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IBU Calculator: Optimize Your Beer Bitterness Levels
This IBU calculator helps you estimate the bitterness of your beer to achieve the perfect balance for your homebrew.
International Bitterness Units (IBU) Calculator
This calculator is designed to help homebrewers determine the International Bitterness Units (IBU) of their beer recipes. IBUs measure the bitterness brought by hops in beer, which is a key component
of beer flavor.
How to Use the Calculator
1. Enter the Original Gravity (OG) of your wort. This is a measure of the sugar concentration before fermentation starts.
2. Enter the Volume of Wort in liters. This is the amount of liquid you are brewing.
3. Enter the Weight of Hops in grams. This is the total weight of hops used for bitterness.
4. Input the Alpha Acid (AA) percentage of your hops. This indicates the bitterness potential of the hops.
5. Fill in the Boil Time in minutes. This is how long you will boil the hops in the wort.
6. Click ‘Calculate’ to find out the IBU of your recipe.
How the Results are Calculated
The calculation uses the formula for IBUs which takes into account the gravity of the wort, the volume of the wort, the weight of the hops, the alpha acid percentage, and the boiling time to
determine the utilization of the hops. The formula applies factors for bigness, which accounts for gravity, and another for boil time.
Limitations of the Calculator
The IBU calculator provides an estimated value based on the inputs provided. It does not account for specific hop utilization rates that may vary with different brewing conditions or systems. The
calculator also assumes a simple linear relationship between time and utilization, which may not be accurate for all boil durations or hop varieties. | {"url":"https://madecalculators.com/ibu-calculator/","timestamp":"2024-11-06T20:44:18Z","content_type":"text/html","content_length":"143248","record_id":"<urn:uuid:3a71a02c-2470-4603-911b-c0346ae25e88>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00852.warc.gz"} |
Partition calculator - Ratios of directed line segments calculator
Ratios of directed line segments calculator
Enter points and ratio in the given input boxes and select the type of partition, i.e. internally or externally. Hit the calculate button to find the coordinates of point using partitioning a line
segment calculator.
Partition calculator
Ratios of directed line segments calculator is an online tool that finds the point coordinates dividing the line (internally/externally) joining two points A and B for a given ratio m and n.
Coordinates of a point - Definition
Coordinates of a point are a pair of numbers that state its precise location on a two-dimensional plane. The coordinate plane has two axes at right angles to each other, called the x and y axis.
Formula of coordinates dividing a line segment
Coordinates dividing line internally
Coordinates dividing line externally
x[1], x[2], y[1], y[2] are end point of a line segment,
m and n are external and internal ratios.
How to find ratio of a directed line segment?
Let’s find out coordinates of points dividing a line segment joining two points with an example.
Find the point coordinates that divides a directed line segment internally having endpoints (2, 4), (3, 6) with a ratio of 5:6.
Step 1: Identify the values.
x[1] = 2, x[2] = 3, y[1] = 4, y[2] = 6, m = 3, n = 1
Step 2: Place the values in equation of internal ratio of directed line segment.
= (mx[2]+nx[1]/m+n, my[2]+ny[1]/m+n)
= (3×3+1×2/3+1, 3×6+1×4/3+1)
= (11/4, 22/4)
Hence, the points (11/4, 22/4) divides the line joining points (2, 4), (3, 6) with a ratio of 5:6.
Use external ratio segment formula for external points coordinates or use our segment partition calculator above to automate your calculations. | {"url":"https://www.allmath.com/points-coordinates.php","timestamp":"2024-11-07T03:53:47Z","content_type":"text/html","content_length":"49016","record_id":"<urn:uuid:0479459a-76c2-4ac7-8d1f-2f12df3c90e8>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00689.warc.gz"} |
K-Means Clustering
kmeans {clustlearn} R Documentation
K-Means Clustering
Perform K-Means clustering on a data matrix.
max_iterations = 10,
initialization = "kmeans++",
details = FALSE,
waiting = TRUE,
data a set of observations, presented as a matrix-like object where every row is a new observation.
centers either the number of clusters or a set of initial cluster centers. If a number, the centers are chosen according to the initialization parameter.
max_iterations the maximum number of iterations allowed.
initialization the initialization method to be used. This should be one of "random" or "kmeans++". The latter is the default.
details a Boolean determining whether intermediate logs explaining how the algorithm works should be printed or not.
waiting a Boolean determining whether the intermediate logs should be printed in chunks waiting for user input before printing the next or not.
... additional arguments passed to proxy::dist().
The data given by data is clustered by the k-means method, which aims to partition the points into k groups such that the sum of squares from points to the assigned cluster centers is minimized. At
the minimum, all cluster centers are at the mean of their Voronoi sets (the set of data points which are nearest to the cluster center).
The k-means method follows a 2 to n step process:
1. The first step can be subdivided into 3 steps:
1. Selection of the number k of clusters, into which the data is going to be grouped and of which the centers will be the representatives. This is determined through the use of the centers
2. Computation of the distance from each data point to each center.
3. Assignment of each observation to a cluster. The observation is assigned to the cluster represented by the nearest center.
2. The next steps are just like the first but for the first sub-step:
1. Computation of the new centers. The center of each cluster is computed as the mean of the observations assigned to said cluster.
The algorithm stops once the centers in step n+1 are the same as the ones in step n. However, this convergence does not always take place. For this reason, the algorithm also stops once a maximum
number of iterations max_iterations is reached.
The initialization methods provided by this function are:
A set of centers observations is chosen at random from the data as the initial centers.
The centers observations are chosen using the kmeans++ algorithm. This algorithm chooses the first center at random and then chooses the next center from the remaining observations with
probability proportional to the square distance to the closest center. This process is repeated until centers centers are chosen.
A stats::kmeans() object.
Eduardo Ruiz Sabajanes, eduardo.ruizs@edu.uah.es
### Voronoi tesselation
voronoi <- suppressMessages(suppressWarnings(require(deldir)))
cols <- c(
### Helper function
test <- function(db, k) {
print(cl <- clustlearn::kmeans(db, k, 100))
plot(db, col = cl$cluster, asp = 1, pch = 20)
points(cl$centers, col = seq_len(k), pch = 13, cex = 2, lwd = 2)
if (voronoi) {
x <- c(min(db[, 1]), max(db[, 1]))
dx <- c(x[1] - x[2], x[2] - x[1])
y <- c(min(db[, 2]), max(db[, 2]))
dy <- c(y[1] - y[2], y[2] - y[1])
tesselation <- deldir(
cl$centers[, 1],
cl$centers[, 2],
rw = c(x + dx, y + dy)
tiles <- tile.list(tesselation)
asp = 1,
add = TRUE,
showpoints = FALSE,
border = "#00000000",
fillcol = cols
### Example 1
test(clustlearn::db1, 2)
### Example 2
test(clustlearn::db2, 2)
### Example 3
test(clustlearn::db3, 3)
### Example 4
test(clustlearn::db4, 3)
### Example 5
test(clustlearn::db5, 3)
### Example 6
test(clustlearn::db6, 3)
### Example 7 (with explanations, no plots)
cl <- clustlearn::kmeans(
clustlearn::db5[1:20, ],
details = TRUE,
waiting = FALSE
version 1.0.0 | {"url":"https://search.r-project.org/CRAN/refmans/clustlearn/html/kmeans.html","timestamp":"2024-11-04T11:00:00Z","content_type":"text/html","content_length":"7005","record_id":"<urn:uuid:b14af243-816b-4cda-8d99-5b413fbf661a>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00277.warc.gz"} |
The Aggregate analysis transforms a time series so that the values are summed either periodically (each year for instance) or continuously, starting from a specified date. Aggregate is useful when
you are working with 'flow' series.
In this analysis, you can define the following settings to determine how the calculation is done:
The window of the calculation is set by the period you choose. If you select 'All' a continuous sum will be performed and you can set a date at which the sum should start.
Selecting this option will express the result as a percentage, so it will calculate the sum and divide by 100.
If you’ve chosen a period other than 'All', you can select 'rolling' to perform the sum on a rolling basis. The window of the rolling sum is the same as the period you have chosen.
In this example, the aggregate analysis is used to calculate an annual rolling sum of the German current account. In other words, the sum is performed on a rolling window of 1 year.
How to calculate fiscal year rolling aggregate on a series from country which doesn't report it like calendar year?
How do I calculate a rolling sum?
There are two main possibilities to calculate a rolling sum:
Set the 'Period' to the desired rolling length, and to not forget to tick the setting 'Rolling'.
You can also use the formula
sum(series, window)
sum(usflof8344, YearsLength(2))
This will calculate a 2-years rolling sum on 'usflof8344'.
For more about formulas and how formula language in Macrobond works see Formula analysis.
How to keep series as is and start aggregating it from certain point in time?
Use formula (on Series list or in Formula analysis):
AggregateSum(CutStart(series, Date(YYYY, MM, DD)))
join(older_series, newer_series, Start(newer_series))
CutStart() will create series from fragment you wish to aggregate. AggregateSum() will cumulate values. Then you need to connect that cumulated fragment with regular series using join(). As in below
join(sek, AggregateSum(CutStart(sek, Date(2024, 4, 2))), Start(AggregateSum(CutStart(sek, Date(2024, 4, 2))))) | {"url":"https://help.macrobond.com/tutorials-training/3-analyzing-data/analyses/transformations/aggregate/","timestamp":"2024-11-14T12:16:10Z","content_type":"text/html","content_length":"58716","record_id":"<urn:uuid:f4dd47b3-c914-4cc7-9027-b6c7ab587495>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00776.warc.gz"} |
A note on cyclotomic Euler systems and the double complex method
Let double-struk f sign be a finite real abelian extension of ℚ. Let M be an odd positive integer. For every squarefree positive integer r the prime factors of which are congruent to 1 modulo M and
split completely in double-struk f sign, the corresponding Kolyvagin class κ[r] ∈ double-struk f sign^× /double-struk f sign ^×M satisfies a remarkable and crucial recursion which for each prime
number ℓ dividing r determines the order of vanishing of κ [r] at each place of double-struk f sign above ℓ in terms of κ[r/ℓ]. In this note we give the recursion a new and universal interpretation
with the help of the double complex method introduced by Anderson and further developed by Das and Ouyang. Namely, we show that the recursion satisfied by Kolyvagin classes is the specialization of a
universal recursion independent of double-struk f sign satisfied by universal Kolyvagin classes in the group cohomology of the universal ordinary distribution à la Kubert tensored with ℤ/Mℤ. Further,
we show by a method involving a variant of the diagonal shift operation introduced by Das that certain group cohomology classes belonging (up to sign) to a basis previously constructed by Ouyang also
satisfy the universal recursion.
Dive into the research topics of 'A note on cyclotomic Euler systems and the double complex method'. Together they form a unique fingerprint. | {"url":"https://experts.umn.edu/en/publications/a-note-on-cyclotomic-euler-systems-and-the-double-complex-method","timestamp":"2024-11-06T09:41:02Z","content_type":"text/html","content_length":"51241","record_id":"<urn:uuid:e229eedf-ccb8-4c15-85ef-fcc785c44f1b>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00397.warc.gz"} |
RISC Activity Database
author = {Clemens G. Raab},
title = {{Using Groebner bases for finding the logarithmic part of the integral of transcendental functions}},
language = {english},
abstract = {We show that Czichowski's algorithm for computing the logarithmic part of the integral of a rational function can be carried over to a rather general class of transcendental
functions. Also an asymptotic bound on the number of field operations needed is given.},
journal = {Journal of Symbolic Computation},
volume = {47},
number = {10},
pages = {1290--1296},
isbn_issn = {ISSN 0747-7171},
year = {2012},
refereed = {yes},
keywords = {Symbolic Integration, Elementary Integral, Differential Algebra, Groebner Basis, Special Functions},
length = {7} | {"url":"https://www3.risc.jku.at/publications/show-bib.php?activity_id=4347","timestamp":"2024-11-12T09:05:45Z","content_type":"text/html","content_length":"3382","record_id":"<urn:uuid:62c4c5c1-30a9-491d-9f47-31795d0c5fc0>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00651.warc.gz"} |
mikeash.com: Friday Q&A 2013-05-17: Let's Build stringWithFormat:
Friday Q&A 2013-05-17: Let's Build stringWithFormat:
Our long effort to rebuild Cocoa piece by piece continues. For today, reader Nate Heagy has suggested building NSString's stringWithFormat: method.
String Formatting
It's hard to get very far in Cocoa without knowing about format strings, but just in case, here's a recap.
stringWithFormat:, as well as other calls like NSLog, take strings that can use special format specifiers of the form %x. The % indicates that it's a format specifier, which reads an additional
argument and adds it to the string. The character after it specifies what kind of data to display. For example:
[NSString stringWithFormat: @"Hello, %@: %d %f", @"world", 42, 1.0]
This produces the string:
This is useful for all sorts of things, from creating user-visible text, to making dictionary keys, to printing debug logs.
Variable Arguments
This method takes variable arguments, which is an odd corner of C. For more extensive coverage of how to write such methods, see my article on vararg macros and functions. Here's a quick recap.
You declare the function or method to take variable arguments by putting ... at the end of the parameter list. For a method, this ends up being slightly odd syntax:
+ (id)stringWithFormat: (NSString *)format, ...;
That , ... thing at the end is actually legal Objective-C.
Once in the method, declare a variable of type va_list to represent the variable argument list. The va_start and va_args macros initialize and clean it up. The va_arg macro will extract one argument
from the list and return it.
As usual, I have posted the code on GitHub. You can view the repository here:
This code supports an extremely limited subset of the full NSString formatting functionality. NSString supports a huge number of specifiers, as well as options such as field width, precision, and
out-of-order arguments. My reimplementation sticks to a basic set that's enough to illustrate what's going on. In particular, it supports:
• %d - int
• %ld - long
• %lld - long long
• %u, %lu, and %llu, for the unsigned variants of the above.
• %f - float
• %s - C strings
• %@ - Objective-C objects
• %% - Output a single % character.
Furthermore, no options are supported.
For my reimplementation, I wrote a function called MAStringWithFormat that does the same thing as [NSString stringWithFormat:]. However, I wrapped the meat of the implementation in a class to
organize the various bits of state needed. That function just makes a va_list for the arguments, instantiates a formatter, and asks it to do the work:
NSString *MAStringWithFormat(NSString *format, ...)
va_list arguments;
va_start(arguments, format);
MAStringFormatter *formatter = [[MAStringFormatter alloc] init];
NSString *result = [formatter format: format arguments: arguments];
return result;
The MAStringFormatter class essentially carries out two tasks in parallel. First, it reads through the format string character-by-character, and secondly, it writes the resulting string. Accordingly,
it has two groups of instance variables. The first group deals with reading through the format string:
CFStringInlineBuffer _formatBuffer;
NSUInteger _formatLength;
NSUInteger _cursor;
CFStringInlineBuffer is a little-known API in CFString that allows for efficiently iterating through the individual characters of a string. Making a function or method call for each character is
slow, so CFStringInlineBuffer allows fetching them in bulk for greater efficiency. The length of the format string is stored to avoid running off the end, and the current position within the format
string is stored in _cursor.
The second group deals with collecting the output of the formatting operation. It consists of a buffer of characters, the current location within that buffer, and its total size:
unichar *_outputBuffer;
NSUInteger _outputBufferCursor;
NSUInteger _outputBufferLength;
This could be implemented using an NSMutableData or NSMutableString, but this is much more efficient. While this code isn't intended to be particularly fast in general, I just couldn't stand the
thought of making each character run through a call to a string object.
MAStringFormatter has a read method which fetches the next character from _formatBuffer, and returns -1 once it reaches the end of the string. There isn't a whole lot to this, just an if check and a
call to CFStringGetCharacterFromInlineBuffer:
- (int)read
if(_cursor < _formatLength)
return CFStringGetCharacterFromInlineBuffer(&_formatBuffer, _cursor++);
return -1;
Writing is a little more complex, because the size of the output string isn't known ahead of time. First, there's a doubleOutputBuffer method that increases the size of the output buffer. If the
buffer is completely empty, it allocates it to hold 64 characters. If it's already allocated, it doubles the size:
- (void)doubleOutputBuffer
if(_outputBufferLength == 0)
_outputBufferLength = 64;
_outputBufferLength *= 2;
Once the new buffer length is computed, a simple call to realloc allocates or reallocates the buffer:
_outputBuffer = realloc(_outputBuffer, _outputBufferLength * sizeof(*_outputBuffer));
Next, there's a write: method, which takes a single unichar and appends it to the buffer. If the write cursor is already at the end of the buffer, it first increases the size of the buffer:
- (void)write: (unichar)c
if(_outputBufferCursor >= _outputBufferLength)
[self doubleOutputBuffer];
Once sufficient storage is assured, it places c at the current cursor position, and advances the cursor:
_outputBuffer[_outputBufferCursor] = c;
The format:arguments: method is the entry point to where the real work gets done. The first thing it does is fill out the format string instance variables using the format argument:
- (NSString *)format: (NSString *)format arguments: (va_list)arguments
_formatLength = [format length];
CFStringInitInlineBuffer((__bridge CFStringRef)format, &_formatBuffer, CFRangeMake(0, _formatLength));
_cursor = 0;
It also initializes the output variables. This isn't necessary, strictly speaking, but leaves open the possibility of reusing a single formatter object:
_outputBuffer = NULL;
_outputBufferCursor = 0;
_outputBufferLength = 0;
After that, it loops through the format string until it runs off the end:
int c;
while((c = [self read]) >= 0)
All format specifiers begin with the '%' character. If c is not a '%', then just write the character directly to the output:
if(c != '%')
[self write: c];
This comparison uses the character literal '%' despite the fact that read deals in unichar. This works because the first 128 Unicode code points map directly to the 128 ASCII characters. When a
unichar contains a %, it contains the same value as the ASCII '%', and the same is true for any other ASCII character. This is terribly convenient when working with ASCII data in NSStrings.
If c is a '%' character, then there's a format specifier to come. What happens at this point depends on what the next character is:
int next = [self read];
If the format specifier is a 'd', then it reads an int from the arguments and passes it to the writeLongLong: method, which handles the actual work of formatting the value into the output. All signed
integers pass through that method. Since long long is the largest signed data type handled, a single method that prints those will work for all signed types:
if(next == 'd')
int value = va_arg(arguments, int);
[self writeLongLong: value];
If the format specifier is 'u', then it does the same thing as above, but with unsigned, and calling through to the writeUnsignedLongLong: method:
else if(next == 'u')
unsigned value = va_arg(arguments, unsigned);
[self writeUnsignedLongLong: value];
Note that int and unsigned are the smallest integer types handled here. There is no code to handle char or short. This is because of C promotion rules for functions that take variable arguments. When
passed as a variable arguments, values of type char or short are promoted to int, and likewise the unsigned variants are promoted to unsigned int. This means that the code for int handles the smaller
data types as well, without any additional work.
If the next character is 'l', then we need to keep reading to figure out what to do:
else if(next == 'l')
next = [self read];
If the character following the 'l' is 'd', then the argument is a long. Follow the same basic procedure as before:
if(next == 'd')
long value = va_arg(arguments, long);
[self writeLongLong: value];
Likewise, if the next character is 'u', it's an unsigned long:
else if(next == 'u')
unsigned long value = va_arg(arguments, unsigned long);
[self writeUnsignedLongLong: value];
If the next character is 'l' again, then we need to read one character further
else if(next == 'l')
next = [self read];
Here, 'd' indicates a long long, and 'u' indicates an unsigned long long. These are handle in the same fashion as before:
if(next == 'd')
long long value = va_arg(arguments, long long);
[self writeLongLong: value];
else if(next == 'u')
unsigned long long value = va_arg(arguments, unsigned long long);
[self writeUnsignedLongLong: value];
That's it for the deep sequence of 'l' variants. Next comes a check for 'f'. In that case, the argument is a `double', and gets passed off to a method built to handle that:
else if(next == 'f')
double value = va_arg(arguments, double);
[self writeDouble: value];
Once again, promotion rules simplify things a bit. When a float is passed as a variable argument, it's promoted to a double, so no extra code is needed to handle float.
If the format specifier is an 's', then the argument is a C string:
else if(next == 's')
const char *value = va_arg(arguments, const char *);
This is simple enough not to need a helper method. It iterates through the string until it reaches the terminating 0, writing each character as it goes. This assumes the string contains only ASCII:
[self write: *value++];
If the format specifier is a '@', then the argument is an Objective-C object:
else if(next == '@')
id value = va_arg(arguments, id);
To find out what to output, ask the value for its description:
NSString *description = [value description];
The length of the description is also handy:
NSUInteger length = [description length];
Now, copy the contents of description into the output buffer. I decided to get a bit fancy here. A simple loop could suffice, perhaps using CFStringInlineBuffer for speed, but I wanted something
nicer. An NSString can put its contents into an arbitrary buffer, so why not ask description to put its contents directly into the output buffer? To do that, the output buffer must first be made
large enough to contain length characters:
while(length > _outputBufferLength - _outputBufferCursor)
[self doubleOutputBuffer];
Doing this in a while loop is mildly inefficient if description is larger than the buffer is already. However, that's an uncommon case, and the code is nicer by being able to share
doubleOutputBuffer, so I decided to use this approach.
Now that the output buffer is sufficiently large, use getCharacters:range: to dump the contents of description into it, putting it at the location of the output cursor:
[description getCharacters: _outputBuffer + _outputBufferCursor range: NSMakeRange(0, length)];
Finally, move the cursor past the newly written data:
_outputBufferCursor += length;
We're nearly to the end. If the character following the '%' is another '%', that's the siganl to write a literal '%' character:
else if(next == '%')
[self write: '%'];
That's the last case handled by this miniature implementation. Once the loop terminates, the resulting unichars are located in _outputBuffer, with _outputBufferCursor indicating the number of
unichars in the buffer. Create an NSString from it and return the new string:
NSString *output = [[NSString alloc] initWithCharactersNoCopy: _outputBuffer length: _outputBufferCursor freeWhenDone: YES];
return output;
Using the NoCopy variant makes this potentially more efficient, and removes the need to manually free the buffer.
That's the basic shell of the formatting code. To complete it, we need the code to print signed and unsigned long longs, and code to print doubles.
unsigned long long
Let's start with the most fundamental helper method, writeUnsignedLongLong:. The others ultimately rely on this one for much of their work.
The algorithm is simple: divide by successive powers of ten, produing a single digit each time. Convert the digit to a unichar and write it.
We'll store the power of ten in a variable called cursor and start it at 1:
- (void)writeUnsignedLongLong: (unsigned long long)value
unsigned long long cursor = 1;
However, what we really want is the power of ten with as many digits as the input number. For example, for 42, we want 10. For 123456, we want 100000. To obtain this, we just keep multiplying cursor
by ten until it has the same number of digits as value, which is easily tested by seeing if value is less than ten times larger than cursor:
while(value / cursor >= 10)
cursor *= 10;
Now we just loop, dividing cursor by ten each time, until we run out of cursor:
The current digit is obtained by dividing value by cursor:
int digit = value / cursor;
To compute the unichar that corresponds with digit, just add the literal '0' character. ASCII (and therefore Unicode) lays out digits sequentially starting with '0', making this easy:
[self write: '0' + digit];
With the digit written, we remove it from value, then move cursor down:
value -= digit * cursor;
cursor /= 10;
And just like that, the value flows into the output. This code even correctly handles zero, due to ensuring that cursor is always at least 1.
long long
The writeLongLong: method is simple. If the number is less than zero, write a '-' and negate the number. For positive numbers, do nothing special. Pass the final non-negative number to
- (void)writeLongLong: (long long)value
unsigned long long unsignedValue = value;
if(value < 0)
[self write: '-'];
unsignedValue = -unsignedValue;
[self writeUnsignedLongLong: unsignedValue];
There's an odd corner case in here. Due to the nature of the two's complement representation of signed integers, the magnitude of the smallest representable long long is one greater than the
magnitude of the largest representable long long on systems we're likely to encounter.
A long long on a typical system can hold numbers all the way down to -9223372036854775808, but only up to 9223372036854775807. This means you can't negate the smallest possible negative number and
get a positive number, because the data type can't hold the appropriate positive number. If you try to negate -9223372036854775808, you get an overflow and undefined behavior, although the result is
usually just -9223372036854775808 again.
However, negation is well defined on all unsigned values, and it has the same bitwise result as negation on the bitwise-equivalent signed values. In other words, -signedLongLong produces the same
bits as -(unsigned long long)signedLongLong. It also works on the bits that make up -9223372036854775808, and produces 9223372036854775808. By moving value into unsignedValue and then negating that,
the above code works around the problem of undefined behavior when negating the smallest representable long long.
Now it's time for the really fun one. Due to the nature of floating-point arithmetic, figuring out how to properly and accurately print the value of a double was pretty tough. I did some research,
even dove into an open-source implementation of printf to see how they did it, but it was so crazy and incomprehensible that I didn't get too far. I finally settled on a technique which works fairly
well, and I think is as accurate as the data type allows, although the output tends to have more digits than it strictly needs.
The first step in solving the problem is to break it into two pieces. I split the double into the integer part and the fractional part, then deal with each one separately. Print each part in base 10,
separate the two with a dot, and done.
The trick, then, is how to print the integer and fractional parts in base 10. I didn't want to use the same technique of successive division that I used for unsigned long long, because I was
concerned that it would lose accuracy. There are integers that can be represented in a double, but where the result of dividing the integer by ten can't be exactly represented in a double. Similarly,
I was afraid that the equivalent successive multiplication by ten for the fractional part would lose precision.
However, dividing or multiplying a double by two to move is always safe, unless it pushes it beyond the limits of what can be represented. If you only do this to push it closer to 1.0, then it will
never lose precision. Furthermore, it's possible to chop off the fractional part of a double without losing precision in the integer part, and vice versa. Put together, these operations allow
extracting information from a double bit by bit, which is enough to compute an integer representation of its integer and fractional parts. With those in hand, the existing writeUnsignedLongLong:
method can be used to print the digits.
With this in mind, I set off. The first step is to check for negative values. If value is negative, write a '-', and negate it:
- (void)writeDouble: (double)value
if(value < 0.0)
[self write: '-'];
value = -value;
Unlike the long long case, there are no double values less than 0.0 that can't be safely and correctly negated, so no shenanigans are needed here.
Next, check for infinity and NaN, and short circuit the whole attempt for them:
if(isinf(value) || isnan(value))
const char *str = isinf(value) ? "INFINITY" : "NaN";
[self write: *str++];
If the number is an actual number, extract the integer and fractional parts.
double intpart = trunc(value);
double fracpart = value - intpart;
With those in hand, call out to helper methods to write those two parts, separated by a dot:
[self writeDoubleIntPart: intpart];
[self write: '.'];
[self writeDoubleFracPart: fracpart];
Integer Part
Writing the integer part is the simpler of the two, conceptually. The strategy is to shift the double value one bit to the right until the value becomes zero. Each bit that's extracted is added to an
unsigned long long accumulator. Once the double becomes zero, the accumulator contains its integer value.
The one tricky part is how to handle the case where the double contains a value that's larger than an unsigned long long can contain. To handle this, whenever the value of the current bit extracted
from the double threatens to overflow, the accumulator is divided by ten to shift it rightwards and allow more room. The total number of shifts is recorded, and the appropriate number of extra zeroes
are printed at the end of the number. Dividing the accumulator by ten loses precision, but the 64 bits of an unsigned long long exceeds the 53 bits of precision in a double, so the lost precision
should not actually result in incorrect output. At the least, while the output may not precisely match the integer value stored in the double, it will be closer to that value than to any other
representable double value, which I'm calling close enough.
In order to know when the accumulator threatens to overflow, the code needs to know the largest power of ten that can be represented in an unsigned long long. This method computes it by just
computing successive powers of ten until it gets close to ULLONG_MAX:
- (unsigned long long)ullongMaxPowerOf10
unsigned long long result = 1;
while(ULLONG_MAX / result >= 10)
result *= 10;
return result;
The writeDoubleIntPart: method starts off by initializing a total variable to zero:
- (void)writeDoubleIntPart: (double)intpart
unsigned long long total = 0;
This is the accumulator that will hold the total computed value so far. It also keeps track of the value of the current bit:
unsigned long long currentBit = 1;
This is multiplied by two each time a bit is extracted from the double, and represents the value of that bit.
The maximum value that can be stored in total before overflow threatens is cached:
unsigned long long maxValue = [self ullongMaxPowerOf10] / 10;
This is one digit less than the maximum representable power of ten, in order to make sure that it can never overflow the accumulator. There is a surplus of 11 bits of precision in the accumulator, so
losing one digit doesn't hurt too much.
The number of times that total and currentBit have been shifted to the right is recorded so that the appropriate number of trailing zeroes can be output later:
unsigned surplusZeroes = 0;
Setup is complete, now it's time to loop until intpart is exhausted:
A bit is extracted from intpart by dividing it by two:
Because intpart contains an integer, dividing it by two produces a number with a fractional part that is either .0 or .5. The .5 case represents a one bit that needs to be added to total. The
presence of .5 is checked by using the fmod function, which computes the remainder when dividing by a number. Using fmod with 1.0 as the second argument just produces the fractional part of the
If the bit is set, then currentBit is added to total, and the .5 is sliced off of intpart using the trunc function:
total += currentBit;
intpart = trunc(intpart);
Next, currentBit is multiplied by two so that it holds the right value for the next bit to be extracted:
If currentBit exceeds maxValue, then both currentBit and total get divided by ten, and surplusZeroes is incremented. Both are rounded when dividing by adding 5 to them first, to aid in preserving as
much precision as possible:
if(currentBit > maxValue)
total = (total + 5) / 10;
currentBit = (currentBit + 5) / 10;
Once intpart is exhausted, total contains an approximation of its original value, and surplusZeroes indicates how many times it got shifted over. First, it prints total:
[self writeUnsignedLongLong: total];
Finally, it prints the appropriate number of trailing zeroes:
for(unsigned i = 0; i < surplusZeroes; i++)
[self write: '0'];
Fractional Part
The basic idea for printing the fractional part is similar to printing the integer part. The difference is that the accumulator can't directly represent the fractional value, because unsigned long
long doesn't do fractions. Instead, it holds the fractional value, scaled up by some large power of ten. For example, 100 might represent 1.0, in which case the value of the first bit in the
fractional part of a double is 50, the second bit is 25, and so forth. The actual numbers used contain a lot more zeroes at the end.
The accumulator for the integer part can overflow, while the accumulator for the fractional part can underflow. If the double contains an extremely small value, the accumulator will end up containing
zero, which is no good. A similar strategy is used to deal with this problem, but in the opposite direction: whenever the accumulator and current bit become too small, they are multiplied by ten, and
an extra leading zero is output.
The method starts off with its accumulator initialized to zero:
- (void)writeDoubleFracPart: (double)fracpart
unsigned long long total = 0;
The value of the current bit is started at the largest power of ten that will fit into an unsigned long long. This represents 1.0, and will be divided by 2 right away to properly represent 0.5 for
the first bit extracted from the double:
unsigned long long currentBit = [self ullongMaxPowerOf10];
The threshold for when the numbers become too small is the maximum representable power of ten, divided by ten. When this value is reached, there's a conceptual leading zero, and it's time to shift
everything over:
unsigned long long shiftThreshold = [self ullongMaxPowerOf10] / 10;
Now it's time for the loop. Keep extracting bits from fracpart until there's nothing left:
fracpart is shifted to the left by one bit, while currentBit is simultaneously shifted to the right:
currentBit /= 2;
fracpart *= 2;
The integer part of the resulting number will be either 1 or 0. If it's 1, the corresponding bit is 1, so add currentBit to total, and chop the 1 off fracpart:
if(fracpart >= 1.0)
total += currentBit;
fracpart -= 1.0;
If both the accumulator and currentBit are below shiftThreshold, it's time to shift everything over, and write a leading zero. Note that the number of shifts doesn't need to be tracked like it did in
the previous method, because the leading zeroes can be written out immediately:
if(currentBit <= shiftThreshold && total <= shiftThreshold)
[self write: '0'];
currentBit *= 10;
total *= 10;
Once the loop exits, there's one more task to be done. total now contains an integer representation of the decimal representation of the fractional part that was passed into the method (whew!), but
with potentially a large number of redundant trailing zeroes. For example, if fracpart contained 0.5, then total now contains 5000000000000000000, but those trailing zeroes shouldn't be printed in
the output. They're removed by just dividing total by ten repeatedly to get rid of trailing zeroes:
while(total != 0 && total % 10 == 0)
total /= 10;
Once that's done, total is ready to print, so it's passed to writeUnsignedLongLong:
[self writeUnsignedLongLong: total];
That's the end of the adventure of printing a double.
stringWithFormat: is an extremely useful method that is, at its heart, a straightforward function that takes variable arguments. There are a ton of subtleties in how to output all of the various data
formats, such as the adventure in printing a double above. There are further complications in supporting all of the various options available in format strings, which the above code doesn't even
address. However, it's ultimately a big loop that looks for '%' format specifiers, and uses va_arg to extract the arguments passed in by the caller. Although stringWithFormat: is considerably more
complex, you now have a basic idea of how it's put together.
That's it for today. Come back next time for more bitwise adventures. Friday Q&A is driven by reader suggesions, so until the next time, please keep sending in your ideas for topics.
Did you enjoy this article? I'm selling whole books full of them! Volumes II and III are now out! They're available as ePub, PDF, print, and on iBooks and Kindle.
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Great article.
All of the various implementations of printing out different types is fun, but if you actually wanted to write your own stringWithFormat:, wouldn't you just handle instances of "%@" and then wrap
: Unfortunately, no. A real
implementation has to handle, among other things, positional parameters and a wide array of options in each format specifier. There's no support in C for modifying the contents of a
on the fly, which would be required for correctly supporting positional parameters when wrapping the existing
-style functions. The options are 1) using something other than % as your custom specifier character, or 2) reimplementing
from the ground up. Apple did the latter in Core Foundation; look at the function
for an incredibly complicated example :).
Bug report: writeDouble will print negative infinity as "INFINITY", even though it's distinct from positive infinity.
Thank you for pointing that out, I've fixed it by moving the negative check to the top, and updated the article.
If you're interested in accurate printing/decoding of floating point numbers you should check out Bruce Dawson's blog: Random Ascii -
It's worthing noting that almost all vendor implementations of printf- functionality deviate from the ISO spec in some minor ways, which can create issues with cross-platform code (same inputs !=
same outputs).
Why not use David Gay's dtoa() function? It is a well tested standard conversion function. Any homegrown replacement is likely to reinvent many long since fixed bugs, or at least will have inferior
for examples of conversion bugs. The articles on this site also give some suggestions on tricky values to test on.
Technically it is easy to print floats -- I published some straightforward code here:
This code can easily be extended to doubles. However making the code efficient without breaking it is very hard. Adding rounding is left as an exercise for the reader.
Why not use a pre-made function? Because this is Let's Build, not Let's Find Some Existing Code and Call It. Otherwise the post could be reduced down to a single sentence that says, "Just call
I can't argue with the "Let's Build" ethos.
Just beware of the significant challenges in trying to make printing of doubles and floats both efficient and correctly rounded. For my "Let's Build" code for printing floats I decided to ignore both
of these, which allowed for much simpler code (and better elucidation) but clearly made my code less generally useful.
Yep, I totally agree, and I'm definitely grateful for the additional information.
This code is definitely not intended for real-world use. After all, Apple already provides a better built-in implementation, so just use theirs....
Very interesting. I am a new reader of your blog and to iOS. I am still trying to build up a appreciation of when to use what approach, hence the following questions..
I am sure that you know that to get the lowest bit you can do intpart & 1 and then shift one place use intpart >>= 1 Since you are "talking" about bits this would be more explicit, was there a reason
you did not do this ?
With the writeUnsignedLongLong why did you not just divide by 10, put the results in a temp buffer and place the reverse of the temp buffer in the output?
thanks for your help.
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Spam and off-topic posts will be deleted without notice. Culprits may be publicly humiliated at my sole discretion.
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Euler's rotation theorem
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Euler's rotation theorem
In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation
about some axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation. Therefore the set of rotations has a group structure, known as a rotation
The theorem is named after Leonhard Euler, who proved it in 1775 by means of spherical geometry. The axis of rotation is known as an Euler axis, typically represented by a unit vector ê. Its product
by the rotation angle is known as an axis-angle. The extension of the theorem to kinematics yields the concept of instant axis of rotation, a line of fixed points.
In linear algebra terms, the theorem states that, in 3D space, any two Cartesian coordinate systems with a common origin are related by a rotation about some fixed axis. This also means that the
product of two rotation matrices is again a rotation matrix and that for a non-identity rotation matrix one eigenvalue is 1 and the other two are both complex, or both equal to −1. The eigenvector
corresponding to this eigenvalue is the axis of rotation connecting the two systems.
Theorema. Quomodocunque sphaera circa centrum suum conuertatur, semper assignari potest diameter, cuius directio in situ translato conueniat cum situ initiali.**
When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position.
Euler's original proof was made using spherical geometry and therefore whenever he speaks about triangles they must be understood as spherical triangles.
To arrive at a proof, Euler analyses what the situation would look like if the theorem were true. To that end, suppose the yellow line in Figure 1 goes through the center of the sphere and is the
axis of rotation we are looking for, and point O is one of the two intersection points of that axis with the sphere. Then he considers an arbitrary great circle that does not contain O (the blue
circle), and its image after rotation (the red circle), which is another great circle not containing O. He labels a point on their intersection as point A. (If the circles coincide, then A can be
taken as any point on either; otherwise A is one of the two points of intersection.)
Now A is on the initial circle (the blue circle), so its image will be on the transported circle (red). He labels that image as point a. Since A is also on the transported circle (red), it is the
image of another point that was on the initial circle (blue) and he labels that preimage as α (see Figure 2). Then he considers the two arcs joining α and a to A. These arcs have the same length
because arc αA is mapped onto arc Aa. Also, since O is a fixed point, triangle αOA is mapped onto triangle AOa, so these triangles are isosceles, and arc AO bisects angle ∠αAa.
Construction of the best candidate point
Let us construct a point that could be invariant using the previous considerations. We start with the blue great circle and its image under the transformation, which is the red great circle as in the
Figure 1. Let point A be a point of intersection of those circles. If A’s image under the transformation is the same point then A is a fixed point of the transformation, and since the center is also
a fixed point, the diameter of the sphere containing A is the axis of rotation and the theorem is proved.
Otherwise we label A’s image as a and its preimage as α, and connect these two points to A with arcs αA and Aa. These arcs have the same length. Construct the great circle that bisects ∠αAa and
locate point O on that great circle so that arcs AO and aO have the same length, and call the region of the sphere containing O and bounded by the blue and red great circles the interior of ∠αAa.
(That is, the yellow region in Figure 3.) Then since αA = Aa and O is on the bisector of ∠αAa, we also have αO = aO.
Proof of its invariance under the transformation
Now let us suppose that O′ is the image of O. Then we know ∠αAO = ∠AaO′ and orientation is preserved,^[1] so O′ must be interior to ∠αAa. Now AO is transformed to aO′, so AO = aO′. Since AO is also
the same length as aO, ∠AaO = ∠aAO. But ∠aAO = ∠AaO′, so ∠AaO = ∠AaO′ and therefore O′ is the same point as O. In other words, O is a fixed point of the transformation, and since the center is also a
fixed point, the diameter of the sphere containing O is the axis of rotation.
Final notes about the construction
Euler also points out that O can be found by intersecting the perpendicular bisector of Aa with the angle bisector of ∠αAO, a construction that might be easier in practice. He also proposed the
intersection of two planes:
• the symmetry plane of the angle ∠αAa (which passes through the center C of the sphere), and
• the symmetry plane of the arc Aa (which also passes through C).
Proposition. These two planes intersect in a diameter. This diameter is the one we are looking for.
Proof. Let us callOeither of the endpoints (there are two) of this diameter over the sphere surface. SinceαAis mapped onAaand the triangles have the same angles, it follows that the triangleOαAis
transported onto the triangleOAa. Therefore the pointOhas to remain fixed under the movement.
Corollaries. This also shows that the rotation of the sphere can be seen as two consecutive reflections about the two planes described above. Points in a mirror plane are invariant under reflection,
and hence the points on their intersection (a line: the axis of rotation) are invariant under both the reflections, and hence under the rotation.
Another simple way to find the rotation axis is by considering the plane on which the points α, A, a lie. The rotation axis is obviously orthogonal to this plane, and passes through the center C of
the sphere.
Given that for a rigid body any movement that leaves an axis invariant is a rotation, this also proves that any arbitrary composition of rotations is equivalent to a single rotation around a new
A spatial rotation is a linear map in one-to-one correspondence with a 3 × 3 rotation matrix R that transforms a coordinate vector x into X, that is Rx = X. Therefore, another version of Euler's
theorem is that for every rotation R, there is a nonzero vector n for which Rn = n; this is exactly the claim that n is an eigenvector of R associated with the eigenvalue 1. Hence it suffices to
prove that 1 is an eigenvalue of R; the rotation axis of R will be the line μn, where n is the eigenvector with eigenvalue 1.
A rotation matrix has the fundamental property that its inverse is its transpose, that is
where I is the 3 × 3 identity matrix and superscript T indicates the transposed matrix.
Compute the determinant of this relation to find that a rotation matrix has determinant ±1. In particular,
A rotation matrix with determinant +1 is a proper rotation, and one with a negative determinant −1 is an improper rotation, that is a reflection combined with a proper rotation.
It will now be shown that a rotation matrix R has at least one invariant vector n, i.e., Rn = n. Because this requires that (R − I)n = 0, we see that the vector n must be an eigenvector of the matrix
R with eigenvalue λ = 1. Thus, this is equivalent to showing that det(R − I) = 0.
for any 3 × 3 matrix A and
(since det(R) = 1) to compute
This shows that λ = 1 is a root (solution) of the characteristic equation, that is,
In other words, the matrix R − I is singular and has a non-zero kernel, that is, there is at least one non-zero vector, say n, for which
The line μn for real μ is invariant under R, i.e., μn is a rotation axis. This proves Euler's theorem.
Equivalence of an orthogonal matrix to a rotation matrix
Two matrices (representing linear maps) are said to be equivalent if there is a change of basis that makes one equal to the other. A proper orthogonal matrix is always equivalent (in this sense) to
either the following matrix or to its vertical reflection:
Then, any orthogonal matrix is either a rotation or an improper rotation. A general orthogonal matrix has only one real eigenvalue, either +1 or −1. When it is +1 the matrix is a rotation. When −1,
the matrix is an improper rotation.
If R has more than one invariant vector then φ = 0 and R = I. Any vector is an invariant vector of I.
Excursion into matrix theory
In order to prove the previous equation some facts from matrix theory must be recalled.
An m × m matrix A has m orthogonal eigenvectors if and only if A is normal, that is, if A†A = AA†.^[2] This result is equivalent to stating that normal matrices can be brought to diagonal form by a
unitary similarity transformation:
and U is unitary, that is,
The eigenvalues α1, ..., αm are roots of the characteristic equation. If the matrix A happens to be unitary (and note that unitary matrices are normal), then
and it follows that the eigenvalues of a unitary matrix are on the unit circle in the complex plane:
Also an orthogonal (real unitary) matrix has eigenvalues on the unit circle in the complex plane. Moreover, since its characteristic equation (an mth order polynomial in λ) has real coefficients, it
follows that its roots appear in complex conjugate pairs, that is, if α is a root then so is α∗. There are 3 roots, thus at least one of them must be purely real (+1 or −1).
After recollection of these general facts from matrix theory, we return to the rotation matrix R. It follows from its realness and orthogonality that we can find a U such that:
If a matrix U can be found that gives the above form, and there is only one purely real component and it is −1, then we define R to be an improper rotation. Let us only consider the case, then, of
matrices R that are proper rotations (the third eigenvalue is just 1). The third column of the 3 × 3 matrix U will then be equal to the invariant vector n. Writing u1 and u2 for the first two columns
of U, this equation gives
If u1 has eigenvalue 1, then φ = 0 and u2 has also eigenvalue 1, which implies that in that case R = E.
Finally, the matrix equation is transformed by means of a unitary matrix,
The columns of U′ are orthonormal. The third column is still n, the other two columns are perpendicular to n. We can now see how our definition of improper rotation corresponds with the geometric
interpretation: an improper rotation is a rotation around an axis (here, the axis corresponding to the third coordinate) and a reflection on a plane perpendicular to that axis. If we only restrict
ourselves to matrices with determinant 1, we can thus see that they must be proper rotations. This result implies that any orthogonal matrix R corresponding to a proper rotation is equivalent to a
rotation over an angle φ around an axis n.
The trace (sum of diagonal elements) of the real rotation matrix given above is 1 + 2 cos φ. Since a trace is invariant under an orthogonal matrix similarity transformation,
it follows that all matrices that are equivalent to R by such orthogonal matrix transformations have the same trace: the trace is a class function. This matrix transformation is clearly an
equivalence relation, that is, all such equivalent matrices form an equivalence class.
In fact, all proper rotation 3 × 3 rotation matrices form a group, usually denoted by SO(3) (the special orthogonal group in 3 dimensions) and all matrices with the same trace form an equivalence
class in this group. All elements of such an equivalence class share their rotation angle, but all rotations are around different axes. If n is an eigenvector of R with eigenvalue 1, then An is also
an eigenvector of ARAT, also with eigenvalue 1. Unless A = I, n and An are different.
Suppose we specify an axis of rotation by a unit vector [x, y, z], and suppose we have an infinitely small rotation of angle Δθ about that vector. Expanding the rotation matrix as an infinite
addition, and taking the first order approach, the rotation matrix ΔR is represented as:
A finite rotation through angle θ about this axis may be seen as a succession of small rotations about the same axis. Approximating Δθ as θ/N where N is a large number, a rotation of θ about the axis
may be represented as:
It can be seen that Euler's theorem essentially states that all rotations may be represented in this form. The product Aθ is the "generator" of the particular rotation, being the vector (x,y,z)
associated with the matrix A. This shows that the rotation matrix and the axis–angle format are related by the exponential function.
One can derive a simple expression for the generator G. One starts with an arbitrary plane (in Euclidean space) defined by a pair of perpendicular unit vectors a and b. In this plane one can choose
an arbitrary vector x with perpendicular y. One then solves for y in terms of x and substituting into an expression for a rotation in a plane yields the rotation matrix R which includes the generator
G = baT − abT.
To include vectors outside the plane in the rotation one needs to modify the above expression for R by including two projection operators that partition the space. This modified rotation matrix can
be rewritten as an exponential function.
Analysis is often easier in terms of these generators, rather than the full rotation matrix. Analysis in terms of the generators is known as the Lie algebra of the rotation group.
It follows from Euler's theorem that the relative orientation of any pair of coordinate systems may be specified by a set of three independent numbers. Sometimes a redundant fourth number is added to
simplify operations with quaternion algebra. Three of these numbers are the direction cosines that orient the eigenvector. The fourth is the angle about the eigenvector that separates the two sets of
coordinates. Such a set of four numbers is called a quaternion.
While the quaternion as described above, does not involve complex numbers, if quaternions are used to describe two successive rotations, they must be combined using the non-commutative quaternion
algebra derived by William Rowan Hamilton through the use of imaginary numbers.
Rotation calculation via quaternions has come to replace the use of direction cosines in aerospace applications through their reduction of the required calculations, and their ability to minimize
round-off errors. Also, in computer graphics the ability to perform spherical interpolation between quaternions with relative ease is of value.
In higher dimensions, any rigid motion that preserves a point in dimension 2n or 2n + 1 is a composition of at most n rotations in orthogonal planes of rotation, though these planes need not be
uniquely determined, and a rigid motion may fix multiple axes.
A rigid motion in three dimensions that does not necessarily fix a point is a "screw motion". This is because a composition of a rotation with a translation perpendicular to the axis is a rotation
about a parallel axis, while composition with a translation parallel to the axis yields a screw motion; see screw axis. This gives rise to screw theory.
• Euler angles
• Euler–Rodrigues parameters
• Rotation formalisms in three dimensions
• Rotation operator (vector space)
• Angular velocity
• Matrix exponential
• Axis–angle representation
Citation Linkopenlibrary.orgOrientation is preserved in the sense that if αA is rotated about A counterclockwise to align with Oa, then Aa must be rotated about a counterclockwise to align with O′a.
Likewise if the rotations are clockwise.
Sep 24, 2019, 4:03 AM
Citation Linkopenlibrary.orgThe dagger symbol † stands for complex conjugation followed by transposition. For real matrices complex conjugation does nothing and daggering a real matrix is the same as
transposing it.
Sep 24, 2019, 4:03 AM
Citation Linkopenlibrary.orgNovi Commentarii academiae scientiarum Petropolitanae 20, 1776, pp. 189–207 (E478)
Sep 24, 2019, 4:03 AM
Citation Link//doi.org/10.4169%2F000298909x47701410.4169/000298909x477014
Sep 24, 2019, 4:03 AM
Citation Linkwww.17centurymaths.comEuler's original text (in Latin) and English translation
Sep 24, 2019, 4:03 AM
Citation Linkdemonstrations.wolfram.comWolfram Demonstrations Project for Euler's Rotation Theorem
Sep 24, 2019, 4:03 AM
Citation Linkdoi.org10.4169/000298909x477014
Sep 24, 2019, 4:03 AM
Citation Linkwww.17centurymaths.comEuler's original text (in Latin) and English translation
Sep 24, 2019, 4:03 AM
Citation Linkdemonstrations.wolfram.comWolfram Demonstrations Project for Euler's Rotation Theorem
Sep 24, 2019, 4:03 AM
Citation Linken.wikipedia.orgThe original version of this page is from Wikipedia, you can edit the page right here on Everipedia.Text is available under the Creative Commons Attribution-ShareAlike
License.Additional terms may apply.See everipedia.org/everipedia-termsfor further details.Images/media credited individually (click the icon for details).
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1 To 20 Times Table Worksheets Free Downloads Multiplication Tables | Multiplication Chart Printable
1 To 20 Times Table Worksheets Free Downloads Multiplication Tables
1 To 20 Times Table Worksheets Free Downloads Multiplication Tables
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Why Do We Use a Multiplication Chart?
A multiplication chart is a layout that demonstrates how to multiply two numbers. It normally includes a top row and a left column. Each row has a number standing for the product of both numbers. You
choose the very first number in the left column, move it down the column, and then select the second number from the top row. The product will certainly be the square where the numbers satisfy.
Multiplication charts are useful for numerous reasons, consisting of helping children discover just how to separate as well as streamline portions. They can additionally aid children find out exactly
how to select an efficient common measure. Since they offer as a continuous suggestion of the trainee’s progression, multiplication charts can also be helpful as workdesk sources. These tools help us
create independent students who recognize the fundamental concepts of multiplication.
Multiplication charts are also useful for aiding trainees memorize their times tables. They help them learn the numbers by decreasing the variety of steps required to finish each operation. One
method for memorizing these tables is to concentrate on a solitary row or column each time, and then relocate onto the following one. Eventually, the whole chart will certainly be committed to
memory. Just like any ability, remembering multiplication tables requires time and also practice.
20×20 Multiplication Chart Blank
Multiplication Grid Chart 20×20 20×20 Multiplication Table
20×20 Multiplication Chart Blank
If you’re searching for 20×20 Multiplication Chart Blank, you’ve concerned the ideal area. Multiplication charts are offered in various formats, including full dimension, half dimension, and also a
selection of cute layouts. Some are vertical, while others include a straight layout. You can likewise discover worksheet printables that consist of multiplication formulas and math realities.
Multiplication charts as well as tables are vital tools for children’s education. These charts are great for usage in homeschool math binders or as class posters.
A 20×20 Multiplication Chart Blank is an useful tool to enhance mathematics truths as well as can aid a youngster discover multiplication quickly. It’s additionally a terrific tool for skip checking
as well as discovering the times tables.
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ISEVEN - Excel docs, syntax and examples
The ISEVEN function in Excel is used to check if a number is even.
number The number you want to check if it is even.
About ISEVEN 🔗
When you need a quick way to determine whether a number is even or not in Excel, turn to the ISEVEN function. This function comes in handy when working with datasets and wanting to categorize numbers
based on their parity – whether they are divisible by 2 without a remainder. It simplifies the process of identifying even numbers within your data, aiding in various analytical and computational
tasks where the number's divisibility by 2 is a key factor.
Examples 🔗
If you want to check if the number 14 is even, use the formula: =ISEVEN(14). This will return TRUE since 14 is an even number.
To verify if 77 is an even number, enter: =ISEVEN(77). This will result in FALSE as 77 is an odd number.
The ISEVEN function only checks if a number is even and returns TRUE or FALSE based on the evaluation. It can be utilized in combination with other functions for conditional formatting, data
validation, or logical calculations within Excel spreadsheets.
Questions 🔗
What does the ISEVEN function return?
The ISEVEN function returns TRUE if the provided number is even, and FALSE if it is odd.
Can I use the ISEVEN function with non-numeric values?
No, the ISEVEN function is designed to work with numeric values only. If you try to use non-numeric values, it will result in an error or return FALSE.
How can I combine the ISEVEN function with other functions in Excel?
You can combine the ISEVEN function with logical functions like IF, AND, or OR to create more complex formulas based on even or odd number conditions. For instance, you can use it within an IF
function to perform specific actions based on whether a number is even or not.
Related functions 🔗
Leave a Comment | {"url":"https://spreadsheetcenter.com/excel-functions/iseven/","timestamp":"2024-11-15T04:35:21Z","content_type":"text/html","content_length":"27725","record_id":"<urn:uuid:6b2f7eee-3f79-4947-bc91-a563184a32be>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00606.warc.gz"} |
Eclipse Predictions and Earth's Rotation
Fred Espenak
When Sir Isaac Newton first published his revolutionary theory of gravitation in the Principia (1687), it laid the ground work for the prediction of planetary motion throughout the solar system.
Edmund Halley played a pivital role in motivating Newton to develop this mathematical description of gravity. In fact, Halley even financed much of the Principia's publication costs.
Halley was quite curious about the orbits of the planets. Using Newton's Principia, Halley calculated orbits for the comets of 1531, 1607, and 1682 and discovered that they must be successive returns
of the same object. He correctly predicted that the comet would return in 1758 and it has been known as Halley's Comet ever since. He also devised a method to determine Earth's distance from the Sun
using rare transits of Venus across the Sun's disk.
Although not as well known, Halley also made important scientific contributions in his studies of eclipses. He is credited with the first eclipse map showing the path of the Moon's shadow across
England during the upcoming total eclipse of 1715. He also rediscovered the Saros cycle of 18 years plus 10 or 11 days (depending on the number of intervening leap years) over which eclipses seem to
repeat. The Saros was used by Chaldeans and Babylonians (and later, the Greeks) for simple lunar eclipse predictions but it was unknown in Halley's day. Using Newton's Theory of the Moon's Motion (or
TMM) and the Saros cycle, Halley made a series of calculations to identify ancient eclipses in the literature. But Halley soon encountered a problem. The eclipse paths he predicted were shifted with
respect to the historical records. Either the Moon was accelerating in its orbit or Earth's rotation rate was slowing down (i.e. - length of the day was increasing). Although both are actually true,
Halley correctly identified the increasing length of the day as the primary culprit. It took another 300 years to understand why.
Earth's Rotation
ocean tides
are casued by the gravitational pull of the Moon and, to a lesser extent, the Sun. But as the tides are attracted to the Moon, the oceans appear to rise and fall while Earth rotates beneath them.
This tidal friction gradually transfers angular momentum from Earth to the Moon. Earth looses energy and slows down while the Moon gains the energy and consequently its orbital period and distance
from Earth increase.
The Moon's average distance from Earth is increasing by 3.8 centmeters per year. Such a precise value is possible due to the Apollo laser reflectors which the astronauts left behind during the lunar
landing missions. Eventually, the Moon's distance will increase so much that it will be to far away to produce total eclipses of the Sun (See: Extinction of Total Solar Eclipses).
In comparison, the secular change in the rotation rate of Earth currently increases the length of day by 2.3 milliseconds per century. While this amount may seem astonishingly small, its accumulated
effects have important consequences. In one century, Earth looses about 40 seconds, while in one millennium, the planet is over one hour "behind schedule." Astronomers use the quantity delta-T to
describe this time difference.
Unfortunately, Earth's rotation is not slowing down at a uniform rate. Non-tidal effects of climate (global warming, polar ice caps and ocean depths) and the dynamics of Earth's molten core make it
impossible to predict the exact value of delta-T in the remote past or distant future.
Good values of delta-T only exist sometime after the invention of the telescope (1610). Careful analysis of telescopic timings of stellar occultations by the Moon permits the direct measurement of
delta-T during this time period. Prior to the 1600's, values of delta-T must rely on historical records of the naked eye observations of eclipses and occultations. Such observations are rare in the
literature and of coarse precision. Ê
Stephenson and collaborators have made a number of important contributions concerning Earth's rotation during the past several millennia. In particular, they have identified hundreds of eclipse and
occultation observations in early European, Middle Eastern and Chinese annals, manuscripts, canons and records. In spite of their relatively low precision, these data represent our only record of the
value of delta-T during the past several millennia.
In particular, Stephenson and Morrison (1984) have fit hundreds of records with simple polynomials to achieve a best fit for describing the value of delta-T from 700 BCE to 1600 CE. An abbreviated
table of their results is as follows:
Year delta-T Longitude
(sec) Shift
1500 BCE 39610 = 11h 00m 165.0°
1000 BCE 27364 = 07h 36m 114.0°
500 BCE 17444 = 04h 51m 72.7°
1 BCE 9848 = 02h 44m 41.0°
500 CE 4577 = 01h 16m 19.1°
1000 CE 1625 = 00h 27m 6.8°
1500 CE 275 = 00h 05m 1.1°
Note: BCE (Before Common Era) and CE (Common Era) are secular alternatives for the terms BC and AD, respectively.
For more information, see Year Dating Conventions.
Take special note of the column labeled "Longitude Shift." This is the amount that an eclipse path must be shifted in order to take into account the cumulative effects of delta-T. The historical
eclipse and occultation records for Stephenson and Morrison (1984) only extends back to about 700 BCE. Thus, any values of delta-T before this time must either be 1) a direct extrapolation from known
values, or 2) based on theoretical models of purely tidal breaking of Earth's rotation. The best available solution is probably to combine both of the above methods when looking into the distant past
(before 1000 BCE), but the uncertainties grow so rapidly that no meaningful results can be obtained earlier than about 2000 BCE.
Stephenson and Houlden (1986) estimate the uncertainties in the adopted values of delta-T as follows:
Year Uncertainty Uncertainty
(Time) (Longitude)
1500 BCE ~900 sec ~4°
400 BCE ~420 sec ~2°
1000 CE ~80 sec 20' (0.33°)
1600 CE 30 sec 7.5' (0.13°)
1700 CE 5 sec 75"
1800 CE 1 sec 15"
1900 CE 0.1 sec 1.5"
The uncertainty in delta-T means that reliable eclipse paths prior to about 1500 BCE are not possible. Similarly, all future values of delta-T are simple extrapolations of current values and trends.
Such estimates are prone to growing uncertainty as one extrapolates further and further into the furure. By the year 3000 CE, the value of delta-T could be on the order of one hour with an
extrapolated uncertainty of about ten minutes or several degrees in longitude.
In more recent work, Stephenson (1997) has made improvements in his analysis of delta-T using additional historical records.
