- related one given category with another.This doesn't have to be co related directly but more of the comparison
df = pd.DataFrame(np.random.rand(10, 4), columns=["a", "b", "c", "d"])
Histogram is similar to bar only difference this contains multiple sub versions in a small buckets call bins which actually tells the width of these bars over a single axis.
df = pd.DataFrame(
"a": np.sqrt(np.random.randn(1000) + 1),
"b": np.random.randn(1000),
columns=["a", "b", "c"],
This plot is used to check the anomalies in the data through scatter plot with the co relation between two variables.We also use it check types of models which is under fitting model or overfitting
model of machine learning.
df = pd.DataFrame(np.random.rand(100, 2),
columns =['a', 'b'])
df.plot.scatter(x ='a', y ='b');
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Earth's oblateness
1 Introduction
The rotating Earth is oblate, that is, it is slightly ‘flat’ in the North Pole–South Pole direction, compared to the slightly ‘bulging’ Equator. This is the result of the hydrostatic balance between
the dominant gravitational force, which wants to pull the Earth into a spherically symmetric configuration, and the centrifugal force due to Earth's rotation, which wants to expel mass away from the
rotating axis but in the end only manages to modify the Earth into an slightly oblate body.
Quantitatively, this oblateness is about 1 part in 300, which is very close (but see below) the ratio of the centrifugal force on the equator to the gravitational force. This is by far the Earth's
largest deviation from a spherically symmetric body. There are certain thermodynamic, but secondary, processes that cause departures from the rotational-hydrostatic equilibrium. Sustained by the
internal heat engine and manifested as external gravity anomalies reflecting lateral heterogeneity of internal mass distribution, these deviations are in general no more than parts per million in
relative terms.
Yet all these deviations are not static or constant; they change with time. The rotation of the Earth is itself changing over geological time, and the aforementioned mass heterogeneities also vary on
timescales upwards from millions of years. On more human timescales, there operates a myriad of dynamic processes that involve mass redistribution in or on the Earth, from tides to atmosphere–ocean
circulations, to internal phenomena like earthquakes, post-glacial rebound and core flows. For the Earth, these changes are typically on the order of parts per billion at the largest [9,24]. The
present article is a story about the oblateness in particular and how and why it changes with time, where we examine the geophysical implications.
2 The Earth's oblateness parameters and their inter-relationships
As long as the Earth is a 3-D body, we shall use the word oblateness to describe its off-spherical shape. Traditionally, the term ‘flatness’, or ‘ellipticity’, has been used; these names are
imprecise because the Earth, of course, is not ‘flat’, and it is a 2-D geometric object only when we try to draw it on paper.
There are several parameters in use to describe the oblateness; each one has its significance depending on the application in question. For simplicity let's for the moment assume the Earth is axially
symmetric, or a body of revolution and so essentially a 2-D body, which is a good approximation.
By equating the general expression of the spherical harmonic expansion of the external gravitational potential field V with that of the mass distribution of a body, one concludes that the spherical
harmonic coefficients, or Stokes coefficients, of V are simply normalized multi-poles of the density function of the body [4,5]. When specialized, the degree-2 Stokes coefficients are related to the
body's inertia tensor elements through a set of equations known as generalized MacCullagh's formulas. In particular, the degree-2 zonal (order 0) Stokes coefficient is given by:
Symbol namesake in honor of Sir Harold Jeffreys [19] and called the oblateness coefficient for the Earth, $J2$ has the physical meaning of the difference of the axial or polar (greatest) moment of
inertia C with the equatorial (least) moment of inertia A, normalized by $Ma2$ where M and a are the Earth's mass and mean Equatorial axis, respectively. Its corresponding term in the harmonic
expansion of V is the dominant term next to the ‘monopole’ term representing the total mass [4].