Year delta-T Longitude
(sec) Shift
500 BCE 16800 = 04h 40m 70.0°
1 BCE 10600 = 02h 57m 44.2°
500 CE 5700 = 01h 35m 23.7°
1000 CE 1600 = 00h 27m 6.7°
1500 CE 180 = 00h 03m 0.8°
If a computer program does not incorporate these values into its calculations, then the resultant eclipse paths will contain longitude shifts inconsistent with the best estimates for delta-T.
References for Rotation and Delta-T
Dickey, J.O., "Earth Rotation Variations from Hours to Centuries", in: I. Appenzeller (ed.), Highlights of Astronomy: Vol. 10 (Kluwer Academic Publishers, Dordrecht/Boston/London, 1995), pp. 17-44.
Meeus, J., "The Effect of Delta T on Astronomical Calculations", Journal of the British Astronomical Association, 108 (1998), 154-156.
Morrison, L.V. and Ward, C. G., "An analysis of the transits of Mercury: 1677-1973", Mon. Not. Roy. Astron. Soc., 173, 183-206, 1975.
Spencer Jones, H., "The Rotation of the Earth, and the Secular Accelerations of the Sun, Moon and Planets", Monthly Notices of the Royal Astronomical Society, 99 (1939), 541-558.
Stephenson, F.R. & Morrison, L.V., "Long-Term Changes in the Rotation of the Earth: 700 BC to AD 1980", Philosophical Transactions of the Royal Society of London, Ser. A, 313 (1984), 47-70.
Stephenson F.R and Houlden M.A., Atlas of Historical Eclipse Maps: East Asia 1500 BD - AD 1900, Cambridge Univ.Press., 1986.
Stephenson, F.R. & Morrison, L.V., "Long-Term Fluctuations in the Earth's Rotation: 700 BC to AD 1990", Philosophical Transactions of the Royal Society of London, Ser. A, 351 (1995), 165-202.
Stephenson F.R., Historical Eclipses and Earth's Rotation , Cambridge Univ.Press, 1997.
• Delta T - NASA Eclipse Home Page
• Delta T - Felix Verbelen (Belgium)
• Delta T - Robert van Gent (The Netherlands)
• Delta-T - IERS Rapid Service/Prediction Center | {"url":"https://eclipse.gsfc.nasa.gov/SEhelp/rotation.html","timestamp":"2024-11-13T22:08:37Z","content_type":"application/xhtml+xml","content_length":"14570","record_id":"<urn:uuid:8701f724-a7af-4e6c-a824-359e88d383b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00351.warc.gz"} |
Essential Discrete Mathematics for Computer Science
• ISBN : 9780691179292
• 출판사 : Princeton
• 출판일 : 20190319
• 저자 : Harry Lewis
● Finally, a book that covers discrete mathematics the way I like to teach it to students of computer science. Saad Mneimneh, Hunter College, City University of New York Lewis and Zax give us a nice
introduction to the essential concepts of discrete mathematics that any computer scientist should know. Their book presents a rigorous treatment of the important results, but it also goes beyond
A more intuitive approach to the mathematical foundation of computer sciencex0DDiscrete mathematics is the basis of much of computer science, from algorithms and automata theory to combinatorics and
graph theory. This textbook covers the discrete mathematics that every…
Essential Discrete Mathematics for Computer Science | {"url":"https://bookmana.github.io/posts/307981328/","timestamp":"2024-11-10T14:40:57Z","content_type":"text/html","content_length":"14401","record_id":"<urn:uuid:bba57d98-f1e7-4993-8f3c-b3ccc860a6a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00186.warc.gz"} |
Stellar Wind -- Parker Wind Theory
Next: Virial Analysis Up: Super- and Subsonic Flow Previous: Steady State Flow under   Contents
Stellar winds are observed around various type of stars. Early type (massive) stars have large luminosities; the photon absorbed by the bound-bound transition transfers its outward momentum to the
gas. This line-driven mechanism seems to work around the early type stars. On the other hand, acceleration mechanism of less massive stars are thought to be related to the coronal activity or dust
driven mechanism (dusts absorb the emission and obtain outward momentum from the emission).
Here, we will see the identical mechanism in 2.8.1 and 2.8.2 works to accelerate the wind from a star. Consider a steady state and ignore 2.1) gives
where we used
This leads to
The equation of motion (2.2) is as follows:
where we used 2.96) and (2.97) we obtain
where 2.89) and (2.93). That is, the fact that the rhs of equation (2.98) is positive corresponds to increasing the cross-section
Figure 2.7: Right-hand side of equation (2.98) is plotted against the distance from the center
For simplicity, we assume the gas is isothermal. The rhs of equation (2.98) varies shown in Figure 2.7. Therefore, near to the star, the flow acts as the cross-section of nozzle is decreasing and far
from the star it does as the cross-section is increasing. This is the same situation that the gas flows through the Laval nozzle.
Using a normalized distance 2.98) becomes
This is rewritten as
we obtain the solution of equation (2.99) as
This gives how the Mach number
is a function only depending on
is a function only depending on
Since the minima of
1. If 2.9.
2. If 2.9.
3. If 2.9.
Out of the two solutions of
Next: Virial Analysis Up: Super- and Subsonic Flow Previous: Steady State Flow under   Contents Kohji Tomisaka 2007-07-08 | {"url":"http://th.nao.ac.jp/MEMBER/tomisaka/Lecture_Notes/StarFormation/3/node41.html","timestamp":"2024-11-02T13:54:54Z","content_type":"text/html","content_length":"19500","record_id":"<urn:uuid:7ec6a5e0-bba9-4c2d-9a57-62423116ded1>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00268.warc.gz"} |
Understanding Coin Toss Probabilities: A Mathematical Exploration
Written on
Chapter 1: Introduction to Coin Toss Probability
A fair coin has an equal chance of landing on heads or tails, specifically a 50% probability for each outcome. Conversely, an unfair coin will not adhere to this balance. Today, we will tackle a
problem that requires determining how many times an unfair coin has been tossed based on the given data.
Before you continue, I encourage you to pause and try solving it yourself with pen and paper. Once you're ready, let's explore the solution!
Section 1.1: Analyzing the Problem
To begin, let's assume there are n coins, out of which 2 result in heads. This indicates that n - 2 coins do not show heads. The selection of 2 coins from n is mathematically represented as:
This expression is commonly referred to as "n choose 2." To calculate the probability that 2 coins will display heads while the remaining n - 2 do not, we derive the following:
From this, we find that the probability is 3/4, derived from 1 - 1/4. Multiplying the aforementioned terms will yield the overall probability of getting 2 heads from n coins.
As a little challenge for you, can you use similar logic to determine the probability of obtaining 3 heads from n coins? Give it a shot!
Section 1.2: Simplifying the Expressions
Did you arrive at the correct solution? Ultimately, we aim to simplify the two expressions and equate them. The formulas are as follows:
• nC2 = n(n-1)/2
• nC3 = n(n-1)(n-2)/6
Mathematically, we can express this as:
Thus, our final answer is 11!
Chapter 2: Rewarding Your Efforts
To celebrate your hard work, here's a little treat for you:
The first video, How Random is a Coin Toss? - Numberphile - YouTube, delves deeper into the randomness behind coin tosses.
Next, we have another enlightening video that breaks down practical examples of coin toss probabilities:
Probability: Toss a Coin 3 Times Example - YouTube provides a clear illustration of how to apply these concepts.
What an intriguing journey into probability! I'm eager to hear your thoughts on this process, so please share them in the comments below.
Lastly, if you're interested in more mathematical challenges, check out the list of the best math puzzles available on Medium!
Math Puzzles
The best math puzzles on Medium
Algebra, Geometry, Calculus, Number Theory, and more
Share this with your friends!
Thank you for taking the time to read this article. If you found it valuable, please give it a clap!
If you appreciate the effort I put into crafting each article, consider buying me a coffee! Your support means the world to me.
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Credit Card Payment Calculator
Download a Credit Card Calculator for Microsoft Excel®
Our new Credit Card Payment Calculator will help you calculate your minimum payment and estimate how long it will take you to pay off your credit card by making either minimum payments or fixed
payments. See below for more information about how to calculate the minimum payment on your credit card.
This calculator will help you realize just how much it really costs to pay the minimum on your credit card. There IS one case in which it might actually be beneficial to only pay the minimum. Read on
below to find out when.
Unlike our debt reduction calculator and credit card payoff calculator, this spreadsheet lets you see how long it will take to pay off your credit card if you make only the minimum payment
Credit Card Payment Calculator
for Excel and OpenOffice
This credit card minimum payment calculator is a simple Excel spreadsheet that calculates your minimum payment, total interest, and time to pay off. It also creates a payment schedule and graphs your
payment and balance over time.
You can now add extra payments into the Payment schedule to see how making occasional extra payments could help you pay off your credit card faster (see the screenshot). You can also choose to make
Fixed Monthly Payments instead of paying the minimum payment.
Update 10/16/2016 (.xlsx version only): I have added an optional 0% Introductory Period so that you can simulate paying off a card or doing a balance transfer to a card offering 0% interest for a
number of months.
How to Calculate the Minimum Credit Card Payment?
The minimum payment on your credit card is usually either a percentage of the current balance (2% - 5%) or a minimum fixed dollar amount (like $15.00), whichever is greater. The minimum payment might
also be defined as the interest plus a percentage of the current balance. Check the fine print on your credit card agreement to determine how your credit card company defines your minimum payment.
This is the minimum possible payment that you could make to avoid having your balance increase. But, if you only only pay the interest month-to-month, you'll never pay off the credit card. The basic
calculation for the monthly interest-only payment is:
(Annual Rate / 12) * Balance
If your interest rate was 18%, then the monthly interest rate would be approximately 18% / 12 = 1.5%.
Percent of Balance
Credit cards are a type of revolving line of credit that don't have a specific amortization period defined. So, to ensure that each payment includes interest plus some portion of the principal, the
minimum payment is defined as a percentage that is greater than the monthly interest rate. This percentage will usually be between 2% and 5%.
Interest plus Percent of Balance
Some credit cards may define the minimum payment as "X% of the balance plus interest" - especially cards where the interest rate is allowed to change. Defining the minimum payment like this ensures
that the credit card payment will always cover interest plus X% of the principal balance.
In the credit card payment calculator, enter the X% in the "Min Payment % of Balance" field and then check the "Plus Interest" box.
Fixed Dollar Amount
When your balance gets low, the "Percent of Balance" calculations might result in a very small minimum payment, and in theory you'd never actually finish paying off the balance. So, there is almost
always a minimum fixed dollar amount, usually about $15.00.
In the credit card calculator, you enter the $15.00 minimum value in "Min Payment for Low Balance" field.
0% Interest Period
Some companies offer 0% interest for a number of months to entice you to sign up for their new card. After the 0% introductory period, the interest rate rises to the normal high rate. You can use the
latest version of this spreadsheet to simulate that scenario, but be aware that missing a payment can cancel the introductory period.
How is Credit Card Interest Calculated?
For credit cards, interest is usually accrued daily or based on the average daily balance, but most credit card calculators estimate the monthly interest by assuming that (1) the balance is constant
and (2) the interest rate is the annual rate divided by 12. This is a pretty good estimate, but probably won't be exactly what you see on your monthly statement.
Minimum Payments vs. Fixed Payments
The credit card payment calculator lets you enter a Fixed Monthly Payment amount. If you do, that amount will override what you have entered in the Min Payment fields. If the fixed payment is the
same as or greater than the first minimum payment, you will generally pay off the credit card much sooner and pay much less interest overall.
Why? If you are only making minimum payments, the minimum payment decreases as the balance decreases, so you aren't paying as much of the principal from month to month. Our credit card calculator can
help you see just how much the difference might be.
When Should I Pay only the Minimum Payment?
There may be extenuating circumstances where you might want to only make a minimum payment (such as lack of money).
There is also a case where it may be mathematically beneficial to pay the minimum. And that is ... if you are using the snowball method to pay off multiple credit cards.
Using the snowball method, you can pay less overall interest and pay off debts faster if you pay off the credit card with the highest interest first and make only minimum payments on the other credit
cards. This assumes that you are allocating a fixed total amount to paying off your debts so that everything left over after making the minimum payments on the other credit cards goes to paying off
the one with the higher interest rate.
Related Resources
• Online Minimum Payment Calculator at Bankrate.com - This is a handy online minimum credit card payment calculator.
• Online Credit Card Payment Calculator at creditcards.com - This is another online calculator to help you calculate the true cost of paying the minimum.
Disclaimer: This spreadsheet and the information on this page is for illustrative and educational purposes only. Results are only estimates. We do not guarantee the results or the applicability to
your unique financial situation. You should seek the advice of qualified professionals regarding financial decisions. | {"url":"https://totalsheets.com/Calculators/credit-card-payment-calculator.html","timestamp":"2024-11-07T04:32:16Z","content_type":"text/html","content_length":"34861","record_id":"<urn:uuid:bf554e57-1389-4a85-b275-f4b438d4fe26>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00597.warc.gz"} |
NCERT SOLUTION OF CLASS 11TH PHYSICS LAW OF MOTION - Wisdom TechSavvy Academy
About Lesson
QUESTIONS FROM TEXTBOOK ( law of motion )
Question 5. 1. Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skilfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer: (a) As the drop of rain is falling with constant speed, in accordance with first law of motion, the net force on the drop of rain is zero.
(b) As the cork is floating on water, its weight is being balanced by the upthrust (equal to.weight of water displaced). Hence net force on the cork is zero.
(c) Net force on a kite skilfully held stationary in sky is zero because it is at rest.
(d) Since car is moving with a constant velocity, the net force on the car is zero.
(e) Since electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on electron is zero.
Question 5. 2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion, .
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction 1 Ignore air resistance.
Answer: (a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms^-2 = 0.5 N.
(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).
(c) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at 45° because
no other acceleration except ‘g’ is acting on pebble.
Question 5. 3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/ h,
(c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
(d) lying on the floor of a train which is accelerating with 1 m s~2, the stone being at rest relative to the train.Neglect air resistance throughout.
Answer: (a) Mass of stone = 0.1 kg
Net force, F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).
(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).
(c) The stone will experience an additional force F’ (along horizontal) i.e.,F = ma = 0.1 x l = 0.1 N
As the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).
(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train.
.•. force acting on the stone, F = ma = 0.1 x 1 = 0.1 N.
It acts along the direction of motion of the train.
Question 5. 4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net
force on the particle (directed towards the centre) is:
(i) T, (ii) T – mv^2/l, (iii) T +mv^2/l, (iv) 0
T is the tension in the string. [Choose the correct alternative].
Answer: (i) T
The net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.
Question 5. 5. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-3. How long does the body take to stop?
Answer: Here m = 20 kg, F = – 50 N (retardation force)
As F = ma
Question 5. 9. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer: Here, m = 20000 kg = 2 x 10^4 kg
Initial acceleration = 5 ms^-2
Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms^-2.
Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms^-2.
Since, thrust = force = mass x acceleration
F = 2 x 10^4 x 14.8 = 2.96 x 10^5 N.
Question 5. 10. A body of mass 0.40 kg moving initially with a constant speed of 10 ms^-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant
the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.
Question 5. 11. A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
Answer: u = 0, a = 2 ms^-2, t 10 s
Using equation, v = u + at, we get
v = 0 + 2 x 10 = 20 ms^-1
(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.
Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.
.-. vx = v = 20 ms^-1
Let us now consider vertical motion which is controlled by force of gravity.
u=0, a = g = 10 ms^-2, t = (11 — 10) s = 1 s
Answer: Let the bob be oscillating as shown in the figure.
(a) When the bob is at its extreme position (say B), then its velocity is zero. Hence on cutting the string the bob will fall vertically downward under the force of its weight F = mg.
(b) When the bob is at its mean position (say A), it has a horizontal velocity of v = 1 ms^-1 and on cutting the string it will experience an acceleration a = g = 10 ms^-2 in vertical downward
direction. Consequently, the bob will behave like a projectile and will fall on ground after describing a parabolic path.
Question 5. 13. A man of mass 70 kg, stands on a weighing machine in a lift, which is moving
(a) upwards with a uniform speed of 10 ms^-1.
(b) downwards with a uniform acceleration of 5 ms^-2.
(c) upwards with a uniform acceleration of 5 ms^-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Answer: Here, m = 70 kg, g = 10 m/s^2
The weighing machine in each case measures the reaction R i.e., the apparent weight.
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.
R = mg = 70 x 10 = 700 N
(b) When the lift moves downwards with a = 5 ms^-2
R = m (g – a) = 70 (10 – 5) = 350 N
(c) When the lift moves upwards with a = 5 ms^-2
R = m (g + a) = 70 (10 + 5) = 1050 N
(d) If the lift were to come down freely under gravity, downward acc. a = g
:. R = m(g -a) = m(g-g) = Zero.
Question 5. 14. Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider
one-dimensional motion only).
law of motion
law of motion
Question 5. 18. Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to
the other?
Answer: Initial momentum of each ball before collision
= 0.05 x 6 kg ms^-1 = 0.3 kg ms^-1
Final momentum of each ball after collision
= – 0.05 x 6 kg ms^-1 = – 0.3 kg ms^-1 Impulse imparted to each ball due to the other
= final momentum – initial momentum = 0.3 kg m s-1 – 0.3 kg ms^-1
= – 0.6 kg ms^-1 = 0.6 kg ms^-1 (in magnitude)
The two impulses are opposite in direction.
law of motion
law of motion
Answer: Here acceleration of conveyor belt a = 1 ms^-2, μ[s]= 0.2 and mass of man m = 65 kg. t As the man is in an accelerating frame, he experiences a pseudo force F[s] = ma as shown
in fig. (a). Hence to maintain his equilibrium, he exerts a force F = – F[s] = ma = 65 x 1 = 65 N in forward direction i.e., direction of motion of belt.
.’. Net force acting on man = 65 N (forward)
As shown in fig. (b), the man can continue to be stationary with respect to belt, if force of friction
law of motion
law of motion
Question 5. 27. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on surrounding air,
(c) force on the helicopter due to the surrounding air,
Answer: Here, mass of helicopter, m[1]= 1000 kg
Mass of the crew and passengers, m[2] = 300 kg upward acceleration, a = 15 ms^-2 and g = 10 ms^-2
(a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers
= m[2] (g + a) = 300 (10 + 15) N = 7500 N
(b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action
F = (m[1]+ m[2]) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N
(c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.
5.28.A stream of water flowing horizontally with a speed of 15 ms^-1 pushes out of a tube of cross sectional area 10^-2 m^2, and hits at a vertical wall nearby. What is the force exerted on the wall
by the impact of water, assuming that it does not rebound?
Ans.In one second, the distance travelled is equal to the velocity v.
Volume of water hitting the wall per second, V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube.
V = 10^-2 m^2 x 15 ms^-1 = 15 x 10^-2 m^3 s^-1
Mass of water hitting the wall per second
= 15 x 10^-2 x 10^3 kg s^-1 = 150 kg s^-1 [v density of water = 1000 kg m^-3] Initial momentum of water hitting the wall per second
= 150 kg s^-1 x 15 ms^-1 = 2250 kg ms^-2 or 2250 N Final momentum per second = 0 Force exerted by the wall = 0 – 2250 N = – 2250 N Force exerted on the wall = – (- 2250) N = 2250 N.
Question 5. 29. Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of
(a) the force on the 7th coin (counted from the bottom) due to all coins above it.
(b) the force on the 7th coin by the eighth coin and
(c) the reaction of the sixth coin on the seventh coin.
Answer: (a) The force on 7th coin is due to weight of the three coins lying above it. Therefore,
F = (3 m) kgf = (3 mg) N
where g is acceleration due to gravity. This force acts vertically downwards.
(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.
F = 2 m + m = (3 m) kg f = (3 mg) N The force acts vertically downwards.
(c) The sixth coin is under the weight of four coins above it.
Reaction, R = – F = – 4 m (kg) = – (4 mgf) N Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.
law of motion
law of motion
law of motion
Answer: In 1st case, man applies an upward force of 25 kg wt.r (same as the weight of the block). According to Newton’s third law of motion, there will be a downward reaction on the floor.
The action on the floor by the man.
= 50 kg wt. + 25 kg wt. = 75 kg wt = 75 kg x 10 m/s^2 = 750 N.
In case II, the man applies a downward force of 25 kg wt. According to Newton’s third law, the reaction is in the upward direction.
In this case, action on the floor by the man
= 50 kg wt – 25 kg wt. = 25 kg wt. = 25 kg x 10 m/s^2 = 250 N.
Therefore, the man should adopt the second method.
Question 5. 33. A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 ms^-2
(b) climbs down with an acceleration of 4 ms^-2
(c) climbs up with a uniform speed of 5 ms^-1
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).
law of motion
law of motion
Question 5. 35. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-1 for
20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer: (a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N Force of friction,F[f]= p mg = 0.18 x 15 x 10 = 27 N. i.e., force experienced by block will be less than the friction.So the block
will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.
(b) The observer moving with trolley has an accelerated motion i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.
Question 5. 36. The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is
0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box). | {"url":"https://free-education.in/courses/class-11th-physics-online-class-for-100-result/lesson/ncert-solution-of-class-11th-physics-law-of-motion/","timestamp":"2024-11-05T18:52:20Z","content_type":"text/html","content_length":"301947","record_id":"<urn:uuid:ec3f8212-a437-4bff-9670-864b0524e8aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00439.warc.gz"} |
Simplex Method Calculator
The simplex method is universal. It allows you to solve any linear programming problems.
Òhe solution by the simplex method is not as difficult as it might seem at first glance.
This calculator only finds a general solution when the solution is a straight line segment.
You can solve your problem or
see examples of solutions that this calculator has made | {"url":"http://reshmat.ru/simplex_method_lpp.html","timestamp":"2024-11-08T18:35:29Z","content_type":"text/html","content_length":"14673","record_id":"<urn:uuid:1fcbf0ea-b2c0-4663-804c-8c6b64709e91>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00317.warc.gz"} |
Da Vinci Code
13-3-2-21-1-1-8-5 O, DRACONIAN DEVIL! OH, LAME SAINT! The Da Vinci Code is one of the most widely read but controversial books of all time. In this book the writer Dan Brown used a very interesting
encryption technique to keep a secret message. It required a great deal of intelligence to decipher the code since there was not enough information available. Here, a similar kind of problem is given
with sufficient clues to solve. In this problem, you will be given a series of numbers, taken from Fibonacci series and a cipher text. Your task is to decipher the text using the decryption technique
described below. Lets follow an example. Any cipher text will consist of two lines. First line is the key which contains some numbers drawn from Fibonacci series. The second line is the actual cipher
text. So, given the following cipher text 13 2 89 377 8 3 233 34 144 21 1 OH, LAME SAINT! the output will be: THE MONA LISA For this problem, assume that the first number in Fibonacci series is 1,
second one is 2 and each subsequent number is found by adding previous two numbers of the series. So, the Fibonacci series is 1,2,3,5,8,13... So, how do we get the string “THE MONA LISA” from the
string “OH, LAME SAINT!”? Here some numbers are drawn from Fibonacci series, given in the first line. The first one is 13 which is the sixth (6th) Fibonacci number in Fibonacci series. So the first
uppercase letter in the cipher text O goes to the sixth (6th) position in the output string. Second number in the input string is 2 which is the second Fibonacci number and thus H goes to second
position in the output string; then comes 89 which is the 10th Fibonacci number, so L which is the third uppercase letter in the cipher goes to the 10th position in the output string. And this
process continues until the cipher text ends and hence we find the string “THE MONA LISA”. Note that only uppercase letters conveys the message; other characters are simply garbage. If a Fibonacci
number is missing in the input sequence then a blank space is put at its position in the output string. As in the above example fourth and ninth Fibonacci numbers 5 and 55 are missing. So, two blank
spaces are inserted in fourth and ninth positions of the output string. But there must not be any trailing spaces. Input Input starts with a line consisting of a single number T. T test cases follow.
Each test case consists of three lines. The first line contains a single positive integer N. In the second line there are N numbers drawn from the Fibonacci series. The numbers will be separated from
each other using spaces. Finally, the third line contains the cipher text to be decrypted.
2/2 Output For each test case, output contains a single line containing the decrypted text. Remember that the decrypted text will contain only uppercase letters. Constraints • Value of any input
Fibonacci number is less than 231. • The length of the cipher text will be at most 100. Sample Input 2 11 13 2 89 377 8 3 233 34 144 21 1 OH, LAME SAINT! 15 34 21 13 144 1597 3 987 610 8 5 89 2 377
2584 1 O, DRACONIAN DEVIL! Sample Output THE MONA LISA LEONARDO DA VINCI | {"url":"https://ohbug.com/uva/11385/","timestamp":"2024-11-03T17:26:04Z","content_type":"text/html","content_length":"4410","record_id":"<urn:uuid:10452c20-0e29-465d-88b2-b1c89c05f8d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00827.warc.gz"} |
(beer) to Perch
Gallon (beer) to Perch Converter
⇅ Switch toPerch to Gallon (beer) Converter
How to use this Gallon (beer) to Perch Converter 🤔
Follow these steps to convert given volume from the units of Gallon (beer) to the units of Perch.
1. Enter the input Gallon (beer) value in the text field.
2. The calculator converts the given Gallon (beer) into Perch in realtime ⌚ using the conversion formula, and displays under the Perch label. You do not need to click any button. If the input
changes, Perch value is re-calculated, just like that.
3. You may copy the resulting Perch value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Gallon (beer) to Perch?
The formula to convert given volume from Gallon (beer) to Perch is:
Volume[(Perch)] = Volume[(Gallon (beer))] × 0.00659371492704826
Substitute the given value of volume in gallon (beer), i.e., Volume[(Gallon (beer))] in the above formula and simplify the right-hand side value. The resulting value is the volume in perch, i.e.,
Calculation will be done after you enter a valid input.
Consider that a brewery produces 50 gallons (beer) of ale in a batch.
Convert this volume from gallons (beer) to Perch.
The volume in gallon (beer) is:
Volume[(Gallon (beer))] = 50
The formula to convert volume from gallon (beer) to perch is:
Volume[(Perch)] = Volume[(Gallon (beer))] × 0.00659371492704826
Substitute given weight Volume[(Gallon (beer))] = 50 in the above formula.
Volume[(Perch)] = 50 × 0.00659371492704826
Volume[(Perch)] = 0.3297
Final Answer:
Therefore, 50 beer gal is equal to 0.3297 per.
The volume is 0.3297 per, in perch.
Consider that a keg holds 30 gallons (beer) of lager.
Convert this capacity from gallons (beer) to Perch.
The volume in gallon (beer) is:
Volume[(Gallon (beer))] = 30
The formula to convert volume from gallon (beer) to perch is:
Volume[(Perch)] = Volume[(Gallon (beer))] × 0.00659371492704826
Substitute given weight Volume[(Gallon (beer))] = 30 in the above formula.
Volume[(Perch)] = 30 × 0.00659371492704826
Volume[(Perch)] = 0.1978
Final Answer:
Therefore, 30 beer gal is equal to 0.1978 per.
The volume is 0.1978 per, in perch.
Gallon (beer) to Perch Conversion Table
The following table gives some of the most used conversions from Gallon (beer) to Perch.
Gallon (beer) (beer gal) Perch (per)
0.01 beer gal 0.00006593715 per
0.1 beer gal 0.00065937149 per
1 beer gal 0.00659371493 per
2 beer gal 0.01318742985 per
3 beer gal 0.01978114478 per
4 beer gal 0.02637485971 per
5 beer gal 0.03296857464 per
6 beer gal 0.03956228956 per
7 beer gal 0.04615600449 per
8 beer gal 0.05274971942 per
9 beer gal 0.05934343434 per
10 beer gal 0.06593714927 per
20 beer gal 0.1319 per
50 beer gal 0.3297 per
100 beer gal 0.6594 per
1000 beer gal 6.5937 per
Gallon (beer)
The beer gallon is a unit of measurement used to quantify liquid volumes, particularly in the brewing industry. It is defined as 128 US fluid ounces, which is equivalent to 3.785 liters.
Historically, the beer gallon was introduced to standardize the measurement of beer for commercial trade and consumption. Today, it remains relevant in the brewing industry and for regulatory
purposes, providing a consistent measure for large volumes of beer and facilitating accurate trade and distribution.
The perch is a unit of measurement used to quantify volume, area, and length, primarily in historical and specific regional contexts. As a volume measure, it is often associated with a cubic
measurement of 1 cubic yard or approximately 0.7646 cubic meters. Historically, the perch was used in land measurement, particularly for timber and stone, and was commonly employed in construction
and trade. Today, while its use has largely declined, the perch is still referenced in some historical contexts and in certain industries where traditional units are preserved.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Gallon (beer) to Perch in Volume?
The formula to convert Gallon (beer) to Perch in Volume is:
Gallon (beer) * 0.00659371492704826
2. Is this tool free or paid?
This Volume conversion tool, which converts Gallon (beer) to Perch, is completely free to use.
3. How do I convert Volume from Gallon (beer) to Perch?
To convert Volume from Gallon (beer) to Perch, you can use the following formula:
Gallon (beer) * 0.00659371492704826
For example, if you have a value in Gallon (beer), you substitute that value in place of Gallon (beer) in the above formula, and solve the mathematical expression to get the equivalent value in | {"url":"https://convertonline.org/unit/?convert=gallon_beer-perch","timestamp":"2024-11-04T17:22:31Z","content_type":"text/html","content_length":"93252","record_id":"<urn:uuid:bee2247f-28e2-4e96-b0c4-5f443ac896d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00102.warc.gz"} |
5,052 research outputs found
An analytical theory of the nonlinear electromagnetic response of a two-dimensional (2D) electron system in the second order in the electric field amplitude is developed. The second-order
polarizability and the intensity of the second harmonic signal are calculated within the self-consistent-field approach both for semiconductor 2D electron systems and for graphene. The second
harmonic generation in graphene is shown to be about two orders of magnitude stronger than in GaAs quantum wells at typical experimental parameters. Under the conditions of the 2D plasmon resonance
the second harmonic radiation intensity is further increased by several orders of magnitude.Comment: 9 pages, 2 figure
We present a formulation for the nonlinear optical response in gapped graphene, where the low-energy single-particle spectrum is modeled by massive Dirac theory. As a representative example of the
formulation presented here, we obtain closed form formula for the third harmonic generation (THG) in gapped graphene. It turns out that the covariant form of the low-energy theory gives rise to a
peculiar logarithmic singularities in the nonlinear optical spectra. The universal functional dependence of the response function on dimension-less quantities indicates that the optical nonlinearity
can be largely enhanced by tuning the gap to smaller values.Comment: http://iopscience.iop.org/0953-8984/labtalk-article/4938
We study a class of integrable non-linear differential equations related to the A.III-type symmetric spaces. These spaces are realized as factor groups of the form SU(N)/S(U(N-k) x U(k)). We use the
Cartan involution corresponding to this symmetric space as an element of the reduction group and restrict generic Lax operators to this symmetric space. The symmetries of the Lax operator are
inherited by the fundamental analytic solutions and give a characterization of the corresponding Riemann-Hilbert data.Comment: 14 pages, 1 figure, LaTeX iopart styl
The wavelength and the propagation length of the edge magnetoplasmons, running along the edge of a two-dimensional electron layer in a semiconductor quantum-well structure are theoretically studied
as a function of frequency, magnetic field, electron density, mobility, and geometry of the structure. The results are intended to be used for analysis and optimization of operation of recently
invented quantum-well microwave spectrometers operating at liquid-nitrogen temperatures (I. V. Kukushkin {\em et al}, Appl. Phys. Lett. {\bf 86}, 044101 (2005)).Comment: 4 pages, including 4 figures.
Accepted for publication in Appl. Phys. Let
A quantum theory of the third-harmonic generation in graphene is presented. An analytical formula for the nonlinear conductivity tensor $\sigma^{(3)}_{\alpha\beta\gamma\delta}(\omega,\omega,\omega)$
is derived. Resonant maxima of the third harmonic are shown to exist at low frequencies $\omega\ll E_F/\hbar$, as well as around the frequency $\omega=2E_F/\hbar$, where $E_F$ is the Fermi energy in
graphene. At the input power of a CO$_2$ laser ($\lambda\approx 10$ $\mu$m) of about 1 MW/cm$^2$ the output power of the third-harmonic ($\lambda\approx 3.3$ $\mu$m) is expected to be $\simeq 50$ W/
cm$^2$.Comment: 5 pages, 3 figure
A portion of the electromagnetic wave spectrum between $\sim 0.1$ and $\sim 10$ terahertz (THz) suffers from the lack of powerful, effective, easy-to-use and inexpensive emitters, detectors and
mixers. We propose a multilayer graphene -- boron-nitride heterostructure which is able to emit radiation in the frequency range $\sim 0.1-30$ THz with the power density up to $\sim 0.5$ W/cm$^2$ at
room temperature. The proposed device is extremely thin, light, flexible, almost invisible and may completely cover the needs of science and technology in the sources of terahertz radiation.Comment:
4 pages, 4 figures, slightly modified version of the previous submissio
We show that all questions raised in the Comment can be easily answered within the theory formulated in the commented papers. It is also shown that the bulk theory promoted by the commenter fails to
explain the most important fundamental features of the discussed phenomenon.Comment: 4 pages, 3 figure
A theory of the nonlinear plasma waves in graphene is developed in the nonperturbative regime. The influence of strong electric fields on the position and linewidth of plasma resonances in the
far-infrared transmission experiments, as well as on the wavelength and the propagation length in the scanning near-field optical microscopy experiments is studied. The theory shows that the fields
of order of a few to a few tens of kV/cm should lead to a red shift and broadening of plasma resonances in the first type and to a reduction of the wavelength and the propagation length in the second
type of experiments.Comment: 6 pages, 3 figure | {"url":"https://core.ac.uk/search/?q=author%3A(Mikhailov%20S%20A)","timestamp":"2024-11-09T13:49:28Z","content_type":"text/html","content_length":"133522","record_id":"<urn:uuid:1cdebe1f-f51e-49c1-b067-0dcabc55ed83>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00217.warc.gz"} |
" used in SRM 12 (Division I Level Two
Division II Level Two)
Class Name: Derivatives
Method Name: calcDerivative
Parameters: String,int,int
Returns: int
Implement a class Derivatives which contains a method calcDerivative. The
method takes a String representing a polynomial, and two ints, k and x. The
method returns the kth derivative of the polynomial evaluated at x.
The first derivative of any term of a polynomial is:
1st derivative of (a*x^n) = (a*n*x^(n-1)).
The first derivative of any polynomial is the sum of the first derivatives of
the polynomial's terms.
The kth derivative of a polynomial is the 1st derivative of the (k-1)th
derivative of the polynomial.
The 0th derivative of a polynomial is the polynomial itself.
The String will be of the form:
There are no spaces.
All the a's are non-negative integers less than 20
All the n's are unique non-negative integers less than 10.
The String is at most 50 characters and at least 5 characters.
Here is the method signature:
public int calcDerivative(String poly,int k,int x);
poly is of the correct form.
k is an integer between 0 and 10, inclusive.
x is an integer between -10 and 10, inclusive.
*If poly="3*x^3+2*x^1+2*x^0", k=1, and x=1:
The first derivative is: 3*3*x^(3-1)+2*1*x^(1-1)+0*2*x^(0-1)=9*x^2+2*x^0.
The first derivative evaluated at x=1 is: 9*1^2+2*1^0=11
So the method returns 11.
*If poly="2*x^5+3*x^2", k=2, and x=-2,
The first derivative is: 5*2*x^(5-1)+3*2*x^(2-1)=10*x^4+6*x^1.
The second derivative is: 10*4*x^(4-1)+6*1*x^(1-1)=40*x^3+6*x^0.
The second derivative evaluated at x=-2 is 40*(-2)^3+6*(-2)^0=-314.
So the method returns -314. | {"url":"http://topcoder.bgcoder.com/print.php?id=50","timestamp":"2024-11-13T09:27:35Z","content_type":"text/html","content_length":"4371","record_id":"<urn:uuid:d78aa035-3a39-439c-ad59-6b81c88d225b>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00656.warc.gz"} |
Example: Gas Release due to Cavitation Within a Resistor
The performance of the enhanced DSHplus components for modeling cavitation effects in resistors is demonstrated using an example simulation.
The flow rate limitation due to vapour cavitation and the subsequent release of gas is analysed by a simple simulation set-up consisting of three pipes and a resistor.
The adjacent video shows the content of this page as a webinar.
Simulation Set-Up
Three pipe elements, a resistor element, a connecting element and two line terminations are used to represent the hydraulic system (red box in the adjacent picture).
At the “open ends” of pipe 1 and pipe 3, pressures (p[Inlet] and p[Outlet]) are prescribed to the system.
Since the pressures may vary with time, signal generating components (orange boxes) are used.
In order to visualise the distribution of flow quantities like pressure or velocity along the pipe network, specialised plotting components are used (green box).
The fluid properties are provided by a fluid property component (blue box). The liquid's vapour pressure p[v] equals 8.61 bar (absolute).
In order to demonstrate how the occurrence of cavitation as well as the gas release and absorption are influenced by the pressure level of the experiment, different parameter sets are evaluated.
At the beginning of the simulation, the pressures at the system inlet and outlet are identical (40 bar) and the gas-free liquid is at rest.
After a short time, a pressure difference of 5 bar between the inlet and outlet pressure is created:
• For parameter set "A", this is achieved by raising the inlet pressure to 45 bar while keeping the outlet pressure at 40 bar (upper diagram in the adjacent figure)
• For parameter set "B", this is achieved by keeping the inlet pressure at 40 bar while lowering the outlet pressure to 35 bar (lower diagram)
Due to the pressure difference, a flow from inlet to outlet is initiated and the pipe is slowly filled with the gas-carrying liquid (α[G0] = 2.0 %).
Since all pressures are larger than the pressure p[α] = 0.2 bar (relative pressure) for which α[G](p) = 0 %, the gas is entirely dissolved in the liquid, i.e. there are no gas bubbles.
After a steady state is reached, the pressure difference is increased further to 30 bar.
• For parameter set "A", this is achieved by raising the inlet pressure to 70 bar while keeping the outlet pressure at 40 bar (upper diagram in the adjacent figure)
• For parameter set "B", this is achieved by keeping the inlet pressure at 40 bar while lowering the outlet pressure to 10 bar (lower diagram). This pressure is very close to the vapour pressure of
the liquid!
The pressure drops across the system are always identical for both parameter sets!
Can we also expect an identical discharge behaviour for both parameter sets?
Results for Parameter Set "A" (40 bar to 70 bar)
Top diagrams (left):
• Current values of pressures prescribed at first and last pipe
Center diagrams (left):
• Current values of flow rate & cavitation index
• Flow rate increases continuously with growing pressure difference since vapour cavitation does not occur (cavitation index C > C[crit] = 2.0)
Bottom diagrams (left):
• Current values of undissolved gas fractions
• No undissolved gas α[G](p) anywhere since p > p[α] everywhere
Top diagrams (right):
• Pressure distribution along the pipe network
• Prominent pressure drop @ resistor
Center diagrams (right):
• Distribution of gas fractions along the pipe network
Bottom diagrams (right):
• Distribution of undissolved gas fractions along the pipe network
• No gas release at resistor
Results for Parameter Set "B" (40 bar to 10 bar)
Top diagrams (left):
• Current values of pressures prescribed at inlet and outlet
Center diagrams (left):
• Current values of flow rate & cavitation index
• Initially, flow rate increases with increasing pressure difference
• After further increase, cavitation index C drops below C[crit]
• Onset of vapour cavitation
• Limitation of low rate (choking)
Bottom diagrams (left):
• Current values of undissolved gas fractions at inlet and outlet
• No undissolved gas at outlet
Top diagrams (right):
• Pressure distribution along the pipe network
• Prominent pressure drop @ resistor
Center diagrams (right):
• Distribution of gas fractions along the pipe network
Bottom diagrams (right):
• Distributions of undissolved gas fractions along the pipe network
• Gas release at resistor
• Due to a small absorption time constant τ[Abs], the gas remains undissolved for longer!
Results for modified Parameter Set "B" (40 bar to 10 bar)
This parameter set is based on parameter set "B". To demonstrate the impact of finite gas absorption rates, the time constant for absorption is increased by a factor of 10.
Top diagrams (left):
• Current values of pressures prescribed at first and last pipe
Center diagrams (left):
• Current values of flow rate & cavitation index
• Initially, flow rate increases with increasing pressure difference
• After further increase, cavitation index C drops below C[crit]
• Onset of vapour cavitation
• Limitation of flow rate (choking)
Bottom diagrams (left):
• Current values of undissolved gas fractions at inlet and outlet
• Undissolved gas at outlet due to larger absorption time constant!
Top diagrams (right):
• Pressure distribution along the pipe network
• Prominent pressure drop @ resistor
Center diagrams (right):
• Distribution of gas fractions along the pipe network
Bottom diagrams (right):
• Distribution of undissolved gas fractions along the pipe network
• Gas release at resistor
• Due to a larger absorption time constant τ[Abs], the gas remains undissolved for longer and can potentially travel further downstream!
Analysis of discharge behaviour
The simulation results are analysed further by plotting the flow rate Q as a function of the root of the pressure difference Δp.
For parameter set "A", the flow rate increases with every increase of the pressure drop (grey curve in the adjacent figure). Because the flow rate is plotted against the root of the pressure drop,
the dependency appears linear.
With the parameter sets "B" and "B modified", the flow rate initially follows the same law. At higher pressure losses (and lower pressures in the "vena contracta"), cavitation occurs and thus a flow
limitation - "choking" - can be observed (dashed red curve).
The pressure drop is the same in both situations!
The resistor behaves differently for both parameter sets since the pressures in parameter set "A" are further away from the vapour pressure p[v]! | {"url":"https://fluidon.com/en/tools/dshplus/piping/cavitation/example-gas-release-within-resistor","timestamp":"2024-11-12T16:22:20Z","content_type":"text/html","content_length":"64952","record_id":"<urn:uuid:751ce3c8-8dd8-48b8-8fda-a6025d0e602f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00110.warc.gz"} |
How To Draw Derivatives
How To Draw Derivatives - Web 0:00 / 31:20. Explore key concepts by building secant and tangent line sliders, or illustrate important calculus ideas like the mean value theorem. We take the
derivative of f(x) to obtain f'(x) = 2x. Web graphing of functions using first and second derivatives the following problems illustrate detailed graphing of functions of one variable using the first
and second derivatives. Using the second derivative can sometimes be a simpler method than using the first derivative.
Mark zeros at the locations of any turning points or stationary inflection points. We can use critical values to. Web 0:00 / 31:20. Problems range in difficulty from average to challenging. We will
use that understanding a. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. This calculus video tutorial explains how to sketch the derivatives of
the parent function using the graph f (x).
How to Sketch the Graph of the Derivative
Explore key concepts by building secant and tangent line sliders, or illustrate important calculus ideas like the mean value theorem. Web 6 years ago hi katelin, since acceleration is the derivative
of velocity, you can plot the slopes of the velocity graph to find the acceleration graph. Mark zeros at the locations of any turning.
Pin on Graphing The Derivative of a Function
Web explore math with our beautiful, free online graphing calculator. Web curve sketching with calculus: Using the second derivative can sometimes be a simpler method than using the first derivative.
Web to sketch the derivative graph of a function: This calculus video tutorial explains how to sketch the derivatives of the parent function using the.
MATH221 Lesson 009B Drawing Derivatives YouTube
Explain the relationship between a function and its first and second derivatives. First, notice that the derivative is equal to 0 when x = 0. First, we learn how to sketch the derivative graph of a
continuous, differentiable function f (x), either given the original function or its graph y=f (x). Web derivative graph rules..
How to Sketch the Graph of the Derivative
If the derivative (which lowers the degree of the starting function by 1) ends up with 1 or lower as the degree, it is linear. 1 2 3 4 1 2 − 2 y x a a 1 2 3 4 1 2 − 2 y x a Logarithm math > ap®︎
calculus ab (2017.
How to Sketch the Graph of the Derivative
Web the second deriviatve is just the derivative of the first derivative. Differentiation allows us to determine the change at a given point. Below are three pairs of graphs. We take the derivative
of f(x) to obtain f'(x) = 2x. 1 2 3 4 1 2 − 2 y x a a 1 2 3.
Draw the Function given Graph of Derivative YouTube
So say we have f(x) = x^2 and we want to evaluate the derivative at point (2, 4). If the derivative (which lowers the degree of the starting function by 1) ends up with 1 or lower as the degree, it
is linear. What do you notice about each pair? Use concavity and inflection points.
Steps to Sketch Graph of Function From Derivative YouTube
If the slope of f (x) is negative, then the. We know from calculus that if the derivative is 0 at a point, then it is a critical value of the original function. Below are three pairs of graphs. Web 6
years ago hi katelin, since acceleration is the derivative of velocity, you can plot.
How to sketch first derivative and Function from graph of second
Let f be a function. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph. Below are three pairs of graphs. Graph functions,
plot points, visualize algebraic equations, add sliders, animate graphs, and more. Sketching a derivative using a function use the following graph.
Sketching the graph of a derivative, (as in d𝑦/d𝑥), A Level Maths, 12th
Where the slope is positive in y’, y” is positive. Unleash the power of differential calculus in the desmos graphing calculator. Web you just take the derivative of that function and plug the x
coordinate of the given point into the derivative. Web curve sketching with calculus: Below are three pairs of graphs. Explain the.
Drawing the Graph of a Derivative YouTube
Web sketching the derivative of a function. What do you notice about each pair? First, notice that the derivative is equal to 0 when x = 0. Web thanks to all of you who support me on patreon. A
linear function is a function that has degree one (as in the highest power of the.
How To Draw Derivatives A linear function is a function that has degree one (as in the highest power of the independent variable is 1). Web 0:00 / 31:20. Let f be a function. ( 14 votes) upvote flag
puspita A function f(x) is said to be differentiable at a if f ′ (a) exists.
Web 6 Years Ago Hi Katelin, Since Acceleration Is The Derivative Of Velocity, You Can Plot The Slopes Of The Velocity Graph To Find The Acceleration Graph.
Web curve sketching with calculus: Problems range in difficulty from average to challenging. It explains how to graph. The derivative function, denoted by f ′, is the function whose domain consists
of those values of x such that the following limit exists:
Unleash The Power Of Differential Calculus In The Desmos Graphing Calculator.
( 14 votes) upvote flag puspita If the slope of f (x) is negative, then the. What do you notice about each pair? Web sketching the derivative of a function.
Sketching A Derivative Using A Function Use The Following Graph Of [Latex]F(X)[/Latex] To Sketch A Graph Of [Latex]F^{\Prime}(X)[/Latex].
Below are three pairs of graphs. The top graph is the original function, f (x), and the bottom graph is the derivative, f’ (x). Web you just take the derivative of that function and plug the x
coordinate of the given point into the derivative. Plot a function and its derivative, or graph the derivative directly.
First, Notice That The Derivative Is Equal To 0 When X = 0.
Mark zeros at the locations of any turning points or stationary inflection points. This video contains plenty of examples and. 1 2 3 4 1 2 − 2 y x what is the graph of its derivative, g ′ ? Web to
sketch the derivative graph of a function:
How To Draw Derivatives Related Post : | {"url":"https://sandbox.independent.com/view/how-to-draw-derivatives.html","timestamp":"2024-11-03T19:18:13Z","content_type":"application/xhtml+xml","content_length":"23430","record_id":"<urn:uuid:a6b8b17e-8cfa-41b7-969e-202500036acd>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00826.warc.gz"} |
Statistical Data Analysis
Both admin recorded data and the honor system have the same flaw: invalid data points. Simply adding or removing one digit of a participant’s data can drastically affect the outcome of the challenge.
Therefore, a system of validation should be put in place to check the reasonableness of every entry. The simplest and most effective method for performing this is to use standard deviation.
Standard deviation is used to determine confidence that a particular data point falls within an ordinary range. By using two standard deviations, you can assume 95% confidence that the value in
question is valid if it falls within the given range.
The first step in finding the standard deviation is finding the mean. To determine the mean, add all of the data points and then divide by the number of data points.
E.g. for a given set of steps walked in a day (1000, 3000, 4000, 5000, 5000, 11000), the mean is:
Mean = (1000 + 3000 + 4000 + 5000 + 5000 + 11000) / 6 = 4833
Next, compute the variance by subtracting each data point by the mean, squaring it and then determining the average.
Variance = ((1000 – 4833)^2 + (3000 – 4833)^2 + (4000 - 4833)^2 + (5000 - 4833)^2 + (5000 - 4833)^2 + (11000 - 4833)^2) / 6
= (14691889 + 3359889 +693889 + 27889 + 27889 + 38031889) / 6
= 9472222
Finally, to compute the standard deviation, take the square root of the variance:
Standard Deviation = √9472222 = 3078
Now that you have the standard deviation, you can use it to determine confidence by computing the upper and lower bounds for your range of numbers. This is accomplished by subtracting the standard
deviation from the mean for the lower bound and adding the standard deviation to the mean for the upper bound. For example:
Lower Bound = 4833 – 3078 = 1755
Upper Bound = 4833 + 3078 = 7911
In a normal distribution, 68% of all values will fall within one standard deviation. In our example, both the 1000 data point and 11000 data point would fall outside of one standard deviation. If we
are checking on every outlier that is reported in our fitness challenge and 32% are considered outliers, we are in for a lot of work. Instead, we should try two standard deviations which will give us
95% confidence that our data is valid. To calculate the upper and lower bounds with two standard deviations, simply multiply the standard deviation by two:
Lower Bound = 4833 – 6155 = -1322
Upper Bound = 4833 + 6155 = 10988
Now, only the 11000 data point barely falls outside of the standard deviation and should be checked out. If you are considering thousands of data points, you may even want to consider using three
standard deviations which would raise confidence to over 99%.
Doing this by hand would require considerable work. Fortunately, spreadsheets can accomplish this with much less effort. Challenge management systems should also provide this analysis automatically.
A sample report from ChallengeRunner.com appears as follows | {"url":"https://challengerunner.com/handbook/Statistical-Data-Analysis","timestamp":"2024-11-13T22:25:17Z","content_type":"text/html","content_length":"34049","record_id":"<urn:uuid:ec2cd2b9-dcb8-415d-9fe0-cdcee7a4371c>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00291.warc.gz"} |
Undetermined Coefficients
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Section 3.9 : Undetermined Coefficients
In this section we will take a look at the first method that can be used to find a particular solution to a nonhomogeneous differential equation.
\[y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\]
One of the main advantages of this method is that it reduces the problem down to an algebra problem. The algebra can get messy on occasion, but for most of the problems it will not be terribly
difficult. Another nice thing about this method is that the complementary solution will not be explicitly required, although as we will see knowledge of the complementary solution will be needed in
some cases and so we’ll generally find that as well.
There are two disadvantages to this method. First, it will only work for a fairly small class of \(g(t)\)’s. The class of \(g(t)\)’s for which the method works, does include some of the more common
functions, however, there are many functions out there for which undetermined coefficients simply won’t work. Second, it is generally only useful for constant coefficient differential equations.
The method is quite simple. All that we need to do is look at \(g(t)\) and make a guess as to the form of \(Y_{P}(t)\) leaving the coefficient(s) undetermined (and hence the name of the method). Plug
the guess into the differential equation and see if we can determine values of the coefficients. If we can determine values for the coefficients then we guessed correctly, if we can’t find values for
the coefficients then we guessed incorrectly.
It’s usually easier to see this method in action rather than to try and describe it, so let’s jump into some examples.
Example 1
Determine a particular solution to \[y'' - 4y' - 12y = 3{{\bf{e}}^{5t}}\]
Show Solution
The point here is to find a particular solution, however the first thing that we’re going to do is find the complementary solution to this differential equation. Recall that the complementary
solution comes from solving,
\[y'' - 4y' - 12y = 0\]
The characteristic equation for this differential equation and its roots are.
\[{r^2} - 4r - 12 = \left( {r - 6} \right)\left( {r + 2} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,{r_1} = - 2,\,\,\,\,{r_2} = 6\]
The complementary solution is then,
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}}\]
At this point the reason for doing this first will not be apparent, however we want you in the habit of finding it before we start the work to find a particular solution. Eventually, as we’ll see,
having the complementary solution in hand will be helpful and so it’s best to be in the habit of finding it first prior to doing the work for undetermined coefficients.
Now, let’s proceed with finding a particular solution. As mentioned prior to the start of this example we need to make a guess as to the form of a particular solution to this differential equation.
Since \(g(t)\) is an exponential and we know that exponentials never just appear or disappear in the differentiation process it seems that a likely form of the particular solution would be
\[{Y_P}\left( t \right) = A{{\bf{e}}^{5t}}\]
Now, all that we need to do is do a couple of derivatives, plug this into the differential equation and see if we can determine what \(A\) needs to be.
Plugging into the differential equation gives
\[\begin{align*}25A{{\bf{e}}^{5t}} - 4\left( {5A{{\bf{e}}^{5t}}} \right) - 12\left( {A{{\bf{e}}^{5t}}} \right) & = 3{{\bf{e}}^{5t}}\\ - 7A{{\bf{e}}^{5t}} & = 3{{\bf{e}}^{5t}}\end{align*}\]
So, in order for our guess to be a solution we will need to choose \(A\) so that the coefficients of the exponentials on either side of the equal sign are the same. In other words we need to choose \
(A\) so that,
\[ - 7A = 3\hspace{0.25in} \Rightarrow \hspace{0.25in}A = - \frac{3}{7}\]
Okay, we found a value for the coefficient. This means that we guessed correctly. A particular solution to the differential equation is then,
\[{Y_P}\left( t \right) = - \frac{3}{7}{{\bf{e}}^{5t}}\]
Before proceeding any further let’s again note that we started off the solution above by finding the complementary solution. This is not technically part the method of Undetermined Coefficients
however, as we’ll eventually see, having this in hand before we make our guess for the particular solution can save us a lot of work and/or headache. Finding the complementary solution first is
simply a good habit to have so we’ll try to get you in the habit over the course of the next few examples. At this point do not worry about why it is a good habit. We’ll eventually see why it is a
good habit.
Now, back to the work at hand. Notice in the last example that we kept saying “a” particular solution, not “the” particular solution. This is because there are other possibilities out there for the
particular solution we’ve just managed to find one of them. Any of them will work when it comes to writing down the general solution to the differential equation.
Speaking of which… This section is devoted to finding particular solutions and most of the examples will be finding only the particular solution. However, we should do at least one full blown IVP to
make sure that we can say that we’ve done one.
Example 2
Solve the following IVP \[y'' - 4y' - 12y = 3{{\bf{e}}^{5t}}\hspace{0.25in}\hspace{0.25in}y\left( 0 \right) = \frac{{18}}{7}\hspace{0.25in}y'\left( 0 \right) = - \frac{1}{7}\]
Show Solution
We know that the general solution will be of the form,
\[y\left( t \right) = {y_c}\left( t \right) + {Y_P}\left( t \right)\]
and we already have both the complementary and particular solution from the first example so we don’t really need to do any extra work for this problem.
One of the more common mistakes in these problems is to find the complementary solution and then, because we’re probably in the habit of doing it, apply the initial conditions to the complementary
solution to find the constants. This however, is incorrect. The complementary solution is only the solution to the homogeneous differential equation and we are after a solution to the nonhomogeneous
differential equation and the initial conditions must satisfy that solution instead of the complementary solution.
So, we need the general solution to the nonhomogeneous differential equation. Taking the complementary solution and the particular solution that we found in the previous example we get the following
for a general solution and its derivative.
\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}} - \frac{3}{7}{{\bf{e}}^{5t}}\\ y'\left( t \right) & = - 2{c_1}{{\bf{e}}^{ - 2t}} + 6{c_2}{{\bf{e}}^{6t}} - \frac
Now, apply the initial conditions to these.
\[\begin{align*}\frac{{18}}{7} = y\left( 0 \right) & = {c_1} + {c_2} - \frac{3}{7}\\ - \frac{1}{7} = y'\left( 0 \right) & = - 2{c_1} + 6{c_2} - \frac{{15}}{7}\end{align*}\]
Solving this system gives \(c_{1} = 2\) and \(c_{2} = 1\). The actual solution is then.
\[y\left( t \right) = 2{{\bf{e}}^{ - 2t}} + {{\bf{e}}^{6t}} - \frac{3}{7}{{\bf{e}}^{5t}}\]
This will be the only IVP in this section so don’t forget how these are done for nonhomogeneous differential equations!
Let’s take a look at another example that will give the second type of \(g(t)\) for which undetermined coefficients will work.
Example 3
Find a particular solution for the following differential equation. \[y'' - 4y' - 12y = \sin \left( {2t} \right)\]
Show Solution
Again, let’s note that we should probably find the complementary solution before we proceed onto the guess for a particular solution. However, because the homogeneous differential equation for this
example is the same as that for the first example we won’t bother with that here.
Now, let’s take our experience from the first example and apply that here. The first example had an exponential function in the \(g(t)\) and our guess was an exponential. This differential equation
has a sine so let’s try the following guess for the particular solution.
\[{Y_P}\left( t \right) = A\sin \left( {2t} \right)\]
Differentiating and plugging into the differential equation gives,
\[ - 4A\sin \left( {2t} \right) - 4\left( {2A\cos \left( {2t} \right)} \right) - 12\left( {A\sin \left( {2t} \right)} \right) = \sin \left( {2t} \right)\]
Collecting like terms yields
\[ - 16A\sin \left( {2t} \right) - 8A\cos \left( {2t} \right) = \sin \left( {2t} \right)\]
We need to pick \(A\) so that we get the same function on both sides of the equal sign. This means that the coefficients of the sines and cosines must be equal. Or,
\[\begin{align*}& \cos \left( {2t} \right)\,: & - 8A & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 0\\ & \sin \left( {2t} \right)\,:& - 16A & = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}A = -
Notice two things. First, since there is no cosine on the right hand side this means that the coefficient must be zero on that side. More importantly we have a serious problem here. In order for the
cosine to drop out, as it must in order for the guess to satisfy the differential equation, we need to set \(A = 0\), but if \(A = 0\), the sine will also drop out and that can’t happen. Likewise,
choosing \(A\) to keep the sine around will also keep the cosine around.
What this means is that our initial guess was wrong. If we get multiple values of the same constant or are unable to find the value of a constant then we have guessed wrong.
One of the nicer aspects of this method is that when we guess wrong our work will often suggest a fix. In this case the problem was the cosine that cropped up. So, to counter this let’s add a cosine
to our guess. Our new guess is
\[{Y_P}\left( t \right) = A\cos \left( {2t} \right) + B\sin \left( {2t} \right)\]
Plugging this into the differential equation and collecting like terms gives,
\[\begin{align*} - 4A\cos \left( {2t} \right) - 4B\sin \left( {2t} \right) - 4\left( { - 2A\sin \left( {2t} \right) + 2B\cos \left( {2t} \right)} \right) - \\ 12\left( {A\cos \left( {2t} \right) + B\
sin \left( {2t} \right)} \right) & = \sin \left( {2t} \right)\\ \left( { - 4A - 8B - 12A} \right)\cos \left( {2t} \right) + \left( { - 4B + 8A - 12B} \right)\sin \left( {2t} \right) & = \sin \left(
{2t} \right)\\ \left( { - 16A - 8B} \right)\cos \left( {2t} \right) + \left( {8A - 16B} \right)\sin \left( {2t} \right) & = \sin \left( {2t} \right)\end{align*}\]
Now, set the coefficients equal
\[\begin{align*} & \cos \left( {2t} \right)\,: &- 16A - 8B & = 0\\ & \sin \left( {2t} \right)\,: & 8A - 16B & = 1\end{align*}\]
Solving this system gives us
\[A = \frac{1}{{40}}\hspace{0.25in}\hspace{0.25in}B = - \frac{1}{{20}}\]
We found constants and this time we guessed correctly. A particular solution to the differential equation is then,
\[{Y_P}\left( t \right) = \frac{1}{{40}}\cos \left( {2t} \right) - \frac{1}{{20}}\sin \left( {2t} \right)\]
Notice that if we had had a cosine instead of a sine in the last example then our guess would have been the same. In fact, if both a sine and a cosine had shown up we will see that the same guess
will also work.