We can express $J2$ in the following form:
$J2=[C/Ma2][(C−A)/C]≡ηH$ (2)
where $η≡C/Ma2$ is a fundamental functional of the Earth's internal structure, and $H≡(C−A)/C$ is called the dynamic oblateness, which can be determined from the observation of the astronomical
precession of the Earth [27]. The Earth's η can be readily determined by knowing the values of $J2$ and H. However, this has not been feasible for other planets because their H's are generally
The ‘geopotential’ field is V, modified by the centrifugal potential, i.e. $V−12a2ω2$, where ω is the angular speed of Earth's rotation. If one approximates the equipotential surface, known as the
geoid, to an oblate spheroid of revolution, then one obtains the geoid oblateness for the Earth which can be given by the Clairaut's first relationship (for a review, see [27]):
where c is the mean axial or polar axis of the (ellipsoidal) Earth, and $m≡a3ω2/GM$ is the ratio of the centrifugal force $aω2$ on the equator to gravity $GM/a2$ (where G is the gravitational
constant). Eq. (3) is based on the first-order theory, which is sufficient for the present purpose for the Earth (for high-order formulas see, e.g., [23]). For planets with much larger m (for
example, Jupiter, see below), the second-order effects become rather significant [29].
Two more concepts of oblateness can be defined at this point: suppose the Earth is under rotational-thermo-hydrostatic equilibrium. A hypothetical hydrostatic geoid oblateness $fH$ representing an
idealized Earth can be defined; to first order $fH$ can be found by [18]:
$fH=(5/2)m/[1+(25/4)(1−1.5η)2]$ (4)
Under such equilibrium, the Earth's geometric surface would conform to and coincide with the geoid, so the geometric oblateness, similarly defines as Eq. (3), is simply equal to f. That is why f is
often referred to as defining the ‘figure’ of the Earth. In reality, due to its heterogeneities, the Earth's geometric oblateness (as properly defined or approximated) would depart slightly from f.
Furthermore, the Earth's true f also departs slightly from $fH$, where the departure signifies interesting geophysical dynamics sustaining non-equilibrium. The lateral heterogeneity and
non-equilibrium configuration of the Earth also manifest themselves in the (relatively small) difference between the two equatorial principal moments of inertia A and B. In the above we have assumed
that the Earth is axially symmetric where $A=B$, which would be the case if the Earth were an otherwise spherically symmetric body subject only to an axial rotation. The real Earth, of course, is not
so (we will further discuss this below.) In fact, the strict definition of $J2$ is $[C−(A+B)/2]/Ma2$, to which Eq. (2) is only an approximation, or valid only for axial-symmetric Earth.
Now let us examine the numerical values. m is known to be $3.46775×10−3$, or 1/288.371, close to 1 part in 300 or 1/300. According to Eq. (3), half of it contributes directly to the geoid oblateness
f. For f, the remaining contribution comes from $32J2$, which, of course, shares the same dynamic origin as m, i.e. Earth's rotation. $J2$ is measured from satellite geodesy (see below) to a high
accuracy, $1.082626×10−3$, about one third of 1/300. So, the two terms in (3) contribute almost the same amount to f, i.e., 50% each, and f itself becomes close to 1/300, at $f=3.35281×10−3$ or 1/
298.257. Finally, the dynamic oblateness H in Eq. (2) is observed to be $3.27379×10−3$ or 1/305.456, again close to 1/300.
Are these matchings in values just fortuitous? From dynamical considerations, one can rightly ‘guess’ that all parameters should be on the order of m, which they indeed are. However, upon closer
examination as follows, they do not necessarily have to have such similar values, so in a sense the latter is fortuitous.
For a reasonable Earth configuration, we should have $H≈f$ because of its moderate sensitivity to the internal density profile, although it is well recognized that H would be somewhat less than f
because of the smaller oblateness of the interior layers (due to smaller centrifugal force) and the higher density toward the center of the Earth (and hence proportionally lower importance in
contributing to the moment of inertia) [27]. Putting this condition into Eqs. (2) and (3), we see the following: the two terms would contribute near-equal shares to f in Eq. (3) and hence all the
values would be close to m or 1/300, only if $η=1/3$.
The interesting, but certainly not out of the ordinary, fact is that, knowing $J2$ and H in Eq. (2), the Earth in reality has $η=0.33069$, indeed almost exactly 1/3! This of course does not have to
be the case, but one does expect a η value somewhat less than 0.4, that of a uniform-density sphere, for a ‘reasonable’ centrally-heavy, terrestrial planet body such as the Earth.
Based on the PREM Earth model (Preliminary Reference Earth Model [15]) derived from seismological data, the Earth should have an estimated hydrostatic $fH$ of 1/299.66 [29], about 0.5% smaller than
the observed. This corresponds to a hydrostatic $J2$ of $1.0722×10−3$, about 1% smaller than the observed. On the other hand, Liu and Chao [21] formulated the relation between A, B and the two Stokes
coefficients of degree 2 and order 2. Using the gravity-observed values for the latter, they get $B−A=7.260×10−6Ma2$, amounting to 69.4-m difference in the equivalent geoidal semi-major axis and
semi-minor axis on the Equator. Although only $∼1/150$ that of $C−A$, this amount is comparable to the non-hydrostatic portion in $C−A$, as pointed out by [17]. They concluded that the
non-hydrostatic portion of the three principal moments of inertia A, B, C only describes a triaxial body and appears to have no preference in orientation. As far as the aforementioned excess
oblateness over the hydrostatic value is concerned, this does not favor the notion that this excess oblateness is a remnant, lagging ‘memory’ of the past, as the Earth slows down due to the tidal
3 Comparative planetology
For a contrast, let us compare the Earth with the giant planet Jupiter. Jupiter has a faster rotation and a much larger mass, and hence larger radius and gravity. Its $m=0.0892=1/11.2$. We can expect
that the geoid oblateness f and the dynamic oblateness H to be similar to m, but not necessarily very close in value. The observed $J2=0.01469$. Adopting second-order formulas [29], which are more
accurate than Eq. (2), $f=0.0649=1/15.4$. Assuming rotational-hydrostatic equilibrium, $η=0.254$ (cf. Eq. (4)), indicating that, not surprisingly, Jupiter is somewhat more centrally-heavy than the
Earth. The derived $H=(1/η)J2=0.0578=1/17.3$. We further expect that the geometric oblateness is the same as f, except possibly for some small departures from rotational-hydrostatic equilibrium.
In another extreme example, let us consider a non-rotating, uniform-density body not under hydrostatic equilibrium (hence the shape sustained by its internal material strength), such as an asteroid.
Then there exists an analytical, but complex, relationship between the spherical harmonic coefficients of gravity and geometrical shape [7]. For the present discussion, let us further assume a
special case where the body is a slightly oblate spheroid. Then, letting $m=0$ in Eq. (2), we have the geoid oblateness $f=32J2$. Since the body is not under hydrostatic equilibrium, the geometric
oblateness is not equal to f; rather, according to Eq. (9) of [7], it equals $53f$. Finally, the dynamic oblateness $H=(1/η)J2$, when $η=25$ (for a uniform spherical body), equals $53f$, the same as
the geometric oblateness, as expected.
4 Consequences of oblateness
We live in the Earth's gravity field, controlled by the dominant monopole term $GM/a2$. We hardly notice any consequence of the oblateness of the Earth (or for that matter the rotational centrifugal
force). However, dynamically, the Earth's oblateness is an essential element in our livelihood – it stabilizes our Earth's rotation. The Earth is ‘bombarded’ all the time by countless geophysical
agents exerting external torques as well as internal torques or mass transports that exchange angular momenta. Yet its rotation axis hardly changes relative to the Earth-fixed geography. This is not
true if it were spherically symmetric: then the crawl of a bug or a firing of a canon, for instance, would completely ‘tumble’ the Earth relative to the (spatially stationary) rotation axis [16,22].
On the other hand, it is well known from classical mechanics that the rotation of a body about its principal axis of the greatest moment of inertia (C) is a stable one. What prevents large shifts of
the rotation axis from happening is the extra oblateness in the form of $C−A$ with the associated extra angular momentum, which is to be overcome by any geophysical agent that tries to shift the
Earth's rotational pole positions. Since the oblateness itself is a consequence of the rotation in the first place, it can be stated that the rotating planet is self-stabilized.