Let’s take a look at the third and final type of basic \(g(t)\) that we can have. There are other types of \(g(t)\) that we can have, but as we will see they will all come back to two types that
we’ve already done as well as the next one.
Example 4
Find a particular solution for the following differential equation. \[y'' - 4y' - 12y = 2{t^3} - t + 3\]
Show Solution
Once, again we will generally want the complementary solution in hand first, but again we’re working with the same homogeneous differential equation (you’ll eventually see why we keep working with
the same homogeneous problem) so we’ll again just refer to the first example.
For this example, \(g(t)\) is a cubic polynomial. For this we will need the following guess for the particular solution.
\[{Y_P}\left( t \right) = A{t^3} + B{t^2} + Ct + D\]
Notice that even though \(g(t)\) doesn’t have a \({t^2}\) in it our guess will still need one! So, differentiate and plug into the differential equation.
\[\begin{align*}6At + 2B - 4\left( {3A{t^2} + 2Bt + C} \right) - 12\left( {A{t^3} + B{t^2} + Ct + D} \right) & = 2{t^3} - t + 3\\ - 12A{t^3} + \left( { - 12A - 12B} \right){t^2} + \left( {6A - 8B -
12C} \right)t + 2B - 4C - 12D & = 2{t^3} - t + 3\end{align*}\]
Now, as we’ve done in the previous examples we will need the coefficients of the terms on both sides of the equal sign to be the same so set coefficients equal and solve.
\[\begin{align*} & {t^3}\,: & - 12A & = 2 & \Rightarrow \hspace{0.25in}A & = - \frac{1}{6}\\ & {t^2}\,: & - 12A - 12B & = 0 & \Rightarrow \hspace{0.25in}B & = \frac{1}{6}\\ & {t^1}\,: & 6A - 8B - 12C
& = - 1 & \Rightarrow \hspace{0.25in}C & = - \frac{1}{9}\\ & {t^0}\,: & 2B - 4C - 12D & = 3 & \Rightarrow \hspace{0.25in}D & = - \frac{5}{{27}}\end{align*}\]
Notice that in this case it was very easy to solve for the constants. The first equation gave \(A\). Then once we knew \(A\) the second equation gave \(B\), etc. A particular solution for this
differential equation is then
\[{Y_P}\left( t \right) = - \frac{1}{6}{t^3} + \frac{1}{6}{t^2} - \frac{1}{9}t - \frac{5}{{27}}\]
Now that we’ve gone over the three basic kinds of functions that we can use undetermined coefficients on let’s summarize.
\(g(t)\) \(Y_{P}(t)\) guess
\(a{{\bf{e}}^{\beta t}}\) \(A{{\bf{e}}^{\beta t}}\)
\(a\cos \left( {\beta t} \right)\) \(A\cos \left( {\beta t} \right) + B\sin \left( {\beta t} \right)\)
\(b\sin \left( {\beta t} \right)\) \(A\cos \left( {\beta t} \right) + B\sin \left( {\beta t} \right)\)
\(a\cos \left( {\beta t} \right) + b\sin \left( {\beta t} \right)\) \(A\cos \left( {\beta t} \right) + B\sin \left( {\beta t} \right)\)
\(n^{\mbox{th}}\) degree polynomial \({A_n}{t^n} + {A_{n - 1}}{t^{n - 1}} + \cdots {A_1}t + {A_0}\)
Notice that there are really only three kinds of functions given above. If you think about it the single cosine and single sine functions are really special cases of the case where both the sine and
cosine are present. Also, we have not yet justified the guess for the case where both a sine and a cosine show up. We will justify this later.
We now need move on to some more complicated functions. The more complicated functions arise by taking products and sums of the basic kinds of functions. Let’s first look at products.
Example 5
Find a particular solution for the following differential equation. \[y'' - 4y' - 12y = t{{\bf{e}}^{4t}}\]
Show Solution
You’re probably getting tired of the opening comment, but again finding the complementary solution first really a good idea but again we’ve already done the work in the first example so we won’t do
it again here. We promise that eventually you’ll see why we keep using the same homogeneous problem and why we say it’s a good idea to have the complementary solution in hand first. At this point all
we’re trying to do is reinforce the habit of finding the complementary solution first.
Okay, let’s start off by writing down the guesses for the individual pieces of the function. The guess for the \(t\) would be
\[At + B\]
while the guess for the exponential would be
Now, since we’ve got a product of two functions it seems like taking a product of the guesses for the individual pieces might work. Doing this would give
\[C{{\bf{e}}^{4t}}\left( {At + B} \right)\]
However, we will have problems with this. As we will see, when we plug our guess into the differential equation we will only get two equations out of this. The problem is that with this guess we’ve
got three unknown constants. With only two equations we won’t be able to solve for all the constants.
This is easy to fix however. Let’s notice that we could do the following
\[C{{\bf{e}}^{4t}}\left( {At + B} \right) = {{\bf{e}}^{4t}}\left( {ACt + BC} \right)\]
If we multiply the \(C\) through, we can see that the guess can be written in such a way that there are really only two constants. So, we will use the following for our guess.
\[{Y_P}\left( t \right) = {{\bf{e}}^{4t}}\left( {At + B} \right)\]
Notice that this is nothing more than the guess for the \(t\) with an exponential tacked on for good measure.
Now that we’ve got our guess, let’s differentiate, plug into the differential equation and collect like terms.
\[\begin{align*}{{\bf{e}}^{4t}}\left( {16At + 16B + 8A} \right) - 4\left( {{{\bf{e}}^{4t}}\left( {4At + 4B + A} \right)} \right) - 12\left( {{{\bf{e}}^{4t}}\left( {At + B} \right)} \right) & = t{{\bf
{e}}^{4t}}\\ \left( {16A - 16A - 12A} \right)t{{\bf{e}}^{4t}} + \left( {16B + 8A - 16B - 4A - 12B} \right){{\bf{e}}^{4t}} & = t{{\bf{e}}^{4t}}\\ - 12At{{\bf{e}}^{4t}} + \left( {4A - 12B} \right){{\bf
{e}}^{4t}} & = t{{\bf{e}}^{4t}}\end{align*}\]
Note that when we’re collecting like terms we want the coefficient of each term to have only constants in it. Following this rule we will get two terms when we collect like terms. Now, set
coefficients equal.
\[\begin{align*} & t{{\bf{e}}^{4t}}\,: & - 12A & = 1 & \Rightarrow \hspace{0.25in}A & = - \frac{1}{{12}}\\ & {{\bf{e}}^{4t}}\,: & 4A - 12B & = 0 & \Rightarrow \hspace{0.25in}B & = - \frac{1}{{36}}\
A particular solution for this differential equation is then
\[{Y_P}\left( t \right) = {{\bf{e}}^{4t}}\left( { - \frac{t}{{12}} - \frac{1}{{36}}} \right) = - \frac{1}{{36}}\left( {3t + 1} \right){{\bf{e}}^{4t}}\]
This last example illustrated the general rule that we will follow when products involve an exponential. When a product involves an exponential we will first strip out the exponential and write down
the guess for the portion of the function without the exponential, then we will go back and tack on the exponential without any leading coefficient.
Let’s take a look at some more products. In the interest of brevity we will just write down the guess for a particular solution and not go through all the details of finding the constants. Also,
because we aren’t going to give an actual differential equation we can’t deal with finding the complementary solution first.
Example 6
Write down the form of the particular solution to \[y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\]
1. \(g\left( t \right) = 16{{\bf{e}}^{7t}}\sin \left( {10t} \right)\)
2. \(g\left( t \right) = \left( {9{t^2} - 103t} \right)\cos t\)
3. \(g\left( t \right) = - {{\bf{e}}^{ - 2t}}\left( {3 - 5t} \right)\cos \left( {9t} \right)\)
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\(g\left( t \right) = 16{{\bf{e}}^{7t}}\sin \left( {10t} \right)\)
Show Solution
So, we have an exponential in the function. Remember the rule. We will ignore the exponential and write down a guess for \(16\sin \left( {10t} \right)\) then put the exponential back in.
The guess for the sine is
\[A\cos \left( {10t} \right) + B\sin \left( {10t} \right)\]
Now, for the actual guess for the particular solution we’ll take the above guess and tack an exponential onto it. This gives,
\[{Y_P}\left( t \right) = {{\bf{e}}^{7t}}\left( {A\cos \left( {10t} \right) + B\sin \left( {10t} \right)} \right)\]
One final note before we move onto the next part. The 16 in front of the function has absolutely no bearing on our guess. Any constants multiplying the whole function are ignored.
\(g\left( t \right) = \left( {9{t^2} - 103t} \right)\cos t\)
Show Solution
We will start this one the same way that we initially started the previous example. The guess for the polynomial is
\[A{t^2} + Bt + C\]
and the guess for the cosine is
\[D\cos t + E\sin t\]
If we multiply the two guesses we get.
\[\left( {A{t^2} + Bt + C} \right)\left( {D\cos t + E\sin t} \right)\]
Let’s simplify things up a little. First multiply the polynomial through as follows.
\[\begin{array}{c}\left( {A{t^2} + Bt + C} \right)\left( {D\cos t} \right) + \left( {A{t^2} + Bt + C} \right)\left( {E\sin t} \right)\\ \left( {AD{t^2} + BDt + CD} \right)\cos t + \left( {AE{t^2} +
BEt + CE} \right)\sin t\end{array}\]
Notice that everywhere one of the unknown constants occurs it is in a product of unknown constants. This means that if we went through and used this as our guess the system of equations that we would
need to solve for the unknown constants would have products of the unknowns in them. These types of systems are generally very difficult to solve.
So, to avoid this we will do the same thing that we did in the previous example. Everywhere we see a product of constants we will rename it and call it a single constant. The guess that we’ll use for
this function will be.
\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right)\cos t + \left( {D{t^2} + Et + F} \right)\sin t\]
This is a general rule that we will use when faced with a product of a polynomial and a trig function. We write down the guess for the polynomial and then multiply that by a cosine. We then write
down the guess for the polynomial again, using different coefficients, and multiply this by a sine.
\(g\left( t \right) = - {{\bf{e}}^{ - 2t}}\left( {3 - 5t} \right)\cos \left( {9t} \right)\)
Show Solution
This final part has all three parts to it. First, we will ignore the exponential and write down a guess for.
\[ - \left( {3 - 5t} \right)\cos \left( {9t} \right)\]
The minus sign can also be ignored. The guess for this is
\[\left( {At + B} \right)\cos \left( {9t} \right) + \left( {Ct + D} \right)\sin \left( {9t} \right)\]
Now, tack an exponential back on and we’re done.
\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {At + B} \right)\cos \left( {9t} \right) + {{\bf{e}}^{ - 2t}}\left( {Ct + D} \right)\sin \left( {9t} \right)\]
Notice that we put the exponential on both terms.
There a couple of general rules that you need to remember for products.
1. If \(g(t)\) contains an exponential, ignore it and write down the guess for the remainder. Then tack the exponential back on without any leading coefficient.
2. For products of polynomials and trig functions you first write down the guess for just the polynomial and multiply that by the appropriate cosine. Then add on a new guess for the polynomial with
different coefficients and multiply that by the appropriate sine.
If you can remember these two rules you can’t go wrong with products. Writing down the guesses for products is usually not that difficult. The difficulty arises when you need to actually find the
Now, let’s take a look at sums of the basic components and/or products of the basic components. To do this we’ll need the following fact.
If \(Y_{P1}(t)\) is a particular solution for
\[y'' + p\left( t \right)y' + q\left( t \right)y = {g_1}\left( t \right)\]
and if \(Y_{P2}(t)\) is a particular solution for
\[y'' + p\left( t \right)y' + q\left( t \right)y = {g_2}\left( t \right)\]
then \(Y_{P1}(t)\) + \(Y_{P2}(t)\) is a particular solution for
\[y'' + p\left( t \right)y' + q\left( t \right)y = {g_1}\left( t \right) + {g_2}\left( t \right)\]
This fact can be used to both find particular solutions to differential equations that have sums in them and to write down guess for functions that have sums in them.
Example 7
Find a particular solution for the following differential equation. \[y'' - 4y' - 12y = 3{{\bf{e}}^{5t}} + \sin \left( {2t} \right) + t{{\bf{e}}^{4t}}\]
Show Solution
This example is the reason that we’ve been using the same homogeneous differential equation for all the previous examples. There is nothing to do with this problem. All that we need to do it go back
to the appropriate examples above and get the particular solution from that example and add them all together.
Doing this gives
\[{Y_P}\left( t \right) = - \frac{3}{7}{{\bf{e}}^{5t}} + \frac{1}{{40}}\cos \left( {2t} \right) - \frac{1}{{20}}\sin \left( {2t} \right) - \frac{1}{{36}}\left( {3t + 1} \right){{\bf{e}}^{4t}}\]
Let’s take a look at a couple of other examples. As with the products we’ll just get guesses here and not worry about actually finding the coefficients.
Example 8
Write down the form of the particular solution to \[y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\]
for the following \(g(t)\)’s.
1. \(g\left( t \right) = 4\cos \left( {6t} \right) - 9\sin \left( {6t} \right)\)
2. \(g\left( t \right) = - 2\sin t + \sin \left( {14t} \right) - 5\cos \left( {14t} \right)\)
3. \(g\left( t \right) = {{\bf{e}}^{7t}} + 6\)
4. \(g\left( t \right) = 6{t^2} - 7\sin \left( {3t} \right) + 9\)
5. \(g\left( t \right) = 10{{\bf{e}}^t} - 5t{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{ - 8t}}\)
6. \(g\left( t \right) = {t^2}\cos t - 5t\sin t\)
7. \(g\left( t \right) = 5{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\cos \left( {6t} \right) - \sin \left( {6t} \right)\)
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\(g\left( t \right) = 4\cos \left( {6t} \right) - 9\sin \left( {6t} \right)\)
Show Solution
This first one we’ve actually already told you how to do. This is in the table of the basic functions. However, we wanted to justify the guess that we put down there. Using the fact on sums of
function we would be tempted to write down a guess for the cosine and a guess for the sine. This would give.
\[\underbrace {A\cos \left( {6t} \right) + B\sin \left( {6t} \right)}_{{\mbox{guess for the cosine}}} + \underbrace {C\cos \left( {6t} \right) + D\sin \left( {6t} \right)}_{{\mbox{guess for the
So, we would get a cosine from each guess and a sine from each guess. The problem with this as a guess is that we are only going to get two equations to solve after plugging into the differential
equation and yet we have 4 unknowns. We will never be able to solve for each of the constants.
To fix this notice that we can combine some terms as follows.
\[\left( {A + C} \right)\cos \left( {6t} \right) + \left( {B + D} \right)\sin \left( {6t} \right)\]
Upon doing this we can see that we’ve really got a single cosine with a coefficient and a single sine with a coefficient and so we may as well just use
\[{Y_P}\left( t \right) = A\cos \left( {6t} \right) + B\sin \left( {6t} \right)\]
The general rule of thumb for writing down guesses for functions that involve sums is to always combine like terms into single terms with single coefficients. This will greatly simplify the work
required to find the coefficients.
\(g\left( t \right) = - 2\sin t + \sin \left( {14t} \right) - 5\cos \left( {14t} \right)\)
Show Solution
For this one we will get two sets of sines and cosines. This will arise because we have two different arguments in them. We will get one set for the sine with just a \(t\) as its argument and we’ll
get another set for the sine and cosine with the 14\(t\) as their arguments.
The guess for this function is
\[{Y_P}\left( t \right) = A\cos t + B\sin t + C\cos \left( {14t} \right) + D\sin \left( {14t} \right)\]
\(g\left( t \right) = {{\bf{e}}^{7t}} + 6\)
Show Solution
The main point of this problem is dealing with the constant. But that isn’t too bad. We just wanted to make sure that an example of that is somewhere in the notes. If you recall that a constant is
nothing more than a zeroth degree polynomial the guess becomes clear.
The guess for this function is
\[{Y_p}\left( t \right) = A{{\bf{e}}^{7t}} + B\]
\(g\left( t \right) = 6{t^2} - 7\sin \left( {3t} \right) + 9\)
Show Solution
This one can be a little tricky if you aren’t paying attention. Let’s first rewrite the function
\[\begin{align*}g\left( t \right) & = 6{t^2} - 7\sin \left( {3t} \right) + 9\hspace{0.25in}{\mbox{as}}\\ g\left( t \right) & = 6{t^2} + 9 - 7\sin \left( {3t} \right)\end{align*}\]
All we did was move the 9. However, upon doing that we see that the function is really a sum of a quadratic polynomial and a sine. The guess for this is then
\[{Y_P}\left( t \right) = A{t^2} + Bt + C + D\cos \left( {3t} \right) + E\sin \left( {3t} \right)\]
If we don’t do this and treat the function as the sum of three terms we would get
\[A{t^2} + Bt + C + D\cos \left( {3t} \right) + E\sin \left( {3t} \right) + G\]
and as with the first part in this example we would end up with two terms that are essentially the same (the \(C\) and the \(G\)) and so would need to be combined. An added step that isn’t really
necessary if we first rewrite the function.
Look for problems where rearranging the function can simplify the initial guess.
\(g\left( t \right) = 10{{\bf{e}}^t} - 5t{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{ - 8t}}\)
Show Solution
So, this look like we’ve got a sum of three terms here. Let’s write down a guess for that.
\[A{{\bf{e}}^t} + \left( {Bt + C} \right){{\bf{e}}^{ - 8t}} + D{{\bf{e}}^{ - 8t}}\]
Notice however that if we were to multiply the exponential in the second term through we would end up with two terms that are essentially the same and would need to be combined. This is a case where
the guess for one term is completely contained in the guess for a different term. When this happens we just drop the guess that’s already included in the other term.
So, the guess here is actually.
\[{Y_P}\left( t \right) = A{{\bf{e}}^t} + \left( {Bt + C} \right){{\bf{e}}^{ - 8t}}\]
Notice that this arose because we had two terms in our \(g(t)\) whose only difference was the polynomial that sat in front of them. When this happens we look at the term that contains the largest
degree polynomial, write down the guess for that and don’t bother writing down the guess for the other term as that guess will be completely contained in the first guess.
\(g\left( t \right) = {t^2}\cos t - 5t\sin t\)
Show Solution
In this case we’ve got two terms whose guess without the polynomials in front of them would be the same. Therefore, we will take the one with the largest degree polynomial in front of it and write
down the guess for that one and ignore the other term. So, the guess for the function is
\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right)\cos t + \left( {D{t^2} + Et + F} \right)\sin t\]
\(g\left( t \right) = 5{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\cos \left( {6t} \right) - \sin \left( {6t} \right)\)
Show Solution
This last part is designed to make sure you understand the general rule that we used in the last two parts. This time there really are three terms and we will need a guess for each term. The guess
here is
\[{Y_P}\left( t \right) = A{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\left( {B\cos \left( {6t} \right) + C\sin \left( {6t} \right)} \right) + D\cos \left( {6t} \right) + E\sin \left( {6t} \right)\]
We can only combine guesses if they are identical up to the constant. So, we can’t combine the first exponential with the second because the second is really multiplied by a cosine and a sine and so
the two exponentials are in fact different functions. Likewise, the last sine and cosine can’t be combined with those in the middle term because the sine and cosine in the middle term are in fact
multiplied by an exponential and so are different.
So, when dealing with sums of functions make sure that you look for identical guesses that may or may not be contained in other guesses and combine them. This will simplify your work later on.
We have one last topic in this section that needs to be dealt with. In the first few examples we were constantly harping on the usefulness of having the complementary solution in hand before making
the guess for a particular solution. We never gave any reason for this other that “trust us”. It is now time to see why having the complementary solution in hand first is useful. This is best shown
with an example so let’s jump into one.
Example 9
Find a particular solution for the following differential equation. \[y'' - 4y' - 12y = {{\bf{e}}^{6t}}\]
Show Solution
This problem seems almost too simple to be given this late in the section. This is especially true given the ease of finding a particular solution for \(g\)(\(t\))’s that are just exponential
functions. Also, because the point of this example is to illustrate why it is generally a good idea to have the complementary solution in hand first we’ll let’s go ahead and recall the complementary
solution first. Here it is,
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}}\]
Now, without worrying about the complementary solution for a couple more seconds let’s go ahead and get to work on the particular solution. There is not much to the guess here. From our previous work
we know that the guess for the particular solution should be,
\[{Y_P}\left( t \right) = A{{\bf{e}}^{6t}}\]
Plugging this into the differential equation gives,
\[\begin{align*}36A{{\bf{e}}^{6t}} - 24A{{\bf{e}}^{6t}} - 12A{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ 0 & = {{\bf{e}}^{6t}}\end{align*}\]
Hmmmm…. Something seems wrong here. Clearly an exponential can’t be zero. So, what went wrong? We finally need the complementary solution. Notice that the second term in the complementary solution
(listed above) is exactly our guess for the form of the particular solution and now recall that both portions of the complementary solution are solutions to the homogeneous differential equation,
\[y'' - 4y' - 12y = 0\]
In other words, we had better have gotten zero by plugging our guess into the differential equation, it is a solution to the homogeneous differential equation!
So, how do we fix this? The way that we fix this is to add a \(t\) to our guess as follows.
\[{Y_P}\left( t \right) = At{{\bf{e}}^{6t}}\]
Plugging this into our differential equation gives,
\[\begin{align*}\left( {12A{{\bf{e}}^{6t}} + 36At{{\bf{e}}^{6t}}} \right) - 4\left( {A{{\bf{e}}^{6t}} + 6At{{\bf{e}}^{6t}}} \right) - 12At{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ \left( {36A - 24A -
12A} \right)t{{\bf{e}}^{6t}} + \left( {12A - 4A} \right){{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ 8A{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\end{align*}\]
Now, we can set coefficients equal.
\[8A = 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,A = \frac{1}{8}\]
So, the particular solution in this case is,
\[{Y_P}\left( t \right) = \frac{t}{8}{{\bf{e}}^{6t}}\]
So, what did we learn from this last example. While technically we don’t need the complementary solution to do undetermined coefficients, you can go through a lot of work only to figure out at the
end that you needed to add in a \(t\) to the guess because it appeared in the complementary solution. This work is avoidable if we first find the complementary solution and comparing our guess to the
complementary solution and seeing if any portion of your guess shows up in the complementary solution.
If a portion of your guess does show up in the complementary solution then we’ll need to modify that portion of the guess by adding in a \(t\) to the portion of the guess that is causing the
problems. We do need to be a little careful and make sure that we add the \(t\) in the correct place however. The following set of examples will show you how to do this.
Example 10
Write down the guess for the particular solution to the given differential equation. Do not find the coefficients.
1. \(y'' + 3y' - 28y = 7t + {{\bf{e}}^{ - 7t}} - 1\)
2. \(y'' - 100y = 9{t^2}{{\bf{e}}^{10t}} + \cos t - t\sin t\)
3. \(4y'' + y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\)
4. \(4y'' + 16y' + 17y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\)
5. \(y'' + 8y' + 16y = {{\bf{e}}^{ - 4t}} + \left( {{t^2} + 5} \right){{\bf{e}}^{ - 4t}}\)
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Show Discussion
In these solutions we’ll leave the details of checking the complementary solution to you.
\(y'' + 3y' - 28y = 7t + {{\bf{e}}^{ - 7t}} - 1\)
Show Solution
The complementary solution is
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}{{\bf{e}}^{ - 7t}}\]
Remembering to put the “-1” with the 7\(t\) gives a first guess for the particular solution.
\[{Y_P}\left( t \right) = At + B + C{{\bf{e}}^{ - 7t}}\]
Notice that the last term in the guess is the last term in the complementary solution. The first two terms however aren’t a problem and don’t appear in the complementary solution. Therefore, we will
only add a \(t\) onto the last term.
The correct guess for the form of the particular solution is.
\[{Y_P}\left( t \right) = At + B + Ct{{\bf{e}}^{ - 7t}}\]
\(y'' - 100y = 9{t^2}{{\bf{e}}^{10t}} + \cos t - t\sin t\)
Show Solution
The complementary solution is
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{10t}} + {c_2}{{\bf{e}}^{ - 10t}}\]
A first guess for the particular solution is
\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right){{\bf{e}}^{10t}} + \left( {Et + F} \right)\cos t + \left( {Gt + H} \right)\sin t\]
Notice that if we multiplied the exponential term through the parenthesis that we would end up getting part of the complementary solution showing up. Since the problem part arises from the first term
the whole first term will get multiplied by \(t\). The second and third terms are okay as they are.
The correct guess for the form of the particular solution in this case is.
\[{Y_P}\left( t \right) = t\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{10t}} + \left( {Et + F} \right)\cos t + \left( {Gt + H} \right)\sin t\]
So, in general, if you were to multiply out a guess and if any term in the result shows up in the complementary solution, then the whole term will get a \(t\) not just the problem portion of the
\(4y'' + y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\)
Show Solution
The complementary solution is
\[{y_c}\left( t \right) = {c_1}\cos \left( {\frac{t}{2}} \right) + {c_2}\sin \left( {\frac{t}{2}} \right)\]
A first guess for the particular solution is
\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \left
( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]
In this case both the second and third terms contain portions of the complementary solution. The first term doesn’t however, since upon multiplying out, both the sine and the cosine would have an
exponential with them and that isn’t part of the complementary solution. We only need to worry about terms showing up in the complementary solution if the only difference between the complementary
solution term and the particular guess term is the constant in front of them.
So, in this case the second and third terms will get a \(t\) while the first won’t
The correct guess for the form of the particular solution is.
\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + t\left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + t\
left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]
\(4y'' + 16y' + 17y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\)
Show Solution
To get this problem we changed the differential equation from the last example and left the \(g(t)\) alone. The complementary solution this time is
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}}\cos \left( {\frac{t}{2}} \right) + {c_2}{{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right)\]
As with the last part, a first guess for the particular solution is
\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \left
( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]
This time however it is the first term that causes problems and not the second or third. In fact, the first term is exactly the complementary solution and so it will need a \(t\). Recall that we will
only have a problem with a term in our guess if it only differs from the complementary solution by a constant. The second and third terms in our guess don’t have the exponential in them and so they
don’t differ from the complementary solution by only a constant.
The correct guess for the form of the particular solution is.
\[{Y_P}\left( t \right) = t{{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \
left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]
\(y'' + 8y' + 16y = {{\bf{e}}^{ - 4t}} + \left( {{t^2} + 5} \right){{\bf{e}}^{ - 4t}}\)
Show Solution
The complementary solution is
\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 4t}} + {c_2}t{{\bf{e}}^{ - 4t}}\]
The two terms in \(g(t)\) are identical with the exception of a polynomial in front of them. So this means that we only need to look at the term with the highest degree polynomial in front of it. A
first guess for the particular solution is
\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]
Notice that if we multiplied the exponential term through the parenthesis the last two terms would be the complementary solution. Therefore, we will need to multiply this whole thing by a \(t\).
The next guess for the particular solution is then.
\[{Y_P}\left( t \right) = t\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]
This still causes problems however. If we multiplied the \(t\) and the exponential through, the last term will still be in the complementary solution. In this case, unlike the previous ones, a \(t\)
wasn’t sufficient to fix the problem. So, we will add in another \(t\) to our guess.
The correct guess for the form of the particular solution is.
\[{Y_P}\left( t \right) = {t^2}\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]
Upon multiplying this out none of the terms are in the complementary solution and so it will be okay.
As this last set of examples has shown, we really should have the complementary solution in hand before even writing down the first guess for the particular solution. By doing this we can compare our
guess to the complementary solution and if any of the terms from your particular solution show up we will know that we’ll have problems. Once the problem is identified we can add a \(t\) to the
problem term(s) and compare our new guess to the complementary solution. If there are no problems we can proceed with the problem, if there are problems add in another \(t\) and compare again.
Can you see a general rule as to when a \(t\) will be needed and when a t^2 will be needed for second order differential equations? | {"url":"https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx","timestamp":"2024-11-14T12:11:07Z","content_type":"text/html","content_length":"128555","record_id":"<urn:uuid:bb26fad5-265d-457c-bc60-66c3e7cecb95>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00267.warc.gz"} |
class pydl.pcomp(x, standardize=False, covariance=False)[source]¶
Bases: object
Replicates the IDL PCOMP() function.
The attributes of this class are all read-only properties, implemented with lazyproperty.
A 2-D array with \(N\) rows and \(M\) columns.
standardizebool, optional
If set to True, the input data will have its mean subtracted off and will be scaled to unit variance.
covariancebool, optional.
If set to True, the covariance matrix of the data will be used for the computation. Otherwise the correlation matrix will be used.
Attributes Summary
coefficients (ndarray) The principal components.
derived (ndarray) The derived variables.
eigenvalues (ndarray) The eigenvalues.
variance (ndarray) The variances of each derived variable.
Attributes Documentation
(ndarray) The principal components. These are the coefficients of derived. Basically, they are a re-scaling of the eigenvectors.
(ndarray) The derived variables.
(ndarray) The eigenvalues. There is one eigenvalue for each principal component.
(ndarray) The variances of each derived variable. | {"url":"https://pydl.readthedocs.io/en/stable/api/pydl.pcomp.html","timestamp":"2024-11-08T15:11:18Z","content_type":"text/html","content_length":"15442","record_id":"<urn:uuid:79f966d3-5afb-4be6-a990-1a799932a588>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00304.warc.gz"} |
Multi-Scale Modeling | Accuracy, Speed & Applications in Continuum Mechanics
Multi-scale modeling
Explore the power of Multi-Scale Modeling in Continuum Mechanics, its balance of accuracy and speed, and diverse applications in technology.
Understanding Multi-Scale Modeling: A Revolution in Continuum Mechanics
Multi-Scale Modeling (MSM) has emerged as a transformative approach in the realm of continuum mechanics, bridging the gap between macroscopic phenomena and their microscopic origins. This methodology
integrates various scales, from atomic to macroscopic levels, offering a comprehensive understanding of material behavior and system dynamics. The essence of MSM lies in its ability to encapsulate
detailed microstructural details into larger-scale models, thus providing insights into the complex interactions governing material properties and system performance.
Accuracy and Speed in Multi-Scale Modeling
One of the critical advantages of MSM is its balance between accuracy and computational efficiency. Traditional continuum models, while useful for large-scale predictions, often overlook microscopic
details essential for accurate material characterization. Conversely, atomistic models, though precise, are computationally intensive and impractical for large systems. MSM adeptly navigates this
trade-off by integrating different modeling techniques, such as Molecular Dynamics (MD), Finite Element Analysis (FEA), and Computational Fluid Dynamics (CFD), to capture the essential physics at
each scale.
Applications in Continuum Mechanics
The versatility of MSM is evident in its wide range of applications in continuum mechanics. In materials science, it enables the prediction of mechanical properties like strength, ductility, and
fracture toughness, considering the underlying microstructure. In fluid dynamics, MSM aids in understanding complex flow phenomena by linking molecular interactions to macroscopic flow
characteristics. Additionally, MSM plays a crucial role in the development of advanced materials, such as high-performance composites and nanomaterials, by facilitating the optimization of their
microstructural attributes.
Key Challenges and Future Directions
Despite its potential, MSM faces several challenges. Ensuring seamless integration between different scales and maintaining accuracy across these scales remain critical concerns. Advances in
computational power and algorithms continue to address these issues, making MSM more accessible and reliable. Furthermore, the development of standardized protocols for MSM implementation in various
fields is crucial for its broader adoption.
In conclusion, Multi-Scale Modeling stands at the forefront of innovation in continuum mechanics, offering a powerful tool for scientists and engineers. Its ability to provide detailed insights while
maintaining computational efficiency opens new horizons in material design, system optimization, and the understanding of complex phenomena.
Advancements in Computational Techniques and Software
Progress in computational methods has significantly enhanced the capabilities of MSM. Novel algorithms and high-performance computing resources have enabled the handling of complex simulations with
greater speed and accuracy. The development of specialized software and open-source tools has democratized access to MSM, allowing researchers and engineers to conduct advanced analyses without
extensive computational resources. These advancements are crucial in tackling large-scale problems that were previously beyond the scope of traditional modeling approaches.
Integration with Machine Learning and Artificial Intelligence
The integration of machine learning (ML) and artificial intelligence (AI) with MSM represents a groundbreaking shift in continuum mechanics. ML algorithms can identify patterns and relationships
within data sets generated by multi-scale simulations, leading to more accurate predictive models. AI can automate the selection of appropriate scales and modeling techniques based on the problem at
hand, significantly enhancing the efficiency of the modeling process. This synergy between MSM, ML, and AI is paving the way for smarter, adaptive models capable of handling unprecedented complexity.
Environmental and Sustainability Applications
MSM is increasingly applied in environmental science and sustainability. It aids in understanding the behavior of materials under environmental stressors, which is crucial in designing sustainable
materials and structures. Moreover, MSM contributes to the development of renewable energy systems, such as solar panels and wind turbines, by optimizing material properties for enhanced performance
and durability under varying environmental conditions.
Conclusion: The Future of Multi-Scale Modeling in Continuum Mechanics
Multi-Scale Modeling has established itself as an indispensable tool in continuum mechanics, offering a unique perspective that combines the microscopic and macroscopic worlds. The future of MSM is
bright, with ongoing advancements in computational power, software development, and the integration of AI and ML. These developments are not only enhancing the accuracy and efficiency of MSM but are
also expanding its application horizons. From designing next-generation materials to addressing global environmental challenges, MSM stands as a cornerstone in the pursuit of scientific and
engineering breakthroughs. Its ability to provide comprehensive insights into complex systems ensures its vital role in shaping the future of technology and sustainability. | {"url":"https://modern-physics.org/multi-scale-modeling/","timestamp":"2024-11-05T17:10:33Z","content_type":"text/html","content_length":"160893","record_id":"<urn:uuid:2fe0ec69-d88a-4b53-adce-9447ba51ced7>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00231.warc.gz"} |
The Mathlingua Language
Mathlingua is a declarative language designed to precisely and concisely describe statements of mathematical definitions, axioms, theorems, and conjectures, such that anyone familiar with reading and
writing mathematical literature can easily learn to read and write Mathlingua text, and content written in Mathlingua has automated checks such as (but not limited to):
• no definition or symbol is used that is not unknown
• no duplicate definitions occur
• no definition is used incorrectly (i.e. the inputs to any definition used are of the correct count and type)
• no statement is ambiguous (i.e. text such as \(a * b\) where the meaning of \(*\) cannot be determined is not allowed)
Why is it needed?
When learning mathematics, books, articles, and encyclopediae can be a great resources that are generally easy to read and write. Sometimes though, the content in these resources can be informal and
sometimes ambiguous when the meaning of some symbols need to be implied by the context.
Next, proof assistants such as Lean, Coq, Isabelle, and others are very formal but have a very steep learning curve and, although they can be used to write proofs that can be verified by computer,
can be very difficult to read and write.
Mathlingua aims to take the best of both approaches. In particular, it is designed to be easy to read and write, be precise and concise, and allow proofs to be expresed, with some checks done. The
language isn't rigid enough to allow proofs to be automatically verified by the system, but has enough structure to allow people to write proofs that can have the checks mentioned above automatically
performed so that humans can focus on checking the logic of the proof.
What is the purpose of the language?
The Mathlingua language is designed to create Mathlore, a free and open knowledgebase of mathematical knowledge to allow anyone access to precise mathematical knowledge.
What does it look like?
To get a feel for the language, the following is a definition of a prime integer:
Describes: p
extends: 'p is \integer'
. 'p != 1'
. not:
. exists: a, b
where: 'a, b is \integer'
. 'a != 1'
. 'b != 1'
. 'p = a * b'
. called: "prime natural number"
that is rendered as:
Describes: \(p\)
extends: \(p\) is integer
· \(p \neq 1\)
· not:
. exists: \(a, b\)
where: \(a, b\) is integer
. \(a \neq 1\)
. \(b \neq 1\)
. \(p = ab\)
· called: prime natural number
Next, the following is an example of a theorem:
given: p, a, b
. 'p is \prime.integer'
. 'a, b is \integer'
if: 'p | a * b' (1)
. anyOf:
. 'p | a'
. 'p | b'
. called: "Euclid's Lemma"
. withoutLossOfGenerality:
. suppose: 'p \coprime.to/ a'
. sequentially:
. notice:
. exists: r, s
where: 'r, s is \integer'
suchThat: 'r*p + s*a = 1'
by: '\bezouts.lemma'
. hence: 'r*p*b + s*a*b = b' (2)
because: "multiply both sides by $b$"
. notice: 'p | r*p*b'
. next: 'p | s*a*b'
by: '\(1)'
because: 'p | a * b'
. thus: 'p | r*p*b + s*a*b'
. hence: 'p | b'
by: '\(2)'
. qed:
Note that the name \prime.integer uses the . character to specify that the prime being described is a prime integer. For a prime element in an arbitrary commutative algebra, a different definition
would be created, perhaps called \prime.element:of{A} that specifies that A must be a \commutative.algebra.
Next, the definition describes how a prime integer is related to an integer by stating that it extends an integer. That is, a prime integer is an integer with additional properties.
Although not shown here, Mathlingua also allows one to describe that something can be viewed as something else, perhaps through a morphism.
Further, not only is the precise mathematical statement of the definition and theorem expressed, but further information, such as what the items are called, is encoded.
Although not shown here, Mathlingua allows for describing a much larger assortment of knowledge associated with a mathematical item, such as how a symbol or expression is written, the item's history,
discoverer(s), importance, informal description, references, relationship to other mathematical concepts, etc. in not only English but in any other written language. | {"url":"https://mathlingua.org/","timestamp":"2024-11-05T10:33:48Z","content_type":"text/html","content_length":"31477","record_id":"<urn:uuid:3d087093-a489-4b80-914d-f2e44977e0d1>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00131.warc.gz"} |
Wrong placement using \displaylimits and unicode-math
8 Dec 2021 8 Dec '21
11:32 p.m.
Hello there, there were some reports regarding the placement of the limits in integrals a few years ago https://tug.org/pipermail/luatex/2015-March/005076.html. It was with regard to display math and
`\limits`. They seem to work fine for me, but I spotted some weird behavior using `\displaylimits` *in in-line math*. In my opinion, the limits of integration are placed way too much on the right.
The expected behavior is to have the same placement as with `\nolimits` in in-line math, and as with `\limits` in display math. The problem doesn't occur with XeLaTeX and it seems to be independent
of a font. Here is a minimal working example comparing different options: ``` latex %! TeX program = lualatex \documentclass{article} \usepackage{unicode-math} \begin{document} \(\int\limits_a^b f(x)
\,\mathrm dx\) \(\int\nolimits_a^b f(x)\,\mathrm dx\) \(\int\displaylimits_a^b f(x)\,\mathrm dx\) \[\int\limits_a^b f(x)\,\mathrm dx\] \[\int\nolimits_a^b f(x)\,\mathrm dx\] \[\int\displaylimits_a^b
f(x)\,\mathrm dx\] \end{document} ```` Here, the third line after `\begin{document}` is the faulty one. Best regards Jakub Kaczor
Last active (days ago)
0 comments
1 participants | {"url":"https://mailman.ntg.nl/archives/list/dev-luatex@ntg.nl/thread/DY3N62VZJAQU2NXAKZH4DQGBT3GG6KV5/?sort=thread","timestamp":"2024-11-08T01:50:12Z","content_type":"text/html","content_length":"17948","record_id":"<urn:uuid:151e612a-1b25-4bb2-bba8-99a2c73662df>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00777.warc.gz"} |
How far does a .45 caliber bullet travel? | [November Updated]
How far does a .45 caliber bullet travel?
How far does a .45 caliber bullet travel?
A .45 caliber bullet can travel up to 1.5 miles when fired at a flat angle.
How far can a .45 caliber bullet travel when fired at an angle?
When fired at an angle, a .45 caliber bullet can travel even further, up to 2 miles.
What factors can affect the distance a .45 caliber bullet travels?
Factors such as muzzle velocity, bullet weight, and wind conditions can all affect the distance a .45 caliber bullet travels.
Can a .45 caliber bullet travel through walls?
Yes, a .45 caliber bullet has the potential to penetrate through multiple interior walls.
What is the effective range of a .45 caliber handgun?
The effective range of a .45 caliber handgun is typically around 50 meters.
Can a .45 caliber bullet travel through body armor?
In most cases, standard body armor is not able to stop a .45 caliber bullet.
Does the type of ammunition affect the distance a .45 caliber bullet travels?
Yes, different types of ammunition, such as hollow points or full metal jackets, can impact the bullet’s trajectory and distance.
What is the maximum effective range of a .45 caliber rifle?
The maximum effective range of a .45 caliber rifle is generally around 200-300 meters.
Can a .45 caliber bullet travel through a car door?
Yes, a .45 caliber bullet has the potential to penetrate through a car door.
How does the barrel length of a firearm affect the distance a .45 caliber bullet travels?
A longer barrel length can generally result in higher muzzle velocity and a longer travel distance for a .45 caliber bullet.
What is the maximum range of a .45 caliber subsonic round?
The maximum range of a .45 caliber subsonic round is typically around 3000 meters.
Does the angle of the shot affect how far a .45 caliber bullet travels?
Yes, the angle of the shot can significantly impact the distance a .45 caliber bullet travels.
Can a .45 caliber bullet travel through a tree trunk?
Yes, a .45 caliber bullet has the potential to penetrate through a tree trunk, depending on the type and size of the tree.
How far can a .45 caliber bullet travel in water?
A .45 caliber bullet can travel up to 5 feet in water before losing its momentum.
Are there any legal restrictions on the use of .45 caliber firearms based on their range?
In some jurisdictions, there may be restrictions on the use of .45 caliber firearms based on their potential range and destructive capabilities.
Leave a Comment | {"url":"https://thegunzone.com/how-far-does-a-45-caliber-bullet-travel/","timestamp":"2024-11-08T15:37:15Z","content_type":"text/html","content_length":"65141","record_id":"<urn:uuid:5553ce48-593b-42db-8ce8-e8b1f100b875>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00351.warc.gz"} |
Orders of Magnitude, Logarithmic Scales, and Decibels
Ultimate Electronics: Practical Circuit Design and Analysis
Easily work with numerical quantities that range over many orders of magnitude. 10 min read
Electrical engineering, to a degree more than many other fields of engineering, features coexisting, interacting quantities over many orders of magnitude. Both a resistor and a resistor are common,
and it is not unusual to find a circuit board that features at least one of each.
Determining if a particular resistor should be is far more important than whether it should be .
Logarithms compress values from many orders of magnitude down to a smaller space. Some of the basic rules about logarithms to remember include:
(The equation is read aloud as: “The log-base-two of sixteen equals four.”)
If the base is not explicitly specified, the context usually determines the intended base. Base is commonly used in engineering and is commonly written as . In mathematics, the natural base is
commonly used and written as . In computer science, base is commonly used and written as . If it’s ever unspecified and unclear, ask.
Taking a logarithm compresses multiplicative differences into additive differences. For example, if we have two functions and , then the logarithms (in any base) of these two functions will only
differ by a constant:
This is a useful way of thinking about lots of electronic systems like amplifiers and filters, which in general apply multiplicative transformations to an input signal. With logarithms, we can think
of that multiplicative transformation as just shifting the value up or down on the y-axis.
The decibels difference between two multiplicatively-related voltage or current signals is defined as:
where A is in dB. In this context, the logarithm is defined as being base 10.
Where does the factor 20 come from? The answer is that this was defined so that a 10X multiplicative increase in power would be equivalent to adding +10dB.
Because power is proportional to the square of voltage (which we’ll discuss in more detail in the Power section), a increase in voltage through an amplifier results in a increase in power, or +20dB.
Since , then:
(To avoid complicating the discussion of decibels at this early stage, we’re going to avoid talking about reference impedance here. As logarithms of negative numbers don’t exist, we’re also generally
talking about the envelope or amplitude of the signal, rather than the signal waveform itself.)
Here are some common decibel values and hints about where they’re from:
It is not unusual to see decibel values outside these bounds: +80 dB represents a power multiplier of and a voltage multiplier of . In general, we can just invert the equations above for A dB:
The values for and are not exact but are common approximations, because:
These exact decibel values for multiplication by and are commonly truncated. That’s why engineers speak colloquially about +3 dB or +6 dB to refer to a power gain of 2 or a voltage gain of 2,
If we connected two amplifiers together in series, each of which had 10x voltage gain, then each would contribute +20dB gain. We can add the decibel values to get +40dB overall voltage gain – a
factor of in voltage and a factor of in power.
Absolute Decibel Units
Any unit specified only as “dB” refers only to a relative factor between signals.
However, it is often convenient to define the denominator as a specific unit-bearing quantity.
For example, “dBm” means “decibels relative to 1 milliwatt.” Therefore, +20 dBm represents “+20dB relative to 1 milliwatt.” As Watts are a unit of power, and +20 dB is a multiplication of 100X in
power, then “+20 dBm” is simply a short way of saying “100 milliwatts.”
In another example, “dBV” means “decibels relative to 1 Volt.” Therefore, -60 dBv represents “-60dB relative to 1 volt.” As -60 dB in voltage is a voltage factor of , then “-60 dBV” is simply a short
way of saying “1 millivolt.”
Be careful to track whether a decibel unit is relative or absolute, and whether the absolute reference represents a power or voltage level.
Decibels Rolloff
In future sections when we talk about frequency response of amplifiers and filters, we will sometimes consider numbers like “-20 dB/decade” or “-6 dB/octave”.
A decade is a 10X increase in frequency, for example from 100 Hz to 1000 Hz. Saying “minus 20 dB per decade” might mean that the output signal at 1000 Hz is only as large in voltage amplitude as the
signal at 100 Hz.
An octave is a 2X increase in frequency, just as it is in music. For example 100 Hz and 200 Hz are one octave apart. Saying “minus 6 dB per octave” might mean that the output signal at 200 Hz is only
as large in voltage amplitude as the signal at 100 Hz.
It turns out that these two examples are actually referring to the same thing:
Let’s prove it. How many octaves are in a decade?
If we start at frequency and double it three times (i.e. go up three octaves), we’ll end up with – almost enough to reach , but not quite. If we double four times (i.e. go up four octaves), we’ll end
up with – more than a decade; we’ve gone too far. The real number of octaves necessary to complete a decade is between three and four, and is shown as above.
If our signal is reduced by -6 dB/octave, and we go up in frequency by 3.322 octaves, the total reduction is:
That -19.932 dB is really precisely -20dB when we account for the fact that -6 dB is only a rounded-off approximation for .
Semi-Logarithmic Plotting
On a semi-log plot, the y-axis is plotted on a logarithmic scale while the x-axis remains linear. This turns functions that are exponentially changing (with any base) into straight lines. For
example, consider the value of a savings account with initial value 1000 growing at 5% per year, with time in years. The value of the account is:
If we take the log,
The logarithm has transformed an exponential (with base 1.05) into a straight line.
For a similar example, plotting the value of a stock or mutual fund should be done on a semi-log plot. (It should also be plotted with dividends reinvested to show total returns, though this is a
separate issue we can’t address here.) Why logarithmic? Because if we plot with a linear scale, it looks like all the growth has happened in just the last few years. If we plot with a logarithmic
scale, it reveals that in fact an investor at roughly any point has roughly doubled their investment value in about a decade. This reflects the idea that in general, in any given year, the company’s
profit (whether retained earnings or distributed as dividends) will be roughly a few percent of its current value.
To see for yourself, load this data, click “Max” to set the maximum timescale from 1970 onwards. The linear graph looks absolutely tiny for the first half of the dataset. Now, click “Edit Graph”,
click “Format”, and check the box next to “Log Scale”; if the horizontal scale resets you may have to click “Max” again. Now, while it’s not smooth or consistent or risk free, you’d see that in fact
the general trend of growth (in a %/year growth sense) has been present for many decades.
Log-Log Plotting
If plotting with a logarithmic scale on the y-axis is powerful, then plotting with a logarithmic scale on both the x-axis and y-axis is really powerful.
Consider our function that we used for computing various approximations in the Algebraic Approximations section. Taking the log of both sides:
This alone doesn’t look very easy to understand, but let’s again apply various approximations in different limiting asymptotes for :
Now let’s define and . These represent the logarithmically-transformed and axes for plotting.
Now we have two very simple equations for lines which are good approximations for in different regimes: the first section has a slope of 0 (a horizontal line), and the second section has a slope of
-1. The approximation line will do badly near the “knee” where the two approximations meet, but will do well everywhere else.
Instead of defining these in terms of we can transform to , and join the lines into a single approximation function (again looking only at positive due to the domain of logarithm):
Finally, we can join these at by seeing that we can take the minimum value of either segment at any given :
This is simply a compact representation of the two segments above, but importantly, it is a representation that the mathematical expression evaluation software inside CircuitLab knows how to handle
directly using the MIN(a,b) notation – see the documentation on expressions for more information.
We’ll show how good the log-log plot is by considering value of from – a factor of 10 billion from bottom to top!
Exercise Click the “circuit” shown above, then click “Simulate”, then “Run DC Sweep”. You’ll see two lines plotted:
The approximation is quite good for and , which corresponds to and . In the middle, neither approximation is excellent, but the error can be quantified now that we know it’s there.
We’ve taken a complicated-looking fraction and turned it into a collection of straight line segments by using two tools: log-log plots and asymptotic approximations. That’s the power of log-log
plotting: it turns rational polynomials (fractions with in the numerator or denominator) into straight line approximations that work over many orders of magnitude.
Note that the second line segment has slope -1, corresponding to . If the original fraction had , the slope would be -2.
Log-log plots and the straight line approximation concept will come up again when we look at filters and amplifiers and frequency response. Log-log is a great fit there because the independent
variable (frequency) varies over many orders of magnitude, as does the dependent variable (amplification of magnitude).
And, as we’ll see, the frequency response of common filters can be described by rational polynomials, so we’ll get straight line segment approximations there too when using these same tools. For
example, a simple RC low-pass filter will have a amplitude response for frequencies above its corner frequency, mapping precisely to the example we’ve just illustrated.
If the concepts of asymptotic approximations, semi-log, and log-log plots are still unfamiliar, you’re encouraged to brush up and explore these concepts on paper and/or in simulation, as they will
come up again and again as we go.
Metric Decimal Unit Prefixes
As electronic quantities span many orders of magnitude, it’s common in electronics to use unit prefixes that help designate orders-of-magnitude at various powers of . These make human-readable
numbers simpler.
The abbreviation for micro is , the Greek lowercase letter mu. However, because this is hard to type into a computer, it’s often written with just the lowercase letter u: , but is spoken aloud as
“twenty two micro Farads” either way.
Most electronics software, including the CircuitLab simulation software, will understand these prefixes if the prefix comes immediately after the digits of the number (with no space in between). For
example, you can enter “2.2k” in the value field for a resistor to get a 2200 ohm resistance. One confusing case is that mega is represented by uppercase “M” and milli is represented by lowercase
“m”; modern software like CircuitLab will interpret these two correctly as written, but some older SPICE-based circuit software dates back to the 1970s and can’t handle the difference between
uppercase and lowercase characters! These older programs instead use “MEG” to denote mega. Be careful that the computer interprets your value correctly.
What’s Next
In the next section, Complex Numbers, we’ll talk about real and imaginary numbers and their connection to circles, trigonometry, and sine waves. (Later, we’ll connect complex numbers with our
algebraic approximations & log-log plotting to explore the frequency domain.)
Robbins, Michael F. Ultimate Electronics: Practical Circuit Design and Analysis. CircuitLab, Inc., 2021, ultimateelectronicsbook.com. Accessed . (Copyright © 2021 CircuitLab, Inc.) | {"url":"https://ultimateelectronicsbook.com/orders-of-magnitude-logarithmic-scales-decibels/","timestamp":"2024-11-09T04:43:20Z","content_type":"text/html","content_length":"269885","record_id":"<urn:uuid:596b1034-01fb-493e-97e4-d486a1a1009d>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00724.warc.gz"} |
Expected value of the future
Expected value of the future
Is the expected value of the future positive or negative? A crucial consideration for the value of extinction risk reduction.
What should we do given that we can't evaluate the vast indirect effects of our actions?
Expected value and fanaticism
Expected value and fanaticism
Is maximising expected value the right approach for trying to do the most good? How should we approach tiny probabilities of doing vast amounts of good? | {"url":"https://library.globalchallengesproject.org/navigation/all-topics/topics/increasing-the-accuracy-of-our-judgements/topics","timestamp":"2024-11-11T05:26:24Z","content_type":"text/html","content_length":"140575","record_id":"<urn:uuid:5e7c88e7-0b20-40a8-8e13-7edb737aae28>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00069.warc.gz"} |
Log Weight Chart
Log Weight Chart - Web in this calculator, you will learn to answer the question how much does a log weigh? and what affects the weight of wood. Download or print a pdf of the green log weight. Web
estimate the weight and density of any green log by selecting the species and measuring the log size. The green log weight chart is a handy quick reference guide to 66 different tree species with
weights. Web calculate the weight of logs based on species, diameter, length and quantity. Web find out how to calculate the weight of green logs using a formula and a chart. Use the green log
weight. Choose from a list of common north. Calculate the merchantable volume and green weight of. Volume and green weight of a log calculator.
12 Best Printable Weight Log Sheet PDF for Free at Printablee
Web find out how to calculate the weight of green logs using a formula and a chart. The green log weight chart is a handy quick reference guide to 66 different tree species with weights. Web the
green log weight chart from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Web.
Doyle Log Scale How To Determine Board Feet In A Log Log home
Calculate the merchantable volume and green weight of. Web the green log weight chart from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Web find out
how to calculate the weight of green logs using a formula and a chart. Web in this calculator, you will learn to answer the question.
Download Monthly Weight Loss Log With Charts Excel Template ExcelDataPro
Volume and green weight of a log calculator. Web estimate the weight and density of any green log by selecting the species and measuring the log size. Web find out how to calculate the weight of
green logs using a formula and a chart. Four diameters for each of the four lengths. Download or print a pdf of the green.
Green Wood Weight Chart
Calculate the merchantable volume and green weight of. Web the green log weight chart from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Use the
green log weight. Download or print a pdf of the green log weight. Web green log weight chart.
Green Log Weight Charts & Calculator Sherrilltree
Use the green log weight. Web find out how to calculate the weight of green logs using a formula and a chart. Web calculate the weight of logs based on species, diameter, length and quantity. The
green log weight chart is a handy quick reference guide to 66 different tree species with weights. Volume and green weight of a log.
12 Best Printable Weight Log Sheet Printablee Com Vrogue
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from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Calculate the merchantable volume.
Printable Doyle Log Scale Chart
Web green log weight chart. Web this table provides the green log weights of the species found in our milling area. Download or print a pdf of the green log weight. Web in this calculator, you will
learn to answer the question how much does a log weigh? and what affects the weight of wood. The green log weight chart.
Weekly weight tracker chart virtpixel
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species and measuring the log size. Four diameters for each of the four lengths. Calculate the merchantable volume and green weight of.
Editable Weekly Weight Tracker Weight Log Weight Loss Etsy
Web in this calculator, you will learn to answer the question how much does a log weigh? and what affects the weight of wood. Calculate the merchantable volume and green weight of. Web the green log
weight chart from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Web find out how.
Green Log Weight Chart AppRecs
Choose from a list of common north. Volume and green weight of a log calculator. Web in this calculator, you will learn to answer the question how much does a log weigh? and what affects the weight
of wood. Download or print a pdf of the green log weight. Web find out how to calculate the weight of green logs.
Download or print a pdf of the green log weight. Web estimate the weight and density of any green log by selecting the species and measuring the log size. Web green log weight chart. Web the green
log weight chart from tcia is a handy quick reference guide to 66 different tree species with weights per cubic foot. Web find out how to calculate the weight of green logs using a formula and a
chart. Calculate the merchantable volume and green weight of. Use the green log weight. Web calculate the weight of logs based on species, diameter, length and quantity. The green log weight chart is
a handy quick reference guide to 66 different tree species with weights. Choose from a list of common north. Four diameters for each of the four lengths. Web in this calculator, you will learn to
answer the question how much does a log weigh? and what affects the weight of wood. Web this table provides the green log weights of the species found in our milling area. Volume and green weight of
a log calculator.
Download Or Print A Pdf Of The Green Log Weight.
Web green log weight chart. Volume and green weight of a log calculator. Web calculate the weight of logs based on species, diameter, length and quantity. Web find out how to calculate the weight of
green logs using a formula and a chart.
Four Diameters For Each Of The Four Lengths.
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guide to 66 different tree species with weights per cubic foot. The green log weight chart is a handy quick reference guide to 66 different tree species with weights.
Choose From A List Of Common North.
Calculate the merchantable volume and green weight of. Web in this calculator, you will learn to answer the question how much does a log weigh? and what affects the weight of wood. Web this table
provides the green log weights of the species found in our milling area.
Related Post: | {"url":"https://poweredbytwente.nl/en/log-weight-chart.html","timestamp":"2024-11-12T18:53:59Z","content_type":"text/html","content_length":"28287","record_id":"<urn:uuid:7f177399-f3d6-4df2-91fa-ee7fe24ad26f>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00712.warc.gz"} |
Different number of steps of Levenberg-Marquardt
I use the Levenberg-Marquardt algorithm with linear inequalities for least squares fitting. The target function has an analytic jacobian, but the optimizer's behavior seems strange to me:
optimization is faster if I use a numerical derivative. I will study the problem in detail and give technical details at the end.
The calculation of the target function with the calculation of jacobian lasts 15 times longer than the calculation without jacobian. And this is good, because the optimization takes place in the
40-dimensional space. So, the optimization should be faster with jacobian. But it's not. With different parameters, optimization lasts 35% longer with an analytic derivative. For an unknown reason,
the optimizer makes more steps with the analytic derivative. I have performed a hundred solutions of the regression problem and got that the version with the numerical derivative makes 181 steps on
the average, with the standard deviation 87. Of 100 times, two stopped because they reached the limit of 500 steps. With analytic jacobian enabled, the optimizer makes 185 steps on average, with a
standard deviation of 166. Out of 100 times, 16 stopped at 500 steps.
I use rough stopping criterion alglib::minlmsetcond(state_, 0.0001, 500) because it is faster. If you put step norm 0.e-12, it gets worse. The numerical derivative makes 206 steps with a standard
deviation of 87 and 3/100 end in step 500. The analytical jacobian takes 202 steps with a standard deviation of 178 and 21/100 ending in step 500.
Why is the number of steps different? Why does analytical jacobian need more steps? 500-iterations stop at the right minimum, by the way. I also checked the jacobian with
alglib::minlmoptguardgradient(), so it works correctly.
Profiling shows that alglib_impl::minlmiteration takes a lot of CPU time. You can expect the optimizer to take more steps, so the target function with the analytic jacobian will take a larger share
than the function with the numerical derivative. But it is not. With a numerical derivative, minlmiteration takes 50% of the CPU, and with an analytic jacobian it takes 60%. And that's weird, too.
I have set bound-box and linear inequality constraints A*x < 0. Is it normal that minlmiteration works so long? In the target function, I fill and solve 130 linear systems AX = BY. 100 of them are
1x1 and the rest are sparse. The largest matrix is A 253x253. I know that free alglib doesn't have vectorization, but does the optimizer should work longer than this target function? At the moment,
one regression solution takes about 1 second, and minlmiteration takes 0.6 sec of it.