A corollary of the above, but on a less dramatic scale, the Earth's dynamic oblateness under the tidal torques exerted by the Moon and Sun gives rise to the astronomical precession of the Earth's
rotation axis in space, and hence is the deciding factor for the precessional period. That in fact is how the dynamic oblateness H is determined. On the same token, H acts as the restoring factor
that prescribes the free wobble, known as the Chandler wobble, of the Earth's polar motion. The period of the Chandler wobble would be $1/H$ days if the Earth were a rigid body, but was found to be
significantly lengthened by the Earth's non-rigidity, or finite elasticity [22].
As stated, the geometrical shape of the Earth largely conforms to the oblate geoid. Therefore, the mean equatorial radius and the mean polar radius of the geoid differ by as much as $a−c=fc∼21 km$.
In particular, the global sea level follows closely this oblate geoid, only undulating on top of the geoid geographically no more than 200 m peak-to-peak and temporally less than 10 m or so. The land
topography undulates up to $∼10 km$, but largely supported isostatically.
As such, the oblateness also affects various geophysical quantities. For example, in the space geodesy enterprise using near-Earth satellites, the oblateness term resides in all Earth surface
geometry that locates the geodetic observatories and altimetric targets. Similarly, the oblateness prevails in the external gravity field that significantly affects the satellite orbits from which
geodetic measurements are made. On the Earth surface, together with the centrifugal force field, the oblateness gives the surface gravity a slight latitudinal dependence which is actually the largest
term in the surface gravity anomaly on the global scale. In another example, the Earth's elastic free oscillation modes (often excited by large earthquakes) see splitting in their otherwise
degenerate characteristic periods due to Earth's oblateness and rotation, completely analogous in the atomic world to the Stark splitting and Zeeman splitting, respectively, as such splitting is
determined by the symmetry properties common to different dynamic systems [2].
5 Historical Notes
Sir Isaac Newton, based on his law of gravitation and force laws, was the first to realize that the Earth under rotational equilibrium should possess a non-vanishing oblateness. The value of $1/f$
favored by him and given in the Principia was 230. Cassini subsequently came up with a negative value, −95, presumably owing to certain systematic errors. The value had evolved [19], since the Peru/
Lapland expedition in the 1740s from a value between 179 and 266, to 301, 295, 297.0, and finally in the early 1950s to Sir Harold Jeffreys' $297.1±0.4$ which is within 0.4% of the modern value
Then came the space age, ushered in by the launch of USSR's Sputnik I spacecraft in October 1957. A month later Sputnik II was launched, and within a few weeks, by monitoring the nodal precession of
its orbit in space, our knowledge of $J2$ grew almost an order of magnitude, to about 0.1% of the modern value. This measurement was arguably one of the very first scientific triumphs of space
Today, after nearly half a century of precise orbit determination of dozens of geodesy-quality satellite orbits around the Earth, the Earth's global gravity field has been solved to harmonic degrees
as high as 120, among which the average $J2$ coefficient has been determined to the accuracy of seven significant figures ($1.082627×10−3$) [20].
Since the 1980s, thanks to the advent of the technique of satellite laser ranging [1], tiny temporal variations around the average value of $J2$ began to be noted. The variation occurs in the last
digit of the above-quoted number and beyond, typically no more than one part in a billion! This will be discussed next.
6 How and why does Earth's $J2$ change?
Mass transports in the atmosphere–hydrosphere–cryosphere–solid Earth–core system (the ‘Earth system’) occur on all temporal and spatial scales for a host of geophysical and climatic reasons [9,24].
According to Newton's gravitational law, such mass transport will cause the gravity field to change with time, producing time-variable gravity signals.
Increasingly refined models for the Earth's static gravity field in terms of spherical harmonic components have been determined by means of decades of precise orbit tracking data of many geodetic
satellites. On top of that, low-degree components of Earth's time-variable gravity have been clearly observed by the space geodetic technique of satellite laser ranging (SLR) [1]. Although tiny in
relative terms (no more than 1 part per billion), these variations signify global-scale mass redistribution in the Earth system.