Technical characteristics:
Alglib::minlmcreatevj() method
Generally speaking, optimization is launched in each thread, but if you leave one thread, nothing changes. All measurements are performed in single-threaded mode.
alglib::minlmsetcond(state, 0.0001, 500);
alglib::minlmsetacctype(state, 1);
If a numerical derivative is used, the step of differentiation is 0.001.
bound-box restrictions and linear inequality A * x < 0.
Optimization space has dimension 40. All coordinates are small and of the same order: 0.001 < x < 100
gcc: -O2 -march-native
The code is not simple, but if needed, it's here:
Running the optimizier:
Setting parameters:
Running the alglib::minlmoptimize()
Thank you in advance for your answer. | {"url":"http://forum.alglib.net/viewtopic.php?f=2&t=4318&view=next","timestamp":"2024-11-10T06:23:15Z","content_type":"application/xhtml+xml","content_length":"18380","record_id":"<urn:uuid:fc111e27-502c-4835-8c81-790d8ad27411>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00283.warc.gz"} |
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Elementary Algebra is a work text that covers the traditional topics studied in a modern elementary algebra course. It is intended for students who:
1. This text can be used in standard lecture or self-paced classes. To help students meet these objectives and to make the study of algebra a pleasant and rewarding experience, Elementary Algebra is
organized as follows.
□ Objectives
Sample Sets
Elementary Algebra contains examples that are set off in boxes for easy reference. The examples are referred to as Sample Sets for two reasons:
1. A parallel Practice Set follows each Sample Set, which reinforces the concepts just learned. The answers to all Practice Sets are displayed with the question when viewing this content
online, or at the end of the chapter in the print version.
Section Exercises
The problems are paired so that the odd-numbered problems are equivalent in kind and difficulty to the even-numbered problems. Answers to the odd-numbered problems are provided with the
exercise when viewed online, or at the back of the chapter in the print version.
Exercises for Review
This section consists of problems that form a cumulative review of the material covered in the preceding sections of the text and is not limited to material in that chapter. The exercises
are keyed by section for easy reference.
Summary of Key Concepts
A summary of the important ideas and formulas used throughout the chapter is included at the end of each chapter. More than just a list of terms, the summary is a valuable tool that
reinforces concepts in preparation for the Proficiency Exam at the end of the chapter, as well as future exams. The summary keys each item to the section of the text where it is
Exercise Supplement
In addition to numerous section exercises, each chapter includes approximately 100 supplemental problems, which are referenced by section. Answers to the odd-numbered problems are
included with the problems when viewed online and in the back of the chapter in the print version.
Proficiency Exam
Each chapter ends with a Proficiency Exam that can serve as a chapter review or a chapter evaluation. The proficiency Exam is keyed to sections, which enables the student to refer back to
the text for assistance. Answers to all Proficiency Exam problems are included with the exercises when viewed online, or in the back of the chapter in the print version.
The writing style is informal and friendly, offering a no-nonsense, straightforward approach to algebra. We have made a deliberate effort not to write another text that minimizes the use
of words because we believe that students can be study algebraic concepts and understand algebraic techniques by using words and symbols rather than symbols alone. It has been our
experience that students at the elementary level are not experienced enough with mathematics to understand symbolic explanations alone; they also need to read the explanation.
We have taken great care to present concepts and techniques so they are understandable and easily remembered. After concepts have been developed, students are warned about common
This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are
presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be
assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
Basic Properties of Real Numbers
The symbols, notations, and properties of numbers that form the basis of algebra, as well as exponents and the rules of exponents, are introduced in Basic Properties of Real Numbers. Each
property of real numbers and the rules of exponents are expressed both symbolically and literally. Literal explanations are included because symbolic explanations alone may be difficult
for a student to interpret.
Basic Operations with Real Numbers
The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a
visual understanding of the meaning of ∣x∣. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and
the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.
Algebraic Expressions and Equations
Operations with algebraic expressions and numerical evaluations are introduced in Algebraic Expressions and Equations. Coefficients are described rather than merely defined. Special
binomial products have both literal symbolic explanation and since they occur so frequently in mathematics, we have been careful to help the student remember them. In each example
problem, the student is “talked” through the symbolic form.
Solving Linear Equations and Inequalities
In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving
applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the “Five-Step Method” of solving applied problems
(discussed in Section 5.6 and Section 5.7).
Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student’s understanding of the
factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be
placed inside the parentheses.
The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the
standard “trial and error” method, and the “collect and discard” method (a method similar to the “ac” method), are presented for factoring trinomials with leading coefficients different
from 1.
Graphing Linear Equations and Inequalities in One and Two Variables
In this chapter the student is shown how graphs provide information that is not always evident from the equation alone. The chapter begins by establishing the relationship between the
variables in an equation, the number of coordinate axes necessary to construct the graph, and the spatial dimension of both the coordinate system and the graph. Interpretation of graphs
is also emphasized throughout the chapter, beginning with the plotting of points. The slope formula is fully developed, progressing from verbal phrases to mathematical expressions. The
expressions are then formed into an equation by explicitly stating that a ratio is a comparison of two quantities of the same type (e.g., distance, weight, or money). This approach
benefits students who take future courses that use graphs to display information.
The student is shown how to graph lines using the intercept method, the table method, and the slope-intercept method, as well as how to distinguish, by inspection, oblique and horizontal/
vertical lines.
A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to
a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since
the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by
using a “freeze frame approach.
The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter
also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.
Roots, Radicals, and Square Root Equations
The distinction between the principal square root of the number x , , and the secondary square root of the number x , , is made by explanation and by example. The simplification of
radical expressions that both involve and do not involve fractions is shown in many detailed examples; this is followed by an explanation of how and why radicals are eliminated from the
denominator of a radical expression. Real-life applications of radical equations have been included, such as problems involving daily output, daily sales, electronic resonance frequency,
and kinetic energy.
Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are
carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2 , and
applied problems from the areas of manufacturing, population, physics, geometry, mathematics (number and volumes), and astronomy, which are solved using the five-step method.
Systems of Linear Equations
Beginning with the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent, and dependent systems and examples to illustrate the applications
for these systems. The substitution method and the addition method of solving a system by elimination are explained, noting when to use each method. The five-step method is again used to
illustrate the solutions of value and rate problems (coin and mixture problems), using drawings that correspond to the actual solution. | {"url":"https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(Ellis_and_Burzynski)/00%3A_Front_Matter/05%3A_Preface","timestamp":"2024-11-04T23:58:14Z","content_type":"text/html","content_length":"140861","record_id":"<urn:uuid:92f54f73-93be-4c27-8033-e116d186048c>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00252.warc.gz"} |
How To Normalize A NumPy Array To Within A Certain Range? - The Citrus Report
How To Normalize A NumPy Array To Within A Certain Range?
NumPy is a fundamental library in Python for scientific computing and data analysis. It provides support to the creation and manipulation of arrays. So, if you are looking to work with large
numerical data sets, you should definitely consider using NumPy. In this post, we will explore how to normalize arrays in NumPy.
What is Normalization?
Normalization is a way of bringing all numerical values in a particular range. The main purpose of normalization is to bring all values in the same scale, so that they are easily comparable. This
becomes particularly important when dealing with large data sets containing different units and ranges of numerical values.
In NumPy, there are two main types of normalization: Min-Max scaling and Z-score normalization. Let’s take a look at both of these methods in more detail.
Min-Max Scaling
Min-Max scaling is a common normalization technique. In this method, you will scale all values to the range of 0 to 1. Suppose you have an array of numerical values in the range of 10 to 20 and you
want to normalize it to a range of 0 to 1. The formula for Min-Max scaling is:
(x - min(x)) / (max(x) - min(x))
This formula will give you the normalized value of x. Here, min(x) and max(x) represent the minimum and maximum values in the array, respectively.
In NumPy, you can use the min() and max() functions to get the minimum and maximum values in an array. Then, you can use the formula we just discussed to normalize the array.
Let’s take a look at an example:
import numpy as np
arr = np.array([10, 15, 20])
min_val = np.min(arr)
max_val = np.max(arr)
normalized_arr = (arr - min_val) / (max_val - min_val)
In the above code, we first created an array with the values 10, 15, and 20. Then, we used the min() and max() functions to get the minimum and maximum values of the array. Next, we used the formula
we discussed earlier to normalize the array, and printed the normalized array.
This will output the following:
[0. 0.5 1. ]
As you can see, all values in the array have been normalized to the range of 0 to 1.
Z-Score Normalization
Z-score normalization is another common normalization technique. In this method, you will scale all values so that they have a mean of 0 and a standard deviation of 1. The formula for Z-score
normalization is:
(x - mean(x)) / standard_deviation(x)
This formula will give you the normalized value of x. Here, mean(x) and standard_deviation(x) represent the mean and standard deviation of the array, respectively.
In NumPy, you can use the mean() and std() functions to get the mean and standard deviation of an array. Then, you can use the formula we just discussed to normalize the array.
Let’s take a look at an example:
import numpy as np
arr = np.array([10, 15, 20])
mean_val = np.mean(arr)
std_val = np.std(arr)
normalized_arr = (arr - mean_val) / std_val
In the above code, we first created an array with the values 10, 15, and 20. Then, we used the mean() and std() functions to get the mean and standard deviation of the array. Next, we used the
formula we discussed earlier to normalize the array, and printed the normalized array.
This will output the following:
[-1.22474487 0. 1.22474487]
As you can see, all values in the array have been normalized to have a mean of 0 and a standard deviation of 1.
NumPy provides several methods to normalize arrays, including Min-Max scaling and Z-score normalization. These techniques are important to bring numerical values to the same scale, making them easily
comparable. In this post, we covered both of these techniques, and how to implement them using NumPy.
If you are looking to work with large numerical data sets, normalization should be an important step in your data pre-processing pipeline. With NumPy, you have powerful tools at your disposal to
normalize arrays in an efficient and effective way.
Leave a Comment | {"url":"https://thecitrusreport.com/questions/how-to-normalize-a-numpy-array-to-within-a-certain-range","timestamp":"2024-11-07T15:34:19Z","content_type":"text/html","content_length":"111999","record_id":"<urn:uuid:33491c6a-d5bd-42ba-b50c-b23c8d257106>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00346.warc.gz"} |
Nebraska Spina Bifida, Inc.
Latex Allergy
Allergy to latex commonly occurs in individuals with spina bifida. It first became evident in the late 1980’s. Experts think that up to 73% of children with spina bifida have a problem with latex.
The reason is unclear but may have to do with too much contact with latex due frequent surgeries, shunt revisions, and other allergies.
Latex can be found in many things,so the FDA requires labeling medical supplies that have natural rubber in them. Other places where latex can be found are:
• bananas passion fruit
• rubber bands avocados
• chestnuts melons
• erasers soles of shoes
• gloves celery
• balloons wheelchair tire inner-tubes
• kiwi condoms
• bandages
Signs of an allergic reaction include:
• watery and itchy eyes a hard time breathing
• sneezing and coughing rash or hives
• swelling of the wind pipe life-threatening collapse
• of blood circulation
You can avoid a latex allergy by staying away from items that contain latex or latex powder. The powder can get into the air and can breathed in by a person or land on their skin. | {"url":"https://www.nebraskaspinabifida.org/what-is-spina-bifida/latex-allergy/","timestamp":"2024-11-09T15:51:44Z","content_type":"text/html","content_length":"34418","record_id":"<urn:uuid:d6854eb7-6c49-4ae2-b1c4-0a8a336fdfb4>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00601.warc.gz"} |
orksheets for 11th Class
Understanding arrangements and factorials
Algebra 2 - Intro to FCP & Factorials
Pre-Cal Quiz Factorials, Permutations & Combinations
Factorials and Operations
Counting Principles and Factorials
Probability Quiz Alg 2 MCR
Factorials & Counting Techniques
FCP, Factorials and Permutations
Permutations, Combinations, Factorials, & Counting Principles
Factorials Practice - 3/3
Evaluating factorials and permutation problems
Probability Exam 1.6, factorials
Algebra 2: Fund Counting Princ, Factorials, Permutations
Arithmetic Sequence, Series, Partial Sums, & Factorials
Factorials/Counting Principle
Explore factorials Worksheets by Grades
Explore Other Subject Worksheets for class 11
Explore printable factorials worksheets for 11th Class
Factorials worksheets for Class 11 are an essential resource for teachers looking to enhance their students' understanding of math, probability, and statistics. These worksheets provide a variety of
problems and exercises that challenge students to apply their knowledge of factorials in different contexts. By incorporating these worksheets into their lesson plans, teachers can help students
build a strong foundation in these crucial mathematical concepts. Additionally, these worksheets can be used as supplementary material for homework assignments or as a review tool before exams. With
factorials worksheets for Class 11, teachers can ensure that their students are well-prepared to tackle more advanced math topics in the future.
Quizizz is an excellent platform for teachers to access a wide range of educational resources, including factorials worksheets for Class 11. This platform offers a variety of interactive quizzes and
games that can help students better understand math, probability, and statistics concepts. Teachers can easily customize these quizzes to align with their lesson plans and learning objectives.
Furthermore, Quizizz provides teachers with valuable insights into their students' performance, allowing them to identify areas where students may need additional support or practice. By
incorporating Quizizz into their teaching strategies, teachers can create an engaging and effective learning environment for their Class 11 students, ensuring they excel in math, probability, and | {"url":"https://quizizz.com/en-in/factorials-worksheets-class-11","timestamp":"2024-11-11T21:03:15Z","content_type":"text/html","content_length":"139245","record_id":"<urn:uuid:738335a4-7a5a-40ab-bf7c-f7419686a677>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00076.warc.gz"} |
Secondary 3 E Math Tuition | The Math Champ
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Sec 3 E Math Tuition
Your affordable, comprehensive Private E Math Tutor in Singapore
Our Teaching Method for Sec 3 E Math
Secondary 3 E Math is the foundation to all math topics moving forward. With a strong foundational grasp of these concepts, it would be significantly easier for students to understand the more
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At the Math Champ, we strive to ensure each student’s weaknesses and create a customised learning path towards improvement. No 2 math students will be learning the exact same way as we strive to
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Application Over Memorisation
We focus more on student’s practical learning and application of these math concepts. We do not believe that memorisation is helpful in scoring well as there may be tricky questions that test
student’s conceptual understanding.
During each math tuition class, dedicate a significant amount of time to understanding the concepts and practice.
Holistic Learning
At our Sec 3 E Math tuition, we believe in a holistic teaching approach that goes beyond just learning math. We focus on building a deep understanding of mathematical concepts, ensuring that students
not only master the syllabus but also develop critical thinking and problem-solving skills.
By integrating real-world applications and encouraging a growth mindset, we aim to inspire a genuine interest in mathematics and help students achieve their full potential.
Our Lesson Syllabus
• Indices
• Quadratic Equations
• Linear Inequalities
• Conditions of Congruence and similarity
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• Arc Lengths, Sector Areas and Radian Measure
• Properties of circles
• Problems in real-world contexts
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Mathematics - Numerical analysis - College School Essays
│Tests of Between-Subjects Effects │
│Dependent Variable: The grade level │
│Source │Type III Sum of Squares │df │Mean Square│F │Sig.│
│Corrected Model│.833^a │1 │.833 │3.169 │.078│
│Intercept │282.133 │1 │282.133 │1072.773│.000│
│StudentType │.833 │1 │.833 │3.169 │.078│
│Error │31.033 │118│.263 │ │ │
│Total │314.000 │120│ │ │ │
│Corrected Total│31.867 │119│ │ │ │
│a. R Squared = .026 (Adjusted R Squared = .018) │
This problem uses the attached file. Based on the following output, please answer this question: is the main effect of Student type significant or not?
F(1,118) = ?, p = .078. Does it show that the average grade level is roughly the same for the two student groups? Please explain all work.
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09:33:53This problem uses the attached file. Based on the following output, please answer this question: is the main
Using the below exhibit formulate a statistical hypothesis appropriate for the consumer
Using the below exhibit formulate a statistical hypothesis appropriate for the consumer group’s purpose, then calculate the mean average miles per gallon. compute the sample variance and sample
standard deviation. determine the most appropriate statistical test using 0.05 significance level.
purchaser miles per gallon Purchaser miles per gallon
1 30.9 13 27
2 24.5 14 26.7
3 31.2 15 31
4 28.7 16 23.5
5 35.1 17 29.4
6 29 18 26.3
7 28.8 19 27.5
8 23.1 20 28.2
9 31 21 28.4
10 30.2 22 29.1
11 28.4 23 21.9
12 29.3 24 30.9
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03:58:39Using the below exhibit formulate a statistical hypothesis appropriate for the consumer
Q15. If one city has a higher CPI than another city, the city with the higher CPI must have a higher cost of living.
Q15. If one city has a higher CPI than another city, the city with the higher CPI must have a higher cost of living.
a. true
b. false
If banks had $10 million in legal reserves, $105 million in checkable deposits, and a 10 percent reserve requirement, they would have to reduce their checkable deposits or increase their reserves.
a. true
b. false
Q17. Nominal wages can be converted into real wages by
a. multiplying the nominal wages by the CPI
b. adding the CPI to the nominal wages
c. subtracting the CPI from the nominal wages
d. dividing the nominal wages by the CPI
Q20. According to the equation of exchange, if total output is 2,000 units, the velocity of money is 5, and the money supply is $1,000, the average price per transaction will be
a. $0.50
b. $2.50
c. $5.00
d. $7.50
Q37. Which of the following Fed actions would increase the level of total bank reserves?
a. buying securities from individuals or businesses
b. selling securities to commercial banks
c. reducing reserve requirements
d. raising the discount rate
Q42. According to the Keynesian analysis, equilibrium occurs at the point where total aggregate expenditure equals total output.
a. true
b. false
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20:13:26Q15. If one city has a higher CPI than another city, the city with the higher CPI must have a higher cost of living.
10. As a result of higher expected inflation, (Points : 1)
10. As a result of higher expected inflation, (Points : 1)
the demand and supply curves for loanable funds both shift to the right and the equilibrium interest rate usually
the demand and supply curves for loanable funds both shift to the left and the equilibrium interest rate usually
the demand curve for loanable funds shifts to the right, the supply curve for loanable funds shifts to the left, and
the equilibrium interest rate usually rises.
the demand curve for loanable funds shifts to the left, the supply curve for loanable funds shifts to the right, and
the equilibrium interest rate usually rises.
the demand and supply curves for loanable funds both shift to the right and the equilibrium interest rate usually rises.
the demand and supply curves for loanable funds both shift to the left and the equilibrium interest rate usually falls.
the demand curve for loanable funds shifts to the right, the supply curve for loanable funds shifts to the left, and the equilibrium interest rate usually rises.
the demand curve for loanable funds shifts to the left, the supply curve for loanable funds shifts to the right, and the equilibrium interest rate usually rises.
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14:02:2910. As a result of higher expected inflation, (Points : 1)
After a number of complaints about its directory assistance
After a number of complaints about its directory assistance, a telephone company
examined samples of calls to determine the frequency of wrong numbers given to callers.
Each sample consisted of 110 calls.
Number of 6 3 5 2 2 6 3 3 5 10 3 2 5 4 6 5
a. Determine 95 percent limits. (Do not round your intermediate calculations. Round your
final answers to 3 decimal places.)
b. Is the process stable (i.e., in control)?
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10:18:04After a number of complaints about its directory assistance
Problem: Given the following printout, answer questions a through e
Problem: Given the following printout, answer questions a through e.
Welch Two Sample t-test
data: math by gender
t = -0.411, df = 187.575, p-value = 0.6816
95 percent confidence interval:
-3.193325 2.092206
sample estimates:
mean in group female mean in group male
52.39450 52.94505
a) State the null and the alternative hypothesis in symbols and words.
b) What is your conclusion regarding the null? (Reject or fail to reject)
c) Write your conclusion within context.
d) Interpret the 95% confidence interval.
e) Find the relevant descriptive on R and show how the t value and confidence interval was calculated? (The data is attached hsb2.csv)
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05:08:21Problem: Given the following printout, answer questions a through e
suppose you know σ and you want an 85% confidence level. What value would you use as z in formula of
suppose you know σ and you want an 85% confidence level. What value would you use as z
in formula of confidence interval for a population mean? (Round your answer to 2 decimal
Value of z [removed]
a. The sample size is 15 and the level of confidence is 95%.
Value of t [removed]
b. The sample size is 24 and the level of confidence is 98%.
Value of t [removed]
c. The sample size is 12 and the level of confidence is 90%.
Value of t
Past surveys reveal that 30% of tourists going to Las Vegas to gamble spend more than
$1,000. The Visitor’s Bureau of Las Vegas wants to update this percentage.
a. The new study is to use the 90% confidence level. The estimate is to be within 1% of
the population proportion. What is the necessary sample size? (Round your answer to
the next whole number.)
Sample size [removed]
b. The Bureau feels the sample size determined above is too large. What can be done to
reduce the sample? Based on your suggestion, recalculate the sample size. (Hint: Use
an allowable error in the range of 0.01 to 0.05) (Round your answer to the next
whole number.)
Sample size [removed]
Forty-nine items are randomly selected from a population of 500 items. The sample mean
is 40 and the sample standard deviation 9.
Develop a 99% confidence interval for the population mean. (Round your answers to 3
decimal places.)
The confidence interval is between [removed] and [removed]
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04:42:03suppose you know σ and you want an 85% confidence level. What value would you use as z in formula of
Below is the profit model spreadsheet for a shoe manufacturer in the month of January. 1)
Below is the profit model spreadsheet for a shoe manufacturer in the month of January. 1) Calculate the revenue for units sold. 2) Calculate the variable cost of production. 3) Calculate the total
Profit Model for January Cost in Dollars
Unit Price 49
Unit Cost 23
Fixed Cost for Production 350,000
Demand 40,000
Unit Price 49
Quantity Sold 38,000
Unit Cost 23
Quantity Produced 38,000
Variable Cost
Fixed Cost 300,000
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04:07:38Below is the profit model spreadsheet for a shoe manufacturer in the month of January. 1)
Pond’s Age-Defying Complex, a cream with alpha hydroxy acid,
Pond’s Age-Defying Complex, a cream with alpha hydroxy acid, advertises that it canreduce wrinkles and improve the skin. In a study published in Archives of Dermatology(recent year), 33 middle-aged
women used a cream with alpha hydroxy acid for 22weeks. At the end of the study period, a dermatologist judged whether each womanexhibited skin improvement. The results for the 33 women are listed
Improved Improved No improvement Improved
No improvement No improvement Improved Improved
Improved Improved Improved Improved
No improvement Improved Improved Improved
No improvement Improved Improved Improved
No improvement Improved No improvement Improved
Improved Improved Improved Improved
Improved No improvement Improved Improved
No improvement
a) MINITAB Output. Enter these data into a MINITAB worksheet all in onecolumn. Use Skin Cream as your header. Select Stat > Basic Statistics > 1-Proportion… Complete the dialog in order to obtain a
97 percent confidenceinterval.
b) Based on the 97% confidence interval for women exhibiting noimprovement, do you have sufficient evidence to conclude that the creamwill improve the skin of more than 60% of middle-aged women?
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02:16:05Pond’s Age-Defying Complex, a cream with alpha hydroxy acid,
In an analysis of hunting by African lions, biologists filmed prey captures
In an analysis of hunting by African lions, biologists filmed prey captures from the safety of their vehicles. Prey captures were then divided into a sequence of events. One of the events is the
stalk, defined as the reduction of predator-prey distance for prey that has been specifically targeted. The investigators identified two types of stalk: (a) “crouching,” — the lion is concealed and
either the lion advances toward the prey or the prey advances (unaware) toward the lion, and (b) “running,” — the lion is less concealed and advances toward the prey in a rapid manner.
Data on lions’ stalks of wildebeests and zebras from a simple random sample of 159 kills appear in the table below.
Characteristic Numeric value
Mean stalking time 31.6 min
Standard deviation of stalk time 16.4 min
Proportion of stalks of the crouching type 0.92
The same monitoring of radio-collared lions over the years suggests that the overall proportion of stalks that are the crouching type is about 0.87. Do the data above provide evidence that for this
population of lions the proportion of crouching stalks of wildebeests and zebras is greater than what was originally thought? Use alpha=0.05
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Mutually Exclusive Events - Definition, Formula, Examples
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Mutually Exclusive Events
When you toss a coin, you either get heads or tails, but there is no other way you could get both results. This is an example of mutually exclusive events. In probability theory, two events are said
to be mutually exclusive events if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events.
Further, if two events are considered disjoint events, then the probability of both events occurring at the same time will be zero. Let us learn more about this concept in this short lesson along
with solved examples.
1. What Are Mutually Exclusive Events?
2. How Do You Calculate Mutually Exclusive Events?
3. Probability of Disjoint (or) Mutually Exclusive Events
4. How Do You Show Mutually Exclusive Events?
5. Do Mutually Exclusive Events Add up to 1?
6. Mutually Exclusive Events Probability Rules
7. Conditional Probability for Mutually Exclusive Events
8. Solved Examples
9. Practice Questions
10. FAQs on Mutually Exclusive Events
What Are Mutually Exclusive Events?
What is the meaning of mutually exclusive events? Mutually exclusive events are the events that cannot occur or happen at the same time. In other words, the probability of the events happening at the
same time is zero.
Example of Mutually Exclusive Events
A student wants to go to school. There are two paths; one that takes him to school and the other one that takes him home. Which path will he choose? He will choose one of the two paths. Obviously, he
can't choose both at the same time. This is an example of a mutually exclusive events.
How Do You Calculate Mutually Exclusive Events?
Mutually exclusive events are events that cannot occur or happen at the same time. The occurrence of mutually exclusive events at the same time is 0. If A and B are two mutually exclusive events in
math, the probability of them both happening together is: P(A and B) = 0. The formula for calculating the probability of two mutually exclusive events is given below:
P(A or B) = P(A) + P(B)
Do you know special symbols are used to show the relation between two sets: The two important relationships between two sets are the intersection of sets and union of sets.
Intersection of sets: The symbol used for the intersection is "\(\cap\)" and "and" is also used. If two sets are there say for example; A = {1, 2, 3} and B = {2, 3, 4}. Then A intersection B is
represented as\( A\cap B\).
A \(\cap\) B = {2, 3}
Union of sets: The symbol used for the union is "\(\cup\) " and "or" is also used. If two sets are there say for example; A = {1, 2, 3} and B = {2, 3, 4}. Then A union B is represented as \(A\cup B
A \(\cup\) B = {1, 2, 3, 4}
Probability of Disjoint (or) Mutually Exclusive Events
The probability of disjoint or mutually exclusive events A and B is written as the probability of the intersection of the events A and B. Probability of Disjoint (or) Mutually Exclusive Events = P (
A \(\cap \) B) = 0. In probability, the specific addition rule is valid when two events are mutually exclusive events. It states that the probability of either event occurring is the sum of
probabilities of each event occurring. If A and B are said to be mutually exclusive events then the probability of an event A occurring or the probability of event B occurring is given as P(A) + P
(B), ,
P (A U B) = P(A) + P(B)
Some of the examples of the mutually exclusive events are:
• When tossing a coin, the event of getting head and tail are mutually exclusive events. Because the probability of getting head and tail simultaneously is 0.
• In a six-sided die, the events “2” and “5” are mutually exclusive events. We cannot get both events 2 and 5 at the same time when we threw one die.
• In a deck of 52 cards, drawing a red card and drawing a club are mutually exclusive events because all the clubs are black.
If the events A and B are not mutually exclusive events, the probability of getting A or B is given as:
P (A U B) = P(A) + P(B) – P (A\(\cap \) B)
How Do You Show Mutually Exclusive Events?
We can use Venn diagrams to show mutually exclusive events. The figures shown below indicate mutually exclusive events and events that are not mutually exclusive events or non-mutually exclusive
events. Note that there is no common element in mutually exclusive events.
Do Mutually Exclusive Events Add up to 1?
We know that mutually exclusive events cannot occur at the same time. The sum of the probability of mutually exclusive events can never be greater than 1 It is always less than 1, until and unless
the same set of events are also exhaustive (at least one of them being true). In this case, the sum of their probability is exactly 1.
Mutually Exclusive Events Probability Rules
In probability theory, two events are mutually exclusive events or disjoint if they do not occur at the same time. A clear case is the set of results of a single coin toss, which can end in either
heads or tails, but not for both. While tossing the coin, both outcomes are collectively exhaustive, which suggests that at least one of the consequences must happen, so these two events collectively
exhaust all the possibilities.
Though, not all mutually exclusive events are commonly exhaustive. For example, the outcomes of 1 and 4 on rolling six-sided dice, are mutually exclusive events (both 1 and 4 cannot come as result at
the same time) but are not collectively exhaustive (it can result in distinct outcomes such as 2,3,5,6). Further, from the definition of mutually exclusive events, the following rules for probability
can be concluded.
• Addition Rule: P (A + B) = 1
• Subtraction Rule: P (A U B)’ = 0
• Multiplication Rule: P (A ∩ B) = 0
There are different varieties of events also. For instance, think of a coin that has a Head on both sides of the coin or a Tail on both sides. It doesn’t matter how many times you flip it, it will
always occur Head (for the first coin) and Tail (for the second coin). If we check the sample space of such an experiment, it will be either { H } for the first coin and { T } for the second one.
Such events have single point in the sample space and are called “Simple Events”. Such kind of two sample events is always mutually exclusive events.
Conditional Probability for Mutually Exclusive Events
Conditional probability is stated as the probability of event A, given that another event B has occurred. For two independent events A and B, the conditional probability of event B given that A has
occurred is denoted by the expression P( B|A) and it is defined using the following equation.
P(B|A)= P (A ∩ B)/P(A)
Let us redefine the above equation using multiplication rule: P (A ∩ B) = 0
P(B|A)= 0/P(A)
So the conditional probability formula for mutually exclusive events is:
P (B | A) = 0
Important Notes
Here are some important things to remember about mutually exclusive events:
1. The probability of an event that cannot happen is 0
2. The probability of an event that is certain to happen is 1
3. The sum of the probabilities of all the elementary events of an experiment is 1
4. The probability of an event is greater than or equal to 0 and less than or equal to 1
Solved Examples
1. Example 1: Daniel is trying to understand mutually exclusive events using dice. Help Daniel understand what is the probability of a dice showing 4 or 5?
There are a total of 6 faces on a die, hence, the total number of outcomes will be 6
The probability of a die showing 4 is P(4) = 1/6
The probability of a die showing 5 isP(5) = 1/6
The probability of getting 4 or 5 is = P(4 or 5)
= P(4 or 5)
= P (4) + P(5)
= (1/6) + (1/6)
= 1 + 1/ 6
= 2/6
= 1/3
Answer: 1/3 will be the answer.
2. Example 2: Benny's teacher is teaching them about mutually exclusive events and gave him a deck of 52 cards and asked him to select a red card or a 6. Find the probability of selecting a red card
or a 6
The probability of getting a Red card = 26/52
The probability of getting a 6 = 4/52
The probability of getting both a Red and a 6 = 2/52
P(R or 6) = P(R) + P(6) - P(R and 6)
= (26/52) + (4/52) - (2/52)
= (30-2/52)
Answer: 7/13 will be the answer.
3. Example 3: Caroline noticed her mother trying to take out the fish to clean the fish tank. She asked her mother, "How many are males and how many are females?" Her mother replied that the tank
contained 5 male fish and 8 female fish. What is the probability that the fish her mother takes out first is a male fish?
This question can be solved easily by using the formula.
Probability of an event = Number of possible outcomes/ Total no of favorable outcomes
No. of male fish = 5
No. of female fish = 8
Total no of fishes
5+8 = 13
The probability that the fish are taken out is a male fish: No of male fish/ Total no of fishes
The probability that the fish are taken out is a male fish 5/13
Answer: Probability is 5/13.
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Breakdown tough concepts through simple visuals.
Math will no longer be a tough subject, especially when you understand the concepts through visualizations.
FAQs on Mutually Exclusive Events
What Does Mutually Exclusive Events Mean?
Mutually exclusive events are a statistical term describing two or more events that cannot happen simultaneously. It is commonly used to describe a situation where the occurrence of one outcome
supersedes the other.
What Is an Example of Mutually Exclusive Events?
Mutually exclusive events are things that can't happen at the same time. For example, you can't run backward and forwards at the same time. The events “running forward” and “running backward” are
mutually exclusive events. Tossing a coin can also give you this type of event.
What Does It Mean To Say Two Things Are Not Mutually Exclusive Events?
The two activities are said to be mutually exclusive events if one cannot exist when the other is true. Not mutually exclusive events means that they can take place at the same time. And we can say
that "The two are not mutually exclusive."
What Is the Formula for Mutually Exclusive Events?
The formula for mutually exclusive events (they can't occur together), is that the (U) union of the two events must be the sum of both, i.e. 0.20 + 0.35 = 0.55.
How Do You Know if A and B are Mutually Exclusive Events?
A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A \(\cap\) B) = 0.
What Does Mutually Exclusive Events Mean in Probability?
In statistics and probability theory, two events are mutually exclusive events if they cannot occur at the same time. The simplest example of mutually exclusive events is a coin toss. A tossed coin
outcome can be either head or tails, but both outcomes cannot occur simultaneously.
Are Dependent Events Mutually Exclusive Events?
Two mutually exclusive events are neither necessarily independent nor dependent.
Download FREE Study Materials
Mutually Exclusive Events Worksheet
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/data/mutually-exclusive-events/","timestamp":"2024-11-07T21:46:52Z","content_type":"text/html","content_length":"237542","record_id":"<urn:uuid:e92d8ed3-5f11-433f-aea3-b8d8ee3ad63b>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00600.warc.gz"} |
How to Calculate Roof Area | ehow.com
If you're planning to put a new roof on your house, it's important to purchase enough matching shingles to complete the job. To do that, you need to accurately calculate the area of your roof.
Calculating area is easy in two dimensions, but nearly all roofs slope up to a peak, which adds a third dimension. Fortunately, you don't have to know complicated geometry to factor in the slope of
your roof. You need only a level and a tape measure to find the "rise" of your roof. After that, just plug in the corresponding factor. You'll find the instructions below, along with the appropriate
factors and formulas.
Step 1
Determine the dimensions of the roof. Multiply the length times the width to calculate the square footage.
Step 2
Determine the slope of the roof. This is expressed in terms of how many inches the roof rises for every foot of horizontal distance. Mark a level 12 inches from one end. Place the other end of the
level at the peak of the roof and hold it level. Measure the distance from the surface of the roof to the bottom of the level at the 12-inch mark. This distance is the rise. For example, if your
measurement is seven inches, the rise is expressed as 7:12 in roofing terminology.
Step 3
Find your "pitch" factor based on the rise you calculated in Step 2. Choose one of the following: If your rise is 3:12, the pitch factor is 1.035. If your rise is 4:12, the pitch factor is 1.055. If
your rise is 5:12, the pitch factor is 1.085. For a rise of 6:12, the pitch factor is 1.12. Finally, if the rise is 7:12, use a pitch factor of 1.16.
Step 4
Multiply the square footage you calculated in Step 1 by the pitch factor you determined in Step 3. The answer is the area of your roof.
In the construction industry, roofs are measured in “squares.” One square equals 100 square feet. Divide the area you calculated in Step 4 by 100. The answer to this is the total number of squares on
your roof for which you need to purchase shingles. | {"url":"https://www.ehow.com/how_5001455_calculate-roof-area.html","timestamp":"2024-11-13T15:05:48Z","content_type":"text/html","content_length":"327700","record_id":"<urn:uuid:a91f4eca-b469-46c4-bcc3-827861551fd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00662.warc.gz"} |
Energy Modeling in PVSketch Mega - PVComplete
Simplified Model
The Simplified model is based on the PVWatts energy model from NREL. (https://pvwatts.nrel.gov/) PVWatts consists of a set of component models to represent the different parts of a photovoltaic
system. PVWatts performs hourly simulations to calculate the electricity produced by the system over a single year. PVWatts assumes that there are 8,760 hours in one year.
The following is a high-level description of the algorithm PVWatts uses to calculate the photovoltaic system’s hourly electrical output:
• Calculate the hourly plane-of-array (POA) solar irradiance from the horizontal irradiance, latitude, longitude, and time in the solar resource data, and from the array type, tilt and azimuth
• Calculate the effective POA irradiance to account for reflective losses from the module cover depending on the solar incidence angle.
• Calculate the cell temperature based on the array type, POA irradiance, wind speed, and ambient temperature. The cell temperature model assumes a module height of 5 meters above the ground and an
installed nominal operating cell temperature (INOCT) of 49°C for the fixed roof mount option (appropriate for approximately 4 inch standoffs), and of 45°C for the other array type options.
• Calculate the array’s DC output from DC system size at a reference POA irradiance of 1,000 W/m², and the calculated cell temperature, assuming a reference cell temperature of 25°C, and
temperature coefficient of power of -0.47%/°C for the standard module type, -0.35%/°C for the premium type, or -0.20%/°C for the thin film type.
• Calculate the system’s AC output from the calculated DC output and system losses and nominal inverter efficiency input (96% by default) with a part-load inverter efficiency adjustment derived
from empirical measurements of inverter performance.
• The Simplified Model allows for the following system loss inputs:
□ PVWatts does not factor in the specific I-V curve of the solar module or inverter efficiency curve. This simplification means that the results are not as accurate for specific equipment as
our Advanced energy model. However, the Simplified model has the advantage of not requiring .PAN or .OND files to run and therefore is a good solution for early stage project design where
final equipment has not yet been specified.
Advanced Model
The Advanced model is based on Canadian Solar’s System Simulator (CASSYS), an open source energy model developed by the module manufacturer. The Advanced model considers a solar project details
description such as arrays, inverters, and modules, site location, and weather conditions on a sub-hourly interval to calculate the state of the system at each step and provide a detailed estimate of
energy flows and losses in the system.
The Advanced model is a more accurate model than is PVWATTS, allowing, for example, to consider the specific IV curve of the solar module and it’s interactions with the inverter input. Both energy
simulations use the same transposition models Hays and Perez.
Comparison with PVSYST
The Advanced model is organized to function similar PVSYST using the same underlying equations, and users can translate between these two models with relatively low effort and to obtain similar
results, usually with 1% of each other.
The standard test condition (STC) parameters for the module are obtained from a .PAN file. Module behavior is calculated for several non-STC operating conditions such as open circuit, fixed voltage,
and maximum point tracking. Values are then converted from module to array level and losses are applied in accordance with user input values.
Shading factors on the beam, diffuse, and ground-reflected components of incident irradiance, based on the sun position throughout the day resulting from a near shading model, are available for
panels arranged in an unlimited rows or a fixed tilt configuration. In the scenario of an unlimited row model, the Advanced energy model neglects edge effects because it assumes such rows are large
enough that edge effects are not significant. This assumption reduces the calculation of the shading factor at different times of the day to a simple geometrical construct, as does PVSYST.
In a paper by Canadian Solar introducing CASSYS, they validated the model through cross-validation between CASSYS and PVSYST, as well as comparisons against measured data. Real world comparisons
between measured and simulated values (once all post-construction conditions and parameters are reflected in the system definition) show that CASSYS provides a reliable basis to estimate the energy
production by a defined system on a sub-hourly basis. In the same paper it is referred that a more thorough study is required to further understand the sources of error, and the fine-tuning steps for
the model inputs. However, simulations that fall within a couple of percent of the actual values are usually considered excellent, as the accuracy of the various instruments used in the measurements
themselves rarely falls below that threshold. The agreement between the tools is found to be -0.35% for the energy predicted over an entire year (CASSYS being the more conservative tool) when the
inputs to all models are closest to each other.
PVSketch Mega applies a hierarchy to available weather data sources to run the Advanced model. First the Advanced model will try to download data from NSRDB first (https://nsrdb.nrel.gov/), and if it
fails for a given location, the model will default to download data from the PVGIS database (https://ec.europa.eu/jrc/en/pvgis).
Key parameters:
• Latitude and longitude
• Inverter model name and performance parameters
• Module model name and performance parameters
• Number of modules per series string
• Number of series strings per array
• Tilt () and azimuth of the array (or tracking angle algorithm for tracked arrays)
• Albedo of the ground (or roof) surface)
• Horizon map showing potential for shading from obstructions
• Irradiance data is reported as three components: direct normal irradiance (DNI), global horizontal irradiance (GHI), and diffuse horizontal irradiance (DHI). | {"url":"https://mail.pvcomplete.com/energy-modeling-in-pvsketch-mega/","timestamp":"2024-11-02T07:41:46Z","content_type":"text/html","content_length":"103548","record_id":"<urn:uuid:59e2c4a8-a7f8-4dab-83df-8f75860adc90>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00557.warc.gz"} |
Mathematics for the Trades : A Guided Approach, Books a la Carte Edition
Hal Saunders; Robert A Carman
NOTE: This edition features the same content as the traditional text in a convenient, three-hole-punched, loose-leaf version. Books a la Carte also offer a great value; this format costs
significantly less than a new textbook. Before purchasing, check with your instructor or review your course syllabus to ensure that you select the correct ISBN. For Books a la Carte editions that
include MyLab(TM) or Mastering(TM), several versions may exist for each title-including customized versions for individual schools-and registrations are not transferable. In addition, you may need a
Course ID, provided by your instructor, to register for and use MyLab or Mastering platforms. For Basic Math, Math for the Trades, Occupational Math, and similar basic math skills courses servicing
trade or technical programs at the undergraduate/graduate level. A solid foundation in the math needed for a wide range of technical and vocational trades Mathematics for the Trades: A Guided
Approach is the leader in trades and occupational mathematics, equipping students with the math skills required for allied health, electrical trades, automotive trades, plumbing, construction, and
many more - particularly in the physical trades. The math concepts are presented completely within the context of practical on-the-job applications, so students can make an impact on the job from day
one. Authentic applications give students relevant, tangible mathematical examples that they are likely to encounter in future careers. Also available with MyLab Math By combining trusted author
content with digital tools and a flexible platform, MyLab Math personalizes the learning experience and improves results for each student. Note: You are purchasing a standalone product; MyLab Math
does not come packaged with this content. Students, if interested in purchasing this title with MyLab Math, ask your instructor to confirm the correct package ISBN and Course ID. Instructors, contact
your Pearson representative for more information. If you would like to purchase both the physical text and MyLab Math, search for: 0135183723 / 9780135183724 Mathematics for the Trades Books a la
Carte Edition Plus MyLab Math -- Title-Specific Access Card Package, 11/e Package consists of: 0134765788 / 9780134765785 Mathematics for the Trades: A Guided Approach, Books a la Carte Edition
0134836138 / 9780134836133 MyLab Math plus Pearson eText - Standalone Access Card - for Mathematics for the Trades: A Guided Approach
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This book is currently reported out of stock for sale, but WorldCat can help you find it in your local library: | {"url":"https://bookchecker.com/0134765788","timestamp":"2024-11-07T20:24:22Z","content_type":"text/html","content_length":"115866","record_id":"<urn:uuid:ac508e14-e061-4b6d-beda-74ed1207fc3d>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00291.warc.gz"} |
Wiener reconstruction of large-scale structure from peculiar velocities
We present an alternative, Bayesian method for large-scale reconstruction from observed peculiar velocities. The method stresses a rigorous treatment of the random errors, and it allows extrapolation
into poorly sampled regions in real space or in k-space. A likelihood analysis is used in a preliminary stage to determine the fluctuation power spectrum, followed by a Wiener filter (WF) analysis to
obtain the minimum-variance mean fields of velocity and mass density. Constrained realizations (CRs) are then used to sample the statistical scatter about the WF mean field. The method is tested
using mock catalogs of the Mark III data, drawn from a simulation that mimics our local cosmological neighborhood. The success of the reconstruction is evaluated quantitatively. With low-resolution
Gaussian smoothing of radius 1200 km s^-1, the reconstruction is of high signal-to-noise ratio (S/N) in a relatively large volume, with small variance about the mean field. A high-resolution
reconstruction, of 500 km s^-1 smoothing, is of reasonable S/N only in limited nearby regions, where interesting new substructure is resolved. The WF/CR method is applied as a demonstration to the
Mark III data. The reconstructed structures are consistent with those extracted from the same velocity data by the POTENT method, and with the structures seen in the distribution of IRAS 1.2 Jy
galaxies. The reconstructed velocity field is decomposed into its divergent and tidal components relative to a cube of side ± 8000 km s^-1 centered on the Local Group. The divergent component is
similar to the velocity field predicted from the distribution of IRAS galaxies. The tidal component is dominated by a bulk flow of 194 ± 32 km s^-1 in the general direction of the Shapley
concentration, and it also indicates a significant quadrupole.
טביעת אצבע
להלן מוצגים תחומי המחקר של הפרסום 'Wiener reconstruction of large-scale structure from peculiar velocities'. יחד הם יוצרים טביעת אצבע ייחודית. | {"url":"https://cris.openu.ac.il/iw/publications/wiener-reconstruction-of-large-scale-structure-from-peculiar-velo","timestamp":"2024-11-14T05:20:07Z","content_type":"text/html","content_length":"53104","record_id":"<urn:uuid:edfd4adf-0ae6-485c-a465-9d5adf5da4b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00369.warc.gz"} |
How To Write Multiplication Sentences For Fourth Grade Math
Stockbyte/Stockbyte/Getty Images
Perhaps the most important skill for fourth graders is that of multiplication. A key way to teach multiplication is via multiplication sentences. Unlike a traditional sentence, multiplication
sentences use numbers and symbols to express a statement. By learning multiplication sentences, fourth graders learn how multiplication and addition relate to each other.
Parts of a Multiplication Sentence
A multiplication sentence consists of two parts: one part is a mathematical expression and the other part is the product. In multiplication, a mathematical expression is the part of the sentence that
comes before the equal sign. The mathematical expression contains the factors and the multiplication symbol. For example, in the sentence "2 x 8 = 16," the "2 x 8" portion is the mathematical
expression. The mathematical expressions doesn't include the answer, which is also known as the product. In the multiplication sentence "2 x 8 = 16," the two and eight are factors and 16 is the
Create Sentences Using Arrays
Before students can learn about multiplication sentences, they must understand the concept of an array. An array consists of a set of numbers or objects arranged in columns and rows — usually on a
grid. This makes it possible to count the number of columns and to multiply the resulting value by the number of rows. By using multiplication, students don't need to manually count each item in the
grid. This forms the basis for multiplication sentences and prepares students for more advanced math. For example, show the students an array that has nine objects in each row, and a total of six
rows. Show them that they can count each individual item in the array, or they can multiply nine times six for a product of 54. For example, the complete sentence looks like "9 x 6 = 54."
Creating Multiplication Sentences
Multiplication sentences serve a crucial function in enabling fourth graders to learn how to use math in a practical way. The ability to construct a multiplication sentence extends beyond the
classroom, by preparing students to calculate large numbers of items. A student who knows how to create his own multiplication sentences can look at a five-by-five grid of items and will know that
the grid contains a total of 25 items. Ask the students to count the number of rows in a picture and then write that number down on their papers. Then, write a multiplication symbol and write the
number of columns after the symbol. In a five-by-six grid, students should write "5 x 6," with "x" as the symbol for multiplication. Once they do this, tell them to write an equal sign and solve the
problem. For example, a correct multiplication sentence for a five-by-six grid of items looks like "5 x 6 = 30."
When to Use Multiplication Sentences
Multiplication sentences only work when the problem contains an equal number of items in each column or row. For example, if you have a group of items with one item in the first row, two in the
second row and three in the fourth row, you must use an addition sentence and add each of the rows together. The addition sentence looks like "1 + 2 + 3 = 6." There is no way to figure that out using
a multiplication sentence. In contrast, if you have two items in each row and three items in each column, then you can use a multiplication sentence to express the complete equation. In this example,
the sentence would look like "2 x 3 = 6." The number two represents the rows in the array, and the number three represents the number of columns.
Create a Sentence From a Word Problem
Word problems always seem to throw students off, but once students understand how to write a multiplication sentence, word problems should be easier for the students. Provide a word problem, such as
"Matt collected a bushel of apples. He has enough apples to place five apples per row six times. How many apples does Matt have? Hurry up and figure out the answer before he eats one." Instruct the
students to draw a picture on a grid to help them visualize the problem, and then apply the same concept you use when creating sentences from a grid. In this example, the student should write the
multiplication sentence as "5 x 6 = 30."
Cite This Article
Martin, Avery. "How To Write Multiplication Sentences For Fourth Grade Math" sciencing.com, https://www.sciencing.com/write-sentences-fourth-grade-math-7839649/. 24 April 2017.
Martin, Avery. (2017, April 24). How To Write Multiplication Sentences For Fourth Grade Math. sciencing.com. Retrieved from https://www.sciencing.com/write-sentences-fourth-grade-math-7839649/
Martin, Avery. How To Write Multiplication Sentences For Fourth Grade Math last modified August 30, 2022. https://www.sciencing.com/write-sentences-fourth-grade-math-7839649/ | {"url":"https://www.sciencing.com:443/write-sentences-fourth-grade-math-7839649/","timestamp":"2024-11-11T14:07:40Z","content_type":"application/xhtml+xml","content_length":"75268","record_id":"<urn:uuid:0fd08296-e5b7-4923-8d71-1f8bf8c07543>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00563.warc.gz"} |
seminars - Coding Theory from the Viewpoint of Lattices
In the first lecture, we introduce the history and basic concepts of (error-correcting) codes. Codes have been used widely in the mobile phones, compact discs, and the big data storage. They were
found by R. Hamming and C. Shannon in the late 1940s. Since then, Coding Theory has become one of the most practical mathematical areas and has interacted with Algebra, Combinatorics, and Number
Theory. Our today’s IT would have been impossible without the theory of codes.
In the second lecture, we describe an interesting connection between codes and lattices. They share common properties. We begin with some basic definitions of codes and lattices. Codes over rings
have been used in the construction of interesting Euclidean or Hermitian lattices. Given a prime p, B. Fine proved that there are exactly three commutative rings with unity of order p 2 and
characteristic p. Using C. Bachoc’s results, we describe that how these rings can be related to certain quotient rings of the ring of algebraic integers of an imaginary quadratic number field. Then
we construct Hermitian lattices from codes over these rings. Shaska, et al. have studied the theta functions of these Hermitian lattices. We generalize the results of Bachoc, Shaska, et al. We
propose some open problems in this direction. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=room&order_type=desc&l=ko&page=81&document_srl=646479","timestamp":"2024-11-13T01:16:33Z","content_type":"text/html","content_length":"49164","record_id":"<urn:uuid:5f298b11-e404-4b60-98f3-301292ecf42c>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00652.warc.gz"} |
Free Correlation 02 Practice Test - 11th Grade - Commerce
Free Correlation 02 Practice Test - 11th Grade - Commerce
Question 1
The price per/kg of chicken and the quantity of chicken purchased per month in a household is tabulated below. Determine the type of correlation.
Price/kg (Rs)Quantity (kg)406505.5603.5703307.5802
Solution : C
The given data can be represented on a scatter chart as shown below.
It can be seen that there is a negative correlation between quantity and price.
Question 2
To find the line of best fit, which of the following is minimized?
Solution : A
The line of best fit is obtained by minimizing the squared errors of the data values i.e. the square of the difference between the predicted and actual values of the dependent variable with
respect to the independent variable.
Question 3
The demand for a good at different price points is given in the below. Find the correlation coefficeint between price and quantity demanded.
Price (P)Quantity (Q)1012020105308540555040
Solution : D
Question 4
If the slope of the line of best fit for two variables is negative, then the correlation between the variables is also negative. State true or false.
Solution : A
A negative correlation indicates that one quantity decreases as the other increases. So, if the slope of the line of best fit is negative, the correlation is also negative. Hence, the given
statement is true.
Question 5
Which of the following is the correct formula for the correlation coefficient, r?
Solution : A
Question 6
Find cov(X,Y) for the following data.
Solution : B
Question 7
The following table contains the cost of 5 motorcycles and their corresponding mileages in (km/litre). Find the correlation coefficient using the shortcut method.
Cost (X)Mileage (Y)500004010000030150000252000001525000010
Solution : D
Let Ax=150000 & hx=50000Let Ay=25 & hy=5
Question 8
Identify the correct statement(s) about the correlation between two variables.
Statement 1: Correlation does not show that there is a causal relationship between two variables
Statement 2: Correlation shows the degree to which variations in one variable explain the variation of the second variable
Neither Statement 1 nor Statement 2
Both Statement 1 and Statement 2
Solution : A and B
Correlation shows the degree to which variations in one variable explain the variation of the second variable. It does not show that there is a causal relationship between two variables.
Question 9
Spearman's rank coefficient helps to understand the type of correlation by looking at non-linear ranked data.
Solution : A
Spearman's rank coefficient helps to understand the type of correlation by looking at non-linear ranked data.
Question 10
Variables in a correlation are called ___
Dependent and independent variables
Solution : C
Variables in a correlation are called co-variables. | {"url":"https://selfstudy365.com/exam/correlation-02-808","timestamp":"2024-11-06T20:43:50Z","content_type":"text/html","content_length":"270509","record_id":"<urn:uuid:42d813d5-0019-4d83-86c2-2638fc020619>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00201.warc.gz"} |
Free module
(Redirected from Free vector space)
Jump to navigation Jump to search
In mathematics, a free module is a module that has a basis – that is, a generating set consisting of linearly independent elements. Every vector space is a free module,^[1] but, if the ring of the
coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules.
Given any set S and ring R, there is a free R-module with basis S, which is called free module on S or module of formal linear combinations of the elements of S.
A free abelian group is precisely a free module over the ring Z of integers.
For a ring ${\displaystyle R}$ and an ${\displaystyle R}$-module ${\displaystyle M}$, the set ${\displaystyle E\subseteq M}$ is a basis for ${\displaystyle M}$ if:
• ${\displaystyle E}$ is a generating set for ${\displaystyle M}$; that is to say, every element of ${\displaystyle M}$ is a finite sum of elements of ${\displaystyle E}$ multiplied by coefficients
in ${\displaystyle R}$; and
• ${\displaystyle E}$ is linearly independent, that is, ${\displaystyle r_{1}e_{1}+r_{2}e_{2}+\cdots +r_{n}e_{n}=0_{M}}$ for ${\displaystyle e_{1},e_{2},\ldots ,e_{n}}$ distinct elements of ${\
displaystyle E}$ implies that ${\displaystyle r_{1}=r_{2}=\cdots =r_{n}=0_{R}}$ (where ${\displaystyle 0_{M}}$ is the zero element of ${\displaystyle M}$ and ${\displaystyle 0_{R}}$ is the zero
element of ${\displaystyle R}$).
A free module is a module with a basis.^[2]
An immediate consequence of the second half of the definition is that the coefficients in the first half are unique for each element of M.
If ${\displaystyle R}$ has invariant basis number, then by definition any two bases have the same cardinality. The cardinality of any (and therefore every) basis is called the rank of the free module
${\displaystyle M}$. If this cardinality is finite, the free module is said to be free of rank n, or simply free of finite rank.
Let R be a ring.
• R is a free module of rank one over itself (either as a left or right module); any unit element is a basis.
• More generally, a (say) left ideal I of R is free if and only if it is a principal ideal generated by a left nonzerodivisor, with a generator being a basis.
• If R is commutative, the polynomial ring ${\displaystyle R[X]}$ in indeterminate X is a free module with a possible basis 1, X, X^2, ....
• For any non-negative integer n, ${\displaystyle R^{n}=R\times \cdots \times R}$, the cartesian product of n copies of R as a left R-module, is free. If R has invariant basis number (which is true
for commutative R), then its rank is n.
• A direct sum of free modules is free, while an infinite cartesian product of free modules is generally not free (cf. the Baer–Specker group.)
Formal linear combinations[edit]
Given a set E and ring R, there is a free R-module that has E as a basis: namely, the direct sum of copies of R indexed by E
${\displaystyle R^{(E)}=\bigoplus _{e\in E}R}$.
Explicitly, it is the submodule of the cartesian product ${\displaystyle \prod _{E}R}$ (R is viewed as say a left module) that consists of the elements that have only finitely many nonzero
components. One can embed E into R^(E) as a subset by identifying an element e with that of R^(E) whose e-th component is 1 (the unity of R) and all the other components are zero. Then each element
of R^(E) can be written uniquely as
${\displaystyle \sum _{e\in E}c_{e}e,}$
where only finitely many ${\displaystyle c_{e}}$ are nonzero. It is called a formal linear combination of elements of E.
A similar argument shows that every free left (resp. right) R-module is isomorphic to a direct sum of copies of R as left (resp. right) module.
Another construction[edit]
The free module R^(E) may also be constructed in the following equivalent way.
Given a ring R and a set E, first as a set we let
${\displaystyle R^{(E)}=\{f:E\to R|f(x)=0{\text{ for all but finitely many }}x\in E\}.}$
We equip it with a structure of a left module such that the addition is defined by: for x in E,
${\displaystyle (f+g)(x)=f(x)+g(x)}$
and the scalar multiplication by: for r in R and x in E,
${\displaystyle (rf)(x)=r(f(x))}$
Now, as an R-valued function on E, each f in ${\displaystyle R^{(E)}}$ can be written uniquely as
${\displaystyle f=\sum _{e\in E}c_{e}\delta _{e}}$
where ${\displaystyle c_{e}}$ are in R and only finitely many of them are nonzero and ${\displaystyle \delta _{e}}$ is given as
${\displaystyle \delta _{e}(x)={\begin{cases}1_{R}\quad {\mbox{if }}x=e\\0_{R}\quad {\mbox{if }}xeq e\end{cases}}}$
(this is a variant of the Kronecker delta.) The above means that the subset ${\displaystyle \{\delta _{e}|e\in E\}}$ of ${\displaystyle R^{(E)}}$ is a basis of ${\displaystyle R^{(E)}}$. The mapping
${\displaystyle e\mapsto \delta _{e}}$ is a bijection between E and this basis. Through this bijection, ${\displaystyle R^{(E)}}$ is a free module with the basis E.
Universal property[edit]
The inclusion mapping ${\displaystyle \iota :E\to R^{(E)}}$ defined above is universal in the following sense. Given an arbitrary function ${\displaystyle f:E\to N}$ from a set E to a left R-module N
, there exists a unique module homomorphism ${\displaystyle {\overline {f}}:R^{(E)}\to N}$ such that ${\displaystyle f={\overline {f}}\circ \iota }$; namely, ${\displaystyle {\overline {f}}}$ is
defined by the formula:
${\displaystyle {\overline {f}}\left(\sum _{e\in E}r_{e}e\right)=\sum _{e\in E}r_{e}f(e)}$
and ${\displaystyle {\overline {f}}}$ is said to be obtained by extending ${\displaystyle f}$ by linearity. The uniqueness means that each R-linear map ${\displaystyle R^{(E)}\to N}$ is uniquely
determined by its restriction to E.
As usual for universal properties, this defines R^(E) up to a canonical isomorphism. Also the formation of ${\displaystyle \iota :E\to R^{(E)}}$ for each set E determines a functor
${\displaystyle R^{(-)}:{\textbf {Set}}\to R-{\mathsf {Mod}},\,E\mapsto R^{(E)}}$,
from the category of sets to the category of left R-modules. It is called the free functor and satisfies a natural relation: for each set E and a left module N,
${\displaystyle \operatorname {Hom} _{\textbf {Set}}(E,U(N))\simeq \operatorname {Hom} _{R}(R^{(E)},N),\,f\mapsto {\overline {f}}}$
where ${\displaystyle U:R-{\mathsf {Mod}}\to {\textbf {Set}}}$ is the forgetful functor, meaning ${\displaystyle R^{(-)}}$ is a left adjoint of the forgetful functor.
Many statements about free modules, which are wrong for general modules over rings, are still true for certain generalisations of free modules. Projective modules are direct summands of free modules,
so one can choose an injection in a free module and use the basis of this one to prove something for the projective module. Even weaker generalisations are flat modules, which still have the property
that tensoring with them preserves exact sequences, and torsion-free modules. If the ring has special properties, this hierarchy may collapse, e.g., for any perfect local Dedekind ring, every
torsion-free module is flat, projective and free as well. A finitely generated torsion-free module of a commutative PID is free. A finitely generated Z-module is free if and only if it is flat.
See local ring, perfect ring and Dedekind ring.
See also[edit]
• Adamson, Iain T. (1972). Elementary Rings and Modules. University Mathematical Texts. Oliver and Boyd. pp. 65–66. ISBN 0-05-002192-3. MR 0345993.
• Keown, R. (1975). An Introduction to Group Representation Theory. Mathematics in science and engineering. 116. Academic Press. ISBN 978-0-12-404250-6. MR 0387387.
• Govorov, V. E. (2001) [1994], "Free module", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4.
External links[edit]
This article incorporates material from free vector space over a set on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License. | {"url":"https://static.hlt.bme.hu/semantics/external/pages/tenzorszorzatok/en.wikipedia.org/wiki/Free_vector_space.html","timestamp":"2024-11-03T23:03:56Z","content_type":"text/html","content_length":"109239","record_id":"<urn:uuid:7026e68f-6541-4c16-bc1c-1e8ce8331c96>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00249.warc.gz"} |
How to use (.) in haskell?
In Haskell, the (.) operator is used to compose functions. It takes two functions as arguments and returns a new function that is the composition of the two functions.
The type signature of (.) is as follows:
1 (.) :: (b -> c) -> (a -> b) -> a -> c
Here, the types a, b, and c can be any types, and -> denotes the function type.
To understand how to use (.) in Haskell, let's consider an example. Suppose we have two functions:
1 add1 :: Int -> Int
2 add1 x = x + 1
4 double :: Int -> Int
5 double x = x * 2
We can compose these two functions using the (.) operator as follows:
1 composedFunction :: Int -> Int
2 composedFunction = add1 . double
Now, composedFunction is a new function that can take an Int as input and output the result of applying add1 after applying double. For example, composedFunction 3 would return 7 (first applying
double to 3, then add1 to the result).
In this example, the (.) operator takes the function add1 (of type Int -> Int) as its left argument, and the function double (of type Int -> Int) as its right argument. It then returns a new function
that is the composition of add1 and double. | {"url":"https://devhubby.com/thread/how-to-use-in-haskell","timestamp":"2024-11-10T15:18:41Z","content_type":"text/html","content_length":"123092","record_id":"<urn:uuid:0026ebc6-363d-4a01-8c7f-c41936bd9609>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00803.warc.gz"} |
44.03 -- Speed of sound in air vs. helium
A signal generator sends a 20-Hz square wave to a loudspeaker driver, at the left end of the glass tube in the photograph above, and provides a synchronous trigger to the oscilloscope. A
microphone at the end of a long metal rod allows you to probe inside the tube, anywhere along its length. The microphone is connected to an input of the oscilloscope, and when the square wave
reaches it, you see the resulting pulse on the oscilloscope trace, at some delay with respect to the trigger. If you measure this delay with the microphone at two different distances from the
speaker, you can then use the distance and delay differences to calculate the speed of sound. You can also fill the tube with helium and repeat the measurement, to compare the speed of sound
in helium to that in air.
With the signal generator producing a square wave, the microphone circuit produces a pulse with the leading edge of each cycle. Placing the microphone right against the output grille of the
speaker shows that there is a roughly 210-μs delay within the speaker housing. Measuring the distance from the flat face of the speaker housing might account for this delay fairly well, but, of
course, measuring from the grille and then subtracting 210 μs from the travel time eliminates it. The best way to eliminate the error, though, is to measure the delay for two well-defined
references some distance apart along the length of the tube, and then to divide this distance by the difference between the two delays.
Sound waves are longitudinal waves created by a disturbance in an elastic medium. The oscillatory motion of the particles transmitting the wave is parallel to the axis along which the wave
travels, as opposed to that in a transverse wave, in which it is perpendicular to the direction in which the wave travels. In this case, the elastic medium is, of course, air or, if you fill the
tube with helium, helium.
To obtain an expression for the speed of sound in a gas, we can imagine the gas occupying a long cylinder of cross-sectional area A, and an oscillating piston at one end of the cylinder sending a
wave through the gas. With each cycle, the cylinder compresses a portion of the gas, increasing the pressure and density in it above the equilibrium pressure and density. As the gas moves away
from the piston, it compresses layers of gas next to it, and a pulse, in the form of a region of high pressure and density, travels down the cylinder. When the piston moves back to its initial
position, the gas in front of it expands, creating a region where the pressure and density are lower than the equilibrium pressure and density. Thus, a sequence of alternating regions of
higher-than-normal pressure and density, compressions, and regions of lower-than-normal pressure and density, rarefactions, travels down the cylinder.
Now we imagine a single pulse, a compression, being sent down the tube, and assume that it has well-defined edges and uniform density and pressure. In the frame of reference of the pulse, as it
travels through the tube, gas is coming toward it at velocity v. If we take a thin slice of the gas, as it enters the region where the pulse is, its leading face is at higher pressure than its
trailing face, so it is compressed and decelerated. The pressure difference across it is dp, and its velocity inside the pulse region is v + dv. (dv is negative.) When it emerges from the other
side of the pulse region, it experiences the reverse pressure difference, expands and is accelerated to its original velocity, v.
When the plug of gas enters the region where the pulse is, then, it feels a force of F = dpA. As the plug travels down the tube, its length is v dt, where dt is how long it takes for the gas plug
to pass a particular point in the cylinder. Its volume is thus Av dt, and its mass is ρ[0]Av dt, where ρ[0] is the equilibrium density of the gas. So by Newton’s second law, F = ma, we have dpA =
(ρ[0]Av dt)(-dv/dt), which we can rearrange to ρ[0]v^2 = (-v dp/dv). In entering the pulse region, the gas plug is compressed from its equilibrium volume V[0] = Av dt by an amount dV = A dv dt.
Rearranging this gives (dV/V[0]) = (A dv dt/Av dt) = (dv/v).
So ρ[0]v^2 = (-v dp/dv) then becomes ρ[0]v^2 = (-V dp/dV). ), The quantity on the right, the ratio of the change in pressure on a body to its fractional change in volume (it is the same as (-dp/(
dV/V)), is called the bulk modulus of elasticity B of the body. It is positive, because the volume changes in the opposite direction to the pressure. In terms of B, the speed of our pulse is v =
Now we need to know what dP/dV equals. Newton, in his calculation of the speed of sound in air, used Boyle’s law (pV = k, or pV = p[0]V[0]). This assumes that the temperature in the gas does not
deviate from its equilibrium temperature. This condition does not hold in our sound wave. In regions of compression, the gas has had work done on it, and so is hotter than before, and in
rarefactions, the gas has expanded and cooled by the same amount that the compression regions have heated. Heat does not have enough time to travel from the compressions to the rarefactions and
thereby keep the gas at its equilibrium temperature. This means that the difference between the equilibrium pressure and the pressure in a compression region is greater than we would expect from
Boyle’s law (as is that between the equilibrium pressure and the pressure in the rarefactions), and this results in a higher velocity. For Newton, this resulted in an error of about 15% in his
calculation. We must use instead the adiabatic gas law, which states that pV^γ = p[0]V[0]^γ, where γ is the heat capacity ratio, C[P]/C[V]. This is the ratio of heat capacity at constant pressure
to heat capacity at constant volume. So p = p[0]V[0]^γV^-γ. Differentiating and then setting V = V[0] gives dp/dV = -γ p[0]V[0]^γV^-(γ-1) and V[0](dp/dV)[0] = -γp[0]. So the bulk modulus for an
ideal gas is γp[0]. Substituting this into the equation in the previous paragraph gives ρ[0]v^2 = γp[0]. The speed of sound in a gas is thus v = √(γp[0]/ρ[0]). For an ideal gas, p[0] = nRT/V.
Since n = m/M, that is, total mass divided by molar mass, and ρ = m/V, ρ[0] = p[0]M/RT, and v = √(γRT/M). For air at STP (0 °C and 1 atm), γ = 1.40, and v = √((1.40 × 8.314 J/mol·K ×273.16 K)/
(2.8967 × 10^-2 kg/mol)) = 331 m/s.
If we call room temperature 298 K, the speed of sound works out to 346 m/s.
For helium, of course, both γ and M are different. γ = 1.67, and M = 4.003 × 10^-3 kg/mol. So for helium the equation above gives v = 973 m/s at STP, and 1,020 m/s at 298 K. Since the speed,
frequency and length of a sound wave are related by the equation ν = v/λ, the frequency of a sound produced by a resonator of a particular length is proportional to the speed of sound in the
particular gas that fills the resonator. This is why when you inhale helium from a balloon, the pitch of your voice rises so dramatically. Since the speed of sound in helium is almost triple that
in air, the frequency of your voice almost triples when your airway is filled with helium.
1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition(New York: John Wiley and Sons, 1977), pp. 434-436, 510-514.
2) Crawford, Frank S., Jr. Waves: Berkeley Physics Course – Volume 3 (San Francisco: McGraw-Hill Book Company, 1968), pp. 165-169. | {"url":"https://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/44.03.html","timestamp":"2024-11-03T12:28:09Z","content_type":"text/html","content_length":"11387","record_id":"<urn:uuid:6860b406-dab8-4867-bb1d-8e8cb553b054>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00120.warc.gz"} |
On Sobolev spaces and density theorems on Finsler manifolds
[1] R. Adams, Sobolev spaces, Academic press, New York, 1975.
[2] H. Akbar-Zadeh, Initiation to Global Finslerian Geometry, North- Holland Mathematical Library, 2006.
[3] T. Aubin, Some nonlinear problems in Riemannian geometry, Springer-Verlag, 1988.
[4] S. Azami, A. Razavi, Existence and uniqueness for a solution of Ricci flow on Finsler manifolds, Int. J. of Geom. Meth. in Mod. Phy., 10(3) (2013) 1-21.
[5] D. Bao, S. S. Chern, Z. Shen, Riemann-Finsler geometry, Springer-Verlag, 2000.
[6] D. Bao, B. Lackey, A Hodge decomposition theorem for Finsler spaces, C. R. Acad. Sci. Paris S´er. I Math., 323(1) (1996) 51-56.
[7] B. Bidabad, On compact Finsler spaces of positive constant curvature C. R. Acad. Sci. Paris S´er. I Math., 349 (2011) 1191-1194.
[8] B. Bidabad, A. Shahi, Harmonic vector fields on Finsler manifolds, C. R. Acad. Sci. Paris S´er. I Math., 354 (2016) 101-106.
[9] B. Bidabad, A. Shahi, M. Yar Ahmadi, Deformation of Cartan curvature on Finsler manifolds, Bull. Korean Math. Soc. 54(6) (2017) 2119-2139.
[10] B. Bidabad, M. Yar Ahmadi, Convergence of Finslerian metrics under Ricci flow, Sci. China Math. 59(4) (2016) 741-750.
[11] Y. Ge, Z. Shen, Eigenvalues, and eigenfunctions of metric measure manifolds, Proc. London Math. Soc., 82(3) (2001) 725-746.
[12] Q. He, Y. Shen, On Bernstein type theorems in Finsler spaces with the volume form induced from the projective sphere bundle, Proc. Amer. Math. Soc., 134(3) (2006) 871-880.
[13] M. Jim´enez-Sevilla, L. Sanchez-Gonzalez, On some problems on smooth approximation and smooth extension of Lipschitz functions on Banach-Finsler manifolds, Nonlinear Anal. 74(11) (2011)
[14] A. Krist´aly, I. Rudas, Elliptic problems on the ball endowed with Funk-type metrics, Nonlinear Anal., 119 (2015) 199-208.
[15] S. Lakzian, Differential Harnack estimates for positive solutions to heat equation under Finsler-Ricci flow, Pacific J. Math., 278(2) (2015) 447-462
[16] H. Mosel, S. Winkelmann, On weakly harmonic maps from Finsler to Riemannian manifolds, Ann. I. H. Poincar´e, 26 (2009) 39-57.
[17] S. B. Myers, Algebras of differentiable functions, Proc. Amer. Math. Soc., 5 (1954) 917-922.
[18] S. Ohta, Nonlinear geometric analysis on Finsler manifolds, European Journal of Math., 3(4) (2017) 916-952.
[19] Z. Shen, Lectures on Finsler geometry, World Scientific, 2001.
[20] N. Winter, On the distance function to the cut locus of a submanifold in Finsler geometry, Ph.D. thesis, RWTH Aachen University, (2010).
[21] Y. Yang, Solvability of some elliptic equations on Finsler manifolds, math.pku.edu.cn preprint, 1-12. | {"url":"https://ajmc.aut.ac.ir/article_3039.html","timestamp":"2024-11-10T15:53:58Z","content_type":"text/html","content_length":"48929","record_id":"<urn:uuid:ab74ea79-4ba2-4359-9470-0061e3c0f0e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00350.warc.gz"} |
Day 24
Advent of Code 2022 - Day 24
Useful Links
Concepts and Packages Demonstrated
defaultdictdataclassFactory methodBreadth First Search (BFS)
Page Navigation
Problem Intro
This one was tough to get right.
We’ve reached a valley that we need to cross. The valley is full of horizontal and vertical blizzards. Our input data represents the valley, and it looks something like this:
In this map:
• Locations marked # are walls of the valley.
• Locations marked . are clear ground that we are allowed to occupy.
• Locations marked with an arrow contain a blizzard. Each minute, every blizzard moves one unit in the direction it is pointing. All blizzards move simultaneously.
• If a blizzard reaches the boundary of the valley, it wraps around and reappears the other side, pointing in the same direction.
• We start in the clear ground . in the top left.
• We need to get to the clear ground . in the bottom right.
Part 1
What is the fewest number of minutes required to avoid the blizzards and reach the goal?
My strategy is as follows:
• Create a MapState class that represents the current location of all the blizzards, and which knows how to return the blizzard state in the subsequent minute.
• Perform a BFS to calculate the shortest route through the ever-changing blizzard map.
First, I’ll use a Point dataclass, as I often do:
class Point():
""" Point x,y which knows how to add another point, and how to return all adjacent (non-diag) points """
x: int
y: int
def __add__(self, other) -> Point:
""" Add other point to this point, returning new point vector """
return Point(self.x + other.x, self.y + other.y)
def adjacent_points(self) -> set[Point]:
return set(self+vector for vector in VECTORS.values())
def __repr__(self):
return f"P({self.x},{self.y})"
This Point class knows how to add a vector to return a new Point, and it uses this addition method to return all of it’s adjacent points (excluding diagonals), i.e. by adding each of four adjacent
vectors to itself.
Now I define the VECTORS dictionary:
VECTORS = {
'>': Point(1, 0),
'v': Point(0, 1),
'<': Point(-1, 0),
'^': Point(0, -1)
Now I create a MapState() class:
class MapState():
""" Store location of blizzards, grid bounds, start, goal, and time. """
def __init__(self, blizzards: dict, grid_dims: tuple, start: Point, goal: Point, t: int) -> None:
self._blizzards: dict[Point, list] = blizzards
self._width = grid_dims[0]
self._height = grid_dims[1]
self._start = start
self._goal = goal
self._time = t
def init_from_grid(cls, grid_input: list[str]):
""" Create a new MapState using an input grid """
grid: list[str] = grid_input
blizzards = defaultdict(list)
for y, row in enumerate(grid[1:-1]): # ignore top and bottom
for x, col in enumerate(row[1:-1]): # ignore left and right
point = Point(x,y)
if col in VECTORS:
height = len(grid) - 2
width = len(grid[0]) - 2
start = Point(0, -1) # 1 above top grid row
goal = Point(width-1, height) # 1 below bottom grid row
return MapState(blizzards, (width, height), start=start, goal=goal, t=0)
def start(self) -> Point:
return self._start
def start(self, point: Point):
self._start = point
def time(self) -> int:
return self._time
def goal(self) -> Point:
return self._goal
def goal(self, point):
self._goal = point
def next_blizzard_state(self) -> MapState:
""" Move blizzards to achieve next blizzard state. There is only one possible next blizzard state """
next_blizzards = defaultdict(list)
for loc, blizzards_here in self._blizzards.items():
for current_bliz in blizzards_here:
next_bliz_x = (loc + VECTORS[current_bliz]).x % self._width
next_bliz_y = (loc + VECTORS[current_bliz]).y % self._height
next_blizzards[Point(next_bliz_x, next_bliz_y)].append(current_bliz)
return MapState(next_blizzards, (self._width, self._height), self._start, self._goal, self.time+1)
def is_valid(self, point: Point) -> bool:
""" Check if the specified point is an allowed position in the current blizzard state. """
if point in (self._start, self._goal): # out of bounds, but allowed
return True
# out of bounds
if not (0 <= point.x < self._width):
return False
if not (0 <= point.y < self._height):
return False
if point in (self._blizzards):
return False
return True
def __str__(self) -> str:
lines = []
for y in range(0, self._height):
line = ""
for x in range (0, self._width):
loc = Point(x,y)
if loc in self._blizzards:
blizzards_here = self._blizzards[loc]
how_many_blizzards = len(blizzards_here)
if how_many_blizzards == 1: # one blizzard here
line += next(bliz for bliz in blizzards_here)
elif how_many_blizzards > 1: # more than one blizzard here
line += str(how_many_blizzards)
line += '.'
return ("\n".join(lines) +
f"\nTime={self.time}, Hash={hash(self)}")
def __repr__(self) -> str:
return f"Time={self.time}, Hash={hash(self)}"
• We create our first MapState object using an init_from_grid() classmethod, passing in grid data which we read from the input data.
□ This creates a defaultdict(list) where any blizzard locations are the keys, and the values are all the blizzards found at this location.
□ Note that when we read in the grid data, we ignore the four edges, which are made up of walls, #.
□ We store the current width and height of the grid, without its walls.
□ We store our start and goal locations.
□ We set the _time to 0 in this first state.
• The class includes a next_blizzard_state() method. It works by:
□ Creating a new defaultdict to store the blizzards in the next state.
□ Iterates through all the current blizzard locations, and adds the appropriate vector (depending on the direction of each blizzard) to the current location, to populate the new defaultdict.
□ Instantiates a new MapState, using this new dict of blizzards, incrementing the time by 1 minute, but otherwise leaving the current MapState attributes untouched.
• It includes an is_valid() method, which allows us to check if any given location is an allowed location in this MapState. To be valid, the location must be within the current bounds, and must not
contain any blizzards. We also allow the _start and _goal locations.
We can now test our blizzard MapState is able to iterate, as follows:
def test_blizzard_states(init_state: MapState, iterations: int):
state = init_state
for _ in range(iterations):
print(state, end="\n\n")
state = state.next_blizzard_state()
def main():
with open(INPUT_FILE, mode="rt") as f:
data = f.read().splitlines()
state = MapState.init_from_grid(data)
test_blizzard_states(state, 10)
The output looks like this:
Good, that looks correct!
Now we’re ready to implement the BFS.
def bfs(state: MapState) -> MapState:
""" BFS, but we're allowed to backtrack.
Our frontier should only contain the current set of allowed next locations. """
start = state.start
goal = state.goal
# Use a set because the many neighbours of the points in our frontier may be the same position
# We don't want to explore the same location twice IN THE SAME ITERATION
frontier = {start}
while goal not in frontier:
state = state.next_blizzard_state()
# reset frontier because we can revisit locations we've been to before
frontier = set(explore_frontier(state, frontier))
return state
def explore_frontier(current_state, frontier):
""" Generator that returns all valid next locations with current blizzard state
from all locations in the frontier. """
for loc in frontier:
for neighbour in loc.adjacent_points():
if current_state.is_valid(neighbour):
yield neighbour
if current_state.is_valid(loc): # staying still may be a valid move
yield loc
It works like this:
• First, we call the bfs() function, passing in the initial MapState.
• Extract the start and goal locations from this MapState.
• Create a set to be our frontier, and add start to it.
• Now enter a while loop that only exits when we’ve found the goal. In this loop:
□ Get the next MapState, i.e. where the blizzards will be in the subsequent minute.
□ Explore all the locations in our frontier. (For the first iteration, this will only be start.)
□ For each location in the frontier, determine which locations are valid next moves. This is a maximum of five locations: the current locations, plus its four neighbour points. For each of
these candidate locations, check if the location is_valid() for the current MapState.
□ Create a new frontier from these valid locations.
Note that unlike a typical BFS, we’re allowed to backtrack here. That’s why we’re not storing all previous visited points in an explored set, as we would typically do. Instead, we’re creating a new
frontier set for each new MapState and the associated current position.
Finally, we can solve for Part 1, like this:
def main():
with open(INPUT_FILE, mode="rt") as f:
data = f.read().splitlines()
# Part 1
state = MapState.init_from_grid(data)
state = bfs(state)
print(f"Part 1: {state.time}")
Part 2
Oh no! One of the elves left his snacks at the entrance to the valley. So we need to go back to the start, retrieve them, and then journey back to the goal. Thus:
What is the fewest number of minutes required to reach the goal, go back to the start, then reach the goal again?
Our total journey is now made up of three legs:
1. From start to goal.
2. From goal back to start.
3. From start to goal again.
But throughout, the blizzards are changing.
This is pretty trivial for us to solve. We just need to continue where we left off, with leg 1 already complete. We just need to:
1. Swap the locations of start and goal, and repeat the BFS.
2. Swap the locations back again, and repeat the BFS again.
In fact, we just need to amend our main() function to look like this:
def main():
with open(INPUT_FILE, mode="rt") as f:
data = f.read().splitlines()
# Part 1
leg_times = []
state = MapState.init_from_grid(data)
state = bfs(state)
print(f"Part 1: Leg time={leg_times[0]}")
# Part 2
# First, swap goal and start, since we need to go back to the start
state.start, state.goal = state.goal, state.start
state = bfs(state)
leg_times.append(state.time - sum(leg_times))
print(f"Part 2: Return leg time={leg_times[-1]}")
state.start, state.goal = state.goal, state.start
state = bfs(state)
leg_times.append(state.time - sum(leg_times))
print(f"Part 2: Last leg time={leg_times[-1]}")
print(f"Part 2: Total time={sum(leg_times)}")
Here’s the final code:
from __future__ import annotations
from collections import defaultdict
from dataclasses import dataclass
from pathlib import Path
import time
SCRIPT_DIR = Path(__file__).parent
# INPUT_FILE = Path(SCRIPT_DIR, "input/sample_input.txt")
INPUT_FILE = Path(SCRIPT_DIR, "input/input.txt")
class Point():
""" Point x,y which knows how to add another point, and how to return all adjacent (non-diag) points """
x: int
y: int
def __add__(self, other) -> Point:
""" Add other point to this point, returning new point vector """
return Point(self.x + other.x, self.y + other.y)
def adjacent_points(self) -> set[Point]:
return set(self+vector for vector in VECTORS.values())
def __repr__(self):
return f"P({self.x},{self.y})"
VECTORS = {
'>': Point(1, 0),
'v': Point(0, 1),
'<': Point(-1, 0),
'^': Point(0, -1)
class MapState():
""" Store location of blizzards, grid bounds, start, goal, and time. """
def __init__(self, blizzards: dict, grid_dims: tuple, start: Point, goal: Point, t: int) -> None:
self._blizzards: dict[Point, list] = blizzards
self._width = grid_dims[0]
self._height = grid_dims[1]
self._start = start
self._goal = goal
self._time = t
def init_from_grid(cls, grid_input: list[str]):
""" Create a new MapState using an input grid """
grid: list[str] = grid_input
blizzards = defaultdict(list)
for y, row in enumerate(grid[1:-1]): # ignore top and bottom
for x, col in enumerate(row[1:-1]): # ignore left and right
point = Point(x,y)
if col in VECTORS:
height = len(grid) - 2
width = len(grid[0]) - 2
start = Point(0, -1) # 1 above top grid row
goal = Point(width-1, height) # 1 below bottom grid row
return MapState(blizzards, (width, height), start=start, goal=goal, t=0)
def start(self) -> Point:
return self._start
def start(self, point: Point):
self._start = point
def time(self) -> int:
return self._time
def goal(self) -> Point:
return self._goal
def goal(self, point):
self._goal = point
def next_blizzard_state(self) -> MapState:
""" Move blizzards to achieve next blizzard state. There is only one possible next blizzard state """
next_blizzards = defaultdict(list)
for loc, blizzards_here in self._blizzards.items():
for current_bliz in blizzards_here:
next_bliz_x = (loc + VECTORS[current_bliz]).x % self._width
next_bliz_y = (loc + VECTORS[current_bliz]).y % self._height
next_blizzards[Point(next_bliz_x, next_bliz_y)].append(current_bliz)
return MapState(next_blizzards, (self._width, self._height), self._start, self._goal, self.time+1)
def is_valid(self, point: Point) -> bool:
""" Check if the specified point is an allowed position in the current blizzard state. """
if point in (self._start, self._goal): # out of bounds, but allowed
return True
# out of bounds
if not (0 <= point.x < self._width):
return False
if not (0 <= point.y < self._height):
return False
if point in (self._blizzards):
return False
return True
def __str__(self) -> str:
lines = []
for y in range(0, self._height):
line = ""
for x in range (0, self._width):
loc = Point(x,y)
if loc in self._blizzards:
blizzards_here = self._blizzards[loc]
how_many_blizzards = len(blizzards_here)
if how_many_blizzards == 1: # one blizzard here
line += next(bliz for bliz in blizzards_here)
elif how_many_blizzards > 1: # more than one blizzard here
line += str(how_many_blizzards)
line += '.'
return ("\n".join(lines) + f"\nTime={self.time}")
def __repr__(self) -> str:
return f"Time={self.time}, Hash={hash(self)}"
def main():
with open(INPUT_FILE, mode="rt") as f:
data = f.read().splitlines()
# Part 1
leg_times = []
state = MapState.init_from_grid(data)
state = bfs(state)
print(f"Part 1: Leg time={leg_times[0]}")
# Part 2
# First, swap goal and start, since we need to go back to the start
state.start, state.goal = state.goal, state.start
state = bfs(state)
leg_times.append(state.time - sum(leg_times))
print(f"Part 2: Return leg time={leg_times[-1]}")
state.start, state.goal = state.goal, state.start
state = bfs(state)
leg_times.append(state.time - sum(leg_times))
print(f"Part 2: Last leg time={leg_times[-1]}")
print(f"Part 2: Total time={sum(leg_times)}")
def test_blizzard_states(init_state: MapState, iterations: int):
state = init_state
for _ in range(iterations):
print(state, end="\n\n")
state = state.next_blizzard_state()
def bfs(state: MapState) -> MapState:
""" BFS, but we're allowed to backtrack.
Our frontier should only contain the current set of allowed next locations. """
start = state.start
goal = state.goal
# Use a set because the many neighbours of the points in our frontier may be the same position
# We don't want to explore the same location twice IN THE SAME ITERATION
frontier = {start}
while goal not in frontier:
state = state.next_blizzard_state()
# reset frontier because we can revisit locations we've been to before
frontier = set(explore_frontier(state, frontier))
return state
def explore_frontier(current_state, frontier):
""" Generator that returns all valid next locations with current blizzard state
from all locations in the frontier. """
for loc in frontier:
for neighbour in loc.adjacent_points():
if current_state.is_valid(neighbour):
yield neighbour
if current_state.is_valid(loc): # staying still may be a valid move
yield loc
if __name__ == "__main__":
t1 = time.perf_counter()
t2 = time.perf_counter()
print(f"Execution time: {t2 - t1:0.4f} seconds")
Here’s the output with my real input data:
Part 1: Leg time=286
Part 2: Return leg time=255
Part 2: Last leg time=279
Part 2: Total time=820
Execution time: 6.9212 seconds | {"url":"https://aoc.just2good.co.uk/2022/24","timestamp":"2024-11-14T12:05:31Z","content_type":"text/html","content_length":"82138","record_id":"<urn:uuid:2b192ed7-e867-41d7-b83b-f75d992ba941>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00580.warc.gz"} |
Permutations and Combinations Homework Help, Questions with Solutions - Kunduz
Permutations and Combinations Questions and Answers
Permutations and Combinations
Question 1 3 Find the greatest monomial that is a factor of the expression 12f g 15fg 3f g o 3fg o 3g o 6fg o 6f g
Permutations and Combinations
Fill in the blanks with the appropriate numbers or expressions related to the following equation V 8 V 5 V 3 54 v 2v 15 a The restrictions on the variable are the smaller number in the first blank b
When you If not enter none none You are correct 5 c The solution set of the equation is You are incorrect Ive the equation do you get any traneous solutions If so enter the value of the extraneous
solution in the blank 13 10 X You are incorrect and 3 X You are incorrect X Enter Enter the solution set notation If there is more than Open in
Permutations and Combinations
7 STEM Connection Hannah is measuring three sheets of metal Sheet A is 1 times as long as Sheet B Sheet C is 2 3 times as long as Sheet A Order the sheets from least length to greatest length ACB 8
Tyler has three dogs of different weights Max Daisy and Charlie 3 as much as 8 Daisy weighs as much as Max Charlie weighs Daisy Who weighs the least How do you know Charlke 9 Hugo is organizing his
books from shortest to tallest His math book is 2 times as tall as his science book His reading book as tall as his science book In what order should Hugo organize his books 7 is muth reading science
10 Extend Your Thinking What will happen to the product if a whole number is multiplied by Thee 82 predect answer 15
Permutations and Combinations
9 Hugo is organizing his books from shortest to tallest His math book is 2 times as tall as his science book His reading book is as tall as his science book In what order should Hugo organize his
books muth reading science 10 Extend Your Thinking What will happen to the product if a whole number is multiplied by DIC answen 15
Permutations and Combinations
X X S2 Fill in the blanks with the requested information pertaining to the equation X X x 9 x 4x 3 a The restricted values are Type your answer here X x 2x 3 Write your response here Type your answer
here Type your answer here Enter the values from smallest to largest and b Solve the equation Enter the solution set of the equation List solutions in order from smallest to largest separated by a
Permutations and Combinations
Use the product rule to simplify 63 16x45 y 63 16x4y5 Simplify your answer Type an exact answer using radicals as needed www
Permutations and Combinations
Add or subtract as indicated 2 7 3 7 6 7 2 7 3 7 6 7 Simplify your answer Type an exact answer using radicals as needed
Permutations and Combinations
Multiply and then simplify if possible 11 6 5 11 6 5 Simplify your answer Type an exact answer using radicals as needed GEN
Permutations and Combinations
Simplify the given expression Write the answer with positive exponents Assume that all variables represent positive numbers y 1 1 2 2 y z 14 1 4 11 2 2 y Simplify your answer Type exponential
notation with positive exponents Use integers or fractions for any numbers in the expression
Permutations and Combinations
Solve the equation and check your solution It is possible that there is NO SOLUTION If y 8 what is the solution 2x 10y 0 Type a response
Permutations and Combinations
37 Construction The roof of a house is to extend up 13 5 feet above the ceiling which is 36 feet across If the edge of the roof is to extend 2 feet beyond the side of the house and 1 5 feet below the
ceiling find the length of one side of the roof
Permutations and Combinations
Consider the lantern shown below a Which semiregular polyhedron does the lantern resemble small rhombicosidodecahedron snub dodecahedron truncated icosahedron icosidodecahedron O truncated
dodecahedron b The lantern has 90 edges and 60 vertices How many faces does the lantern have
Permutations and Combinations
21 It takes the earth 24 hours to make one complete revolution on its axis Through how many degrees does the earth turn in 12 hours
Permutations and Combinations
The graph on the right gives the annual per person retail availability of beef chicken and pork in pounds from 2007 to 2012 What is the approximate annual per person retail availability of beef
chicken and pork in 2012 The annual per person availability of beef in 2012 is approximately 62 pounds The annual per person availability of chicken in 2012 is approximately pounds G Pounds 70 65 60
55 50 45 beef porke 40 35 2007 2008 2009 2010 2011 2012 Year
Permutations and Combinations
Tell whether each figure is convex or concave a b c convex O concave d convex concave convex O concave convex O concave
Permutations and Combinations
Ifferent rays share endpoint A as shown in the figure A a How many different angles are formed by the three rays b Describe the types of angles that are formed by the three rays There are two acute
angles and one right angle There are two obtuse angles and one right angle O There is one obtuse angle and one acute angle There is one acute angle and one right angle There are two acute angles and
two right angles c Describe a special relationship between two of the angles The Select angles Select ales
Permutations and Combinations
6 A football team consists of 23 players three goalkeepers eight defenders seven midfielders and five forwards How many different match lineups are possible if we assume that a lineup must consist of
exactly 11 players and include exactly one goalkeeper four or five defenders at least three midfielders and at least one forward
Permutations and Combinations
3 Let X be the set of all ten letter words that can be obtained by permuting the letters of the word COLLATERAL How many elements does X contain ii How many elements of X begin and end with L iii How
many elements of X contain three neighbouring letters L iv How many elements of X contain at least two neighbouring letters L v How many elements of X contain no neighbouring vowel letters letters A
E O
Permutations and Combinations
1 Let X be the set of all five digit numbers that can be obtained by permu ting the digits of the number 12558 i How many elements does X contain ii How many even numbers are there in X iii How many
odd numbers are there in X
Permutations and Combinations
3 In how many ways can we put 20 identical silver coins into five coloured boxes so that at most 3 coins go into the blue box at least 4 into red and at most 5 into green The remaining boxes are
yellow and black and may contain any number of coins Every box except the red one may also remain empty
Permutations and Combinations
2 Assume that a basketball line up must consist of 5 players 2 guards 1 center and 2 forwards A coach has in total 12 players in the team there are 5 guards 2 centers 4 forwards and additionally
Peter Williams who can play as center or forward How many 5 player subsets of this 12 element set are acceptable line ups In how many of them does Peter Williams appear
Permutations and Combinations
1 Let X be the set of all different eight letter words that can be obtained by permuting the letters of the word DEBUNKED How many elements does X contain In how many of them there are neighbouring
identical letters In how many elements of X there are no neighbouring vowels terminology E U are vowel letters while B D K N are consonant letters 2 Assume that a basketball line up must consist of 5
players 2 guards 1 center and 2 forwards A coach has in total 12 players in the team there are 5 guards 2 centers 4 forwards and additionally Peter Williams who can play as center or forward How many
5 player subsets of this 12 element set are acceptable line ups In how many of them does Peter Williams appear 3 In how many ways can we put 20 identical silver coins into five coloured boxes so that
at most 3 coins go into the blue box at least 4 into red and at most 5 into green The remaining boxes are yellow and black and may contain any number of coins Every box except the red one may also
remain empty 4 How many solutions of the equation a b c d 80 in integers greater than 0 satisfy simultaneously all the following conditions a 30 10 c 40 and a b c d are all even REMARK If your answer
is eg 3 56 you don t have to compute the numerical answer 546875 Just write 3 56 as the final answer
Permutations and Combinations
What is the least common denominator of rational expressions with the given denominators To enter an exponent use the symbol as in 3x 4 for 3x 4a 6a and 10b Type your answer and submit X X Write your
response here
Permutations and Combinations
When factoring the given binomial what is the trinomial factor 8x 125 Select an answer and submit For keyboard navigation use the up down arrow keys to select an answer a b C P 4x 10x 25 4x 10x 25 2x
10x 25 2x 2 10x 25 G
Permutations and Combinations
7 B 9 1 Write each expression in exponential form 7 5n 8 A 2n C 5n 4 Solve each equation Remember to check for extraneous solutions ANSWER SHOULD BE AN INTEGEROR A FRACTION NO POINTS IF YOU ANSWER AS
A DECIMAL 1 1 b 2 6b 6b 8 b Simplify 9 6x 2x A 2 C 12 11 1 B 5n 4 D 5n 5 20 4 4 5 5 4 A B 2x 3x D 2 3 B D 4 5 25 3 5 2 10 10 30x 9 12 6x A 4 105x 5 3x B 27x x 6 C 54 10x 6x 30x D 105 18x 12 12 4 3 3
25 A 3 12 3 2 16 4 4 3 3 5 D 3 5 3 3 B
Permutations and Combinations
7 B 9 1 Write each expression in exponential form 7 5n 8 A 2n C 5n 4 Solve each equation Remember to check for extraneous solutions ANSWER SHOULD BE AN INTEGEROR A FRACTION NO POINTS IF YOU ANSWER AS
A DECIMAL 1 1 b 2 6b 6b 8 b Simplify 9 6x 2x A 2 C 12 11 1 B 5n 4 D 5n 5 20 4 4 5 5 4 A B 2x 3x D 2 3 B D 4 5 25 3 5 2 10 10 30x 9 12 6x A 4 105x 5 3x B 27x x 6 C 54 10x 6x 30x D 105 18x 12 12 4 3 3
25 A 3 12 3 2 16 4 4 3 3 5 D 3 5 3 3 B
Permutations and Combinations
13 2 3 4 A 14 C 2 2 6 13 2 14 6n A 6n C n Write each expression in radical form Simplify 1 16 36n A 6n C 243n 5 15 B Write each expression in exponential form 15 n 5 A 6n C 5n D B n 2 5 2 2 2 2 6 2 B
3n D n D n B 125n D 216n
Permutations and Combinations
Slape 1 4 6 0 FIND THE SLOPE OF THE LINEAR FUNCTION IN EACH BOX YOU TRY SLOPE SLOPE 2 3 3 y y 8x 6 SLOPE SLOPE 20 14 8 1 y 268 196 124 40 20 96 12 60
Permutations and Combinations
Choose h and k such that the system has a no solution b a unique solution and c many solutions X hx 5 2x 4x k OH The system has a unique solution only when h c Select the correct answer below and
fill in the answer box es to complete your choice Type an integer or simplified fraction OA The system has many solutions only when h OB The system has many solutions only when k OC The system has
many solutions only when h OD The system has many solutions only when h O E The system has many solutions only when h OF The system has many solutions only when h OG The system has many solutions
only when k H The system has many solutions only when h and k and k COOP and h is any real number and k is any real number and k and k and k is any real number and h is any real number and k
Permutations and Combinations
Choose the correct answer below OA There is a pivot position in each row of the coefficient matrix The augmented matrix will have nine columns and will not have a row of the form 000000001 so the
system is consistent OB There is at least one row of the coefficient matrix that does not have a pivot position This means the augmented matrix which will have nine columns have a row of the form
000000001 so the system is inconsistent OC There is at least one row of the coefficient matrix that does not have a pivot position This means the augmented matrix which will have nine columns
00000001 so the system could be inconsistent have a row of the form i OD There is a pivot position in each row of the coefficient matrix The augmented matrix will have seven columns and will not have
a row of the form 0000001 so the system is consistent
Permutations and Combinations
Find the general solution of the system whose augmented matrix is given below 1 7 0 1 0 4 0 10 0 8 1 0 00 1 8 3 0 00 0 0 0 Select the correct choice below and if necessary fill in the answer boxes to
complete your answer O C X x 0 X3 is free OB X X is free X4 is free X5 is free x X3 is free X5 is free OD The system is inconsistent
Permutations and Combinations
220 attended the FSU Marist game M 185 attended the FSU Butler game B 190 attended the FSU NC State gandle NCS 100 attended the Marist game and the NC State game 60 attended the NC State game and the
Butler game attended the Butler game and the Marist game 30 attended all three of the games 65 Sketch a Venn diagram and answer the questions below M NOS B U
Permutations and Combinations
15 x x 2 16 x x 1 17 x x 6 18 x x 9 19 xlx 3 x 3 20 x x 4 x 5 21 x x 1 x 6 22 x x 5 x 7
Permutations and Combinations
Survey researchers design and conduct surveys and analyze data Some survey researchers design public opinion surveys Other survey researchers market research analysts design marketing surveys which
examine products or services that consumers want need or prefer Most survey researchers work in research firms polling organizations nonprofits corporations colleges and universities and government
agencies The majority work full time during regular business hours A bachelor s degree may be sufficient for some entry level positions The median annual wage for survey researchers was 49 760 in May
2014 SOURCE Bureau of Labor and Statistics U S Department of Labor Occupational Outlook Handbook 2016 2017 Edition Survey Researchers Surveys for scientific research cover various fields including
government health social sciences and education For example a survey researcher may try to capture information about the prevalence of drug use or disease An anonymous survey of college students was
taken to determine behaviors regarding alcohol cigarettes and illegal drugs The results were as follows 894 drank alcohol regularly 665 smoked cigarettes 192 used illegal drugs 424 drank alcohol
regularly and smoked cigarettes 114 drank alcohol regularly and used illegal drugs 119 smoked cigarettes and used illegal drugs 97 engaged in all three behaviors 309 engaged in none of the behaviors
Permutations and Combinations
A survey was given to 298 people asking whether people like dogs and or cats 186 said they like dogs 123 said they like cats 67 said they don t like cats or dogs How many said they liked both cats
and dogs people liked both cats and dogs
Permutations and Combinations
The Venn diagram here shows the cardinality of each set Use this to find the cardinality of the given set A C n An BC 8 11 7 5 9 14 B 6 a
Permutations and Combinations
Let the Universal Set be S Let A and B are subsets of S Set A contains 29 elements and Set B contains 99 elements Sets A and B have 24 elements in common If there are 24 elements that are in S but
not in A nor B how many elements are in S
Permutations and Combinations
Let the Universal Set S have 150 elements A and B are subsets of S Set A contains 28 elements and Set B contains 88 elements If Sets A and B have 5 elements in common how many elements are in A but
not in B
Permutations and Combinations
The Venn diagram here shows the cardinality of each set Use this to find the cardinality of each given set 13 n A n ANC M 5
Permutations and Combinations
A quarterback throws an incomplete pass The height of the football at time t is modeled by the equation h t 16t2 40t 7 Rounded to the nearest tenth the solutions to equation when h t 0 feet are 0 2 s
and 2 7 s Which solution can be eliminated and why The solution 0 2 s can be eliminated because the pass was not thrown backward The solution 2 7 s can be eliminated because a ball cannot be in the
air for that long due to gravity The solution 2 7 s can be eliminated because the pass was thrown backward The solution 0 2 s can be eliminated because time cannot be a negative value
Permutations and Combinations
In the last several weeks 74 days saw rain and 39 days saw high winds In that same time period 22 days saw both rain and high winds How many days saw either rain or high winds
Permutations and Combinations
Which equation represents the vertical asymptote of the graph Ox 8 Oy 8 7 Ox 0 6 LO 5 3 13 14 19 12 11 10 9 8 7 6 5 4 3 2 1 2 2 3 345 5 6 0 | {"url":"https://kunduz.com/questions/algebra/permutations-and-combinations/?page=2","timestamp":"2024-11-06T13:52:53Z","content_type":"text/html","content_length":"318813","record_id":"<urn:uuid:b5224911-64c2-459a-82d0-aa4a9f9e33da>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00002.warc.gz"} |
Circular algebraic curve
In geometry, a circular algebraic curve is a type of plane algebraic curve determined by an equation F(x, y) = 0, where F is a polynomial with real coefficients and the highest-order terms of F form
a polynomial divisible by x^2 + y^2. More precisely, if F = F[n] + F[n−1] + ... + F[1] + F[0], where each F[i] is homogeneous of degree i, then the curve F(x, y) = 0 is circular if and only if F[n]
is divisible by x^2 + y^2.
Equivalently, if the curve is determined in homogeneous coordinates by G(x, y, z) = 0, where G is a homogeneous polynomial, then the curve is circular if and only if G(1, i, 0) = G(1, −i, 0) = 0. In
other words, the curve is circular if it contains the circular points at infinity, (1, i, 0) and (1, −i, 0), when considered as a curve in the complex projective plane.
Multicircular algebraic curves
An algebraic curve is called p-circular if it contains the points (1, i, 0) and (1, −i, 0) when considered as a curve in the complex projective plane, and these points are singularities of order at
least p. The terms bicircular, tricircular, etc. apply when p = 2, 3, etc. In terms of the polynomial F given above, the curve F(x, y) = 0 is p-circular if F[n−i] is divisible by (x^2 + y^2)^p−i when
i < p. When p = 1 this reduces to the definition of a circular curve. The set of p-circular curves is invariant under Euclidean transformations. Note that a p-circular curve must have degree at least
The set of p-circular curves of degree p + k, where p may vary but k is a fixed positive integer, is invariant under inversion. When k is 1 this says that the set of lines (0-circular curves of
degree 1) together with the set of circles (1-circular curves of degree 2) form a set which is invariant under inversion.
Original source: https://en.wikipedia.org/wiki/Circular algebraic curve. Read more | {"url":"https://handwiki.org/wiki/Circular_algebraic_curve","timestamp":"2024-11-03T12:21:31Z","content_type":"text/html","content_length":"33757","record_id":"<urn:uuid:556a211d-e1f3-4193-9e5d-e6284daaaaf2>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00131.warc.gz"} |
Kinetic modeling of scrape-off layer plasmas
We study the global well-posedness and asymptotic behavior for a semilinear damped wave equation with Neumann boundary conditions, modeling a one-dimensional linearly elastic body interacting with a
rigid substrate through an adhesive material. The key fea ...
World Scientific Publ Co Pte Ltd | {"url":"https://graphsearch.epfl.ch/en/publication/119497","timestamp":"2024-11-03T19:24:43Z","content_type":"text/html","content_length":"110533","record_id":"<urn:uuid:09d196d7-473d-46f9-8a9f-b78878532e60>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00737.warc.gz"} |
About Articles Books Lectures Presentations Glossary Cite page Help? Translate
Core Concepts Possibilities
Actualism Alternative Possibilities are one of the key requirements for the freedom component of
Adequate free will, critically needed for libertarian free will. They allow for what William James
Determinism called open and ambiguous futures. The old page on Harry Frankfurt's denial of the
Agent-Causality Principle of Alternate Possibilities (PAP) can be found here. This Possibilities page is
Alternative now about the philosophical difference between "possibilities" (especially those never
Possibilities realized) and the one realized "actuality." The existential and ontological status of mere
Causa Sui "possibilities" has been debated by philosophers for many centuries. Diodorus Cronus
Causal Closure dazzled his contemporaries in the fourth century BCE with sophisticated logical arguments,
Causalism especially paradoxes, that logically "proved" there could be only one possible future.
Causality Diodorus' Master Argument is a set of propositions designed to show that the actual is the
Certainty only possible and that some true statements about the future imply that the future is
Chance already determined. This follows logically from his observation that if something in the
Chance Not Direct future is not going to happen, it must have been that statements in the past that it would
Cause not happen must have been true. Modern day "actualists" include Daniel Dennett, for whom
Chaos Theory determinism guarantees that the actual outcome is and always was the only possible
The Cogito Model outcome. Dennett, as dazzling as Diodorus, cleverly asks "Change the future? From what to
Compatibilism what?" The ancient philosophers debated the distinction between necessity and contingency
Complexity (between the a priori and the a posteriori). Necessity includes events or concepts that
Comprehensive are logically necessary and physically necessary, contingency those that are logically or
Compatibilism physically possible. In the middle ages and the enlightenment, necessity was often
Conceptual contrasted with freedom. In modern times it is often contrasted with mere chance.
Analysis Causality is often confused with necessity, as if a causal chain requires a deterministic
Contingency necessity. But we can imagine chains where the linked causes are statistical, and modern
Control quantum physics tells us that all events are only statistically caused, even if for large
Could Do Otherwise macroscopic objects the statistical likelihood approaches certainty for all practical
Creativity purposes. The apparent deterministic nature of classical mechanical laws is only an
Default "adequate" determinism, true for macroscopic objects which are large enough, massive
Responsibility enough, to contain so many elementary particles that the indeterministic quantum effects
De-liberation of individual average out. In modern philosophy, modal theorists like David Lewis discuss
Determination counterfactuals that might be true in other "possible worlds." Lewis' work at Princeton
Determination may have been inspired by the work of Princeton scientist Hugh Everett III. Everett's
Fallacy interpretation of quantum mechanics replaces the "collapse" of the wave function with a
Determinism "splitting" of this world into multiple worlds, in each of which everything is completely
Disambiguation determined!
Double Effect
Either Way The Ontological Status of Alternative Possibilities
Emergent Whereas Actualities are physical events involving material bodies, possibilities normally
Determinism have no material content. They are immaterial, like our thoughts and ideas. In particular,
Epistemic Freedom they are predictions about the future in a universe with multiple possible futures.
Ethical Fallacy Actualists (from Diodorus to Dennett) are determinists who believe that the only possible
Experimental future is the one future that will actually happen.
Extreme Quantum Mechanics and Alternative Possibilities
Event Has Many According to the Schrödinger equation of motion, the time evolution of the wave function
Causes describes a "superposition" of possible quantum states. Standard quantum mechanics says
Frankfurt Cases that interaction of the quantum system with other objects causes the system to collapse
Free Choice into one of those possible states, with probability given by the square of the
Freedom of Action "probability amplitude." One very important kind of interaction is a measurement by an
"Free Will" "observer." In standard quantum theory, when a measurement is made, the quantum system is
Free Will Axiom "projected" or "collapsed" or "reduced" into a single one of the system's allowed states.
Free Will in If the system was "prepared" in one of these "eigenstates," then the measurement will find
Antiquity it in that state with probability one (that is, with certainty). However, if the system is
Free Will prepared in an arbitrary state ψ[a], this state can be represented as being in a linear
Mechanisms combination of the system's basic eigenstates φ[n].
Free Will ψ[a] = Σ c[n] | n >.
Requirements where
Free Will Theorem c[n] = < ψ[a] | φ[n] >.
Future Contingency The system ψ[a] is said to be in "superposition" of those basic states φ[n]. The
Hard probability P[n] of its being found in a particular state φ[n] is
Incompatibilism P[n] = < ψ[a] | φ[n] >^2 = c[n]^2.
Idea of Freedom These probabilities and their information content are ontologically similar to our
Illusion of thoughts and ideas — immaterial predictions about future material events. The astonishing
Determinism mathematical accuracy of these predictions about the future might appear to put them in
Illusionism the same category as logical statements and mathematical proofs. But this is not so.
Impossibilism Physical theories are only tested statistically, by comparison to the outcomes of large
Incompatibilism numbers of identical experiments. Scientific theories are not "proven" true or false,
Indeterminacy neither mathematically nor logically by reasoned arguments. Information philosophy goes
Indeterminism "beyond logic and language" to solve great problems in philosophy and physics.
Laplace's Demon Possibilities and the Existence of Particle Properties in Quantum Mechanics
Liberty of When a quantum measurement is made on a system in a superposition of states, which state
Indifference the system collapses into is quantum random. The means that the particular state did not
Libet Experiments exist before the measurement. There is no objective reality as Albert Einstein hoped. As
Luck the Copenhagen interpretation claimed, the property is brought into existence by the
Master Argument measurement. It creates new information in the universe.
Libertarianism For Teachers
Moral Necessity
Moral Stanford Encyclopedia of Philosophy
Responsibility Wikipedia
Moral Sentiments
Mysteries For Scholars
Necessity Chapter 3.7 - The Ergod Chapter 4.2 - The History of Free Will
Noise Part Three - Value Part Five - Problems
Nonlocality Normal | Teacher | Scholar
Paradigm Case
Random When?/
Rational Fallacy
Same Circumstances
Science Advance
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Free Will
Mental Causation
James Symposium
Possibilities and the Existence of Particle Properties in Quantum Mechanics
Chapter 3.7 - The Ergod Chapter 4.2 - The History of Free Will
Part Three - Value Part Five - Problems | {"url":"https://informationphilosopher.com/freedom/possibilities.html","timestamp":"2024-11-10T08:39:40Z","content_type":"text/html","content_length":"105244","record_id":"<urn:uuid:71595065-fd67-48b3-96aa-f1df20c21937>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00648.warc.gz"} |
Third law of thermodynamics
Third law of thermodynamics
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Functions of state
) †
Pressure / Volume †
Chemical potential / Particle no. †
(† Conjugate variables)
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Work · Heat
Material properties
$T$ $\partial S$
Specific heat capacity $c=$ $N$ $\partial T$
$1$ $\partial V$
Compressibility $\beta=-$ $V$ $\partial p$
Thermal expansion $\alpha=$ $1$ $\partial V$
$V$ $\partial T$
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Free energy
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von Helmholtz
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· Gibbs ·
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The third law of thermodynamics is a statistical law of nature regarding entropy:
The entropy of a perfect crystal approaches zero as the absolute temperature approaches zero.
For other materials, the residual entropy is not necessarily zero,
The third law was developed by the chemist Walther Nernst during the years 1906-1912, and is therefore often referred to as Nernst's theorem or Nernst's postulate. The third law of thermodynamics
states that the entropy of a system at absolute zero is a well-defined constant. This is because a system at zero temperature exists in its ground state, so that its entropy is determined only by the
degeneracy of the ground state. It means that "it is impossible by any procedure, no matter how idealised, to reduce any system to the absolute zero of temperature in a finite number of operations".
An alternative version of the third law of thermodynamics as stated by Gilbert N. Lewis and Merle Randall in 1923:
If the entropy of each element in some (perfect) crystalline state be taken as zero at the absolute zero of temperature, every substance has a finite positive entropy; but at the absolute zero of
temperature the entropy may become zero, and does so become in the case of perfect crystalline substances.
This version states not only ΔS will reach zero at 0 kelvins, but S itself will also reach zero as long as the crystal has a ground state with only one configuration. Some crystals form defects which
causes a residual entropy. This residual entropy disappears when the kinetic barriers to transitioning to one ground state are overcome.
With the development of statistical mechanics, the third law of thermodynamics (like the other laws) changed from a fundamental law (justified by experiments) to a derived law (derived from even more
basic laws). The basic law from which it is primarily derived is the statistical-mechanics definition of entropy for a large system:
$S = k_B \log \, \Omega \$
where S is entropy, k[B] is the Boltzmann constant, and $\Omega$ is the number of microstates consistent with the macroscopic configuration.
In simple terms, the third law states that the entropy of a perfect crystal approaches zero as the absolute temperature approaches zero. This law provides an absolute reference point for the
determination of entropy. The entropy determined relative to this point is the absolute entropy.
The entropy of a perfect crystal lattice as defined by Nernst's theorem is zero (provided that its ground state is unique, whereby ln(1)k = 0).
An example of a system which does not have a unique ground state is one containing half-integer spins, for which time-reversal symmetry gives two degenerate ground states (an entropy of ln(2) k,
which is negligible on a macroscopic scale). Some crystalline systems exhibit geometrical frustration, where the structure of the crystal lattice prevents the emergence of a unique ground state.
Ground-state helium (unless under pressure) remains liquid.
In addition, glasses and solid solutions retain large entropy at 0K, because they are large collections of nearly degenerate states, in which they become trapped out of equilibrium. Another example
of a solid with many nearly-degenerate ground states, trapped out of equilibrium, is ice Ih, which has "proton disorder".
For the third law to apply strictly, the magnetic moments of a perfectly ordered crystal must themselves be perfectly ordered; indeed, from an entropic perspective, this can be considered to be part
of the definition of "perfect crystal". Only ferromagnetic, antiferromagnetic, and diamagnetic materials can satisfy this condition. Materials that remain paramagnetic at 0K, by contrast, may have
many nearly-degenerate ground states (for example, in a spin glass), or may retain dynamic disorder (a spin liquid ). | {"url":"https://ftp.worldpossible.org/endless/eos-rachel/RACHEL/RACHEL/modules/wikipedia_for_schools/wp/t/Third_law_of_thermodynamics.htm","timestamp":"2024-11-06T15:06:26Z","content_type":"text/html","content_length":"26396","record_id":"<urn:uuid:29b2df5c-c26e-486d-81b7-87bf0a3fbb47>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00601.warc.gz"} |
Ball Mill Rotate Revolve
Rotate Ball Mills - yvons-fotostudio.nl
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Are Two Lines Perpendicular To The Same Line Parallel - Graphworksheets.com
Graphing Parallel And Perpendicular Lines Worksheet – Line Graph Worksheets will help you understand how a line graph functions. There are many types of line graphs and each one has its own purpose.
Whether you’re teaching a child to read, draw, or interpret line graphs, we have worksheets for you. Make a line graph Line … Read more | {"url":"https://www.graphworksheets.com/tag/are-two-lines-perpendicular-to-the-same-line-parallel/","timestamp":"2024-11-11T05:05:07Z","content_type":"text/html","content_length":"47199","record_id":"<urn:uuid:a06a014b-a520-4b17-8348-e76853c080c4>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00020.warc.gz"} |
purpose integer arithmetic
The predicates in this section provide more logical operations between integers. They are not covered by the ISO standard, although they are‘part of the community’and found as either library or
built-in in many other Prolog systems.
Low and High are integers, High ≥Low. If Value is an integer, Low ≤Value ≤High. When Value is a variable it is successively bound to all integers between Low and High. If High is inf or infinite^
121We prefer infinite, but some other Prolog systems already use inf for infinity; we accept both for the time being. between/3 is true iff Value ≥Low, a feature that is particularly interesting
for generating integers from a certain value.
True if Int2 = Int1 + 1 and Int1 ≥. At least one of the arguments must be instantiated to a natural number. This predicate raises the domain error not_less_than_zero if called with a negative
integer. E.g. succ(X, 0) fails silently and succ(X, -1) raises a domain error.^122The behaviour to deal with natural numbers only was defined by Richard O'Keefe to support the common
count-down-to-zero in a natural way. Up to 5.1.8, succ/2 also accepted negative integers.
True if Int3 = Int1 + Int2. At least two of the three arguments must be instantiated to integers.
This predicate is a shorthand for computing both the Quotient and Remainder of two integers in a single operation. This allows for exploiting the fact that the low level implementation for
computing the quotient also produces the remainder. Timing confirms that this predicate is almost twice as fast as performing the steps independently. Semantically, divmod/4 is defined as below.
divmod(Dividend, Divisor, Quotient, Remainder) :-
Quotient is Dividend div Divisor,
Remainder is Dividend mod Divisor.
Note that this predicate is only available if SWI-Prolog is compiled with unbounded integer support. This is the case for all packaged versions.
True when Root ** N + Remainder = I. N and I must be integers.^123This predicate was suggested by Markus Triska. The final name and argument order is by Richard O'Keefe. The decision to include
the remainder is by Jan Wielemaker. Including the remainder makes this predicate about twice as slow if Root is not exact. N must be one or more. If I is negative and N is odd, Root and Remainder
are negative, i.e., the following holds for I < 0:
% I < 0,
% N mod 2 =\= 0,
N, I, Root, Remainder),
IPos is -I,
N, IPos, RootPos, RemainderPos),
Root =:= -RootPos,
Remainder =:= -RemainderPos. | {"url":"https://cliopatria.swi-prolog.org/swish/pldoc/man?section=logic-int-arith","timestamp":"2024-11-02T09:37:58Z","content_type":"text/html","content_length":"9182","record_id":"<urn:uuid:f3178b6e-bde2-4750-aa93-0f1a5f1631d3>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00124.warc.gz"} |
Plate Buckling Checks with ABS/DNV Rules | SDC Verifier
Recognition of Structural Members Enables Plate Buckling Checks According to ABS / DNV Rules Directly in General FEA Software
Plate buckling strength is an important aspect of offshore steel construction design. We will show how the problem of both general FEA analysis (strength evaluation, displacement, and deflection
checks) and plate buckling check according to ABS or DNV rules is solved with the help of FEA and code checking tools. The most complicated task in performing a plate buckling check for a big
structure, like a complete ship design, on a general Finite Element Analysis model is to define a big amount of plates and the dimensions of these plates to be verified.