In particular, the lowest-degree zonal harmonic is Earth's oblateness coefficient $J2$, whose temporal variation was the first to be detected among all gravity components. A ‘secular’ decrease in
$J2$ (over the observed quarter century) was first identified from the SLR satellite nodal precession acceleration. Its main excitation source has since been attributed to the post-glacial rebound
(PGR) of the solid Earth [26,30], and for additional secondary causes [3]. Subsequently, many studies reported strong seasonal as well as weaker non-seasonal signals, primarily in $J2$, but of late
also in the next-lowest harmonics [13] and geocenter. The prominent seasonal $J2$ signals (with primarily annual amplitude $∼3×10−10$) have been correlated with mass transports in the atmosphere,
oceans, and land hydrology [8,11,25].
Such was the case until around the turn of the century beginning in 1998, when the SLR data began to reveal that Earth's $J2$ had suddenly deviated significantly from the PGR secular decreasing trend
(at about $−2.8×10−11 yr−1$). This ‘1998 anomaly’ embarked on a reverse, increasing trend over the following years, before quieting back down to the ‘normal’ decreasing trend. This was reported by
[10,12]. Fig. 1a shows an updated time series of the SLR-observed $J2$, using SLR data from up to nine satellites, with more satellites becoming available with time [12]. Note that the relevant
18.6-yr lunar-driven ocean tide amplitude was set to the value recovered in a 21-yr comprehensive solution for the secular zonal rates, low-degree static and annual terms, and the 18.6-yr and the
much smaller 9.3-yr lunar tides. Fig. 1(b) is the same time series but after the removal of (i) the atmospheric contribution calculated according to the global NCEP reanalysis data assuming an
inverted-barometer effect, and (ii) the least-squares fit of the remaining seasonal signals, which are attributable to (the poorly known) seasonal mass redistribution in the oceans and land
hydrology. The PGR slope (the solid line) and the 1998 anomaly are clearly evident.
Fig. 1
A number of possible causes for the 1998 anomaly was speculated by Cox and Chao [12], including oceanic water mass redistribution, melting of polar ice sheets and high-latitude glaciers, global sea
level rise, and material flow in the fluid core. Dickey et al. [14] emphasized and demonstrated the importance of the melting of high-latitude glaciers. Chao et al. [10] report an oceanographic event
that took place in the extratropic North + South Pacific basins that was found to match remarkably well with the time evolution of the $J2$ anomaly; the phenomenon appears to be part of the Pacific
Decadal Oscillation immediately following the episode of the 1997–1998 El Niño.
The difficulty in identifying the definite cause(s) in the above w.r.t. $J2$ stems from the extremely low geographical resolution of the zonal harmonic function in question, namely the degree-2
Legendre function. Thus, a positive $J2$ anomaly only tells us that a net transport of mass from higher latitudes to lower latitudes (across the nodal latitude of the degree-2 Legendre function,
namely ±35.3°) has occurred, in either or both Northern and Southern Hemispheres. For example, an equivalent of as much as 3000 km^3 of water melted from Greenland and spread into the oceans would be
needed to produce the first half of the $J2$ anomaly where the relative change is $+7×10−11$ per year; but we have no way of telling without other ancillary evidences or observations. On the other
hand, the space gravity mission of GRACE (launched in March 2002, with an expected lifetime of over 5 yr), using the satellite-to-satellite tracking (SST) technique, is yielding gravity information
at much higher geographical resolution than the SLR-based information. For example, GRACE is able to detect centimeter level water-height-equivalent mass changes over an area of about 1000 km across
from month to month [28]. However, considering the relatively weak sensitivity of the SST to the longest-wavelength (lowest-degree) gravity components, GRACE's utility in measuring the variation of
$J2$ in particular awaits to be seen. The same applies to future gravity missions of GOCE (using gravity gradiometer) and other SST measurements in ‘follow-on’ gravity missions under planning.