For the precise FEA analysis, the model has to have a fine mesh with small enough finite elements to guarantee the correct results. But at the same time, each plate field must be treated as one
separate structural member for plate buckling checks. With the help of SDC Verifier it is possible to break the boundaries of general FEA Analysis and enable the code checking directly in Simcenter
3D, Femap, Ansys, etc. by enabling the automatic recognition of structural items. The recognition of plates, stiffeners, and girders is based on mesh connectivity and can be performed on any
structure which is built with 2D or, in some cases, even 3D elements. The structural members are defined automatically and mesh independently. This allows an engineer to have a model with fine mesh
for precise results of the general finite element analysis and a list of structural members for code checking.
This article was originally posted in SNAMES 40th Annual Journal.
Plates are commonly used in the design of ships, offshore structures, aircraft, civil, and other engineering structures. Each plate should be verified as it influences the strength and stability of
the whole construction. There are two main failure modes of a plated structural item that can lead to sudden damage: material failure and structural instability, which is called buckling. Most plated
structures are capable of carrying tensile loading, but may be poor in resisting compressive forces. Usually, buckling effects take place suddenly and may lead to severe or even catastrophic
structural failure.
That’s why it is very important to understand the buckling capacities of the plates to avoid a collapse of the complete structure. Buckling analysis of the structure with a general FEA solver may
seem like a quick and easy solution, since it provides a buckling load factor – a ratio of the buckling load to the currently applied load. As a result, you get a value of the factor that will cause
buckling failure in case of multiplication of the load value with it. But this analysis result would be only for a panel that will fail first, which does not guarantee that the rest of the structure
is safe.
This is where it becomes necessary to verify according to the industry standards. A lot of these documents already contain verification procedures or recommended practices for the plate buckling
analysis. Here are some of the codes that are commonly used in the industry:
• DNV RP-C201 Buckling Strength of Plated Structures;
• DNV CN30 Buckling Strength Analysis of Bars and Frames, and Spherical Shells;
• ABS Plate Buckling and Ultimate Strength Assessment for Offshore Structures;
• Eurocode 3 – design of steel structures – Part 1-5: Plated structural elements.
In the case of code checks according to the standards, an engineer is able to perform the calculation of the value of utilization factor for every plate as a result. It is possible to do the
calculation for each load or combination to ensure that the whole structure is safe. This procedure is usually not very simple since a lot of factors, characteristics, and coefficients should be
taken into account.
With a certain knowledge, it’s possible to do the check according to the standard by hand. But in this case, an engineer is able to verify one plate under one loading condition at a time. Though, a
typical offshore structure or ship design consists of thousands or even millions of plates and hundreds of load combinations. This is when automation is a must. When it comes to the code checking in
CAE, there are two ways:
• To run the general finite element analysis for the design, which is mandatory to understand the behavior of the structure and obtain the results of stresses, displacements, forces, other outputs.
And then to perform the Standard verification of important details with scripts, spreadsheets, or hand calculation.
• To use the general FEA analysis for the design and dedicated software for the code checking.
Both of these methods have certain drawbacks. Spreadsheet or hand calculation analysis is time-consuming, and it’s easy to miss a failure, since not always the most stressed or longest plates are the
most subjected to buckling. Using dedicated code checking software is more precise, but since it requires having each structural member defined, it is necessary to build another model for code
checking. Discernibly, double modeling leads to an increase in the overall completion time of the project, since every update and modification has to be done twice.
Automatic Recognition
Since checks are done on structural items and not on finite elements, the best solutions for both execution time and accuracy of the results would be to use the extension for general FEA software
(Ansys, Femap, Simcenter 3D) that is capable of the automatic recognition of the structural members mesh independently. SDC Verifier is software that follows this methodology. The best solution to
avoid double work is to have the same environment for both General FEA and code checks.
Stiffened Panel Finder — is a tool to automatically recognize sections, panels, plates, stiffeners, and girders, and dimensions of these structural members. The detection is based on mesh
connectivity and can be performed on any structure which is built with 2D (plate or shell elements) for plate members and both 1D (beams) or 2D finite elements for stiffeners and girders.
Recognized structural members
Detection is made automatically and meshes independently. This brings to an engineer the opportunity to have a model with fine mesh for precise results of the general finite element analysis and use
the same model for calculation of Eurocode, ABS, or DNV plate buckling checks. At the first stage, Sections are defined by the global or custom coordinates. All the elements that lay in one plane (of
course, with a certain angle of deviation which could be defined by the user in settings) are defined as sections. This allows detection of, for example, Frames, Decks, and Longitudinal sections of
the ship. Hull is also automatically recognized as a custom section.
Frames of the ship automatically recognized by the coordinates
The next step is to define the plates on these sections; plates are also recognized automatically with borders at sections intersection, stiffeners, girders, or any other members perpendicular to the
sections. A user always has control over recognition to add/remove or split the members manually. But if the mesh is fine enough, there is no need for manual interaction with the recognized
structural members. Recognition is completely mesh-independent, any plate of the studied FEA model can consist of hundreds or even thousands of finite elements for precise stress analysis, and it
will still be defined as one structural member for plate buckling checks.
Automatic recognition of the plates defines the following parameters for the code check: length and widths of the plate, direction, amount of edges, material type, and thickness. The analysis is
based on stresses in each finite element of the plate or on the plate average stress.
Plates and Stiffeners recognized on a section
Verification procedure from the user point of view
Despite the fact that material properties, forces, stresses are defined in the FEA program and plate dimensions and types are automatically recognized, some parameters still should be defined by the
user. For example, DNV RP-C201 Plate/Stiffener Buckling (2010) requires user input for a characteristic called Resulting Material Factor. During the analysis procedure, buckling resistance will be
divided by this factor. By default, this factor is 1.15, but an engineer may change this value taking into account the type of structure or consequences of failure.
It is also possible to define a thickness factor that allows to increase/decrease all plate thicknesses quickly without re-solving the model. For example, a thickness factor of 1.2 means a thickness
increase of 20%, which leads to a stress decrease.
An important decision has to be made about what stresses to use. It is possible to use plate average stress; this will result in one buckling factor result on each plate. A more conservative approach
is to use stresses of every element for the analysis; in this case, the maximum buckling factor from all the elements of a plate would be presented as a resulting buckling factor of the whole plate.
Plate Average Stress options.
Elemental Stress options
If the option to use plate average stress is turned off, then there are two options to define elemental stress: average element stress or minimum element midplane stress (which is maximum compressive
Basically, the parameters and decisions described above are the only engineer’s input in case of automatic code checking. The rest of the calculation is done by a code checking program: standard
outputs of the FEA solver and parameters of the model are used as variables for formulas to define plate buckling factor as a result. The benefit of SDC Verifier as a code checking tool is also that
all the formulas are open and refer to the standards, so it is easy to follow the calculation procedure, possible to find the source of the problem quickly, and even modify the existing formulas if
customization of the checks is necessary. It is also possible to see the intermediate results values.
Software completely follows the verification procedure of the selected standard. At the first step of the code check, plate length, width, and thickness are retrieved from the recognition, and
compressive Stresses Sx, Sy, and Sxy are calculated in plate direction. Then the Slenderness and Buckling resistance for both X and Y directions are checked. Every formula is open and has a
description, names of the intermediate variables. For example, the slenderness formula (used to calculate the buckling resistance in the X direction of the plate) from the DNV check is represented
Different types of variables are highlighted with different colors, and description refers to the formula from the standard. In the final step, Buckling factors are calculated for X, Y, and XY
(Shear) direction, as well as Maximum overall directional and combined Buckling Factors.
Results of the Automated Plate Buckling Checks
Result tables
As a result of this automated verification procedure, the user will get a Buckling factor for every plate of the whole structure in minutes, rather than days spent with spreadsheets or hand
calculations. Moreover, the calculation could be done for multiple load combinations and envelope groups of loads. This means that the results of the analysis, which are typically presented in
detailed buckling factor tables for every section/plate, are automatically prepared for each loading condition.
A wide variety of tables are available, and results can be presented over any load or selection. The extreme table type shows the maximum value for the complete selection, and Expand table type
presents the value for every item of this selection to be quite extensive.
In addition to the buckling factor, the following parameters results can also be listed in the table:
• Plate Width;
• Plate Thickness;
• Sx in plate direction;
• Sy in plate direction;
• Sxy in plate directions;
• Equivalent Stress.
Overall Buckling Factor results in a table
The interface of the tables allows presenting not only the final results but also the calculation details – all the formulas with intermediate resulting values of the parameters used for the
calculation. This provides an engineer with an extra instrument to control the calculation and leaves much less room for an error.
Result plots
The graphical interface of the FEA programs is used to visualize the buckling factor or any other output (including the recognition details) values for any user-defined selection.
This provides a user with full control of the view, including the positioning of the model, plotting style, legend settings. Views are stored and can be used to present the results of general FEA
analysis, as well as code checking results, for any load or selection.
Buckling Factor results on a plot
Automatic reporting
Since the calculation core allows to get the results for individual loads and load combinations, and SDC Verifier has an interface to present the results with tables and plots, it’s easy to prepare
an automated template-based structure for the report generation. Typically, the report contains the following parts:
• Model Setup – information about materials, properties, loads and boundary conditions, the basis of calculation, formulas used for the analysis (automatically added because of the open interface
of the code-checking software);
• Results – automatically sorted by load, or by selection results of finite element analysis and verification according to standards;
• Summary – a short explanation of the main results and comparison with the allowable values.
Automation of the reporting process helps to save time on the repetitive documentation routine. It also reduces the deadline pressure: since the report structure is set, there is no need to create a
new report in case of modifications or design changes. The engineer has only to update the model, run the calculation, and regenerate the report.
The code checking approach described in this article brings to marine designers and naval architects the understanding of alternatives for the usual code checking workflow and describes the ways to
save time on routine and repetitive tasks by automating the verification for a complete model in a single CAE environment.
In addition to the time-saving benefits, usage of code checking extensions for the General FEA programs allows to:
• Check the quality of modeling with the help of recognition tools.
• Understand the behavior of a studied structure, by analyzing all possible loading conditions and defining the governing ones.
• Analyze the critical parameters for the checks.
• Quickly improve the design. By using the thickness factors and modifying the plate dimensions. Or with the help of powerful editors in the general FEA tools and instant update of the simulation
data code-checking extension.
• Compare different design approaches of loading conditions in one user-friendly CAE environment.
• Timoshenko, S. P. and Gere, J. Theory of Elastic Stability, 2nd edition, McGraw-Hill, 1961
• Recommended Practice. Det Norske Veritas. DNV-RP-C201 Buckling Strength of Plated Structures. October 2010. | {"url":"https://sdcverifier.com/articles/recognition-of-structural-members-enables-plate-buckling-checks-according-to-abs-dnv-rules-directly-in-general-fea-software/","timestamp":"2024-11-04T11:03:25Z","content_type":"text/html","content_length":"183986","record_id":"<urn:uuid:0dd0786a-3671-4a62-a09c-9f86684bf1a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00414.warc.gz"} |
2,192 research outputs found
We introduce the category Pstem[n] of n-stems, with a functor P[n] from spaces to Pstem[n]. This can be thought of as the n-th order homotopy groups of a space. We show how to associate to each
simplicial n-stem Q an (n+1)-truncated spectral sequence. Moreover, if Q=P[n]X is the Postnikov n-stem of a simplicial space X, the truncated spectral sequence for Q is the truncation of the usual
homotopy spectral sequence of X. Similar results are also proven for cosimplicial n-stems. They are helpful for computations, since n-stems in low degrees have good algebraic models
Recommended from our members
This paper describes an algorithm for locating stationary points of n-forms. Use is made of the associated n-linear form, the stationary points of which are seen to coincide with those of the n-form.
Conditions of convergence are established using the concept of Liapunov stability, and it is seen that the scheme can always be made to converge to the global maximum of the n-form over unit vectors
We develop a general theory of cosimplicial resolutions, homotopy spectral sequences, and completions for objects in model categories, extending work of Bousfield-Kan and Bendersky-Thompson for
ordinary spaces. This is based on a generalized cosimplicial version of the Dwyer-Kan-Stover theory of resolution model categories, and we are able to construct our homotopy spectral sequences and
completions using very flexible weak resolutions in the spirit of relative homological algebra. We deduce that our completion functors have triple structures and preserve certain fiber squares up to
homotopy. We also deduce that the Bendersky-Thompson completions over connective ring spectra are equivalent to Bousfield-Kan completions over solid rings. The present work allows us to show, in a
subsequent paper, that the homotopy spectral sequences over arbitrary ring spectra have well-behaved composition pairings.Comment: Published by Geometry and Topology at http://www.maths.warwick.ac.uk
We formulate and prove a chain rule for the derivative, in the sense of Goodwillie, of compositions of weak homotopy functors from simplicial sets to simplicial sets. The derivative spectrum dF(X) of
such a functor F at a simplicial set X can be equipped with a right action by the loop group of its domain X, and a free left action by the loop group of its codomain Y = F(X). The derivative
spectrum d(E o F)(X)$ of a composite of such functors is then stably equivalent to the balanced smash product of the derivatives dE(Y) and dF(X), with respect to the two actions of the loop group of
Y. As an application we provide a non-manifold computation of the derivative of the functor F(X) = Q(Map(K, X)_+).Comment: Published by Geometry and Topology at http://www.maths.warwick.ac.uk/gt/
The category of I-spaces is the diagram category of spaces indexed by finite sets and injections. This is a symmetric monoidal category whose commutative monoids model all E-infinity spaces. Working
in the category of I-spaces enables us to simplify and strengthen previous work on group completion and units of E-infinity spaces. As an application we clarify the relation to Gamma-spaces and show
how the spectrum of units associated with a commutative symmetric ring spectrum arises through a chain of Quillen adjunctions.Comment: v3: 43 pages. Minor revisions, accepted for publication in
Algebraic and Geometric Topolog
We study the category of discrete modules over the ring of degree zero stable operations in p-local complex K-theory. We show that the p-local K-homology of any space or spectrum is such a module,
and that this category is isomorphic to a category defined by Bousfield and used in his work on the K-local stable homotopy category (Amer. J. Math., 1985). We also provide an alternative
characterisation of discrete modules as locally finitely generated modules.Comment: 19 page
For each n\geq 1 we introduce two new Segal-type models of n-types of topological spaces: weakly globular n-fold groupoids, and a lax version of these. We show that any n-type can be represented up
to homotopy by such models via an explicit algebraic fundamental n-fold groupoid functor. We compare these models to Tamsamani's weak n-groupoids, and extract from them a model for (k-1)connected
n-typesComment: Added index of terminology and notation. Minor amendments and added details is some definitions and proofs. Some typos correcte
Given a complete Heyting algebra we construct an algebraic tensor triangulated category whose Bousfield lattice is the Booleanization of the given Heyting algebra. As a consequence we deduce that any
complete Boolean algebra is the Bousfield lattice of some tensor triangulated category. Using the same ideas we then give two further examples illustrating some interesting behaviour of the Bousfield
lattice.Comment: 10 pages, update to clarify the products occurring in the main constructio
We apply the Dwyer-Kan theory of homotopy function complexes in model categories to the study of mapping spaces in quasi-categories. Using this, together with our work on rigidification from [DS1],
we give a streamlined proof of the Quillen equivalence between quasi-categories and simplicial categories. Some useful material about relative mapping spaces in quasi-categories is developed along
the way | {"url":"https://core.ac.uk/search/?q=authors%3A(Bousfield)","timestamp":"2024-11-06T20:49:52Z","content_type":"text/html","content_length":"143673","record_id":"<urn:uuid:e0a6bdb4-39fe-4a9d-b140-8016b2bf85d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00087.warc.gz"} |
The Elements of Geometry
From inside the book
Page 147 ... angle of the other are to each other as the products of the sides including the equal angles . Describe an isosceles triangle equal in area to a given triangle and having its vertical
angle equal to one angle of the given triangle . 11 ...
Page 152 ... angle of the other are to each other as the products of the sides including the equal angles , prove that the bisector of an angle of a triangle . divides the opposite side into parts
which are proportional to the sides adjacent to them ...
BOOK IV 163
Polyhedral Angles 173
Polyhedrons 177
BOOK VI 197
Figures of Revolution 211
Popular passages
... if two triangles have two sides of one equal, respectively, to two sides of the other...
The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. B' ADC A' D' C' Hyp. In triangles
ABC and A'B'C', ZA = ZA'. To prove = — • A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.
A sphere is a solid bounded by a surface, all the points of which are equally distant from a point within called the center.
Prove that, if from a point without a circle a secant and a tangent be drawn, the tangent is a mean proportional between the whole secant and the part without the circle.
If four quantities are in proportion, they are in proportion by inversion; that is, the second term is to the first as the fourth is to the third. Let...
If two triangles have two angles and the included side of one equal respectively to two angles and the included side of the other, the triangles are congruent.
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... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section.
Bibliographic information | {"url":"https://books.google.com.jm/books?id=WCUAAAAAYAAJ&vq=%22angle+of+the+other+are+to+each+other+as+the+products+of+the%22&dq=editions:UOM39015063898350&lr=&output=html&source=gbs_navlinks_s","timestamp":"2024-11-11T06:45:54Z","content_type":"text/html","content_length":"66664","record_id":"<urn:uuid:82e0c1e5-5fa0-47f5-89b0-24e676889a1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00705.warc.gz"} |
The Procedure of Balancing the Chemical Equations
We need to add the Stoichiometric coefficient in the chemical equations. The main reason behind balancing the equation is the Law of Conservation of Energy. You need to balance the same number of
atoms on both sides of the chemical equations. Balancing chemical equations calculator automatically in the chemical equation. When students are balancing the chemical equation. The calculatored have
provided the simple procedure to balance the chemical equation.
You need to identify the reactant and the product in the equation, then in the next step write down the number of atoms of the reactants and the products. Then you need to understand the coefficient
rules, and in the last step, you can add the coefficient in the chemical equation. Use calculatored.com to balance the chemical equation by the simple procedure.
In this article, we are highlighting the fact how we can explain the process of balancing the chemical equations:
1: Identify your reactants and the product:
For balancing the chemical equation, it should be clear to us. The left side of the chemical equation is the reactants and the right side of the equation is the products. The balance equation
calculator helps us to identify the reactants and the products in the equations:
Fe + O2 Fe2O3
In this equation the Iron and Oxygen are reactants and the Iron oxide is the product of the reaction. Balancing equations calculator balances the chemical equation according to the required number of
atoms in the equation.
Al + O2 Al2O3
In this equation the Aluminum and Oxygen are reactants and the Aluminum oxide is a product of the reaction. Balance chemical equations calculator balances the chemical equation according to the
required number of atoms in the equation.
2: Count the number of atoms:
Now count the number of atoms on both sides of the chemical question. Now consider, in this case, there is 1 atom of Aluminum, 2 atoms of Oxygen, and 1 molecule of Aluminum oxide and it contains 2
atoms of Aluminum and 3 atoms of oxygen.
Al + O2 Al2O3
You can observe there is no balance of the equation, we have 1 atom of the Aluminum on the reactant side and 2 atoms of the Aluminum on the product side. Oxygen has 2 atoms on the reactant, Chemical
equation calculator simply balances the chemical equation; we don’t need to count the atoms on both sides of the equation.
3:Rules of the coefficients:
There are certain rules you need to know when balancing the chemical equation. The main problem for the students, they are not able to understand the rules of the coefficient. Balancing chemical
equations calculator also helps to understand the coefficient rules for the chemical equations:
1. Students need to understand they can only add the coefficients of the reactants and products. When you see this equation then we only add the coefficient of the Aluminum and the Oxygen, and for
the aluminum oxide, but we are not going to change the subscript of the ”O2” or the Al2O3. The balancing chemical equations calculator automatically going to change the number of coefficients of
the elements to balance the equation.
Al + O2 Al2O3
2. We can add the whole number in the chemical equation, as it is not possible to use the fractions, or decimal values in the chemical equation.
3. Balancing chemical equations calculator helps to add the coefficients in a manner that on the reactant side the number of atoms becomes equal to the product side number of the same item. Consider
the Al + O2 Al2O3, when we are balancing the coefficients, we use a strategy to balance the number of atoms of the Aluminum(Al) on both sides. The same goes for the Oxygen, as we can see in this
condition, there are 2 atoms of Oxygen on the reactant side and 3 atoms on the product side. The balancing chemical equations calculator is simple to use, as you only have to enter the value of
the reactant and the product.
4. The other thing, which is an especially critical thing to understand the coefficient applies totally to the same formula 2Al2O3, it means now there are “4” atoms of Aluminum(Al) and “6” atoms of
4:Balancing the chemical equation:
We balance the chemical equation, in the following steps:
2Al + O2 Al2O3
Balance the atoms of the Aluminum:
2Al + O2 Al2O3
Now we can see there are “2” atoms of Aluminum on the reactant and the product side.
Balance the atoms of the Oxygen:
When we are balancing the atoms of the oxygen, we add “3” on the coefficient side and “2” on the reactant side. This would make the number of atoms “6” on both sides of the equation.
2Al + 3O2 2Al2O3
The balanced equation:
Now we can see the Aluminum balance is disturbed, but it is simple to adjust, just see how many atoms of the Aluminum are on the product side. It is “4”, so just simply add the coefficient of the
number “4” on the reactant side. The final chemical equation is as follows:
4Al + 3O2 2Al2O3
Balancing chemical equations calculator directly balances the chemical equation, but you need to understand the manual balancing of the chemical equation.
Students do find the chemical equations difficult to balance, as they have not developed a simple understanding of the rules of balancing the equation. A simple method to generate the concepts of
balancing the equation, Learn the rules of balancing the chemical equation. Try to balance the chemical equation by applying these rules. When you practice, it will become easy for you to balance the
reactants and the products on both sides of the chemical equations. The balancing chemical equations calculator is also helpful in learning the concepts of balancing chemical equations. The concepts
are not difficult at all, the main thing is we try to balance the equation without any understanding of the concepts.
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Lesson: Factorising quadratics (Part 2) | Oak National Academy
Switch to our new maths teaching resources
Slide decks, worksheets, quizzes and lesson planning guidance designed for your classroom.
Lesson details
Key learning points
1. In this lesson, we will further develop your ability to factorise quadratics by spotting factors of terms.
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
6 Questions
Fill in the gap: ________ is the opposite of expanding.
Factorise x² + 11x + 24.
Factorise x² + 14x + 24.
Factorise x² + 8x + 16.
A quadratic expression x² + bx + 20 can be factorised. Find all the possible values for b when b is positive.
7 Questions
Fill in the gap: ________ is the opposite of expanding.
Factorise x² + 14x + 24.
Cala says that you cannot factorise x² -9x + 8 because there does not exist a pair of factors for 8 that add to give a negative number. Is she correct?
Yes, Cala is correct, there DOES NOT exist a pair of factors of 8 that add to give a negative number.
Factorise x² - 3x - 10.
Factorise x² - 9x + 14.
A quadratic expression x² + bx - 20 can be factorised. Find all the possible values for b when b is negative. | {"url":"https://www.thenational.academy/teachers/lessons/factorising-quadratics-part-2-6cw6cr","timestamp":"2024-11-04T19:02:09Z","content_type":"text/html","content_length":"272176","record_id":"<urn:uuid:e43ff9ac-da47-4020-a1a2-ab2c54990eb4>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00779.warc.gz"} |
Excel to Python: SLOPE function - A Complete Guide
How to Use Excel's SLOPE Function in Pandas
The SLOPE function in Excel determines the slope of the line of best fit through a set of points. This is particularly useful in analyzing linear relationships between two variables.
This page explains how to use pandas to calculate the SLOPE of a linear relationship, similar to Excel's SLOPE function.
Implementing the Linear Slope Calculation function in Pandas#
To replicate the SLOPE function in pandas, you typically use linear regression methods or custom calculations. Here are some common implementations:
Basic SLOPE Calculation#
The basic SLOPE calculation in pandas involves using linear regression methods to determine the slope between two series of data.
In Excel, you might use =SLOPE(B2:B10, A2:A10). The pandas equivalent involves fitting a linear model to the data.
# Import the linregress function from scipy.stats
from scipy.stats import linregress
slope, _, _, _, _ = linregress(df['ColumnX'], df['ColumnY'])
Calculating intercept, standard error, p-value, and r-value#
In addition to the slope, you can also calculate the intercept, standard error, p-value, and r-value of the linear regression model.
These additional parameters can help you better understand the relationship between the two variables and are returned by default from the linregress function.
The intercept tells you the value of the dependent variable when the independent variable is zero.
The standard error is the standard deviation of the estimate of the slope. It's a measure of the accuracy of the predictions.
The p-value is the probability that the slope is zero. In other words, it's the probability that the two variables are not related given the observations you provided.
The r-value is the correlation coefficient. It's a measure of how closely the two variables are related. It ranges from -1 to 1, with 1 indicating a perfect positive correlation and -1 indicating a
perfect negative correlation.
# Import the linregress function from scipy.stats
from scipy.stats import linregress
slope, intercept, r_value, p_value, std_err = linregress(df['ColumnX'], df['ColumnY'])
Calculate slope of price change over days#
The SLOPE function can also be used to determine trends in financial indicators over time. For example, you can use it to calculate the slope of a stock price over a period of time.
This is similar to using the SLOPE function in Excel, but pandas has more flexibility in handling date ranges.
import pandas as pd
import numpy as np
from scipy.stats import linregress
# Sample data
data = {
'Date': pd.date_range(start='2023-01-01', periods=5, freq='D'),
'Value': [100, 105, 110, 115, 120]
df = pd.DataFrame(data)
# Convert dates to ordinal numbers
df['Date_ordinal'] = df['Date'].apply(lambda x: x.toordinal())
# Calculate slope
slope, intercept, r_value, p_value, std_err = linregress(df['Date_ordinal'], df['Value'])
Common mistakes when using SLOPE in Python#
While implementing the SLOPE function in pandas, there are several pitfalls that you should be aware of. Here are some common mistakes and their solutions.
Incorrect Data Types#
Calculating SLOPE with incompatible data types in pandas can lead to errors. Ensure that your data columns are numeric.
If your data is in string format, you can convert it to numeric using the to_numeric function.
df['ColumnX'] = pd.to_numeric(df['ColumnX'], errors='coerce')
df['ColumnY'] = pd.to_numeric(df['ColumnY'], errors='coerce')
Understanding the Linear Slope Calculation Formula in Excel#
The SLOPE function in Excel calculates the slope of the linear regression line through a given set of data points.
=SLOPE(known_y's, known_x's)
SLOPE Excel Syntax
Parameter Description Data Type
known_y's The dependent data points range of numbers
known_x's The independent data points range of numbers
Formula Description Result
=SLOPE(B2:B10, A2:A10) Calculate the slope of the linear regression line for the data in ranges A2:A10 and B2:B10 Calculated slope value
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2020-11-15 ASERT
layout: specification
title: ASERT Difficulty Adjustment Algorithm (aserti3-2d)
date: 2020-08-17
category: spec
activation: 1605441600
version: 0.6.3
author: freetrader, Jonathan Toomim, Calin Culianu, Mark Lundeberg, Tobias Ruck
Activation of a new new difficulty adjustment algorithm 'aserti3-2d' (or 'ASERT' for short) for the November 2020 Bitcoin Cash upgrade. Activation will be based on MTP, with the last pre-fork block
used as the anchor block.
• To eliminate periodic oscillations in difficulty and hashrate
• To reduce the difference in profitability between steady miners and those who switch to mining other blockchains.
• To maintain average block intervals close to the 10 minute target.
• To bring the average transaction confirmation time close to target time.
Technical background
The November 2017 Bitcoin Cash upgrade introduced a simple moving average as difficulty adjustment algorithm. This change unfortunately introduced daily periodic difficulty oscillations, which
resulted in long confirmation times followed by a burst of rapid blocks. This harms the user experience of Bitcoin Cash, and punishes steady hashrate miners.
Research into the family of difficulty algorithms based on an exponential moving average (EMA) resulted in ASERT (Absolutely Scheduled Exponentially Rising Targets) [1], which has been developed by
Mark Lundeberg in 2019 and fully described by him in 2020. An equivalent formula was independently discovered in 2018 by Jacob Eliosoff and in 2020 by Werner et. al [6].
ASERT does not have the same oscillations as the DAA introduced in the November 2017 upgrade and has a range of other attractive qualities such as robustness against singularities [15] without a need
for additional rules, and absence of accumulation of rounding/approximation errors.
In extensive simulation against a range of other stable algorithms [2], an ASERT algorithm performed best across criteria that included:
• Average block times closest to an ideal target time of 600 seconds.
• Average transaction confirmation times closest to the target time.
• Reducing the advantage of non-steady mining strategies, thereby maximizing the relative profitability of steady mining.
Terms and conventions
• Fork block: The first block mined according to the new consensus rules.
• Anchor block: The parent of the fork block.
Target computation
The current block's target bits are calculated by the following algorithm.
The aserti3-2d algorithm can be described by the following formula:
next_target = anchor_target * 2**((time_delta - ideal_block_time * (height_delta + 1)) / halflife)
• anchor_target is the unsigned 256 bit integer equivalent of the nBits value in the header of the anchor block.
• time_delta is the difference, in signed integer seconds, between the timestamp in the header of the current block and the timestamp in the parent of the anchor block.
• ideal_block_time is a constant: 600 seconds, the targeted average time between blocks.
• height_delta is the difference in block height between the current block and the anchor block.
• halflife is a constant parameter sometimes referred to as 'tau', with a value of 172800 (seconds) on mainnet.
• next_target is the integer value of the target computed for the block after the current block.
The algorithm below implements the above formula using fixed-point integer arithmetic and a cubic polynomial approximation to the 2^x term.
The 'target' values used as input and output are the compact representations of actual 256-bit integer targets as specified for the 'nBits' field in the block header.
Python-code, uses Python 3 syntax:
def next_target_aserti3_2d(
anchor_height: int, # height of the anchor block.
anchor_parent_time: int, # timestamp (nTime) of the parent of the anchor block.
anchor_bits: int, # 'nBits' value of the anchor block.
current_height: int, # height of the current block.
current_time: int, # timestamp of the current block.
) -> int: # 'target' nBits of the current block.
ideal_block_time = 600 # in seconds
halflife = 172_800 # 2 days (in seconds)
radix = 2**16 # 16 bits for decimal part of fixed-point integer arithmetic
max_bits = 0x1d00_ffff # maximum target in nBits representation
max_target = bits_to_target(max_bits) # maximum target as integer
anchor_target = bits_to_target(anchor_bits)
time_delta = current_time - anchor_parent_time
height_delta = current_height - anchor_height # can be negative
# `//` is truncating division (int.__floordiv__) - see note 3 below
exponent = time_delta - ideal_block_time * (height_delta + 1) // halflife
# Compute equivalent of `num_shifts = math.floor(exponent / 2**16)`
num_shifts = exponent >> 16
exponent = exponent - num_shifts * radix
factor = ((195_766_423_245_049 * exponent +
971_821_376 * exponent**2 +
5_127 * exponent**3 +
2**47) >> 48) + radix
next_target = anchor_target * factor
# Calculate `next_target = math.floor(next_target * 2**factor)`
if num_shifts < 0:
next_target >>= -num_shifts
# Implementations should be careful of overflow here (see note 6 below).
next_target <<= num_shifts
next_target >>= 16
if next_target == 0:
return target_to_bits(1) # hardest valid target
if next_target > max_target:
return max_bits # limit on easiest target
return target_to_bits(next_target)
Note 1: The reference implementations make use of signed integer arithmetic. Alternative implementations may use strictly unsigned integer arithmetic.
Note 2: All implementations should strictly avoid use of floating point arithmetic in the computation of the exponent.
Note 3: In the calculation of the exponent, truncating integer division [7, 10] must be used, as indicated by the // division operator (int.__floordiv__).
Note 5: The convenience functions bits_to_target() and target_to_bits() are assumed to be available for conversion between compact 'nBits' and unsigned 256-bit integer representations of targets.
Examples of such functions are available in the C++ and Python3 reference implementations.
Note 6: If a limited-width integer type is used for current_target, then the << operator may cause an overflow exception or silent discarding of most-significant bits. Implementations must detect and
handle such cases to correctly emulate the behaviour of an unlimited-width calculation. Note that if the result at this point would exceed radix * max_target then max_bits may be returned
Note 7: The polynomial approximation that computes factor must be performed with 64 bit unsigned integer arithmetic or better. It will overflow a signed 64 bit integer. Since exponent is signed, it
may be necessary to cast it to unsigned 64 bit integer. In languages like Java where long is always signed, an unsigned shift >>> 48 must be used to divide by 2^48.
The ASERT algorithm will be activated according to the top-level upgrade spec [3].
Anchor block
ASERT requires the choice of an anchor block to schedule future target computations.
The first block with an MTP that is greater/equal to the upgrade activation time will be used as the anchor block for subsequent ASERT calculations.
This corresponds to the last block mined under the pre-ASERT DAA rules.
Note 1: The anchor block is the block whose height and target (nBits) are used as the 'absolute' basis for ASERT's scheduled target. The timestamp (nTime) of the anchor block's parent is used.
Note 2: The height, timestamp, and nBits of this block are not known ahead of the upgrade. Implementations MUST dynamically determine it across the upgrade. Once the network upgrade has been
consolidated by sufficient chain work or a checkpoint, implementations can simply hard-code the known height, nBits and associated (parent) timestamp this anchor block. Implementations MAY also
hard-code other equivalent representations, such as an nBits value and a time offset from the genesis block.
REQ-ASERT-TESTNET-DIFF-RESET (testnet difficulty reset)
On testnet, an additional rule will be included: Any block with a timestamp that is more than 1200 seconds after its parent's timestamp must use an nBits value of max_bits (0x1d00ffff).
Rationale and commentary on requirements / design decisions
1. Choice of anchor block determination
Choosing an anchor block that is far enough in the past would result in slightly simpler coding requirements but would create the possibility of a significant difficulty adjustment at the
The last block mined according to the old DAA was chosen since this block is the most proximal anchor and allows for the smoothest transition to the new algorithm.
2. Avoidance of floating point calculations
Compliance with IEEE-754 floating point arithmetic is not generally guaranteed by programming languages on which a new DAA needs to be implemented. This could result in floating point
calculations yielding different results depending on compilers, interpreters or hardware.
It is therefore highly advised to perform all calculations purely using integers and highly specific operators to ensure identical difficulty targets are enforced across all implementations.
3. Choice of half-life
A half-life of 2 days (halflife = 2 * 24 * 3600), equivalent to an e^x-based time constant of 2 * 144 * ln(2) or aserti3-415.5, was chosen because it reaches near-optimal performance in
simulations by balancing the need to buffer against statistical noise and the need to respond rapidly to swings in price or hashrate, while also being easy for humans to understand: For every 2
days ahead of schedule a block's timestamp becomes, the difficulty doubles.
4. Choice of approximation polynomial
The DAA is part of a control system feedback loop that regulates hashrate, and the exponential function and its integer approximation comprise its transfer function. As such, standard guidelines
for ensuring control system stability apply. Control systems tend to be far more sensitive to differential nonlinearity (DNL) than integral nonlinearity (INL) in their transfer functions. Our
requirements were to have a transfer function that was (a) monotonic, (b) contained no abrupt changes, (c) had precision and differential nonlinearity that was better than our multi-block
statistical noise floor, (d) was simple to implement, and (e) had integral nonlinearity that was no worse than our single-block statistical noise floor.
A simple, fast to compute cubic approximation of 2^x for 0 <= x < 1 was found to satisfy all of these requirements. It maintains an absolute error margin below 0.013% over this range [8]. In
order to address the full (-infinity, +infinity) domain of the exponential function, we found the 2**(x + n) = 2**n * 2**x identity to be of use. Our cubic approximation gives the exactly correct
values f(0) == 1 and f(1) == 2, which allows us to use this identity without concern for discontinuities at the edges of the approximation's domain.
First, there is the issue of DNL. Our goal was to ensure that our algorithm added no more than 25% as much noise as is inherent in our dataset. Our algorithm is effectively trying to estimate the
characteristic hashrate over the recent past, using a 2-day (~288-block) half-life. Our expected exponential distribution of block intervals has a standard deviation (stddev) of 600 seconds. Over
a 2-day half-life, our noise floor in our estimated hashrate should be about sqrt(1 / 288) * 600 seconds, or 35.3 seconds. Our chosen approximation method is able to achieve precision of 3
seconds in most circumstances, limited in two places by 16-bit operations:
`172800 sec / 65536 = 2.6367 sec`
Our worst-case precision is 8 seconds, and is limited by the worst-case 15-bit precision of the nBits value. This 8 second worst-case is not within the scope of this work to address, as it would
require a change to the block header. Our worst-case step size is 0.00305%,[11] due to the worst-case 15-bit nBits mantissa issue. Outside the 15-bit nBits mantissa range, our approximation has a
worst-case precision of 0.0021%. Overall, we considered this to be satisfactory DNL performance.
Second, there is the issue of INL. Simulation testing showed that difficulty and hashrate regulation performance was remarkably insensitive to integral non-linearity. We found that even the use
of f(x) = 1 + x as an approximation of 2**x in the aserti1 algorithm was satisfactory when coupled with the 2**(x + n) = 2^n * 2^x identity, despite having 6% worst-case INL.[12, 13] An
approximation with poor INL will still show good hashrate regulation ability, but will have a different amount of drift for a given change in hashrate depending on where in the [0, 1) domain our
exponent (modulo 1) lies. With INL of +/- 1%, for any given difficulty (or target), a block's timestamp might end up being 1% of 172800 seconds ahead of or behind schedule. However, out of an
abundance of caution, and because achieving higher precision was easy, we chose to aim for INL that would be comparable to or less than the typical drift that can be caused by one block. Out of a
2-day half-life window, one block's variance comprises:
`600 / 172800 = 0.347%`
Our cubic approximation's INL performance is better than 0.013%,[14] which exceeds that requirement by a comfortable margin.
5. Conversion of difficulty bits (nBits) to 256-bit target representations
As there are few calculations in ASERT which involve 256-bit integers and the algorithm is executed infrequently, it was considered unnecessary to require more complex operations such as doing
arithmetic directly on the compact target representations (nBits) that are the inputs/output of the difficulty algorithm.
Furthermore, 256-bit (or even bignum) arithmetic is available in existing implementation and used within the previous DAA. Performance impacts are negligible.
6. Choice of 16-bits of precision for fixed-point math
The nBits format is comprised of 8 bits of base_256 exponent, followed by a 24-bit mantissa. The mantissa must have a value of at least 0x008000, which means that the worst-case scenario gives
the mantissa only 15 bits of precision. The choice of 16-bit precision in our fixed point math ensures that overall precision is limited by this 15-bit nBits limit.
7. Choice of name
The specific algorithm name 'aserti3-2d' was chosen based on:
□ the 'i' refers to the integer-only arithmetic
□ the '3' refers to the cubic approximation of the exponential
□ the '2d' refers to the 2-day (172800 second) halflife
Implementation advice
Implementations must not make any rounding errors during their calculations. Rounding must be done exactly as specified in the algorithm. In practice, to guarantee that, you likely need to use
integer arithmetic exclusively.
Implementations which use signed integers and use bit-shifting must ensure that the bit-shifting is arithmetic.
Note 1: In C++ compilers, right shifting negative signed integers is formally unspecified behavior until C++20 when it will become standard [5]. In practice, C/C++ compilers commonly implement
arithmetic bit shifting for signed numbers. Implementers are advised to verify good behavior through compile-time assertions or unit tests.
Reference implementations
Test vectors
Test vectors suitable for validating further implementations of the aserti3-2d algorithm are available at:
alternatively at:
and alternatively at:
Thanks to Mark Lundeberg for granting permission to publish the ASERT paper [1], Jonathan Toomim for developing the initial Python and C++ implementations, upgrading the simulation framework [9] and
evaluating the various difficulty algorithms.
Thanks to Jacob Eliosoff, Tom Harding and Scott Roberts for evaluation work on the families of EMA and other algorithms considered as replacements for the Bitcoin Cash DAA, and thanks to the
following for review and their valuable suggestions for improvement:
• Andrea Suisani (sickpig)
• BigBlockIfTrue
• Fernando Pellicioni
• imaginary_username
• mtrycz
• Jochen Hoenicke
• John Nieri (emergent_reasons)
• Tom Zander
[1] "Static difficulty adjustments, with absolutely scheduled exponentially rising targets (DA-ASERT) -- v2", Mark B. Lundeberg, July 31, 2020
[2] "BCH upgrade proposal: Use ASERT as the new DAA", Jonathan Toomim, 8 July 2020
[3] Bitcoin Cash November 15, 2020 Upgrade Specification.
[4] https://en.wikipedia.org/wiki/Arithmetic_shift
[5] https://en.cppreference.com/w/cpp/language/operator_arithmetic
[6] "Unstable Throughput: When the Difficulty Algorithm Breaks", Sam M. Werner, Dragos I. Ilie, Iain Stewart, William J. Knottenbelt, June 2020
[7] "Different kinds of integer division", Harry Garrood, blog, 2018
[8] Error in a cubic approximation of 2^x for 0 <= x < 1
[9] Jonathan Toomim adaptation of kyuupichan's difficulty algorithm simulator: https://github.com/jtoomim/difficulty/tree/comparator
[10] "The Euclidean definition of the functions div and mod", Raymond T. Boute, 1992, ACM Transactions on Programming Languages and Systems (TOPLAS). 14. 127-144. 10.1145/128861.128862
[11] http://toom.im/bch/aserti3_step_size.html
[12] f(x) = (1 + x)/2^x for 0<x<1, WolframAlpha.
[13] https://github.com/zawy12/difficulty-algorithms/issues/62#issuecomment-647060200
[14] http://toom.im/bch/aserti3_approx_error.html
[15] https://github.com/zawy12/difficulty-algorithms/issues/62#issuecomment-646187957
This specification is dual-licensed under the Creative Commons CC0 1.0 Universal and GNU All-Permissive licenses. | {"url":"https://documentation.cash/protocol/forks/2020-11-15-asert.html","timestamp":"2024-11-04T07:57:51Z","content_type":"text/html","content_length":"28395","record_id":"<urn:uuid:d9d78690-46d7-43d6-8cc5-3edcc9a70a82>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00708.warc.gz"} |
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Optional parameter: coordinate system
Optional parameter: coordinate system#
When the minimal example on the previous page is passed to the World Builder, the World Builder fills in anything else it needs with default values. The default values can be viewed in the Parameter
listings part of this manual in The World Builder file.
One of the optional parameters which can most fundamentally change what the resulting world looks like is the /coordinate system. Its default value is cartesian. Reading this document for the first
(or even 10th time) is overwhelming. If you want to learn more about /coordinate system, open the drop box below for a detailed explanation about how to read this section of the manual.
Coordinate system Parameter listings explanation
Parameter listings is a very big collapsible tree in which each heading contains its path. / is the root of the tree, and, for example, /coordinate system is an object directly on the root. You can
collapse objects below /coordinate system by clicking the upward arrow on the right, and click it again (now pointing downward) to expand it back. Each object contains its description, type, default
value and additional specifications. Here you see the default value for /coordinate system is set to cartesian. Hence, if your model is in Cartesian coordinates, you don’t need to do anything!
Since coordinate system is an object, we need to write something like "coordinate system": {} in our input file.
The next dropbox has the path /coordinate system/oneOf. This is the only entry beneath the /coordinate system dropbox. The oneOf subpath indicates you need to choose one object out of a few options.
These are listed in the oneOf box. The paths of these dropboxes in the oneOf dropbox are numbered and unfortunately not named. This means we will have to inspect each to see what they actually
In this case there are two dropboxes. Let’s look at path number 1 (/coordinate system/oneOf/1). We can see that this is an object which has one required key (parameter): model. The documentation
already states that this is a Cartesian coordinate system. When we look into the model dropbox (/coordinate system/oneOf/1/model), we see that there is an enum with only one possible value: cartesian
This is a very long winded way to say that if we want a Cartesian coordinate system, we need to provide a coordinate system object which has a key called model and has an enum value cartesian:
"coordinate system": {"model":cartesian}.
3 "coordinate system": {"model":"cartesian"},
Now see if you can use the documentation how to create a spherical model.
If you managed to do this in one go, great, well done! If you just tried changing cartesian to spherical, you will get an error. Take a good look at the error message. The error message indicates
that a required parameter is missing. /coordinate system/oneOf/2 requires a key depth method. Please see Constant angle in spherical domain issue for more information on why this is needed. The last
option in this section is radius. This key has a default value, so it is optional. The default is set to 6371000.0, which should be fine for most models.
So a spherical model with a user defined radius of 1.0 would look like this:
3 "coordinate system":{"model":"spherical", "depth method":"begin segment", "radius":1.0},
Our previous minimal example looks like this:
2 "version":"1.1",
3 "features":[ ]
We can be more explicit and add one line setting it to the default value. However, there is no difference between this one and the previous code block.
2 "version":"1.1",
3 "coordinate system":{"model":"cartesian"},
4 "features":[ ]
If you want to have a spherical model, please see Constant angle in spherical domain issue first. An input file for a spherical model would like something like this:
2 "version":"1.1",
3 "coordinate system":{"model":"spherical", "depth method":"begin segment"},
4 "features":[ ]
This should be a good default spherical coordinate system. For more information on how to derive this from the parameter listing and what the options are, please expand and read the dropbox above.
In the following sections we will continue using the Cartesian coordinate system. We will also show this in Spherical models, where we show how easy it is to switch between Cartesian and spherical
coordinate systems in our finished subduction example.
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The Atlas of Irish Mathematics: Monaghan (Feb 2019)
The Atlas of Irish Mathematics: Monaghan (Feb 2019)
In past bimonthly Atlas of Irish Mathematics blogs with regional focus, we've shone a light on people associated with Donegal, Wexford, Armagh, Limerick, Westmeath, Mayo, Belfast, Wicklow, Kerry and
Galway. This time, it's Monaghan.
Included below are men who played key roles in the early days of TCD, a famous writer who mentored Seamus Heaney, several Maynooth people of note, and three often overlooked people from the almost
100 Irish or Irish-based people who contributed problems or solutions to the Educational Times in the second half of the 1800s (a future blog will highlight all of them).
Comments, additions and corrections are, as always, welcome. As are more photographs of the forgotten faces from the past. Last updated 6 July 2024.
Thanks to Paul Greaney (NUIG), Olivia Bree (SPD), Tony O'Farrell (MU) & Ciarán P. Mac an Bhaird (MU), ET Trigg and Eddie O'Kane for valuable input.
1. William Clement (1707-1782) was born in Carrickmacross, Monaghan, and was educated at TCD (BA, 1726, MA 1731), where he taught for almost half a century, and also served as vice-provost. He
lectured in botany, physics, and medicine at various stages of his career, and was Donegall Lecturer of mathematics from 1750 to 1759.
TCD / Bio
2. Patrick Flood (1827?-1902) was born in Monaghan. Nothing is known about his education. He was a teacher in Wexford, Tipperary, Clare and Dublin. Over a three decade period, he contributed to
the Educational Times.
Educational Times / 1901 Census / L /
3. Physicist and mathematician Francis Lennon (1838-1920) was born in November in Tyholland, Monaghan, near the Armagh border, and was educated at Clogher. He was ordained in 1862 and appointed
professor at the diocesan seminary St Macartan's. Following the death of Nicholas Callan in 1864, he became professor of natural philosophy in Maynooth, a position he held until 1911. He served
terms as chair of the Dublin Scientific Club and examiner for mathematics for the Intermediate Education Board. He has 2 books to his name: the revised and improved 1872 edition of André Darré's
Elements of Geometry with Both Plane and Spherical Trigonometry, and The Elements of Plane and Spherical Geometry (1875).
Grave / Prize / 1911 census
4. Francis Tarleton (1841-1920) is believed to have been born in Tyholland, east of Monaghan town. He was educated at TCD (BA 1861, MA 1865) where he was elected a fellow in 1866. He was called
to the bar in 1868, but spent his entire career as an academic and administrator at TCD. He occupied the chair of natural philosophy from 1890 to 1901, treating physics as a branch of maths. He
also served as bursar, senior dean, and vice provost, and was awarded an honorary ScD (1891). He authored books on dynamics and the mathematical theory of attraction, and contributed to the
Educational Times. He colourfully viewed Einstein's theory of relativity as being in the same category as Bolshevism.
TCD / Nature / Educational Times 1 / Educational Times 2 / 1901 Census / 1911 Census
5. William Stoops (1844?-1919) was born in Castleblayney, Monaghan, and was educated at Queen's College Cork (BA 1874). At first he taught at the Coleraine Academic Institute, then in 1881 he
became the headmaster at the Newry Intermediate School, Down, where he stayed until 1917. He contributed to the Educational Times.
Obit 1 / Obit 2 / Educational Times / 1901 Census / 1911 Census
6. Francis Rountree (1859?-1944) was born in Cavan and was educated at TCD (BA 1884, MA 1889). His career included spells teaching at the Grammar School in Cork, and in Rockcorry, Monaghan.
Links / 1911 Census
7. James Mills Stoops (1867-1941) was born 25 September in Alsmeed, Muckno Parish, Cremorne, northeast of Castleblayney, Monaghan, a nephew of William Stoops above. He was educated at Queen's
College Belfast (BSc 1891). At first he taught at Victoria College in Belfast, contributed to the Educational Times, and authored solutions for some Blackie's Guides to the Intermediate
Mathematics. In 1899 he matriculated at the University of London, and the rest of his career seems to have been spent as a clergyman in Australia and New Zealand. He taught maths at Emmanuel
College in Brisbane from its 1912 opening. (Thanks to Kerry Mahony for finding the death date.)
Books / Educational Times
8. Doctor William A. Stoops (1869-1938) was born 20 February in Mullyash just outside Castleblayney, Monaghan. He was educated at Queen's Belfast (BSc 1897, MD 1903??) and practiced medicine in
England. He was one of the main mourners at the funeral of William above who died in 1919, and may have been a cousin of James above.
1901 Census
10. John Troughton (1902-1975) was born 24 May in Clones, Monaghan. He was educated at TCD (BA in mental and moral philosophy 1924, LLB 1929), being awarded a maths scholarship in 1925. He spent
his career as a barrister in the British Civil Service in Kenya, Uganda and Swaziland.
1911 Census / Who Was Who / Link
11. Writer Michael McLaverty (1904-1992) was born 5 July in Carrickmacross, Monaghan, was grew up there and in Belfast. He was educated at QUB (BSc 1927, MSc by thesis 1933), and taught maths and
physics at St John's PED for decades before ending his career as headmaster at St Thomas. (Thanks to Eddie O'Kane.)
Wikipedia / Hidden Gems / Ricorso
12. Eoghan Rushe (1917-2001) was born 10 June in Corbane, Corduff, northwest of Carrickmacross, Monaghan. He was educated at UCD (BSc 1939) and later taught at Belcamp College in Dublin, where he
rose to the rank of vice principal. (Thanks to Bernie Ruth & Joe Callan for valuable input here.)
13. Roy Nelson (1920-??) was born 26 June in Corvally, northeast of Monaghan town. He taught maths at UCD 1956-1959. More information is welcome. (He is not the Open University presenter of the
same name who previously taught computer science at QUB in the 1960s.)
14. Paul Carragher (1927-1999) was born 28 June born near Annyalla, Castleblayney, Monaghan. He joined the civil service for a few years, and was then educated at TCD (BAI 1955, MA 1960). He
worked for 2 years as an engineer in Scotland and England, and then taught maths at Izmir Koleji in Turkey. Moving to Canada in 1958, he taught at Memorial University in Newfoundland until 1966,
then for a year at TCD. Most of the rest of his career (1968-1993) was spent at the University of New Brunswick, also in Canada. He received his doctorate from TCD (1978) for a thesis on "Heat
Transfer on Continuous Solid Surfaces" done under Lawrence Crane.
DIAS / ResearchGate
15. Physicist Thomas (Gerry) McGreevy (1929-2015) was born 25 January in Belfast, and grew up mostly in Togan, southwest of Monaghan town. He was educated at Maynooth (BSc 1949) and then worked at
UCG for a while. He earned his PhD at UCD circa 1957, and was on the staff at Maynooth from then to 1982, serving as registrar along the way. He spent his later years as a parish priest in
Monaghan and Donegal.
Irish Times
15B. Sean Clerkin (1932-2016) was born in Monaghan and was educated at Maynooth (1953 BSc). Following ordination, he taught maths at St Michael's College in Enniskillen (1957-1976), before
engaging in parish work in Mullanarockan, Tydavnet, Monaghan.
RIP / 1976 / Parish
16. Dermot Marron was born in Clones, Monaghan and educated at QUB (BSc 1993, PhD 1997). His thesis on "Spittability in Ordered Sets and in Ordered Spaces" was done under Brian McMaster. He
published some papers in topology, and then pursued an actuarial career in Dublin.
Allied Risk /
17. Ciarán P. Mac an Bhaird was born 9 July in Lough Egish, southwest of Castleblayney, Monaghan, and was educated entirely at Maynooth (BSc 1998, MSc 2000, PhD 2007), where he has also spent his
career. His doctorate on "Gauss' Method for the Determination of Cyclotomic Numbers" was done under Pat McCarthy. He authored the book Primality Testing and Gaussian Sums (Logic Press, 2005)
based on his master's thesis on "An Introduction to the Mathematics Involved in the APR Primality Test" which was done with the same advisor. His current interests include maths education, the
history of mathematics, and algebraic number theory.
18. Biostatistician Siobhán Connolly (Connolly-Kernan) was born in Castleblayney, Monaghan, and was educated at Maynooth (BSc 2012) and TCD (PhD 2016). Her thesis on "Investigating the Parental
Role in Autism Spectrum Disorder" was done with Eleisa Heron. She now lectures at DKIT.
DKIT / LinkedIn
19. Jack McDonnell was born in Castleblayney, Monaghan, and was educated entirely at Maynooth (BSc 2013, MSc 2014, PhD 2018+). His thesis on "Predicting Grass Growth at Farm Level to Adapt to
Changing and Volatile Weather Conditions" was done under Caroline Brophy & Deirdre Hennessy. After a postdoc at the Met Service he joined the staff at DKIT. | {"url":"https://www.mathsireland.ie/blog/2019_02_cm","timestamp":"2024-11-04T10:34:33Z","content_type":"application/xhtml+xml","content_length":"68836","record_id":"<urn:uuid:6bf30227-2377-4fa7-ad9b-fc06c95859ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00887.warc.gz"} |
US10076312B2 - Filter assembly for medical image signal and dynamic decimation method using same
- Google Patents
US10076312B2 - Filter assembly for medical image signal and dynamic decimation method using same - Google Patents
Filter assembly for medical image signal and dynamic decimation method using same Download PDF
Publication number
US10076312B2 US10076312B2 US15/551,684 US201515551684A US10076312B2 US 10076312 B2 US10076312 B2 US 10076312B2 US 201515551684 A US201515551684 A US 201515551684A US 10076312 B2 US10076312 B2 US
United States
Prior art keywords
medical image
image signal
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Tai-kyong Song
Hyungil Kang
Jeeun Kang
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Hansono Co Ltd
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Hansono Co Ltd
Sogang University Research Foundation
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Assigned to Sogang University Research Foundation reassignment Sogang University Research Foundation ASSIGNMENT OF ASSIGNORS INTEREST (SEE DOCUMENT FOR DETAILS). Assignors: KANG, Hyungil, KANG,
Jeeun, SONG, TAI-KYONG
Publication of US20180035981A1 publication Critical patent/US20180035981A1/en
Assigned to HANSONO CO. LTD reassignment HANSONO CO. LTD ASSIGNMENT OF ASSIGNORS INTEREST (SEE DOCUMENT FOR DETAILS). Assignors: Sogang University Research Foundation
Application granted granted Critical
Publication of US10076312B2 publication Critical patent/US10076312B2/en
Active legal-status Critical Current
Anticipated expiration legal-status Critical
☆ A—HUMAN NECESSITIES
☆ A61—MEDICAL OR VETERINARY SCIENCE; HYGIENE
☆ A61B—DIAGNOSIS; SURGERY; IDENTIFICATION
☆ A61B8/00—Diagnosis using ultrasonic, sonic or infrasonic waves
☆ A61B8/08—Detecting organic movements or changes, e.g. tumours, cysts, swellings
☆ A—HUMAN NECESSITIES
☆ A61—MEDICAL OR VETERINARY SCIENCE; HYGIENE
☆ A61B—DIAGNOSIS; SURGERY; IDENTIFICATION
☆ A61B8/00—Diagnosis using ultrasonic, sonic or infrasonic waves
☆ A61B8/52—Devices using data or image processing specially adapted for diagnosis using ultrasonic, sonic or infrasonic waves
☆ A61B8/5269—Devices using data or image processing specially adapted for diagnosis using ultrasonic, sonic or infrasonic waves involving detection or reduction of artifacts
☆ A—HUMAN NECESSITIES
☆ A61—MEDICAL OR VETERINARY SCIENCE; HYGIENE
☆ A61B—DIAGNOSIS; SURGERY; IDENTIFICATION
☆ A61B8/00—Diagnosis using ultrasonic, sonic or infrasonic waves
☆ A61B8/52—Devices using data or image processing specially adapted for diagnosis using ultrasonic, sonic or infrasonic waves
☆ A61B8/5207—Devices using data or image processing specially adapted for diagnosis using ultrasonic, sonic or infrasonic waves involving processing of raw data to produce diagnostic data,
e.g. for generating an image
☆ G—PHYSICS
☆ G01—MEASURING; TESTING
☆ G01S—RADIO DIRECTION-FINDING; RADIO NAVIGATION; DETERMINING DISTANCE OR VELOCITY BY USE OF RADIO WAVES; LOCATING OR PRESENCE-DETECTING BY USE OF THE REFLECTION OR RERADIATION OF RADIO
WAVES; ANALOGOUS ARRANGEMENTS USING OTHER WAVES
☆ G01S15/00—Systems using the reflection or reradiation of acoustic waves, e.g. sonar systems
☆ G01S15/88—Sonar systems specially adapted for specific applications
☆ G01S15/89—Sonar systems specially adapted for specific applications for mapping or imaging
☆ G01S15/8906—Short-range imaging systems; Acoustic microscope systems using pulse-echo techniques
☆ G01S15/8977—Short-range imaging systems; Acoustic microscope systems using pulse-echo techniques using special techniques for image reconstruction, e.g. FFT, geometrical transformations,
spatial deconvolution, time deconvolution
☆ G—PHYSICS
☆ G01—MEASURING; TESTING
☆ G01S—RADIO DIRECTION-FINDING; RADIO NAVIGATION; DETERMINING DISTANCE OR VELOCITY BY USE OF RADIO WAVES; LOCATING OR PRESENCE-DETECTING BY USE OF THE REFLECTION OR RERADIATION OF RADIO
WAVES; ANALOGOUS ARRANGEMENTS USING OTHER WAVES
☆ G01S7/00—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00
☆ G01S7/52—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00 of systems according to group G01S15/00
☆ G01S7/52017—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00 of systems according to group G01S15/00 particularly adapted to short-range imaging
☆ G01S7/52023—Details of receivers
☆ G01S7/52025—Details of receivers for pulse systems
☆ G01S7/52026—Extracting wanted echo signals
☆ G—PHYSICS
☆ G01—MEASURING; TESTING
☆ G01S—RADIO DIRECTION-FINDING; RADIO NAVIGATION; DETERMINING DISTANCE OR VELOCITY BY USE OF RADIO WAVES; LOCATING OR PRESENCE-DETECTING BY USE OF THE REFLECTION OR RERADIATION OF RADIO
WAVES; ANALOGOUS ARRANGEMENTS USING OTHER WAVES
☆ G01S7/00—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00
☆ G01S7/52—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00 of systems according to group G01S15/00
☆ G01S7/52017—Details of systems according to groups G01S13/00, G01S15/00, G01S17/00 of systems according to group G01S15/00 particularly adapted to short-range imaging
☆ G01S7/52046—Techniques for image enhancement involving transmitter or receiver
☆ H—ELECTRICITY
☆ H03—ELECTRONIC CIRCUITRY
☆ H03H—IMPEDANCE NETWORKS, e.g. RESONANT CIRCUITS; RESONATORS
☆ H03H17/00—Networks using digital techniques
☆ H03H17/02—Frequency selective networks
☆ H03H17/06—Non-recursive filters
☆ H03H17/0621—Non-recursive filters with input-sampling frequency and output-delivery frequency which differ, e.g. extrapolation; Anti-aliasing
☆ H03H17/0635—Non-recursive filters with input-sampling frequency and output-delivery frequency which differ, e.g. extrapolation; Anti-aliasing characterized by the ratio between the
input-sampling and output-delivery frequencies
☆ H03H17/065—Non-recursive filters with input-sampling frequency and output-delivery frequency which differ, e.g. extrapolation; Anti-aliasing characterized by the ratio between the
input-sampling and output-delivery frequencies the ratio being integer
☆ H03H17/0664—Non-recursive filters with input-sampling frequency and output-delivery frequency which differ, e.g. extrapolation; Anti-aliasing characterized by the ratio between the
input-sampling and output-delivery frequencies the ratio being integer where the output-delivery frequency is lower than the input sampling frequency, i.e. decimation
□ the present invention relates to a technique of receiving and processing a medical image signal and, more particularly, a filter assembly for adaptively processing variation in bandwidth of
an ultrasound image signal received from a probe according to depth of an image, and a dynamic decimation method using the same.
□ Medical imaging technology is a diagnosis technique of visually representing muscles, tendons, and many internal organs, to capture their size, structure, and pathologic lesions with
real-time tomographic images, based on an ultrasound or photoacoustic means. Medical imaging is also used to visualize fetuses during a periodic checkup or in an emergency situation.
Ultrasound has been used to image the interior of the human body for at least 50 years and has become one of the most widely used diagnostic tools in modern medicine. The ultrasound technique
is low in cost and easy in mobility, relative to magnetic resonance imaging (MRI) or X-ray computed tomography (CT).
□ MRI magnetic resonance imaging
□ CT X-ray computed tomography
□ the principle of ultrasound imaging is as follows. First, an ultrasound image is made by bringing a measurement object into contact with a probe and receiving ultrasound reflected by
generation of ultrasound waves. If ultrasound is generated, an ultrasound wave passes into a medium within a very short time and the ultrasound wave is reflected upon passing between two
media having different acoustic impedances. In the ultrasound imaging technique, such a reflection wave is measured and a distance is calculated based on the time until reflection sound
returns back, thereby achieving imaging.
□ the technical objects that can be achieved through the present invention are designed to solve inefficiency of a conventional filter structure for implementing dynamic decimation, in which
filter length increases in proportion to a decimation ratio and thus there are wasted filters and multipliers that are not used according to dynamic variation of the decimation ratio and
solve overhead of hardware cost and the amount of calculations necessary for multiplication at a high data rate because all multipliers are positioned after an expander.
□ a filter assembly for a medical image signal including an expander configured to receive the medical image signal and up-sample the medical image signal; and a decimation filter including an
integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the received medical image signal by dynamically updating an impulse
response and perform decimation on the up-sampled signal according to a decimation ratio.
□ MACs multiplier accumulators
□ the decimation filter may calculate a partial sum, which is the sum of coefficients of a k-th (wherein k is a positive integer) location of a polyphase filter, through each MAC.
□ Each MAC may include a shift register configured to receive and store coefficients of a polyphase filter, a multiplier configured to multiply the coefficients stored in the shift register by
the up-sampled signal, a summer configured to cumulatively sum the multiplied results, and a decimator configured to decimate the summed result.
□ a frequency band of the received signal may be determined by attenuation caused by depth of an object of the medical image signal and filter coefficients for calculating different cutoff
frequencies according to the depth may be supplied to the MAC through the shift register to control a signal-to-noise ratio of the medical image.
□ Each MAC may include a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation in filter length.
□ a filter assembly for a medical image signal including a decimation filter including an integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to
bandwidth of the medical image signal by dynamically updating an impulse response and perform decimation on the received signal according to a decimation ratio, wherein the decimation filter
determines a filter coefficient adjusted by an interval of an integer number to up-sample the medical image signal and supplies the filter coefficient to each MAC.
□ MACs multiplier accumulators
□ the medical image signal supplied to the decimation filter may not be previously up-sampled and operate at a low frequency relative to a previously up-sampled signal.
□ the decimation filter may determine, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as to perform a partial sum calculation of the signal through the
MAC except for a zero padding part out of an output of the expander.
□ the decimation filter may calculate a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through the MAC.
□ the MAC may include a shift register configured to receive and store coefficients of a polyphase filter; a multiplier configured to multiply the coefficients stored in the shift register by
the medical image signal; a summer configured to cumulatively sum the multiplied results; and a decimator configured to perform decimation on the summed result.
□ a method of decimating a medical image signal including receiving the medical image signal; selecting a filter coefficient for changing a cutoff frequency according to bandwidth of the
medical image signal in consideration of a decimation ratio; supplying the selected filter coefficient to a partial sum calculator including an integer number of multiplier accumulators
(MACs); and performing, by the partial sum calculator, dynamic decimation on the received medical image signal, using the selected filter coefficient, wherein the filter coefficient is
determined in consideration of an interval of an integer number to up-sample the received medical image signal.
□ MACs multiplier accumulators
□ the medical image signal supplied to the partial sum calculator may not be previously up-sampled and operate at a low frequency relative to a previously up-sampled signal.
□ the selecting the filter coefficient may include determining, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as to calculate a partial sum through
each MAC except for a zero padding part out of an output of the expander.
□ the partial sum calculator may calculate a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through each MAC.
□ the performing dynamic decimation may includes multiplying, by the MAC, the medical image signal by coefficients of a polyphase filter stored in a shift register by use of a multiplier;
cumulatively summing, by the MAC, the multiplied results by use of a summer; and decimating, by the MAC, the summed result by use of a decimator.
□ the MAC may include a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation in filter length.
□ Embodiments of the present invention require relatively fewer hardware resources and less power consumption upon performing dynamic decimation by implementing a fixed number of multipliers by
a polyphase filter structure and can achieve ultra-slimness of an ultrasound imaging system by adaptively applying a cutoff frequency of a filter in order to raise SNR.
□ FIG. 1 is a diagram for explaining a decimation filter used in a medical imaging system through which embodiments of the present invention are implemented.
□ FIG. 2 is a diagram for explaining problems occurring upon dynamically performing decimation through a finite impulse response (FIR) filter.
□ FIR finite impulse response
□ FIG. 3 is a diagram for explaining an example of implementing a decimation filter using a polyphase structure instead of the FIR filter of FIG. 2 and problems occurring in such an example.
□ FIG. 4 is a diagram illustrating a decimation filter assembly implementing partial sums of phase filters using multiplier accumulators (MACs) according to an embodiment of the present
□ MACs multiplier accumulators
□ FIG. 5 is a block diagram illustrating in detail a partial sum calculator for calculating a k-th (wherein k is a positive integer) partial sum in the decimation filter assembly of FIG. 4
according to an embodiment of the present invention.
□ FIGS. 6A and 6B are diagrams for explaining a structure in which an expander is removed from the decimation filter assembly of FIG. 4 .
□ FIG. 7 is a block diagram for explaining a structure in which an expander is removed according to another embodiment of the present invention.
□ FIG. 8 is a flowchart illustrating a decimation method of a medical image signal using a dynamic decimation filter according to still another embodiment of the present invention.
□ a filter assembly for a medical image signal includes an expander configured to receive the medical image signal and up-sample the medical image signal; and a decimation filter including an
integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the medical image signal by dynamically updating an impulse response and
perform decimation on the up-sampled signal according to a decimation ratio.
□ MACs multiplier accumulators
□ N denotes the length of a filter
□ D denotes a delay
□ L denotes expansion increasing rate of an expander for up-sampling
□ M denotes a decimation ratio
□ FIG. 1 is a diagram for explaining a decimation filter used in a medical imaging system through which embodiments of the present invention are implemented.
□ a dynamic filter is used to maximize SNR as frequency bandwidth decreases and is mainly implemented by a decimation filter used generally to match the data rate of the echo signal to a
screen. Accordingly, such a dynamic decimation filter should be capable of performing decimation on an arbitrary fractional decimation factor M/L and should be capable of dynamically updating
an impulse response thereof.
□ a decimation filter 110 adjusts the number of samples when the number of samples of a baseband signal is larger than the number of samples to be displayed on a screen through an ultrasound
imaging system 120 .
□ an input signal x(n) is integer-fold up-sampled ( 111 ), signal-processed using a coefficient ( 112 ) of a decimation filter, and decimated ( 113 ) at a ratio of M, thereby generating an
output signal y(n).
□ the echo signal in medical ultrasound imaging is affected by depth-dependent attenuation and the bandwidth of the echo signal differs according to depth. Therefore, it is necessary to
dynamically update a cutoff frequency with respect to each depth in order to maximize SNR. That is, the cutoff frequency varies with bandwidth using the dynamic filter.
□ the length N of the filter is proportional to the cutoff frequency and the decimation ratio.
□ FIG. 2 is a diagram for explaining problems occurring upon dynamically performing decimation through a finite impulse response (FIR) filter.
□ FIR finite impulse response
□ the filter length N is equal to the number of multipliers included in the filter and, due to characteristics of the dynamic decimation filter structure, a hardware filter should be
implemented for a maximum decimation ratio.
□ the multipliers operate at a high operation frequency of L times and at a high data rate of L times. Therefore, the amount of calculations per unit time increases and thus overhead increases
in terms of the amount of calculations of a block and hardware cost.
□ FIG. 3 is a diagram for explaining an example of implementing a decimation filter using a polyphase structure instead of the FIR filter of FIG. 2 and problems occurring in such an example.
□ a dynamic decimation filter using a polyphase filter may be used.
□ M-fold decimation is performed after performing L-fold expansion.
□ the filter of FIG. 3 has the same length as the filter of FIG. 2 but a data rate at which multipliers operate is lowered by M times.
□ embodiments of the present invention described hereinbelow propose a filter structure which has all functions of the above-described dynamic decimation filter and simultaneously can be
efficiently implemented without waste of hardware and operation resources. That is, an efficient arbitrary factional decimation structure only using K multiplier accumulators is proposed to
use restricted hardware complexity regardless of L and K.
□ Equation 1 An equation of a general FIR-based decimation filter is defined as follows.
□ M/L an arbitrary fractional decimation factor
□ Equation 1 a signal w(n) which is L-fold up-sampled with respect to an input x(n) is given as indicated in Equation 1.
□ Equation 2 A procedure of filtering the FIR as shown in FIG. 2 with respect to such an input is indicated as Equation 2.
□ h(j) denotes a coefficient of a given FIR filter and N denotes the length of the filter.
□ Equation 2 may be summarized as Equation 3 upon changing a dynamic decimation filter structure to a polyphase structure in which outputs of all polyphase filters are added as illustrated in
FIG. 3 .
□ FIG. 4 is a diagram illustrating a structure in which partial sums P 0 to P k-1 are added based on k-th coefficients with respect to M polyphase filters of FIG. 3 .
□ FIG. 4 is a diagram illustrating rearrangement of the polyphase filters of FIG. 3 through multiplier accumulators (MACs) according to an embodiment of the present invention. It may be
appreciated that the polyphase filters are implemented through a plurality of P 0 -MAC, P 1 -MAC, . . . , P k -MAC connected in parallel to each other.
□ MACs multiplier accumulators
□ Equation 3 indicates that each partial sum P k (n) in FIG. 4 is obtained by multiplying M consecutive pairs of input samples by filter coefficients at a data rate of L ⁇ f x and cumulatively
summing the multiplied results. Since each partial sum is calculated at a period of M samples through Equation 3, it will be appreciated that the partial sum can be implemented by a single
MAC. Outputs of the MACs are summed to produce an output y(n) as indicated by Equation 4.
□ Each MAC that calculates each partial sum P k (n) is represented as P k -MAC and all P k -MAC units receive the same data set. This is implemented by eliminating delays between adjacent
filter blocks illustrated in FIG. 3 and, instead, outputs of the MAC units are implemented as a delayed sum as indicated in Equation 4.
□ the filter assembly of FIG. 4 includes an expander for receiving a medical image signal and up-sampling the medical image signal and a decimation filter that includes an integer number of
MACs, changes a cutoff frequency according to bandwidth of the medical image signal by dynamically updating an impulse response, and performs decimation on the up-sampled signal according to
a decimation ratio.
□ the decimation filter calculates a partial sum, which is the sum of coefficients of a k-th (wherein k is a positive integer) location of a polyphase filter, using each MAC.
□ calculation can be performed using only K (where K is a positive integer) multipliers using the MAC, as opposed to the multipliers used to match the filter length as in FIGS. 2 and 3 .
Therefore, if only the length of the coefficients supplied to the MAC is matched as the filter length varies, arbitrary fractional decimation can be performed only by a limited number of
□ the amount of ultrasound reception signals varies according to depth of an image. For example, if the amount of reception data used to image a specific depth is 1,536, 1,024 samples are
required to display the specific depth at a resolution of 640 ⁇ 480. Therefore, 3/2-fold decimation is needed.
□ a filter coefficient is updated in the shift register by calculating a different cutoff frequency according to depth, a filter having a maximum SNR can be constructed.
□ FIG. 5 is a block diagram illustrating in detail a partial sum calculator 400 for calculating a k-th (wherein k is a positive integer) partial sum in the decimation filter assembly of FIG. 4
according to an embodiment of the present invention.
□ the partial sum calculator 400 includes an expander 410 and a P k -MAC 420 .
□ Each P k -MAC cumulatively sums filter calculations with respect to an L-fold up-sampled input signal and outputs the summed result according to a decimation ratio.
□ M filter coefficient calculations are performed by one multiplier.
□ a filter coefficient of a shift register is adaptively applied according to the cutoff frequency.
□ the P k -MAC 420 includes a shift register 421 for receiving and storing coefficients of a polyphase filter, a multiplier 422 for multiplying the coefficients stored in the shift register 421
by the up-sampled signal, a summer and register 423 for cumulatively summing the multiplied results, and a decimator 424 for performing M-fold decimation on the summed result.
□ the frequency band of the reception signal is determined by attenuation caused by the depth of an object of the medical image signal.
□ the P k -MAC 420 calculates different cutoff frequencies according to the depth and supplies the filter coefficients to the shift register 421 , thereby controlling an SNR of the medical
image. Particularly, since the P k -MAC 420 includes a fixed number of multipliers 422 regardless of the decimation ratio, the waste of multipliers used according to variation in the filter
length is prevented.
□ FIG. 6 is a diagram for explaining a structure in which an expander is removed from the decimation filter assembly of FIG. 4 .
□ (a) of FIG. 6 corresponds to the filter structure introduced in FIGS. 4 and 5 .
□ a medical image signal input to a Pk-MAC is a signal which is L-fold up-sampled by an expander 610 . That is, it may be appreciated that the structure of (a) of FIG. 6 demands L-fold
expansion with respect to input data.
□ a filter assembly for the medical image signal illustrated in (b) of FIG. 6 includes a decimation filter which includes an integer number of MACs, changes a cutoff frequency according to
bandwidth of the medical image signal by dynamically updating an impulse response, and performs decimation on the received signal according to a decimation ratio.
□ the decimation filter selects filter coefficients adjusted by an interval of an integer number in order to up-sample the medical image signal and supplies the selected filter coefficients to
the MAC.
□ the medical image signal supplied to the decimation filter of (b) of FIG. 6 is not previously up-sampled and operates at a low frequency relative to a previously up-sampled signal.
□ the decimation filter determines the filter coefficients such that the MAC may perform a partial sum calculation of a signal except for a zero padding part out of the output of the expander.
That is, if the filter coefficients are adjusted to values other than 0 out of the output of the expander in consideration of the L-fold expander and are adaptively supplied to the MAC to
match frequency bandwidth, the expander may be eliminated and a cutoff frequency matching input frequency bandwidth may be adaptively applied.
□ the decimation filter of (b) of FIG. 6 also calculates a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through the
□ the MAC includes a shift register for receiving and storing the coefficients of the polyphase filter, a multiplier for multiplying the coefficients stored in the shift register by the medical
image signal, a summer for cumulatively summing the multiplied results, and a decimator for performing decimation on the summed result.
□ the decimation filter assembly of (b) of FIG. 6 requires fewer hardware resources and less power consumption by operating at a relatively low frequency and can prevent waste of multipliers
according to variation of the filter length by using a fixed number of multipliers regardless of a decimation ratio.
□ FIG. 7 is a block diagram illustrating a decimation filter assembly in which an expander is removed according to another embodiment of the present invention.
□ a partial sum of a phase filter structure is implemented using MACs.
□ l denotes the smallest value of p and ⁇ circumflex over (n) ⁇ denotes the largest value of q
□ X ( n ) [ x ( ⁇ circumflex over ( n ) ⁇ ) x ( ⁇ circumflex over (n) ⁇ 1) . . . x ( ⁇ circumflex over (n) ⁇ M n +1)] T
□ H k ( l ) [ h ( kM+l ) h ( kM+l+L ) . . . h ( kM+l ( M n ⁇ 1) L )] T [Equation 6]
□ FIG. 8 is a flowchart illustrating a decimation method of a medical image signal using a dynamic decimation filter according to still another embodiment of the present invention. Since the
decimation method includes a procedure corresponding to each configuration of (b) of FIG. 6 described earlier, each process will be briefly described focusing on a time cause-and-effect
relation of operations to avoid a repeated description.
□ step S 810 a medical image signal is received.
□ a filter coefficient for changing a cutoff frequency is selected according to bandwidth of the medical image signal in consideration of a decimation ratio.
□ the filter coefficient in order to remove an expander, is desirably determined in consideration of an interval of an integer number for up-sampling the medical image signal received in step S
810 .
□ the medical image signal supplied to a partial sum calculator is not previously up-sampled and operates at a low frequency relative to a previously up-sampled signal.
□ step S 830 the determined filter coefficient is supplied to the partial sum calculator including an integer number of MACs.
□ the partial sum calculator calculates a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, using the MAC.
□ the MAC includes a fixed number of multipliers regardless of the decimation ratio, thereby preventing waste of multipliers according to variation in filter length.
□ step S 840 dynamic decimation is performed on the received medical image signal using the filter coefficients supplied by the partial sum calculator. More specifically, in step S 840 , the
MAC multiplies the medical image signal by the coefficients of the polyphase filter stored in a shift register, by use of a multiplier, and cumulatively sums the multiplied results, and
decimates the summed result using a decimator.
□ the method for performing decimation on a medical image signal in processing a digital signal may be implemented as code that can be written in a computer-readable recording medium and thus
read by a computer system.
□ the computer-readable recording medium may be any type of recording device in which data that can be read by the computer system is stored.
□ Examples of the computer-readable recording medium include a ROM, a RAM, a CD-ROM, a magnetic tape, a floppy disk, optical data storage, and a carrier wave (e.g., data transmission over the
□ the computer-readable recording medium can be distributed over computer systems connected to a network so that computer-readable code is written thereto and executed therefrom in a
decentralized manner.
□ Functional programs, code, and code segments to realize the embodiments herein can be construed by one of ordinary skill in the art.
□ a fixed number of multipliers is implemented as a polyphase filter structure using a MAC. Therefore, since relatively few hardware resources and less power consumption are needed upon
performing dynamic decimation and a cutoff frequency of a filter is adaptively applied to raise SNR, ultra-slimness of an ultrasound imaging system can be achieved.
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The present invention relates to a filter assembly for a medical image signal and a dynamic decimation method using the same. The filter assembly includes a decimation filter that includes an
integer number of multiplier accumulators (MACs), changes a cut-off frequency depending on a bandwidth of the medical image signal received through a dynamic impulse response update, and performs
a decimation with respect to the received signal according to a decimation ratio, wherein the decimation filter determines a filter coefficient corresponding to an integer interval so as to
up-sample the received medical image signal and supplies the filter coefficient to the MACs.
This application is a § 371 national stage entry of International Application No. PCT/KR2013/013122, filed on Dec. 3, 2015 which claims priority to South Korean Patent Application No.
10-2015-0023981, filed on Feb. 17, 2015, the entire contents of which are incorporated herein by reference.
TECHNICAL FIELD
The present invention relates to a technique of receiving and processing a medical image signal and, more particularly, a filter assembly for adaptively processing variation in bandwidth of an
ultrasound image signal received from a probe according to depth of an image, and a dynamic decimation method using the same.
BACKGROUND ART
Medical imaging technology is a diagnosis technique of visually representing muscles, tendons, and many internal organs, to capture their size, structure, and pathologic lesions with real-time
tomographic images, based on an ultrasound or photoacoustic means. Medical imaging is also used to visualize fetuses during a periodic checkup or in an emergency situation. Ultrasound has been
used to image the interior of the human body for at least 50 years and has become one of the most widely used diagnostic tools in modern medicine. The ultrasound technique is low in cost and easy
in mobility, relative to magnetic resonance imaging (MRI) or X-ray computed tomography (CT).
The principle of ultrasound imaging is as follows. First, an ultrasound image is made by bringing a measurement object into contact with a probe and receiving ultrasound reflected by generation
of ultrasound waves. If ultrasound is generated, an ultrasound wave passes into a medium within a very short time and the ultrasound wave is reflected upon passing between two media having
different acoustic impedances. In the ultrasound imaging technique, such a reflection wave is measured and a distance is calculated based on the time until reflection sound returns back, thereby
achieving imaging.
In such ultrasound imaging, an echo signal returning from a target object attenuates according to depth and thus bandwidth of the signal varies. To improve a signal-to-noise ratio (SNR) caused by
variation in bandwidth, a signal processing procedure is needed. An overview of ultrasound signal processing is given in the prior art document proposed below.
PRIOR ART DOCUMENT
Korean Patent Publication No. 10-2011-0022440, published on Mar. 7, 2011, Sogang University Research Foundation
DETAILED DESCRIPTION OF THE INVENTION Technical Problems
The technical objects that can be achieved through the present invention are designed to solve inefficiency of a conventional filter structure for implementing dynamic decimation, in which filter
length increases in proportion to a decimation ratio and thus there are wasted filters and multipliers that are not used according to dynamic variation of the decimation ratio and solve overhead
of hardware cost and the amount of calculations necessary for multiplication at a high data rate because all multipliers are positioned after an expander.
Technical Solutions
According to an aspect of the present invention, provided herein is a filter assembly for a medical image signal, including an expander configured to receive the medical image signal and
up-sample the medical image signal; and a decimation filter including an integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the
received medical image signal by dynamically updating an impulse response and perform decimation on the up-sampled signal according to a decimation ratio.
The decimation filter may calculate a partial sum, which is the sum of coefficients of a k-th (wherein k is a positive integer) location of a polyphase filter, through each MAC.
Each MAC may include a shift register configured to receive and store coefficients of a polyphase filter, a multiplier configured to multiply the coefficients stored in the shift register by the
up-sampled signal, a summer configured to cumulatively sum the multiplied results, and a decimator configured to decimate the summed result.
A frequency band of the received signal may be determined by attenuation caused by depth of an object of the medical image signal and filter coefficients for calculating different cutoff
frequencies according to the depth may be supplied to the MAC through the shift register to control a signal-to-noise ratio of the medical image.
Each MAC may include a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation in filter length.
In another aspect of the present invention, provided herein is a filter assembly for a medical image signal, including a decimation filter including an integer number of multiplier accumulators
(MACs), configured to change a cutoff frequency according to bandwidth of the medical image signal by dynamically updating an impulse response and perform decimation on the received signal
according to a decimation ratio, wherein the decimation filter determines a filter coefficient adjusted by an interval of an integer number to up-sample the medical image signal and supplies the
filter coefficient to each MAC.
The medical image signal supplied to the decimation filter may not be previously up-sampled and operate at a low frequency relative to a previously up-sampled signal.
The decimation filter may determine, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as to perform a partial sum calculation of the signal through the MAC
except for a zero padding part out of an output of the expander.
The decimation filter may calculate a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through the MAC.
The MAC may include a shift register configured to receive and store coefficients of a polyphase filter; a multiplier configured to multiply the coefficients stored in the shift register by the
medical image signal; a summer configured to cumulatively sum the multiplied results; and a decimator configured to perform decimation on the summed result.
In another aspect of the present invention, provided herein is a method of decimating a medical image signal, including receiving the medical image signal; selecting a filter coefficient for
changing a cutoff frequency according to bandwidth of the medical image signal in consideration of a decimation ratio; supplying the selected filter coefficient to a partial sum calculator
including an integer number of multiplier accumulators (MACs); and performing, by the partial sum calculator, dynamic decimation on the received medical image signal, using the selected filter
coefficient, wherein the filter coefficient is determined in consideration of an interval of an integer number to up-sample the received medical image signal.
The medical image signal supplied to the partial sum calculator may not be previously up-sampled and operate at a low frequency relative to a previously up-sampled signal.
The selecting the filter coefficient may include determining, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as to calculate a partial sum through each
MAC except for a zero padding part out of an output of the expander.
The partial sum calculator may calculate a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through each MAC.
The performing dynamic decimation may includes multiplying, by the MAC, the medical image signal by coefficients of a polyphase filter stored in a shift register by use of a multiplier;
cumulatively summing, by the MAC, the multiplied results by use of a summer; and decimating, by the MAC, the summed result by use of a decimator.
The MAC may include a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation in filter length.
Advantageous Effects
Embodiments of the present invention require relatively fewer hardware resources and less power consumption upon performing dynamic decimation by implementing a fixed number of multipliers by a
polyphase filter structure and can achieve ultra-slimness of an ultrasound imaging system by adaptively applying a cutoff frequency of a filter in order to raise SNR.
DESCRIPTION OF DRAWINGS
FIG. 1 is a diagram for explaining a decimation filter used in a medical imaging system through which embodiments of the present invention are implemented.
FIG. 2 is a diagram for explaining problems occurring upon dynamically performing decimation through a finite impulse response (FIR) filter.
FIG. 3 is a diagram for explaining an example of implementing a decimation filter using a polyphase structure instead of the FIR filter of FIG. 2 and problems occurring in such an example.
FIG. 4 is a diagram illustrating a decimation filter assembly implementing partial sums of phase filters using multiplier accumulators (MACs) according to an embodiment of the present invention.
FIG. 5 is a block diagram illustrating in detail a partial sum calculator for calculating a k-th (wherein k is a positive integer) partial sum in the decimation filter assembly of FIG. 4
according to an embodiment of the present invention.
FIGS. 6A and 6B are diagrams for explaining a structure in which an expander is removed from the decimation filter assembly of FIG. 4.
FIG. 7 is a block diagram for explaining a structure in which an expander is removed according to another embodiment of the present invention.
FIG. 8 is a flowchart illustrating a decimation method of a medical image signal using a dynamic decimation filter according to still another embodiment of the present invention.
110: decimation filter
120: ultrasound imaging system
400: partial sum calculator including MAC
111, 210, 410, 610: expander
420: MAC
421: shift register
422: multiplier
423: register for summer
113, 424: decimator
BEST MODE FOR CARRYING OUT THE INVENTION
A filter assembly for a medical image signal according to an embodiment of the present invention includes an expander configured to receive the medical image signal and up-sample the medical
image signal; and a decimation filter including an integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the medical image signal by
dynamically updating an impulse response and perform decimation on the up-sampled signal according to a decimation ratio.
MODE FOR INVENTION
Prior to a description of embodiments of the present invention, necessity and technical problems of a dynamic decimation filter will be briefly introduced and then a technical means adopted by
the embodiments of the present invention in order to solve these problems will be proposed. Hereinbelow, among symbols represented in the description and drawings of the present invention, ‘N’
denotes the length of a filter, ‘D’ denotes a delay, ‘L’ denotes expansion increasing rate of an expander for up-sampling, and ‘M’ denotes a decimation ratio.
FIG. 1 is a diagram for explaining a decimation filter used in a medical imaging system through which embodiments of the present invention are implemented.
In medical ultrasound imaging, since an ultrasound signal is affected by frequency-dependent and depth-dependent attenuation while passing through soft tissues, center frequency and frequency
bandwidth of an echo signal decrease with depth. In this case, a dynamic filter is used to maximize SNR as frequency bandwidth decreases and is mainly implemented by a decimation filter used
generally to match the data rate of the echo signal to a screen. Accordingly, such a dynamic decimation filter should be capable of performing decimation on an arbitrary fractional decimation
factor M/L and should be capable of dynamically updating an impulse response thereof.
Referring to (a) of FIG. 1, a decimation filter 110 adjusts the number of samples when the number of samples of a baseband signal is larger than the number of samples to be displayed on a screen
through an ultrasound imaging system 120. In (a) of FIG. 1, it is assumed that 512 samples are needed per centimeter when 40 MHz-sampling is performed. Then, 2,048 to 10,240 samples are needed
with respect to an image of 4 to 20 cm. Therefore, 1,024 samples are actually required for a display resolution of 640×480.
Referring to (b) of FIG. 1, in order to adjust the number of samples, an input signal x(n) is integer-fold up-sampled (111), signal-processed using a coefficient (112) of a decimation filter, and
decimated (113) at a ratio of M, thereby generating an output signal y(n).
Meanwhile, as described above, the echo signal in medical ultrasound imaging is affected by depth-dependent attenuation and the bandwidth of the echo signal differs according to depth. Therefore,
it is necessary to dynamically update a cutoff frequency with respect to each depth in order to maximize SNR. That is, the cutoff frequency varies with bandwidth using the dynamic filter.
Typically, the length N of the filter is proportional to the cutoff frequency and the decimation ratio.
FIG. 2 is a diagram for explaining problems occurring upon dynamically performing decimation through a finite impulse response (FIR) filter. The structure of an M/L-fold dynamic decimation filter
is shown in FIG. 2.
In implementing a dynamic decimation filter structure using the FIR filter (when M>L), the length N of the filter increases in proportion to increase in the decimation ratio M. That is, a
condition of N=KM may be assumed. Referring to FIG. 2, it may be appreciated that the filter length N is equal to the number of multipliers included in the filter and, due to characteristics of
the dynamic decimation filter structure, a hardware filter should be implemented for a maximum decimation ratio. In this case, the number of multipliers used varies dynamically. For example, if a
used part 220 includes (M+1) multipliers, multipliers (h(M+1) to h(N−1)=0) of the other part 230 are unused and wasted. That is, if a maximum value of M is large, the filter requires excessive
multipliers and, if the value of M is small in a dynamic decimation processing procedure, many multipliers are wasted, thereby causing inefficiency.
In addition, since all multipliers are provided after an L-fold expander 210, the multipliers operate at a high operation frequency of L times and at a high data rate of L times. Therefore, the
amount of calculations per unit time increases and thus overhead increases in terms of the amount of calculations of a block and hardware cost.
FIG. 3 is a diagram for explaining an example of implementing a decimation filter using a polyphase structure instead of the FIR filter of FIG. 2 and problems occurring in such an example.
To efficiently improve the FIR filter introduced in FIG. 2, a dynamic decimation filter using a polyphase filter may be used. When the decimation filter shown in FIG. 3 is used, M-fold decimation
is performed after performing L-fold expansion. Then, as compared with the dynamic decimation filter structure using the FIR filter, the filter of FIG. 3 has the same length as the filter of FIG.
2 but a data rate at which multipliers operate is lowered by M times.
However, even in this case, since multipliers proportional to the length of the filter are needed, there is waste of multipliers when the value of M is small. For example, if a part 310 which is
used in a dynamic decimation process is small relative to a part 320 which is not used in the dynamic decimation process, inefficiency may occur in using hardware and resources.
Accordingly, embodiments of the present invention described hereinbelow propose a filter structure which has all functions of the above-described dynamic decimation filter and simultaneously can
be efficiently implemented without waste of hardware and operation resources. That is, an efficient arbitrary factional decimation structure only using K multiplier accumulators is proposed to
use restricted hardware complexity regardless of L and K. Hereinafter, the embodiments of the present invention will be described in detail with reference to the attached drawings.
An equation of a general FIR-based decimation filter is defined as follows. When an arbitrary fractional decimation factor is represented as M/L, a signal w(n) which is L-fold up-sampled with
respect to an input x(n) is given as indicated in Equation 1.
$w ( n ) = ( x ( n / L ) , n = pL ( p : integer ) 0 , otherwise 0 [ Equation 1 ]$
Herein, n is a sampling index. A procedure of filtering the FIR as shown in FIG. 2 with respect to such an input is indicated as Equation 2.
$y ( n ) = ∑ j = 0 N - 1 h ( j ) · w ( nM - j ) [ Equation 2 ]$
Herein, h(j) denotes a coefficient of a given FIR filter and N denotes the length of the filter.
An equation for a dynamic decimation structure using the polyphase filter will now be defined. First, Equation 2 may be summarized as Equation 3 upon changing a dynamic decimation filter
structure to a polyphase structure in which outputs of all polyphase filters are added as illustrated in FIG. 3.
$y ( n ) = ∑ k = 0 K - 1 H ( k ) · W ( n - k ) = ∑ k = 0 K - 1 ∑ m = 0 M - 1 h ( kM + m ) · w ( ( n - k ) M - m ) H ( k ) = [ h ( kM ) h ( kM + 1 ) … h (
kM + M - 1 ) ] W ( n ) = [ w ( nM ) w ( nM - 1 ) … w ( nM - ( M - 1 ) ) ] T [ Equation 3 ]$
FIG. 4 is a diagram illustrating a structure in which partial sums P[0 ]to P[k-1 ]are added based on k-th coefficients with respect to M polyphase filters of FIG. 3.
FIG. 4 is a diagram illustrating rearrangement of the polyphase filters of FIG. 3 through multiplier accumulators (MACs) according to an embodiment of the present invention. It may be appreciated
that the polyphase filters are implemented through a plurality of P[0]-MAC, P[1]-MAC, . . . , P[k]-MAC connected in parallel to each other.
Equation 3 indicates that each partial sum P[k](n) in FIG. 4 is obtained by multiplying M consecutive pairs of input samples by filter coefficients at a data rate of L·f[x ]and cumulatively
summing the multiplied results. Since each partial sum is calculated at a period of M samples through Equation 3, it will be appreciated that the partial sum can be implemented by a single MAC.
Outputs of the MACs are summed to produce an output y(n) as indicated by Equation 4.
$y ( n ) = ∑ k = 0 K - 1 P k ( n - k ) = ∑ k = 0 K - 1 ∑ m = 0 M - 1 w ( ( n - k ) M - m ) · h ( kM + m ) P k ( n ) = ∑ m = 0 M - 1 w ( nM - m ) · h ( kM + m ) , k = 0
, 1 , … , K - 1 [ Equation 4 ]$
Each MAC that calculates each partial sum P[k](n) is represented as P[k]-MAC and all P[k]-MAC units receive the same data set. This is implemented by eliminating delays between adjacent filter
blocks illustrated in FIG. 3 and, instead, outputs of the MAC units are implemented as a delayed sum as indicated in Equation 4.
In summary, the filter assembly of FIG. 4 includes an expander for receiving a medical image signal and up-sampling the medical image signal and a decimation filter that includes an integer
number of MACs, changes a cutoff frequency according to bandwidth of the medical image signal by dynamically updating an impulse response, and performs decimation on the up-sampled signal
according to a decimation ratio. The decimation filter calculates a partial sum, which is the sum of coefficients of a k-th (wherein k is a positive integer) location of a polyphase filter, using
each MAC.
More specifically, when the sum of coefficients of a k-th location of each polyphase filter is a partial sum P[k], this partial sum is implemented through a MAC expressed as P[k]-MAC. Only a
limited number of multipliers is used regardless of a decimation ratio M and the coefficients are supplied to the MAC through a shift register (not shown) to perform calculation.
Thus, calculation can be performed using only K (where K is a positive integer) multipliers using the MAC, as opposed to the multipliers used to match the filter length as in FIGS. 2 and 3.
Therefore, if only the length of the coefficients supplied to the MAC is matched as the filter length varies, arbitrary fractional decimation can be performed only by a limited number of
The amount of ultrasound reception signals varies according to depth of an image. For example, if the amount of reception data used to image a specific depth is 1,536, 1,024 samples are required
to display the specific depth at a resolution of 640×480. Therefore, 3/2-fold decimation is needed. To efficiently perform decimation through a dynamic decimation filter having a filter length 32
, M=3, and L=2, a sampling frequency of the reception signal is supplied to input of the filter and is supplied to 8 (K=8) P-MACs.
In addition, since a frequency band of a reception signal decreases according to depth, if a filter coefficient is updated in the shift register by calculating a different cutoff frequency
according to depth, a filter having a maximum SNR can be constructed.
FIG. 5 is a block diagram illustrating in detail a partial sum calculator 400 for calculating a k-th (wherein k is a positive integer) partial sum in the decimation filter assembly of FIG. 4
according to an embodiment of the present invention. The partial sum calculator 400 includes an expander 410 and a P[k]-MAC 420.
Each P[k]-MAC cumulatively sums filter calculations with respect to an L-fold up-sampled input signal and outputs the summed result according to a decimation ratio. Thus, M filter coefficient
calculations are performed by one multiplier. In this case, a filter coefficient of a shift register is adaptively applied according to the cutoff frequency.
More specifically, the P[k]-MAC 420 includes a shift register 421 for receiving and storing coefficients of a polyphase filter, a multiplier 422 for multiplying the coefficients stored in the
shift register 421 by the up-sampled signal, a summer and register 423 for cumulatively summing the multiplied results, and a decimator 424 for performing M-fold decimation on the summed result.
The frequency band of the reception signal is determined by attenuation caused by the depth of an object of the medical image signal. The P[k]-MAC 420 calculates different cutoff frequencies
according to the depth and supplies the filter coefficients to the shift register 421, thereby controlling an SNR of the medical image. Particularly, since the P[k]-MAC 420 includes a fixed
number of multipliers 422 regardless of the decimation ratio, the waste of multipliers used according to variation in the filter length is prevented.
An additional embodiment for improving performance of such a dynamic decimation filter structure in terms of an operation frequency will now be proposed.
FIG. 6 is a diagram for explaining a structure in which an expander is removed from the decimation filter assembly of FIG. 4. (a) of FIG. 6 corresponds to the filter structure introduced in FIGS.
4 and 5.
Referring to (a) of FIG. 6, a medical image signal input to a Pk-MAC is a signal which is L-fold up-sampled by an expander 610. That is, it may be appreciated that the structure of (a) of FIG. 6
demands L-fold expansion with respect to input data.
To improve this, in (b) of FIG. 6, the expander 610 is removed and an L-fold up-sampling procedure is performed in the Pk-MAC so that a signal process procedure performed at a lower frequency may
be derived. To this end, in (b) of FIG. 6, filter coefficients adjusted by an interval of L are supplied to a shift register and then is supplied to a multiplier.
To this end, a filter assembly for the medical image signal illustrated in (b) of FIG. 6 includes a decimation filter which includes an integer number of MACs, changes a cutoff frequency
according to bandwidth of the medical image signal by dynamically updating an impulse response, and performs decimation on the received signal according to a decimation ratio. The decimation
filter selects filter coefficients adjusted by an interval of an integer number in order to up-sample the medical image signal and supplies the selected filter coefficients to the MAC.
Particularly, the medical image signal supplied to the decimation filter of (b) of FIG. 6 is not previously up-sampled and operates at a low frequency relative to a previously up-sampled signal.
More specifically, in consideration of an integer-fold expander for up-sampling, the decimation filter determines the filter coefficients such that the MAC may perform a partial sum calculation
of a signal except for a zero padding part out of the output of the expander. That is, if the filter coefficients are adjusted to values other than 0 out of the output of the expander in
consideration of the L-fold expander and are adaptively supplied to the MAC to match frequency bandwidth, the expander may be eliminated and a cutoff frequency matching input frequency bandwidth
may be adaptively applied.
Obviously, the decimation filter of (b) of FIG. 6 also calculates a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, through
the MAC. As described with reference to FIG. 5, the MAC includes a shift register for receiving and storing the coefficients of the polyphase filter, a multiplier for multiplying the coefficients
stored in the shift register by the medical image signal, a summer for cumulatively summing the multiplied results, and a decimator for performing decimation on the summed result.
In this way, the decimation filter assembly of (b) of FIG. 6 requires fewer hardware resources and less power consumption by operating at a relatively low frequency and can prevent waste of
multipliers according to variation of the filter length by using a fixed number of multipliers regardless of a decimation ratio.
FIG. 7 is a block diagram illustrating a decimation filter assembly in which an expander is removed according to another embodiment of the present invention. To efficiently design filter length
proportional to a decimation ratio M and the amount of calculations caused by an expander, a partial sum of a phase filter structure is implemented using MACs.
More specifically, an original input sequence x(n) is used as input of a Pk-MAC unit. Since it is assumed that decimation is performed, a non-zero value is necessarily present between M elements
of w(n) of Equation 3 when M>L. In this case, w(nM−p)=x(q) (0≤p≤M−1) is satisfied (only when nM−p=qL and w(nM−m)=0 for m≠p). When l denotes the smallest value of p and {circumflex over (n)}
denotes the largest value of q, l=nM mod L and {circumflex over (n)}=└nM/L┘. That is, x({circumflex over (n)})=w(nM−l) indicates the first non-zero value of W(n). Therefore, if M[n ]non-zero
values are present in W(n), M[n ]is the largest value of j satisfying l+(j−1)L≤M−1 and may be indicated as in Equation 5.
$M n = ⌊ M - 1 - l L ⌋ + 1 [ Equation 5 ]$
Since w(nM−l−jL)=x({circumflex over (n)}−j), an input data vector and a filter coefficient supplied to all MAC units are represented as in Equation 6 according to a Pk-MAC.
X(n)=[x({circumflex over (n)})x({circumflex over (n)}−1) . . . x({circumflex over (n)}−M [n]+1)]^T,
H [k](l)=[h(kM+l)h(kM+l+L) . . . h(kM+l(M [n]−1)L)]^T[Equation 6]
As illustrated in FIG. 7, MAC units output K partial sums P[k](n)=X(n)·H[k](l) (k=0, 1, . . . K−1). These partial sums are delayed through a delay chain as in P[k](n−k) (k=0, 1, . . . K−1) and
are summed to produce the output y(n) of Equation 4. Since X(n) has M[n ]elements, the Pk-MAC outputs a partial sum after M[n ]samples are input as illustrated by an M[n]-fold decimator at an
output stage of each MAC. If L and M are relatively prime, the sum of L consecutive values of M[n ]having L repeated patterns is equal to M, i.e., M[n]+M[n−1]+ . . . +M[n−L+1]=M. Therefore, it
can be appreciated that a decimation factor is M/L since the proposed structure provides L outputs per M input samples.
FIG. 8 is a flowchart illustrating a decimation method of a medical image signal using a dynamic decimation filter according to still another embodiment of the present invention. Since the
decimation method includes a procedure corresponding to each configuration of (b) of FIG. 6 described earlier, each process will be briefly described focusing on a time cause-and-effect relation
of operations to avoid a repeated description.
In step S810, a medical image signal is received.
In step S820, a filter coefficient for changing a cutoff frequency is selected according to bandwidth of the medical image signal in consideration of a decimation ratio. In this case, in order to
remove an expander, the filter coefficient is desirably determined in consideration of an interval of an integer number for up-sampling the medical image signal received in step S810. To this
end, in step S820 for selecting the filter coefficient, it is desirable to select the filter coefficient such that a partial sum through a MAC is calculated except for a zero padding part out of
the output of the expander in consideration of integer-fold expansion.
Accordingly, the medical image signal supplied to a partial sum calculator is not previously up-sampled and operates at a low frequency relative to a previously up-sampled signal.
In step S830, the determined filter coefficient is supplied to the partial sum calculator including an integer number of MACs. Herein, the partial sum calculator calculates a partial sum, which
is the sum of coefficients of a k-th (where k is a positive integer) location of a polyphase filter, using the MAC. The MAC includes a fixed number of multipliers regardless of the decimation
ratio, thereby preventing waste of multipliers according to variation in filter length.
In step S840, dynamic decimation is performed on the received medical image signal using the filter coefficients supplied by the partial sum calculator. More specifically, in step S840, the MAC
multiplies the medical image signal by the coefficients of the polyphase filter stored in a shift register, by use of a multiplier, and cumulatively sums the multiplied results, and decimates the
summed result using a decimator.
Meanwhile, the method for performing decimation on a medical image signal in processing a digital signal according to the foregoing exemplary embodiments may be implemented as code that can be
written in a computer-readable recording medium and thus read by a computer system. The computer-readable recording medium may be any type of recording device in which data that can be read by
the computer system is stored.
Examples of the computer-readable recording medium include a ROM, a RAM, a CD-ROM, a magnetic tape, a floppy disk, optical data storage, and a carrier wave (e.g., data transmission over the
Internet). The computer-readable recording medium can be distributed over computer systems connected to a network so that computer-readable code is written thereto and executed therefrom in a
decentralized manner. Functional programs, code, and code segments to realize the embodiments herein can be construed by one of ordinary skill in the art.
While the present invention has been particularly shown and described with reference to exemplary embodiments thereof, those skilled in the art will appreciate that the present invention may be
embodied in other specific forms than those set forth herein without departing from the spirit and essential characteristics of the present invention. The above detailed description is therefore
to be construed in all aspects as illustrative and not restrictive. The scope of the invention should be determined by reasonable interpretation of the appended claims and all changes coming
within the equivalency range of the invention are within the scope of the invention.
According to the above-described embodiments of the present invention, a fixed number of multipliers is implemented as a polyphase filter structure using a MAC. Therefore, since relatively few
hardware resources and less power consumption are needed upon performing dynamic decimation and a cutoff frequency of a filter is adaptively applied to raise SNR, ultra-slimness of an ultrasound
imaging system can be achieved.
Claims (18)
The invention claimed is:
1. A filter assembly for a medical image signal, comprising:
an expander configured to receive the medical image signal and up-sample the medical image signal; and
a decimation filter including an integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the received medical image signal by
dynamically updating an impulse response and perform decimation on the up-sampled signal according to a decimation ratio.
2. The filter assembly according to claim 1, wherein the decimation filter calculates a partial sum, which is the sum of coefficients of a k-th (wherein k is a positive integer) location of a
polyphase filter, through each MAC.
3. The filter assembly according to
claim 1
, wherein each MAC includes:
a shift register configured to receive and store coefficients of a polyphase filter;
a multiplier configured to multiply the coefficients stored in the shift register by the up-sampled signal;
a summer configured to cumulatively sum the multiplied results; and
a decimator configured to decimate the summed result.
4. The filter assembly according to claim 3, wherein a frequency band of the received signal is determined by attenuation caused by depth of an object of the medical image signal and filter
coefficients for calculating different cutoff frequencies according to the depth are supplied to the MAC through the shift register to control a signal-to-noise ratio of the medical image.
5. The filter assembly according to claim 1, wherein each MAC includes a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation
in filter length.
6. A filter assembly for a medical image signal, comprising:
a decimation filter including an integer number of multiplier accumulators (MACs), configured to change a cutoff frequency according to bandwidth of the medical image signal by dynamically
updating an impulse response and perform decimation on the received signal according to a decimation ratio,
wherein the decimation filter determines a filter coefficient adjusted by an interval of an integer number to up-sample the medical image signal and supplies the filter coefficient to each MAC.
7. The filter assembly according to claim 6, wherein the medical image signal supplied to the decimation filter is not previously up-sampled and operates at a low frequency relative to a
previously up-sampled signal.
8. The filter assembly according to claim 6, wherein the decimation filter determines, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as to perform a
partial sum calculation of the signal through the MAC except for a zero padding part out of an output of the expander.
9. The filter assembly according to claim 6, wherein the decimation filter calculates a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a
polyphase filter, through the MAC.
10. The filter assembly according to
claim 6
, wherein the MAC includes:
a shift register configured to receive and store coefficients of a polyphase filter;
a multiplier configured to multiply the coefficients stored in the shift register by the medical image signal;
a summer configured to cumulatively sum the multiplied results; and
a decimator configured to perform decimation on the summed result.
11. The filter assembly according to claim 10, wherein a frequency band of the received signal is determined by attenuation caused by depth of an object of the medical image signal and filter
coefficients for calculating different cutoff frequencies according to the depth are supplied to the MAC through the shift register to control a signal-to-noise ratio of the medical image.
12. The filter assembly according to claim 6, wherein the MAC includes a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation
in filter length.
13. A method of decimating a medical image signal, comprising:
receiving the medical image signal;
selecting a filter coefficient for changing a cutoff frequency according to bandwidth of the medical image signal in consideration of a decimation ratio;
supplying the selected filter coefficient to a partial sum calculator including an integer number of multiplier accumulators (MACs); and
performing, by the partial sum calculator, dynamic decimation on the received medical image signal, using the selected filter coefficient,
wherein the filter coefficient is determined in consideration of an interval of an integer number to up-sample the received medical image signal.
14. The method according to claim 13, wherein the medical image signal supplied to the partial sum calculator is not previously up-sampled and operates at a low frequency relative to a previously
up-sampled signal.
15. The method according to claim 13, wherein the selecting the filter coefficient includes determining, in consideration of an integer-fold expander for up-sampling, the filter coefficient so as
to calculate a partial sum through each MAC except for a zero padding part out of an output of the expander.
16. The method according to claim 13, wherein the partial sum calculator calculates a partial sum, which is the sum of coefficients of a k-th (where k is a positive integer) location of a
polyphase filter, through each MAC.
17. The method according to
claim 13
, wherein the performing dynamic decimation includes:
multiplying, by the MAC, the medical image signal by coefficients of a polyphase filter stored in a shift register by use of a multiplier;
cumulatively summing, by the MAC, the multiplied results by use of a summer; and
decimating, by the MAC, the summed result by use of a decimator.
18. The method according to claim 13, wherein the MAC includes a fixed number of multipliers regardless of the decimation ratio to prevent waste of multipliers used according to variation in
filter length.
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Theory of representations 1
Code Completion Credits Range
01TR1 ZK 2 2+0
Course guarantor:
Basic knowledge about representations of groups, with emphasize given to finite groups.
basic mathematical analysis and linear algebra course (01MAN, 01MAA2-01MAA4, 01LAL, 01LAA2), algebra course (01ALGE)
Syllabus of lectures:
1. The notion of group and its representation. Irreducible representations. Schur's lemma.
2. Direct sum and direct product of representations.
3. Representation characters, orthogonality, Burnside theorem.
4. Character tables.
5. Representations of permutation group.
6. Induced representations, normal subgroups, projective representations.
Syllabus of tutorials:
Study Objective:
Knowledge: basic notions and procedures in representations of finite groups, basic outlook in construction methods.
Skills: explicit construction of representations and character tables of given finite group, analysis of given representation (irreducibility)
Study materials:
Key references:
1. B. Steinberg: Representation Theory of Finite Groups: An Introductory Approach, Springer, 2011
2. J. P. Serre: Linear Representations of Finite Groups, Springer, 2012
Recommended references:
3. B. Simon: Representations of Finite and Compact Groups, AMS, 1996
4. A. Wilson: Modular Representation Theory of Finite Groups, Scitus Academics LLC, 2016
Time-table for winter semester 2024/2025:
Burdík Č.
Wed 16:00–17:50
(lecture parallel1)
Trojanova 13
Time-table for summer semester 2024/2025:
Time-table is not available yet
The course is a part of the following study plans: | {"url":"https://bilakniha.cvut.cz/en/predmet5561206.html","timestamp":"2024-11-05T13:50:29Z","content_type":"text/html","content_length":"21322","record_id":"<urn:uuid:47d2b8c0-7836-44e3-a2c8-fc371ff314bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00652.warc.gz"} |
Some thoughts about epsilon and delta
By Ben Blum-Smith, Contributing Editor
The calculus has a very special place in the 20th century’s traditional course of mathematical study. It is a sort of fulcrum: both the summit toward which the whole secondary curriculum strives, and
the fundamental prerequisite for a wide swath of collegiate and graduate work, both in mathematics itself and in its applications to the sciences, economics, engineering, etc.^[1] At its heart is the
notion of the limit, described in 1935 by T. A. A. Broadbent as the critical turning point:
The first encounter with a limit marks the dividing line between the elementary and the advanced parts of a school course. Here we have not a new manipulation of old operations, but a new
operation; not a new trick, but a new idea.^[2]
Humanity’s own collective understanding of this “new idea” was hard-earned. The great length of the historical journey toward the modern definition in terms of $\epsilon$ and $\delta$ mirrors the
well-known difficulty students have with it. Although it is the foundation of calculus, it is common to push the difficulty of this definition off from a first calculus course onto real analysis.
Indeed, mathematicians have been discussing the appropriate place for the full rigors of this definition in the calculus curriculum for over 100 years.^[3]
There is also a rich vein in the mathematics education research literature studying students’ engagement with the $\epsilon$–$\delta$ definition. Researchers have examined student difficulties coming
from its multiple nested quantifiers^[4] as well as its great distance from the less formal notions of limit with which students typically enter its study,^[5] and have also made an effort to chart
the paths they take toward a full understanding.^[6]
This blog post is a contribution to this conversation, analyzing in detail three learners’ difficulties with $\epsilon$ and $\delta$.^[7] If there is a take-home message, it is to respect the
profound subtlety of this definition and the complexity of the landscape through which students need to move as they learn to work with it.
Some history
Many readers will be familiar with the long struggle to find a rigorous underpinning for the calculus of Newton and Leibniz, leading to the modern definition of the limit in terms of $\epsilon$ and $
\delta$. In this section, I excerpt a few episodes, which will become important in the later discussion of student thought. Readers already familiar with the history of the subject are welcome to
skim or skip this section.
While Newton and Leibniz published their foundational work on (what we now call) derivatives and integrals in the late 17th century, they based these ideas not on the modern limit, but on notions
that look hand-wavy in retrospect.^[8] To Leibniz, the derivative, for example, was a ratio of “infinitesimal” quantities — smaller than finite quantities, but not zero. To Newton, it was an
“ultimate ratio”, the ratio approached by a pair of quantities as they both disappear. Both authors would calculate the derivative of $x^n$ via a now-familiar manipulation: augment $x$ by a small
amount $o$; correspondingly, $x^n$ augments to $(x+o)^n = x^n + nox^{n-1} + \dots + o^n$. The ratio of the change in $x^n$ to the change in $x$ is $nox^{n-1}+\dots+o^n : o$, or $nx^{n-1}+\dots+o^
{n-1}:1$. At this point, they would differ in their explanation of why you can ignore all the terms involving $o$ in this last expression: for Leibniz, it is because they are infinitesimal, and for
Newton, it is because they all vanish when the augmentation of $x$ is allowed to vanish.
A famous critique of both of these lines of reasoning was leveled in 1734 by the British philosopher and theologian Bishop George Berkeley, arguing that since to form the ratio of $nox^{n-1}+\dots+o^
n$ to $o$ in the first place, it was necessary to assume $o$ is nonzero, it is strictly inconsistent to then decide to ignore it.
Hitherto I have supposed that $x$ flows, that $x$ hath a real Increment, that $o$ is something. And I have proceeded all along on that Supposition, without which I should not have been able to
have made so much as one single Step. From that Supposition it is that I get at the Increment of $x^n$, that I am able to compare it with the Increment of $x$, and that I find the Proportion
between the two Increments. I now beg leave to make a new Supposition contrary to the first, i.e., I will suppose that there is no Increment of $x$, or that $o$ is nothing; which second
Supposition destroys my first, and is inconsistent with it, and therefore with every thing that supposeth it. I do nevertheless beg leave to retain $nx^{n-1}$, which is an Expression obtained in
virtue of my first Supposition, which necessarily presupposeth such Supposition, and which could not be obtained without it: All which seems a most inconsistent way of arguing…
It was a long journey from the state of the art in the early 18th century, to which Berkeley was responding, to the modern reformulation of calculus on the basis of the $\epsilon$–$\delta$ limit. The
process took well over a century. I will summarize this story by quoting somewhat telegraphically from William Dunham’s book The Calculus Gallery,^[9] from which I first learned it.
Berkeley penned the now famous question:
… They are neither finite quantities nor quantities infinitely small, nor yet nothing. May we not call them the ghosts of departed quantities?
… Over the next decades a number of mathematicians tried to shore up the shaky underpinnings… pp. 71-72
… Cauchy’s “limit-avoidance” definition made no mention whatever of attaining the limit, just of getting and staying close to it. For him, there were no departed quantities, and Berkeley’s ghosts
disappeared… p. 78
… If his statement seems peculiar, his proof began with a now-familiar ring, for Cauchy introduced two “very small numbers” $\delta$ and $\epsilon$… p. 83
… We recall that Cauchy built his calculus upon limits, which he defined in these words:
When the values successively attributed to a variable approach indefinitely to a fixed value, in a manner so as to end by differing from it by as little as one wishes, this last is called the
limit of all the others.
To us, aspects of this statement, for instance, the motion implied by the term “approach,” seem less than satisfactory. Is something actually moving? If so, must we consider concepts of time and
space before talking of limits? And what does it mean for the process to “end”? The whole business needed one last revision.
Contrast Cauchy’s words with the polished definition from the Weierstrassians:
$\lim_{x\to a} f(x) = L$ if and only if, for every $\epsilon > 0$, there exists a $\delta > 0$ so that, if $0 < |x-a| < \delta$, then $|f(x) - L| < \epsilon$.
Here nothing is in motion, and time is irrelevant. This is a static rather than dynamic definition and an arithmetic rather than a geometric one. At its core, it is nothing but a statement about
inequalities. pp. 129-130
The Weierstrassian definition (i.e., the modern one!) allows the manipulation to which Berkeley objected to be carried to completion without ever asking $o$ to be zero.
Missing the point completely
The core of this blog post is a discussion of three learners’ encounters with the $\epsilon$–$\delta$ limit, seeking to illuminate some of the subtle challenges that can arise. I begin with my own
I “did well” in my college real analysis class, by which I mean that my instructor (a well-known analyst at a major research university) concluded on the basis of my written output that I had
mastered the material. However, I walked away from the course with a subtle but important misunderstanding of the $\epsilon$–$\delta$ definition that was not visible in my written work and so went
entirely undetected by my instructor, and, for many years, by myself as well.
From my previous experience with calculus, I had concluded that you can often identify the value toward which a function is headed, even if the function is undefined “when you get there.” To the
extent I had a definition of limit, this was it: the value toward which the function is headed.
When I studied real analysis as an undergraduate, I found the class easy, including the $\delta$–$\epsilon$ work. I mean, if $L$ is where $f(x)$ is headed as $x\to c$, then sure, for any $\epsilon$
-neighborhood around $L$, there is going to be a $\delta$-neighborhood around $c$ that puts $f(x)$ inside the $\epsilon$-neighborhood. But I related to the notion that $f(x)$ is headed toward $L$ as
conceptually prior to this $\epsilon$–$\delta$ game. The latter seemed like fancy window-dressing to me, possibly useful post-hoc if you need an error estimate. I did not understand that it was a
definition — that it was supposed to be constitutive of the meaning of limit. So, I completely missed the point! But I want to stress that you would not have known this from my written work, and of
course, I didn’t know it either.