7 Relationship between Earth's rotation and $J2$ change
As stated, the Earth's oblateness arises from its rotation; the rotational-hydrostatic relationship, to first order, is given in Eq. (4), where the oblateness is proportional to m, which is in turn
proportional to $ω2$. Therefore,
For example, the Earth's secular spin-down due to the tidal braking would lead to a secular ‘rounding’ of the Earth (barring possible temporal retardation under viscosity), thus decreasing $J2$.
Numerically, at the tidal-braking rate of $ω˙=−6.5×10−22 rads−2$, that decreasing rate of $J2$ is about $−6.1×10−13 yr−1$, contributing only 2% of the observed decreasing rate of $J2$ (see above).
On the other hand, any change in $J2$ will cause ω to change, as dictated by the conservation of angular momentum for the Earth. For instance, a decreasing $J2$ means a faster spin (analogous to a
spinning skater pulling arms closer to the body), and vice versa. That effect can be shown to be [6]:
where the coefficient 2.01 is evaluated from Earth parameters. For example, the decrease in $J2$, at the rate of $−6.1×10−13 yr−1$ due to the tidal braking of ω given above, will in turn feedback to
cause ω to increase, but only by as little as $2.8×10−24 rads−2$, or $∼−0.053 μs$ in the equivalent length-of-day per year. That is completely negligible in today's measurement. On the other hand,
the observed $J2$ rate of change $−2.8×10−11 yr−1$ (see above) presumably speeds up the Earth rotation by $−2.4 μs$ in the equivalent length-of-day per year, which is still negligible.
8 Epilogue
Although numerically small, the oblateness is a fundamental property of the Earth under stable rotation. Its existence and cause, its dynamical and geometrical consequences, its values and departures
from idealized models, and its temporal evolution due to mass transports in the Earth system are all fascinating topics in geophysics, which reveal insights towards the understanding of the structure
and dynamical behavior of the Earth. The measurement and monitoring of the Earth's oblateness have been a triumph as well as a scientific target of the modern space geodesy. As one sees deeper and
finer into the Earth's oblateness, there is little doubt that the Earth will surprise and further fascinate us with a continuing story unfolding with time.
This paper is completed under the support of the NASA Solid Earth program. I am grateful to Christopher Cox for providing Fig. 1. | {"url":"https://comptes-rendus.academie-sciences.fr/geoscience/articles/en/10.1016/j.crte.2006.09.014/","timestamp":"2024-11-08T20:12:05Z","content_type":"text/html","content_length":"109237","record_id":"<urn:uuid:3ea5b020-0cd6-40d4-996e-5b44c0bba74b>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00696.warc.gz"} |
Calculating a logarithmic mean for noise data
Some specialized types of aggregation are not available via Processing & Logic. In these cases, processing data externally in a Google sheet is an option, and the general steps for doing so are
covered in this article.
One specific type of aggregation is used to calculate an average for logarithmic values, such as noise data. This article will describe how to perform this aggregation using a Google sheet. The
method used comes from this blog post which uses Excel to perform the calculations.
Problem statement
Noise data is measured in decibels, which are expressed on a logarithmic scale. This means that a simple arithmetic average (such as the AVERAGE aggregate which is available via Processing & Logic)
is not a suitable way to find the mean value of a set of noise data.
In basic terms, an appropriate logarithmic average can be achieved by calculating the anti-log of each decibel value, finding the average of all the anti-log values, and then finally calculating the
log of that average. And by controlling how many anti-log values are included in the average calculation, this can be achieved in a "rolling" way by having a moving window of raw values included in
each iteration.
Raw data
The sample noise data used in the following example will comprise 10 hours of 1-minute values. The average calculation will include 30 minutes of raw data, and produce one output every minute. The
timestamp of each average value will be at the start of the 30-minute window, just like the built-in eagle.io aggregates (which in turn are derived from OPC Unified Architecture specifications). This
means that the raw values between 00:00 (inclusive) and 00:30 (exclusive) will be aggregated and the resulting logarithmic average will have a timestamp of 00:00. The window will then move to include
raw values between 00:01 (inclusive) and 00:31 (exclusive), with the logarithmic average having a timestamp of 00:01, and so on.
Making the raw data public
The first step for external processing is to make the the data public. In this case we only need a single parameter, so in the security configuration of the parameter, enable public access, and set
the correct timezone (this should be the same as the data source timezone). The time format should be left as the default of YYYY-MM-DDTHH:mm:ss.SSSZ for best compatibility with the Google sheet.
This will provide a public URL displaying the raw noise data in text form:
Getting the raw data into a Google sheet
Now we create a new Google sheet, and in cell A1 import the parameter data using the importdata formula containing the public URL:
Calculating the logarithmic average using the Google sheet
This is where the actual calculations occur, and there are 3 steps which will use 3 columns, C through E. Note that the raw data starts on row 4 of column B, so the first window of 30 values to be
averaged will come from B4 though B33.
Calculation 1: divide the raw data by 10 and anti-log the result
In cell C4, enter =10^(B4/10) and fill down:
Calculation 2: average the first 30 anti-logged values
In cell D4, enter =average(C4:C33) and fill down:
Calculation 3: log the averaged results
In cell E4, enter =10*log(D4) and fill down:
The values in column E are the logarithmic averages, i.e. E4 contains the log average of noise values from B4 though B33, E5 contains the log average of noise values from B5 though B34, E6 contains
the log average of noise values from B6 though B35, and so on.
Now that the calculations are complete, add a label in E2 and units in E3 so it's easier to identify when the data is used in eagle.io:
Getting the averaged results back into eagle.io
From the File menu select Share -> Publish to web:
Select Sheet1, with a format of Comma-separated values, and click Publish:
A URL will be generated; copy this so it can be used in eagle.io:
In eagle.io, create a new data source with a transport of Download via HTTP:
Enter the URL that was just created by the Google sheet:
In the parser configuration, set the time format to be YYYY-MM-DDTHH:mm:ss.SSSZ, the labels row to be 2, the units row to be 3, and disable columns 2, 3 and 4 (because we only need the average data
in column 5):
This will create a new parameter named LAS10 log avg 30-min:
Charting the result
Now we can create a chart to compare the raw noise data in the LAS10 parameter with the averaged data in the LAS10 log avg 30-min parameter:
The averaged data is clearly less spiky, and hopefully more useful to decision makers, than the original raw noise data.
External calculation with Google sheets can be used for a huge variety of useful purposes; even if you don't need to calculate log averages, you can adopt the general approach shown above and
substitute your own calculations, aggregations or formulas as required.
0 comments
Article is closed for comments. | {"url":"https://help.eagle.io/hc/en-au/articles/4770947081231-Calculating-a-logarithmic-mean-for-noise-data","timestamp":"2024-11-11T03:32:58Z","content_type":"text/html","content_length":"36586","record_id":"<urn:uuid:8ead35f9-7ad6-4192-9037-7883a7934e11>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00434.warc.gz"} |
How Alviss AI’s Bayesian Approach Enhances Marketing Mix Modeling Accuracy – Alviss AI Blog
Image from Freepik
If you’ve worked in marketing long enough, you’ve probably heard that Marketing Mix Modeling (MMM) is half science, half art. Traditional MMM can help companies make smarter choices about where to
spend their budgets, but it’s not exactly foolproof. The more complex the data, the harder it gets to understand what’s actually driving results. Most models try to fit a lot of moving parts into one
framework, and that leads to all kinds of uncertainty. And if you’re working with a lot of uncertainty, you’re bound to get hit with surprises—some good, but often, expensive.
At Alviss AI, we take a different approach. Instead of just trying to squeeze every factor into a single predictive model, we use something called Bayesian modeling. And this isn’t just a fancy
statistical trick. Bayesian modeling lets us embrace uncertainty rather than ignore it. By doing so, we get more accurate insights, which are exactly what marketing teams need to avoid guesswork and
start making decisions based on what’s most likely to be true.
Embracing Uncertainty
The way most traditional MMM models work is that they take a bunch of data from different marketing channels and fit them into a single equation. Then, the model says something like, “Based on this
data, we estimate that TV drives 20% of your sales, digital drives 40%, and so on.” But let’s be honest—no one is ever 100% sure of those numbers. And what if a new data source or a competitor’s
unexpected move shakes up the whole market? Traditional models don’t adapt well to that. They get locked into their initial assumptions, and you’re left guessing how reliable the numbers are.
Bayesian models are different. Instead of just making a single guess about how much impact each marketing channel has, Bayesian models give us a range of possible answers and the probability of each.
This gives marketing teams a clearer picture of what’s likely and what’s not, so they can make decisions with confidence, even when the landscape changes.
Why Probabilities Matter
Here’s where Bayesian modeling shines. Let’s say you’re launching a big campaign and want to know how much to invest in different channels. With a Bayesian model, you don’t just get a hard number
saying, “spend $500K on digital.” Instead, you get a range that says something like, “if you spend between $400K and $600K on digital, there’s a 90% chance of hitting your target.” Now you have
something you can work with, something that reflects the real-world uncertainty of marketing.
In traditional models, there’s usually just one set of answers: “Digital has a 40% impact, TV has a 20% impact,” and so on. But in the real world, things are messier. Channels influence each other,
customer behavior shifts, and data isn’t always precise. A Bayesian approach accounts for all this. It doesn’t just assume that digital will always have the same impact or that TV spending will
produce predictable results. Instead, it shows you the most likely outcomes based on the data you have—and keeps those probabilities updated as new data comes in.
Better Decisions, Better Outcomes
So why does any of this matter? Because when you’re working with better insights, you make better decisions. Marketing budgets are rarely flexible, and every dollar counts. With a Bayesian model,
you’re not just placing your bets on a single number or relying on past data to predict the future. You’re looking at the probabilities, weighing the risks, and making decisions that are based on
what’s most likely to work now, not last year.
At Alviss AI, we’ve built Bayesian models that are flexible and transparent. That means our clients can see exactly where the model is less certain and where it’s rock solid. It’s like having a
compass that doesn’t just point north but also tells you if there’s a storm ahead. Instead of following an outdated map, you’re getting directions in real time, adapting to changes as they happen.
The Value of Transparency
There’s another benefit to Bayesian modeling that often goes unnoticed: transparency. Because Bayesian models offer a range of possible outcomes, they’re inherently more open about where the data is
strong and where it’s shaky. Traditional models tend to hide these uncertainties, presenting a single answer as if it were gospel. But in marketing, there’s rarely a single “right” answer. Every
market is a mix of shifting trends, customer behaviors, and unpredictable competitors.
With Alviss AI, we wanted to create a model that shows its work. If there’s high uncertainty around a specific channel’s impact, you’ll know about it. If the data points to a likely outcome, you’ll
see that too. This transparency gives teams confidence in their decisions, knowing that they’re not just guessing. They’re working with a model that’s honest about its limitations and clear about its
Building a Smarter MMM
Using Bayesian modeling doesn’t just make Marketing Mix Modeling more accurate; it makes it smarter. The world of marketing is fast, and the smartest decisions come from understanding probabilities,
not just hard predictions. With Alviss AI, we’re giving marketing teams the tools to see what’s most likely, what’s possible, and what’s risky.
When you have a model that accounts for uncertainty, you’re no longer working in the dark. You’re working with something that respects the reality of marketing—that every decision has some risk, but
with the right insights, you can make that risk work in your favor. That’s the kind of clarity that lets you invest confidently and get results that matter.
This post is part of a 6 part series called “Mastering Marketing Effectiveness with In-Housed MMM”. The posts are outlined below. | {"url":"https://blog.alviss.io/posts/MasteringMMM/Why%20In-Housing%20Marketing%20Mix%20Modeling.html","timestamp":"2024-11-11T18:11:50Z","content_type":"application/xhtml+xml","content_length":"37069","record_id":"<urn:uuid:fc169b30-a510-4d77-b0ff-fa026a3bfa77>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00106.warc.gz"} |