I went on to become a high school math teacher. In the years that followed, I did detect certain limitations in my understanding of limits. For example, I noticed that I didn’t have adequate tools
for thinking about if and when the order of two different limit processes could be safely interchanged. But it did not cross my mind that the place to start in thinking clearly about this was a tool
I had already been given.
After a few years, I began teaching an AP Calculus course. About 3/4 of the way through my first time teaching it, my student Harry^[10] said to me after class, “You know this whole class is based on
a paradox, right?” He proceeded to give me what I now recognize as essentially Bishop Berkeley’s critique. At the time, it did not occur to me to reach for epsilon and delta. Instead, I responded
like an 18th century mathematician, trying to convince him that the terminus of an unending process is something it’s meaningful to talk about. I hadn’t really understood what the problem was. Of
course, Harry left unsatisfied.
The pieces finally came together for me the next year, when I read Dunham’s Calculus Gallery, quoted above. I remember the shift in my understanding: ooooohhhhhh. The $\epsilon$‘s and $\delta$‘s are
not an addendum to, or a gussying-up of, the idea of identifying where an unending process is headed. They are replacing this idea! It was a revelation to reread the definition from this new point of
view. Calculus does not need the infinitesimal! I immediately wished I had a do-over with Harry, whose dissatisfaction I hadn’t comprehended enough to be able to speak to.
I concluded from this that a complete understanding of the $\epsilon$–$\delta$ definition includes an understanding of what it’s for.
But it’s not the same thing at all
Having come to this conclusion, in my own teaching of real analysis I’ve made a great effort to make clear the problem that $\epsilon$ and $\delta$ are responding to. In one course, I began with a
somewhat in-depth study of Berkeley’s critique of the 18th century understanding of the calculus, in order to then be able to offer $\epsilon$ and $\delta$ as an answer to that critique.
In doing this, I ran into a new challenge. To illustrate, I’ll focus on the experience of a student named Ty. Ty arrived in my course having already developed a fairly strong form of a more
intuitive, 17th-18th century understanding of the limit; essentially the Newtonian one, much like the understanding that had carried me myself through all my calculus coursework. He quickly made
sense of Berkeley’s objection, so he was able to see that this understanding was not mathematically satisfactory. I was selling the $\epsilon$–$\delta$ definition as a more satisfactory substitute.
However, Ty objected that important aspects of his understanding of the limit (what Tall and Vinner called his concept image^[11]) were not captured by this new definition. In particular, what had
happened to the notion that the limit was something toward which the function was, or even should have been, headed? The $\epsilon$–$\delta$ definition of $\lim_{x\to a}f(x)$ studiously avoids the
point $a$ “at which the limit is taken,” even speculatively. To Ty, it was the $\epsilon$–$\delta$ definition that was, pun intended, missing the point.
Of course, this studious avoidance is precisely how the Weierstrassian definition gets around Berkeley’s objection. The Newtonian “ultimate ratio” and the Leibnizian “infinitesimal” both ask us to
imagine something counterfactual, or at least pretty wonky. This is exactly what made them hard for Berkeley to swallow, and as I learned from Dunham’s book, the great virtue of $\epsilon$ and $\
delta$ is that they give us a way to uniquely identify the limit that does not ask us to engage in such a trippy flight of fancy that may or may not look sane in the light of day.
But, at the same time, something is lost.^[12] What I learned from Ty is that this loss is pedagogically important to acknowledge.^[13]
One vs. many
Another subtle difficulty in working with the $\epsilon$–$\delta$ definition is revealed when you use it to try to prove something. I think what I am about to describe is a general difficulty
students encounter in learning the methods and conventions of proof-writing, but I speculate that it may be particularly acute with respect to the present topic. Consider this (utterly standard)
proof that if $f,g$ are functions of $x$ such that $\lim_{x\to a}f = L$ and $\lim_{x\to a}g = M$, and $h=f+g$, then $\lim_{x\to a} h = L+M$:
Let $\epsilon > 0$ be given. Because $\lim_{x\to a} f = L$, there exists $\delta_1 > 0$ such that $0 < |x - a| < \delta_1$ implies $|f - L| < \epsilon / 2$. Similarly, because $\lim_{x\to a} g =
M$, there exists $\delta_2 > 0$ such that $0 < |x - a| < \delta_2$ implies $|g-M| < \epsilon/2$.
Take $\delta = \min(\delta_1,\delta_2)$.
Then for values of $x$ satisfying $0 < |x - a| < \delta$, it follows from the triangle inequality and the definition of $h$ that
$|h - (L+M)| = |f - L + g - M| \leq |f-L| + |g-M| < \epsilon / 2 + \epsilon / 2 = \epsilon$.
Since $\epsilon > 0$ was arbitrary, we can conclude that $\lim_{x\to a} h = L+M$.
Here is a surprisingly rich question: is the $\epsilon$ in the proof one number, or many numbers?
On one way of looking at it, of course it is only one number: $\epsilon$ is fixed at the outset of the proof. Indeed, if $\epsilon$ were allowed to be more than one thing, equations like $\epsilon /
2 + \epsilon / 2 = \epsilon$ would be meaningless. More subtly, we usually speak about $\epsilon$ as a single fixed quantity when we justify the existence of $\delta_1,\delta_2$ in terms of the
definition of the limit: we know $\delta_1$ exists because by the definition of the limit, for any $\varepsilon$ there is a $\delta$, so in particular, there is a $\delta_1$ for $\epsilon / 2$, etc.
Note the “in particular”: we produce $\delta_1$ from the definition by specializing it.
But on another way of looking at it, of course $\epsilon$ is many numbers. Indeed, it must represent every positive number, otherwise how can it be used to verify the definition for all $\epsilon>0$?
The singular, fixed $\epsilon$ with which we work in the proof is a sort of chimera: it actually represents all positive numbers at once. That we think of it as a single number is just a
psychological device to allow us to focus in a productive way on what we are doing with all these numbers.
This dual nature of $\epsilon$ in the above was driven home for me by working with Ricky. Fast and accurate with calculations and algebraic manipulations, Ricky was thrown for a loop by real
analysis, which was her first proof-based class, and in particular by the $\epsilon$–$\delta$ proofs. After a lot of work, she had mastered the definition itself. But in trying to write the proofs,
she found the lilting refrain for all $\epsilon > 0$… to be a kind of siren song, leading her astray. She was constantly re-initializing $\epsilon$ with this phrase, so that reading her work, there
were 3 different meanings for $\epsilon$ by the end. “Look at how the proof works,” I would say, referring to the proof of $\lim h = L+M$ above. “You don’t need $|f-L|$ to be less than any old $\
epsilon$. You need it to be less than the particular $\epsilon$ that you are using for $h$.” “What do you mean the particular $\epsilon$ I am using?” she would ask. “I am trying to prove it works for
all $\epsilon$!”
Ricky’s difficulty has led me to a much greater appreciation of the subtle and profound abstraction involved any time an object is introduced into a proof with the words “fix an arbitrary…” In a
sense, this is nothing more — nor less! — than the abstraction at the heart of a student’s first encounter with algebra: if we imagine an unspecified number $x$, and proceed to have a conversation
about it, our conversation applies simultaneously at once to all the values $x$ could have taken, even if we were imagining it the whole time as “only one thing.”^[14] But I don’t think I appreciated
the great demand that “fix an arbitrary…” proofs in general, and $\delta$–$\epsilon$ proofs in particular, place on this abstraction. The mastery of it that is needed here goes far beyond what is
needed to get you through years of pushing $x$ around.
Conclusion: respect the subtlety
I offer the above anecdotes primarily as grist for reflection about learning, and especially about the nature of the particular landscape students tread as they encounter $\epsilon$ and $\delta$.^[15
] But I would like to articulate some lessons and reminders that I myself draw from them:
(1) A complete understanding of a concept might require to go beyond mastery of its internal structure and its downstream implications, to include an understanding of its purpose, i.e. the situation
it was designed to respond to.
(2) Work that successfully responds to the standard set of prompts may still conceal important gaps in understanding, as mine did in my undergraduate real analysis class. More generally, do not
assume because a student is “strong” that they have command of any particular thing.
(3) Conversely, take student thought seriously, even when it looks/sounds wrong. Ricky and Ty were producing unsuccessful work for very mathematically rich reasons; I learned something worthwhile by
taking the time to understand what each of them was getting at. Harry’s issue, which I didn’t take seriously at the time, could have pushed my own understanding of calculus forward — in fact, it did,
albeit belatedly.
Finally, I hope the combination of these anecdotes with the history above serves as a reminder both of the magnitude of the historical accomplishment crystallized in the Weierstrassian $\epsilon$–$\
delta$ definition of the limit, and of the corresponding profundity of the journey students take toward its mastery.
Notes and references
[1] There is an important contemporary argument that calculus’ pride of place in the curriculum should be ceded to statistics. (For example, see the TED talk by Arthur Benjamin.) That debate is
beyond the scope of this blog post.
[2] The First Encounter with a Limit. The Mathematical Gazette, Vol. 19, No. 233 (1935), pp. 109-123. (link [jstor])
[3] In addition to the 1935 Mathematical Gazette article quoted above, see, e.g., E. J. Moulton, The Content of a Second Course in Calculus, AMM Vol. 25, No. 10 (1918), pp. 429-434 (link [jstor]); E.
G. Phillips, On the Teaching of Analysis, The Mathematical Gazette Vol. 14, No. 204 (1929), pp. 571-573 (link [jstor]); N. R. C. Dockeray, The Teaching of Mathematical Analysis in Schools, The
Mathematical Gazette Vol. 19, No. 236 (1935), pp. 321-340 (link [jstor]); H. Scheffe, At What Level of Rigor Should Advanced Calculus for Undergraduates be Taught?, AMM Vol. 47, No. 9 (1940), pp.
635-640 (link [jstor]). I thank Dave L. Renfro for all of these references.
[4] E.g., E. Dubinsky and O. Yiparaki, On student understanding of AE and EA quantification, Research in Collegiate Mathematics Education IV, 8 (2000), pp. 239-289 (link). In this and the next two
notes, the literature cited only scratches the surface.
[5] E.g., D. Tall and S. Vinner, Concept image and concept definition in mathematics with particular reference to limits and continuity, Educational Studies in Mathematics Vol. 12 (1981), pp. 151-169
(link), S. R. Williams, Models of Limit Held by College Calculus Students, Journal for Research in Mathematics Education Vol. 22, No. 3 (1991), pp. 219-236 (link [jstor]), and C. Swinyard and E.
Lockwood, Research on students’ reasoning about the formal definition of limit: An evolving conceptual analysis, Proceedings of the 10th annual conference on research in undergraduate mathematics
education, San Diego State University, San Diego, CA (2007) (link).
Findings about students’ informal understandings of limits that generate friction with their study of $\epsilon$ and $\delta$ include that they are often dynamic/motion-based (like Newton), or
infinitesimals-based (like Leibniz), and meanwhile, they are also often characterized by a “forward” orientation from $x$ to $f(x)$ — “If you bring $x$ close to $a$, it puts $f(x)$ close to $L$.”
This is in contrast with the $\epsilon$–$\delta$ definition’s “backward” orientation from $f(x)$ to $x$ — “To make $f(x)$$\epsilon$-close to $L$, you have to find a $\delta$ to constrain $x$.”
[6] E.g., J. Cottrill, E. Dubinsky, D. Nichols, K. Schwingendorf, K. Thomas, D. Vidakovic, Understanding the Limit Concept: Beginning with a Coordinated Process Scheme, Journal of Mathematical
Behavior Vol. 15, pp. 167-192 (1996), Swinyard and Lockwood op. cit. (which responds to Cottrill et. al.), and C. Nagle, Transitioning from introductory calculus to formal limit conceptions, For the
Learning of Mathematics Vol. 33, No. 2 (2013), pp. 2-10 (link).
[7] To avoid ambiguity, the learners referred to here are myself and the students I below call Ty and Ricky. The student I call Harry illustrates a difficulty one might have without the $\epsilon$–$\
delta$ limit.
[8] The brief account I am about to give represents an orthodox view of the history of calculus, see for example J. V. Grabiner, Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus,
The American Mathematical Monthly Vol. 90, No. 3 (1984), pp. 185-194 (link). This orthodoxy is not without its detractors, e.g., B. Pourciau, Newton and the Notion of Limit, Historia Mathematica No.
28 (2001), pp. 18-30 (link) or H. Edwards, Euler’s Definition of the Derivative, Bulletin of the AMS Vol. 44, No. 4 (2007), pp. 575-580 (link).
Readers interested in more comprehensive accounts of the history of the $\epsilon$–$\delta$ limit can consult Judith Grabiner’s monograph The Origins of Cauchy’s Rigorous Calculus, MIT Press,
Cambridge, MA (1981) and William Dunham’s The Calculus Gallery: Masterpieces from Newton to Lebesgue, Princeton Univ. Press, Princeton, NJ (2005). A very interesting-looking new book on the topic is
David Bressoud’s Calculus Reordered: A History of the Big Ideas, Princeton Univ. Press, Princeton, (2019) (link), which takes a broader view, looking at the development of integration,
differentiation, series, and limits across multiple millennia and continents, and viewing the limit as a sort of culmination driven by the needs of research mathematicians in the 19th century.
Bressoud’s book also considers questions of pedagogy in relation to this history.
[9] Dunham, op. cit. (see previous footnote).
[10] All names of students are pseudonyms.
[11] D. Tall and S. Vinner. Concept image and concept definition in mathematics with particular reference to limits and continuity. Educational Studies in Mathematics Vol. 12, No. 2 (1981), pp.
151-169. (link)
[12] Relatedly, recovering that which was lost from calculus when $\epsilon$ and $\delta$ superseded the Leibnizian infinitesimals is often given as the rationale behind Abraham Robinson’s
development of nonstandard analysis.
[13] This observation is related to the body of research indicated in note [5]. I think it is subtly different though. As I understand that research, the theme is the difficulties students have with
the $\epsilon$–$\delta$ definition due to “interference” from their more informal understandings of limits and derivatives. In contrast, my focus here is on a difficulty Ty had not because of
“interference,” but rather because he recognized (perhaps more clearly than I did) that this new definition is not actually doing the same thing, so if it was being sold it as a substitute, he was
not buying.
[14] To help Ricky contextualize what she needed to do for the $\epsilon$ proof in terms of things she already understood, I asked her to consider this proof that every square number exceeds by one
the product of the two integers consecutive with its square root:
Let $x$ be any integer. Then
$(x + 1) (x - 1) = x^2 - x + x - 1$
$= x^2 + 0 - 1$
$= x^2 - 1$,
so any square number $x^2$ is one more than the product of $x+1$ and $x-1$.
“I think of the $x$ in this proof as every number,” she said. “But you have to relate to it as a single number during the calculation itself,” I replied. “Otherwise, how do you know that $-x + x = 0$
[15] I first encountered the metaphor of a “landscape of learning” attendant to particular mathematical topics in the writings of Catherine Twomey Fosnot and Maarten Dolk.
6 Responses to Some thoughts about epsilon and delta
1. All great points. Thanks for this piece, Ben. For anyone who is interested, I wrote a series of columns on student difficulties with limits in my Launchings columns for July, August, and
September of 2014: Beyond the Limit I, II, and III. And I have a chapter on Limits as the Algebra of Inequalities in the new book Calculus Reordered: A History of the Big Ideas, in which I draw
on the work of Judith Grabiner.
□ Thanks, David! I learned of your new book from Al Cuoco during the editorial phase, and pointed to it in note [8]. I would like to know if you feel it is being correctly characterized there;
I was only able to read the beginning by press time.
For readers interested in the Launchings columns mentioned by David, which are extremely relevant to the present piece, here are the URLS (unfortunately this blog doesn’t support links in the
2. On the “one vs many”: there is an approach to explaining the definition of limit as an adversarial game: I pick an epsilon, you have to respond with a delta. In this approach, it is clear that in
every instance of a game, you have to deal with only one epsilon (the one I picked); but to have a strategy you have to be able to deal with whatever epsilon I throw at you.
□ This is great. Since the point of the blog post is respect for the subtlety of the difficulties, though, allow me to take this opportunity to stay on message: the adversarial paradigm was
*already how Ricky understood the definition.* Indeed, it helped her attain sufficient mastery of the definition to be able to state it with a feeling that she understood all its parts.
But the difficulty discussed above came after this, revealing itself in the context of work on specific proofs. I’m speculating here, but perhaps one way to see it is that she was struggling
with the idea of a uniform strategy; or else with the notion that a uniform strategy can be described in terms of a single (but generic) epsilon.
For what it’s worth, the analogy mentioned in note [14] did shift something for her, because she felt she understood how the algebraic proof worked.
3. The question of whether epsilon is one positive real number or all of them is a real sticking point. One helpful device that I enjoy using in my teaching is Susanna Epp’s use of a “generic
particular” to prove a universally quantified statement:
Method of Generalizing from the Generic Particular
To show that every element of a set satisfies a certain property, suppose x is a particular but arbitrarily chosen element of the set, and show that x satisfies the property.
(from Discrete Mathematics with Applications by Susanna S. Epp)
Anecdotally, I observed that my students with an understanding of “generic particular” were able to produce more coherent proofs of universally quantified statements, and to understand the proofs
process in deeper way.
□ Seems like you already found your way to my related post (https://blogs.ams.org/matheducation/2020/05/20/the-things-in-proofs-are-weird-a-thought-on-student-difficulties/) on the strangeness
of the generic particular! (I’ve never heard this phrase before, thank you for introducing it to me!)
I’m extremely interested in the question of how students develop an understanding of this idea and what steps instructors can take to support this development (cf. that other post). The quote
from Susanna Epp you give directs students to the right path to take, but there’s also the question of how a student gets convinced that that’s the right path. The tack I took with Ricky,
described in note [14], did make some progress with that particular learner, but this is just one trick that worked a little bit, in one context. I’m interested in developing a more
comprehensive map of the landscape of learning (in the sense the phrase is used by Cathy Fosnot in her “Young Mathematicians at Work” books) involved in developing this understanding. Excited
to be in conversation with you about this!
This entry was posted in Faculty Experiences, Mathematics Education Research, Student Experiences and tagged calculus, continuity, definitions, delta, difficulty, epsilon, limit, proof, real analysis
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Online Trigonometry Calculators and Solvers
Easy to use online trigonometry calculators and solvers for various topics in trigonometry are presented. These may be used to practice and explore the properties of trigonometric functions
numerically in order to gain deep understanding of these functions.
Inverse Trigonometric Functions
Trigonometric Ratios
Hyperbolic Functions
More Math Calculators and Solvers. | {"url":"https://www.analyzemath.com/trigonometry-calculators.html","timestamp":"2024-11-01T22:13:23Z","content_type":"text/html","content_length":"22742","record_id":"<urn:uuid:df241898-3a42-4554-beee-2982dca18d42>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00267.warc.gz"} |
Calculate the mean of the following data
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now, we first find the class mark x[i]of each class and then proceed as follows
Therefore, mean
Hence, mean of the given data is 12.93. | {"url":"https://philoid.com/question/29409-calculate-the-mean-of-the-following-data","timestamp":"2024-11-05T16:09:40Z","content_type":"text/html","content_length":"31926","record_id":"<urn:uuid:14b36cb6-4439-4271-b7f7-6469559e9abb>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00266.warc.gz"} |
old Entry Requirements
Please carefully read the information below about entry requirements and supporting documentation before you apply to ensure your application has the best chance for success.
Degree Requirements
Good honours degree (Upper-second class honours degree 2.1 or equivalent). Your undergraduate degree should be in Philosophy, Politics and Economics or a closely related programme with significant
components from each subject area.
MA PPE: Tripartite
Economics Requirement
You should have achieved a good standard (2:1 or higher) in undergraduate modules in microeconomics and macroeconomics at an intermediate level or above (at 40 hours of lectures across intermediate
or final years). Module choice in Economics will restricted according to the strength of your background in Economics to a set list. Students without intermediate-level econometrics will only have
access to a restricted list of Economics optional modules. Whilst formal techniques are taught as part of the Economics component of MA in PPE pathways with Economics, prior training in these areas
of mathematics and statistics is required. All Tripartite students are also required to take the 2-week pre-sessional course in Mathematics and Statistics.
Philosophy Requirement
You should have taken at least 25% of one year of study in intermediate Philosophy modules.
MA PPE: Bipartite: Economics and Philosophy
Economics Requirement
You should have achieved a good standard (2:1 or higher) in undergraduate modules in microeconomics and macroeconomics at an intermediate level or above (at 40 hours of lectures across intermediate
or final years). Module choice in Economics will be restricted according to the strength of your background in Economics to a set list. Econometrics (at 40 hours of lectures or equivalent) at an
intermediate level or above would also be a strong advantage. Bipartite students without intermediate-level econometrics will only have access to a restricted list of Economics optional modules.
In order to take the full selected set of MA in PPE Economics optional modules, students must have obtained a good standard (2:1 or higher) in undergraduate courses in mathematics and economic
statistics (at least the equivalent of a year-long module or two term/semester–long modules, and economic statistics should include econometrics):
• Calculus, functions of several variables, partial derivatives, constrained optimization using Lagrange multipliers, matrix algebra and linear equations.
• Probability theory, distribution theory (binomial, normal and associated distributions), sampling theory, statistical inference, interval estimation, hypothesis testing (means and variances),
least squares regression.
Where this has not been obtained, students may take Quantitative Methods: Econometrics A (30 credits) in Term 1 to enable further Economics optional module choices. Where this has not been taken only
a restricted optional module catalogue will be available. Whilst formal techniques are taught as part of the Economics component of MA in PPE pathways with Economics, prior training in these areas of
mathematics and statistics is required. All MA in PPE Bipartite: Economics and Philosophy students are also required to take the 2-week pre-sessional course in Mathematics and Statistics.
Philosophy Requirement
You should have taken at least 25% of one year of study in intermediate Philosophy modules.
MA PPE: Bipartite: Politics and Economics
Economics Requirement
You should have achieved a good standard (2:1 or higher) in undergraduate modules in microeconomics and macroeconomics at an intermediate level or above (at 40 hours of lectures across intermediate
or final years). Module choice in Economics will be restricted according to the strength of your background in Economics to a set list. Econometrics (at 40 hours of lectures or equivalent) at an
intermediate level or above would also be a strong advantage. Bipartite students without intermediate-level econometrics will only have access to a restricted list of Economics optional modules.
In order to take the full selected set of MA in PPE Economics optional modules, students must have obtained a good standard (2:1 or higher) in undergraduate courses in mathematics and economic
statistics (at least the equivalent of a year-long module or two term/semester–long modules, and economic statistics should include econometrics):
• Calculus, functions of several variables, partial derivatives, constrained optimization using Lagrange multipliers, matrix algebra and linear equations.
• Probability theory, distribution theory (binomial, normal and associated distributions), sampling theory, statistical inference, interval estimation, hypothesis testing (means and variances),
least squares regression. Where this has not been obtained, students may take Quantitative Methods: Econometrics A (30 credits) in Term 1 to enable further Economics optional module choices.
Where this has not been taken only a restricted optional module catalogue will be available. Whilst formal techniques are taught as part of the Economics component of MA in PPE pathways with
Economics, prior training in these areas of mathematics and statistics is required. All MA in PPE Bipartite: Politics and Economics students are also required to take the 2-week pre-sessional course
in Mathematics and Statistics.
MA PPE: Bipartite: Politics and Philosophy
Philosophy Requirement
You should have taken at least 25% of one year of study in intermediate Philosophy modules.
Further Requirements
English Language Fluency and Further Requirements | {"url":"https://warwick.ac.uk/fac/soc/ppe/prospective/masters/entryrequirementsold/","timestamp":"2024-11-06T05:41:24Z","content_type":"text/html","content_length":"37445","record_id":"<urn:uuid:a19f701d-0c62-415b-b2b9-1c079df41dd9>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00851.warc.gz"} |
NBM Card v4.2 - MDL
NBM v4.2 Text Bulletin Card
There are five NBM text bulletins, which cover different time scales and different elements. The products are as follows:
(*Actual forecast hours change according to cycle. Listed times are for 00Z and 12Z cycles.):
Product Name Product Type Time step Forecast Hours covered
NBH Hourly 1-Hourly Hours 1-25
NBS Short 3-Hourly Hours 6-72*
NBE Extended 12-Hourly Hours 24-192*
NBX Super-Extended 12-Hourly Hours 204-264* (continuation of NBE)
NBP Probabilistic (Extended period) 12-Hourly Hours 24-228*
Example bulletins for each of the types (NBH,NBS,NBE,NBX,NBP) are listed below. At the bottom of the page, additional stations from different locations are displayed to show expected data at
different regions (and at land vs. marine stations).
The bulletins consist of the following:
• The first line always consists of: the station call(short name), bulletin name, forecast date, and forecast cycle.
• The following 2-3 lines show the forecast valid date/time and/or model forecast hour. All forecast dates and times are listed in UTC time.
• The lines following the date/time consist of various weather elements. These elements vary by bulletin type, region, and sometimes cycle.
• Large values: any value which would be printed as >998 will be printed as 998 (except MSLP: see element description below for details)
• Large negative values: Any element with a displayed value < -98 will be displayed as -98.
Missing Data:
For all Elements, a value of -99 indicates missing data. If data are missing for all forecast hours, the line for that element will not be printed.
Changes from NBM v4.1 Text Bulletins
Users should take note of the following changes from the NBM v4.1 products:
• Deterministic 10-m wind speed and 10-m wind gust replaced with the mean of the distribution of probabilistic wind speed/gust calculated from the quantile mapping technique.
• Station IDs and locations have been updated (see NBM v4.2 station table: https://vlab.noaa.gov/web/mdl/nbm-stations-v4.2 )
For additional information about NBM weather elements, see the Element Key
NBH Example Bulletin (CONUS, land station)
Data Key
Date/time lines
• UTC = forcast valid hour (in UTC) (date as in line 1, unless past 23Z - then the following date is valid)
• TMP = temperature, degrees F
• TSD = standard deviation of temperature, degrees F
• DPT = dew point temperature, degrees F
• DSD = dew point temperature standard deviation, degrees F
• SKY = sky cover, percent
• SSD = sky cover standard deviation, percent
• WDR = wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
□ oceanic-only stations only have WDR for cycles=0,7,12,19
• WSP = wind speed, knots (note: processes are slightly different between CONUS and other regions)
• WSD = wind speed standard deviation, knots
• GST = wind gust, knots (note: processes are slightly different between CONUS and other regions)
• GSD = wind gust standard deviation, knots
• P01 = 1-hour PoP, percent
• P06 = 6-hour PoP, percent
• Q01 = 1-hour QPF, 1/100 inches
• T01 = 1-hour thunderstorm probability, percent
• PZR = conditional probability of freezing rain, percent
• PSN = conditional probability of snow, percent
• PPL = conditional probability of sleet / ice pellets, percent
• PRA = conditional probability of rain, percent
• S01 = 1-hour snow amount, 1/10 inches
• SLV = snow level, 100s feet MSL
• I01 = 1-hour ice amount, 1/100 inches
• CIG = ceiling height, 100s feet (>5000 ft reported to the nearest 1000 ft; -88=unlimited)
• MVC = probability of ceiling MVFR flight conditions (ceiling <= 3000 ft), percent
• IFC = probability of ceiling IFR flight conditions (ceiling < 1000 ft), percent
• LIC = probability of ceiling LIFR flight conditions (ceiling < 500 ft), percent
• LCB = lowest cloud base, 100s feet (>5000 ft reported to the nearest 1000 ft; -88=unlimited)
• VIS = visibility, 1/10th miles (rounded to nearest mile for values >= 1 mile) [Note: values before approx. 2019-06-01 are in full miles]
• MVV = probability of visibility MVFR flight conditions (visibility <= 5 miles), percent
• IFV = probability of visibility IFR flight conditions (visibility < 3 miles), percent
• LIV = probability of visibility LIFR flight conditions (visibility < 1 mile), percent
• MHT = mixing height, 100s feet AGL
• TWD = transport wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
• TWS = transport wind speed, knots (<0.5 knots will be listed as 0 (calm))
• HID = Haines Index (unitless)
• SOL = instantaneous solar radiation, 10s W/m^2 (ex: 8=80 W/m2; non-zero values < 10 = 1)
NBS Example Bulletin (CONUS, land station)
Data Key
Date/time lines
• DT (or blank) = forcast valid date (in UTC)
• UTC = forcast valid hour (in UTC)
• FHR = model forecast hour (number of hours forward from forecast date/cycle)
• TXN = 18-hour maximum and minimum temperatures, degrees F.
□ Min is between 00Z-18Z and reported at 12Z (except Guam)
□ Max is between 12z(current day)-06Z(next day) and reported at 00z(following day) (except Guam)
□ Guam stations report Tmax at 12z and Tmin at 00z.
• XND = standard deviation of maximum or minimum temperature, degrees F
• TMP = temperature, degrees F
• TSD = standard deviation of temperature, degrees F
• DPT = dew point temperature, degrees F
• DSD = dew point temperature standard deviation, degrees F
• SKY = sky cover, percent
• SSD = sky cover standard deviation, percent
• WDR = wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
□ oceanic-only stations only have WDR for cycles=0,7,12,19
• WSP = wind speed, knots (<0.5 knots will be listed as 0 (calm))
• WSD = wind speed standard deviation, knots
• GST = wind gust, knots
• GSD = wind gust standard deviation, knots
• P06 = 6-hour PoP, percent
• P12 = 12-hour PoP, percent
• Q06 = 6-hour QPF, 1/100 inches
• Q12 = 12-hour QPF, 1/100 inches
• DUR = Duration of precipitation, hours
• T03 = 3-hour thunderstorm probability, percent
• T06 = 6-hour thunderstorm probability, percent
• T12 = 12-hour thunderstorm probability, percent
• PZR = conditional probability of freezing rain, percent
• PSN = conditional probability of snow, percent
• PPL = conditional probability of sleet / ice pellets, percent
• PRA = conditional probability of rain, percent
• S06 = 6-hour snow amount, 1/10 inches
• SLV = snow level, 100s feet MSL (rounded to nearest 1000 ft for values > 10,000 ft)
• I06 = 6-hour ice amount, 1/100 inches
• CIG = ceiling height, 100s feet (>5000 ft reported to the nearest 1000 ft; -88=unlimited)
• IFC = probability of ceiling IFR flight conditions (ceiling < 1000 ft), percent
• LCB = lowest cloud base, 100s feet (>5000 ft reported to the nearest 1000 ft; -88=unlimited)
• VIS = visibility, 1/10th miles up to 10 miles (rounded to nearest mile for values >= 1 mile; VIS > 10 miles = 100)
• IFV = probability of visibility IFR flight conditions (visibility < 3 miles), percent
• MHT = mixing height, 100s feet AGL
• TWD = transport wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
• TWS = transport wind speed, knots (<0.5 knots will be listed as 0 (calm))
• HID = Haines Index (unitless)
• SOL = instantaneous solar radiation, 10s W/m^2 (ex: 8=80 W/m2; non-zero values < 10 = 1)
• SWH = significant wave height, feet (marine and some near-water stations only)
(For additional information about NBM weather elements, see the Element Key)
NBE Example Bulletin (CONUS, land station)
Data Key
Date/time lines
• (second line) = forcast valid dates (in UTC)
• UTC = forcast valid hour (in UTC)
• FHR = model forecast hour (number of hours forward from forecast date/cycle)
• TXN = 18-hour maximum and minimum temperatures, degrees F.
□ Min is between 00Z-18Z and reported at 12Z (except Guam)
□ Max is between 12z(current day)-06Z(next day) and reported at 00z(following day) (except Guam)
□ Guam stations report Tmax at 12z and Tmin at 00z.
• XND = standard deviation of maximum or minimum temperature, degrees F
• TMP = temperature, degrees F
• TSD = standard deviation of temperature, degrees F
• DPT = dew point temperature, degrees F
• DSD = dew point temperature standard deviation, degrees F
• SKY = sky cover, percent
• SSD = sky cover standard deviation, percent
• WDR = wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
□ oceanic-only stations only have WDR for cycles=0,7,12,19
• WSP = wind speed, knots (<0.5 knots will be listed as 0 (calm))
• WSD = wind speed standard deviation, knots
• GST = wind gust, knots
• GSD = wind gust standard deviation, knots
• P12 = 12-hour PoP, percent
• Q12 = 12-hour QPF, 1/100 inches
• Q24 = 24-hour QPF, 1/100 inches
• DUR = Duration of precipitation, hours (CONUS only)
• T12 = 12-hour thunderstorm probability, percent
• PZR = conditional probability of freezing rain, percent
• PSN = conditional probability of snow, percent
• PPL = conditional probability of sleet / ice pellets, percent
• PRA = conditional probability of rain, percent
• S12 = 12-hour snow amount, 1/10 inches
• SLV = snow level, 100s feet MSL
• I12 = 12-hour ice amount, 1/100 inches
• S24 = 24-hour snow amount, 1/10 inches
• SOL = 12-hour mean++ solar radiation, 10s W/m^2 (++for hours in period with sunlight present)
• SWH = significant wave height, feet (marine and some near-water stations only)
(For additional information about NBM weather elements, see the Element Key)
NBX Example Bulletin (CONUS, land station)
Data Key
Date/time lines
• (second line) = forcast valid dates (in UTC)
• UTC = forcast valid hour (in UTC)
• FHR = model forecast hour (number of hours forward from forecast date/cycle)
• TXN = 18-hour maximum and minimum temperatures, degrees F.
□ Min is between 00Z-18Z and reported at 12Z (except Guam)
□ Max is between 12z(current day)-06Z(next day) and reported at 00z(following day) (except Guam)
□ Guam stations report Tmax at 12z and Tmin at 00z.
• XND = standard deviation of maximum or minimum temperature, degrees F
• TMP = temperature, degrees F
• TSD = standard deviation of temperature, degrees F
• DPT = dew point temperature, degrees F
• DSD = dew point temperature standard deviation, degrees F
• SKY = sky cover, percent
• SSD = sky cover standard deviation, percent
• WDR = wind direction, nearest tens degrees (northerly=36; easterly=9; calm=00)
□ oceanic-only stations only have WDR for cycles=0,7,12,19
• WSP = wind speed, knots (<0.5 knots will be listed as 0 (calm))
• WSD = wind speed standard deviation, knots
• GST = wind gust, knots
• GSD = wind gust standard deviation, knots
• P12 = 12-hour PoP, percent
• P24 * = 24-hour PoP, percent (13Z only)
□ For 13Z only: extended P24 displayed for hours ~264-370
• Q12 = 12-hour QPF, 1/100 inches
• Q24 = 24-hour QPF, 1/100 inches
□ For 13Z only: extended QPF from QMD Mean 24-hour QPF displayed for hours ~264-370
• DUR = Duration of precipitation, hours (CONUS only for NBX)
• PZR = conditional probability of freezing rain, percent
• PSN = conditional probability of snow, percent
• PPL = conditional probability of sleet / ice pellets, percent
• PRA = conditional probability of rain, percent
• S12 = 12-hour snow amount, 1/10 inches
• SLV = snow level, 100s feet MSL
• I12 = 12-hour ice amount, 1/100 inches
• S24 = 24-hour snow amount, 1/10 inches
• SOL = 12-hour mean++ solar radiation, 10s W/m^2 (++for hours in period with sunlight present)
• SWH = significant wave height, feet (marine and some near-water stations only)
□ SWH is only available in NBX for the following cycles: 0,1,2,6,7,8,12,13,14,18,19,20
(For additional information about NBM weather elements, see the Element Key)
Note on NBP Data Availability
Data for all elements in NBP is not available for all cycles. All elements are available for 01z, 07z, 13z and 19z. Some data is available for 00z and 12z. All other cycles data is limited. See
entries below for more details.
NBP Example Bulletin (CONUS, land station)
Data Key
Date/time lines
• (second line) = forcast valid dates (in UTC)
• UTC = forcast valid hour (in UTC)
• FHR = model forecast hour (number of hours forward from forecast date/cycle)
About Probabilistic Elements
The weather elements for NBP are probabilistic percentiles. For X percentile, there is an X% probability that the weather element will be EQUAL TO or BELOW the value listed (or a (100-X) % chance the
value will be above what is listed). So for 10th percentile, there is a 10% chance the value will be at or below the number listed, and a 90% chance it will be above. For additional information about
NBM weather elements, see the Element Key
Begin Key
• TXNMN = QMD Mean minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNSD = QMD Standard Deviation minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNP1 = QMD 10th percentile minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNP2 = QMD 25th percentile minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNP5 = QMD 50th percentile minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNP7 = QMD 75th percentile minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
• TXNP9 = QMD 90th percentile minimum/maximum temperature, F. Minimum is listed at 12z, and Maximum is listed at 00z.
□ QMD Temperature data is currently only available for CONUS and Alaska domains.
• WSPP1 = 10th percentile wind speed, knots.
• WSPP2 = 25th percentile wind speed, knots.
• WSPP5 = 50th percentile wind speed, knots.
• WSPP7 = 75th percentile wind speed, knots.
• WSPP9 = 90th percentile wind speed, knots.
• W24P1 * = 10th percentile 24-hour maximum sustained wind speed exceedance, knots.
• W24P2 * = 25th percentile 24-hour maximum sustained wind speed exceedance, knots.
• W24P5 * = 50th percentile 24-hour maximum sustained wind speed exceedance, knots.
• W24P7 * = 75th percentile 24-hour maximum sustained wind speed exceedance, knots.
• W24P9 * = 90th percentile 24-hour maximum sustained wind speed exceedance, knots.
□ 24-hour Max Wind data is currently only available for CONUS domain.
• G24P1 * = 10th percentile 24-hour wind gust exceedance, knots.
• G24P2 * = 25th percentile 24-hour wind gust exceedance, knots.
• G24P5 * = 50th percentile 24-hour wind gust exceedance, knots.
• G24P7 * = 75th percentile 24-hour wind gust exceedance, knots.
• G24P9 * = 90th percentile 24-hour wind gust exceedance, knots.
□ 24-hour Wind Gust data is currently only available for CONUS domain.
• Q24P1 = 10th percentile 24-hour QPF, 1/100 inch.
• Q24P2 = 25th percentile 24-hour QPF, 1/100 inch.
• Q24P5 = 50th percentile 24-hour QPF, 1/100 inch.
• Q24P7 = 75th percentile 24-hour QPF, 1/100 inch.
• Q24P9 = 90th percentile 24-hour QPF, 1/100 inch.
• S24P1 = 10th percentile 24-hour snowfall accumulation, 1/10 inch.
• S24P2 = 25th percentile 24-hour snowfall accumulation, 1/10 inch.
• S24P5 = 50th percentile 24-hour snowfall accumulation, 1/10 inch.
• S24P7 = 75th percentile 24-hour snowfall accumulation, 1/10 inch.
• S24P9 = 90th percentile 24-hour snowfall accumulation, 1/10 inch.
• I24P1 = 10th percentile 24-hour flat ice accumulation, 1/100 inch.
• I24P2 = 25th percentile 24-hour flat ice accumulation, 1/100 inch.
• I24P5 = 50th percentile 24-hour flat ice accumulation, 1/100 inch.
• I24P7 = 75th percentile 24-hour flat ice accumulation, 1/100 inch.
• I24P9 = 90th percentile 24-hour flat ice accumulation, 1/100 inch.
• SLPP1 = 10th percentile mean sea level pressure, mb.
• SLPP2 = 25th percentile mean sea level pressure, mb.
• SLPP5 = 50th percentile mean sea level pressure, mb.
• SLPP7 = 75th percentile mean sea level pressure, mb.
• SLPP9 = 90th percentile mean sea level pressure, mb.
□ Any SLP value 1000 mb or higher does not show the thousands value. (ex: 1000 mb = 000; 987 mb = 987).
(For additional information about NBM weather elements, see the Element Key) | {"url":"https://vlab.noaa.gov/web/mdl/nbm-textcard-v4.2","timestamp":"2024-11-05T20:20:20Z","content_type":"text/html","content_length":"176856","record_id":"<urn:uuid:7dcf08e9-73ef-4237-a078-6034e9aeadad>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00670.warc.gz"} |
How To Generate Passive Income Without A Corpus Right Now? - Capitalworx
[vc_row][vc_column][vc_empty_space][/vc_column][/vc_row][vc_row][vc_column][vc_column_text]After my last post on “The Secret to Build a Comfortable Retirement Income or Passive Income“, I received
questions from some of my concerned readers. They said, “We don’t have that kind of money to invest right now to generate passive income. What should we do?”
Well, this post is for you.
If you don’t have the initial corpus to invest, then you should be focusing on creating that corpus. This is what some people call “planning for retirement”. However, I think that retirement can come
in quite early, if you plan well and execute diligently.[/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][vc_empty_space][/vc_column][/vc_row][vc_row][vc_column][vc_empty_space][/vc_column][/
vc_row][vc_row][vc_column][vc_column_text]Let’s start with the amount of money you would like every month when you retire.
Or maybe you want to generate passive income in addition to your primary source of income. I am assuming the period to be 15 years, after which you will retire.
1. Let’s say this amount is Rs 50000/- per month today. In 15 years time, this amount will be equivalent to Rs 137950/- (considering 7% annual inflation). This inflated amount is our x.
2. So then we need 75x to achieve our plan. (If you don’t understand this statement, read my earlier post). 75×137950 is equal to Rs 1,03,46,250/- which is our target amount.
3. To achieve this target amount, you would require an SIP of Rs 16883/- for 15 years (considering a compounded return of 14% per annum)
4. Once this target amount is achieved, you can proceed to the SWP (systematic withdrawal plan) for passive income, as outlined in my last post.
Read my last post here.[/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][vc_empty_space][/vc_column][/vc_row][vc_row][vc_column][vc_empty_space][/vc_column][/vc_row][vc_row][vc_column]
[vc_column_text]Consider another illustration:
1. Let’s say I want to generate a passive second income of Rs 10000/- per month in 10 years time. The inflated amount would come to Rs 19671/- (assuming 7% inflation). This amount is our x.
2. To generate a passive second income, I need a corpus of 75x, which is, 75×19671 = Rs 14,75,325/-
3. To achieve this target amount in 10 years time, I would need an SIP of Rs 5630/- per month (considering a compounded return of 14% per annum)
4. Once this target amount is achieved, you can proceed to the SWP (systematic withdrawal plan) for generating a second income, as outlined in my last post.
Read my last post here: The Secret to Build a Comfortable Retirement Income or Passive Income
Let me know your thoughts. Contact me here.
How To Generate Passive Income Without A Corpus Right Now? | {"url":"https://www.capitalworx.in/2017/09/26/generate-passive-income-dont-corpus-right-now/","timestamp":"2024-11-07T13:01:36Z","content_type":"text/html","content_length":"66704","record_id":"<urn:uuid:aca0fab6-5452-4c50-badb-b82973ee7572>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00474.warc.gz"} |
Collected Works of William P. Thurston with Commentary: I. Foliations, Surfaces and Differential Geometrysearch
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Collected Works of William P. Thurston with Commentary: I. Foliations, Surfaces and Differential Geometry
Softcover ISBN: 978-1-4704-7472-0
Product Code: CWORKS/27.1.S
List Price: $125.00
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AMS Member Price: $100.00
eBook ISBN: 978-1-4704-6833-0
Product Code: CWORKS/27.1.E
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Softcover ISBN: 978-1-4704-7472-0
eBook: ISBN: 978-1-4704-6833-0
Product Code: CWORKS/27.1.S.B
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Collected Works of William P. Thurston with Commentary: I. Foliations, Surfaces and Differential Geometry
Softcover ISBN: 978-1-4704-7472-0
Product Code: CWORKS/27.1.S
List Price: $125.00
MAA Member Price: $112.50
AMS Member Price: $100.00
eBook ISBN: 978-1-4704-6833-0
Product Code: CWORKS/27.1.E
List Price: $125.00
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• Collected Works
Volume: 27; 2022; 759 pp
MSC: Primary 57; 53
William Thurston's work has had a profound influence on mathematics. He connected whole mathematical subjects in entirely new ways and changed the way mathematicians think about geometry,
topology, foliations, group theory, dynamical systems, and the way these areas interact. His emphasis on understanding and imagination in mathematical learning and thinking are integral elements
of his distinctive legacy.
This four-part collection brings together in one place Thurston's major writings, many of which are appearing in publication for the first time. Volumes I–III contain commentaries by the Editors.
Volume IV includes a preface by Steven P. Kerckhoff.
Volume I contains William Thurston's papers on foliations, mapping classes groups, and differential geometry.
Graduate students and researchers interested in geometric topology, geometric group theory, low-dimensional topology, and dynamical systems of rational maps.
This item is also available as part of a set:
□ Cover
□ Title page
□ Copyright page
□ Contents
□ Preface
□ Acknowledgments
□ Foliations
□ Commentary: Foliations
□ Foliations of three-manifolds which are circle bunldes
□ Anosov flows and the fundamental group
□ Noncobordant foliations of 𝑆³
□ Some remarks on foliations
□ Foliations and groups of diffeomorphisms
□ A generalization of the Reeb stability theorem
□ The theory of foliations of codimension greater than one
□ Foliated bundles, invariant measures and flat manifolds
□ The theory of folations of codimension greater than one
□ On the existence of contact forms
□ A local construction of foliations for three-manifolds
□ On the construction and classification of foliations
□ Existence of codimension-one foliations
□ Polynomial growth in holonomy groups of foliations
□ Anosov flows on new three manifolds
□ A norm for the homology of 3-manifolds
□ Contact structures and foliations on 3-manifolds
□ Confoliations
□ Three-manifolds, foliations and circles, I. Preliminary version
□ Three-manifolds, foliations and circles II. The transverse asymptotic geometry of foliations
□ Surfaces and mapping class groups
□ Commentary: Surfaces and mapping class groups
□ A presentation for the mapping class group of a closed orientable surface
□ New proofs of some results on Nielsen
□ On the geometry and dynamics of diffeomorphisms of surfaces
□ Earthquakes in 2-dimensional hyperbolic geometry
□ Minimal stretch maps between hyperbolic surfaces
□ Non-continuity of the action of the modular group of Bers’ boundary of Teichmuller space
□ Differential geometry
□ Commentary: Differential geometry
□ Some simple examples of symplectic manifolds
□ Characteristic numbers of 3-manifolds
□ Transformation groups and natural bundles
□ Manifolds with canonical coordinate charts: Some examples
□ Pinching constants for hyperbolic manifolds
□ Hyperbolic 4-manifolds and conformally flat 3-manifolds
□ Shapes of polyhedra and trinagulations of the sphere
□ On three-dimensional space groups
□ Other titles in this series
□ Back Cover
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Volume: 27; 2022; 759 pp
MSC: Primary 57; 53
William Thurston's work has had a profound influence on mathematics. He connected whole mathematical subjects in entirely new ways and changed the way mathematicians think about geometry, topology,
foliations, group theory, dynamical systems, and the way these areas interact. His emphasis on understanding and imagination in mathematical learning and thinking are integral elements of his
distinctive legacy.
This four-part collection brings together in one place Thurston's major writings, many of which are appearing in publication for the first time. Volumes I–III contain commentaries by the Editors.
Volume IV includes a preface by Steven P. Kerckhoff.
Volume I contains William Thurston's papers on foliations, mapping classes groups, and differential geometry.
Graduate students and researchers interested in geometric topology, geometric group theory, low-dimensional topology, and dynamical systems of rational maps.
This item is also available as part of a set:
• Cover
• Title page
• Copyright page
• Contents
• Preface
• Acknowledgments
• Foliations
• Commentary: Foliations
• Foliations of three-manifolds which are circle bunldes
• Anosov flows and the fundamental group
• Noncobordant foliations of 𝑆³
• Some remarks on foliations
• Foliations and groups of diffeomorphisms
• A generalization of the Reeb stability theorem
• The theory of foliations of codimension greater than one
• Foliated bundles, invariant measures and flat manifolds
• The theory of folations of codimension greater than one
• On the existence of contact forms
• A local construction of foliations for three-manifolds
• On the construction and classification of foliations
• Existence of codimension-one foliations
• Polynomial growth in holonomy groups of foliations
• Anosov flows on new three manifolds
• A norm for the homology of 3-manifolds
• Contact structures and foliations on 3-manifolds
• Confoliations
• Three-manifolds, foliations and circles, I. Preliminary version
• Three-manifolds, foliations and circles II. The transverse asymptotic geometry of foliations
• Surfaces and mapping class groups
• Commentary: Surfaces and mapping class groups
• A presentation for the mapping class group of a closed orientable surface
• New proofs of some results on Nielsen
• On the geometry and dynamics of diffeomorphisms of surfaces
• Earthquakes in 2-dimensional hyperbolic geometry
• Minimal stretch maps between hyperbolic surfaces
• Non-continuity of the action of the modular group of Bers’ boundary of Teichmuller space
• Differential geometry
• Commentary: Differential geometry
• Some simple examples of symplectic manifolds
• Characteristic numbers of 3-manifolds
• Transformation groups and natural bundles
• Manifolds with canonical coordinate charts: Some examples
• Pinching constants for hyperbolic manifolds
• Hyperbolic 4-manifolds and conformally flat 3-manifolds
• Shapes of polyhedra and trinagulations of the sphere
• On three-dimensional space groups
• Other titles in this series
• Back Cover
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Algorithms For Automatic And Robust Registration Of 3D Head Scans
CVMP 2008
Algorithms For Automatic And Robust Registration
Of 3D Head Scans
Two methods for registering laser-scans of human heads and transforming them to a new semantically consistent topology defined by a user-provided template mesh are described. Both algorithms are
stated within the Iterative Closest Point framework. The first method is based on finding landmark correspondences by iteratively registering the vicinity of a landmark with a re-weighted error
function. Thin-plate spline interpolation is then used to deform the template mesh and finally the scan is resampled in the topology of the deformed template. The second algorithm employs a morphable
shape model, which can be computed from a database of laser-scans using the first algorithm. It directly optimizes pose and shape of the morphable model. The use of the algorithm with PCA mixture
models, where the shape is split up into regions each described by an individual subspace, is addressed. Mixture models require either blending or regularization strategies, both of which are
described in detail. For both algorithms, strategies for filling in missing geometry for incomplete laser-scans are described. While an interpolation-based approach can be used to fill in small or
smooth regions, the model-driven algorithm is capable of fitting a plausible complete head mesh to arbitrarily small geometry, which is known as "shape completion". The importance of regularization
in the case of extreme shape completion is shown.
Keywords: nonrigid registration, ICP, geometry interpolation, morphable head models, shape completion, 3D face processing
Subjects: 3D-Scanner, Computational Geometry, Head
Laser scanners are a popular tool for acquiring detailed 3D models of human heads. However, the meshes generated by the scanners typically have a topology that reflects the operation of the scanner
and is unsuitable for many applications. The data-set used in this work, for example, has a cylindrical grid topology with vertices of the form (ϕ[i], z[i], r[i]) where ϕ[i] are regularly spaced
angles, z[i] are regularly spaced vertical offsets and r[i] are varying radii. In this paper, we describe two methods for semantically consistent registration and topology transfer of laser-scanned
human heads that do not require manual intervention. The algorithms employ a user-provided template mesh which specifies the target topology. From arbitrary head scans they compute meshes with the
following properties:
● All generated meshes have the same topology, i.e. that of the reference mesh. Different heads only vary in vertex locations.
● The mesh topology is semantically interpretable, i.e. topologically equivalent vertices in different heads have the same "meaning" such as tip of the nose, center of upper lip, etc.
The first algorithm proposed is purely geometric and computes the registration without prior knowledge. The second algorithm involves a semantic shape model of the human head, which is a variant of
the popular morphable head model of Blanz and Vetter [ BV99 ]. While the model-driven algorithm is more robust than the geometric one, it presupposes a database of head meshes that are already
registered in a common, semantically interpretable topology. Therefore, the geometric algorithm can be used to generate the database from which the model is computed. Both algorithms build on an
error minimization approach computed in the well known Iterative Closest Points framework.
We also address the use of the registration algorithms with incomplete data, which is often an issue when laser-scanners or other surface reconstruction methods are used. For the geometric algorithm,
we propose an interpolation method for filling in deficient areas that are either small or smooth. The model-driven algorithm can be used to fill in even huge missing parts with complicated geometry
in a plausible fashion. In fact, with proper regularization, it is capable of performing "shape completion", i.e. computing plausible geometry that matches arbitrarily small pieces of given data.
The paper is structured as follows. In section 2 we give an overview of related work in the areas of ICP, morphable shape models and registration of head meshes. In section 3, we briefly recapitulate
the algebra of first order ICP and introduce some notation. The geometric registration approach is described in section 4 and the model-driven scheme is treated in section 5.
Figure 1. The reference mesh used for topology conversion with landmarks for the first algorithm stage.
Both algorithms use the Iterative Closest Points optimization scheme which was first introduced by Besl and MacKay [ BM92 ]. ICP is typically used to compute a rigid transform that aligns two or more
partially overlapping meshes e.g. laser-scans of a larger scene such that the overlapping parts match as good as possible according to some error function. Common error functions are distances of
closest points or point-to-plane distances (e.g. Chen and Medioni [ CM92 ]). The optimal transform depends on matching the right points in both meshes; to match the right points, however, the optimal
transform must be known. Therefore, ICP assumes that the closest points match; the transform induced by these correspondences is computed and applied and the procedure is iterated. For a rigid
transformation, ICP is guaranteed to converge to a minimum of the error function [ BM92 ]. However, this minimum may be local and thereby the meshes must be roughly aligned before ICP is invoked to
find a good match.
There is a huge body of literature on ICP and numerous variants and optimizations have been proposed, aiming mostly at improving stability and speed of convergence. Rusinkiewicz and Levoy [ RL01 ]
give an overview and, more recently, Mitra et al. [ MGPG04 ] introduced a general numeric optimization framework based on second order approximants which subsumes several previous approaches.
The ICP algorithm is neither limited to matching precisely overlapping meshes nor to the estimation of rigid transformations. In fact, any kind of transformation that is uniquely determined by a set
of corresponding points can be estimated with ICP. However, as is the case for many optimization algorithms, the degrees of freedom of the transformation correlate with convergence: The more degrees
of freedom there are, the more likely ICP is to converge to an undesired local minimum. Particularly relevant in the context of this work are non-rigid variants of ICP. These algorithms face the
problem deforming the template in order to minimize the registration error while keeping its overall appearance. To this end, deformation is often controlled by a model that mimics the behavior of an
elastic material. Examples of non-rigid ICP schemes are Haehnel et al. [ HTB03 ] who combine ICP optimization with a coarse to fine approach, or Amberg et al. [ ARV07 ] who minimize a nonrigid
matching error function similar to that of Allen et al. ([ ACP03 ], see section 2.3) in a stepwise optimal fashion. Our model-driven algorithm (section 5) can be seen as a nonrigid ICP scheme with a
data-driven deformation model; it was first published in [ SE09 ] where we compare it to other approaches and address its computational complexity.
"Morphable models" are orthogonal subspace models of 3D shape (and sometimes texture) of a certain object domain. In order to compute a morphable shape model, a database of meshes sharing the same
semantically consistent topology is required. This allows to treat the geometry of a mesh with N vertices as a single measurement in a 3N-dimensional data-space. The morphable model is obtained by
performing Principal Component Analysis (PCA) on the database in this space. By omitting eigenvectors with low eigenvalues a low-dimensional subspace covering the most dominant variations in the
dataset is obtained. For complex shapes, a single linear subspace typically cannot cover sufficient variability. Therefore, the geometry is often split up into regions and each region is described by
its own subspace model. Typically, regions overlap at their borders in order to obtain a smooth overall mesh in the end. Thereby, the morphable model becomes a PCA mixture model.
Morphable models have been used most successfully in the domain of the human head and face, where they were introduced as "morphable head models" by Blanz and Vetter [ BV99 ]. A wide number of
applications in vision and graphics of faces can be driven or supported by morphable models, including, for example, face recognition [ AKV08, BV03 ], face tracking [ PF03, Rom05 ] and several
varieties of 3D reconstruction, e.g. from uncalibrated video [ Fua00 ] or from stereo images [ ABF07 ]. While examples besides faces are rare, a morphable model of teeth is described in Blanz et al.
[ BMVS04 ].
Head and face registration algorithms can roughly be divided into methods that exploit texture information and purely geometric approaches. On the side of geometric methods, some specialize
exclusively on heads and faces while others try to solve the general problem of registering arbitrary surfaces by nonrigid transformations. The algorithms proposed in this paper belong to the class
of head-specific purely geometric approaches. Several algorithms were introduced by the morphable model community since a database of registered meshes is required to compute a morphable model.
On the side of texture-based methods, Blanz and Vetter [ BV99, VB98 ] use an algorithm based on optical flow. Traditionally used to estimate 2D image deformation, optical flow is extended to use 2D
texture as well as local 3D surface properties of the face in order to compute a matching deformation. However, the authors themselves [ BV99 ] as well as Paterson and Fitzgibbon [ PF03 ] report the
method to be unreliable on "exotic" faces. Thus Paterson and Fitzgibbon [ PF03 ] manually annotate 30 landmark points in the texture and use radial basis function interpolation to compute a matching
warp. 3D points are found by inverse-warping the texture coordinates associated with the 3D points.
Purely geometric head registration is used by Kähler et al. [ KHYS02 ] to animate facial expressions of laser-scanned heads with an anatomical animation model. Their method is based on iteratively
subdividing a rough face mesh and aligning the new vertices to a reference model. The initial model for the subdivision is obtained from manually placed landmarks. Allen et al. [ ACP03 ] register
full human body scans by using a generic nonlinear optimizer on an error function penalizing distances of closest points and dissimilarities between transforms of nearby points, thereby controlling
the rigidity of the overall transformation. The approach requires some manually annotated landmarks to improve convergence. No landmarks are required by Anguelov et al. [ ASP05a ], who formulate the
registration problem in the framework of a Markov random field. A variety of local and relational cost measures are incorporated in the field and the registration is computed by loopy belief
Geometric methods were also developed in the field of database retrieval and 3D face recognition. They typically aim at retrieving a stored prototype face from a database which is used to determine
the identity of the scanned subject. To this end, Haar and Veltkamp [ tHV08 ] use curves originating from the nose in all directions. Nair and Cavallaro [ NC08 ] use ICP on incomplete faces in order
to achieve invariance to facial expressions. These techniques, however, are designed to retrieve a model of the same person rather than registering multiple faces in a semantically valid way.
Issues of filling in incomplete data, which are addressed for both algorithms described in this work, are treated in several papers. Knothe et al. [ KRV06 ] use Local Feature Analysis, a localized
variant of PCA, to fit a model to highly sparse data. Blanz et al. [ BMVS04 ] have a probabilistic approach for dealing with sparse fitting targets, where we propose a geometric regularization
(sections 5.2 and 5.3). Anguelov et al. [ ASP05b ] perform shape completion on full body scans in arbitrary pose. Note that in this work, the issue of filling in smaller holes is dealt with in the
context of the geometric algorithm, while shape completion and fitting to extremely sparse data is treated in the context of the model-driven approach.
Both algorithms treated in this paper are best described on the basis of the linearized ICP algorithm. In the following, we briefly recapitulate the algebra of this approach and introduce some
Let rot(θ) denote a rotation matrix for a vector θ of three Euler angles. Further, be t ∈ ^3 a translation vector and s an anisotropic scale factor. Be p[i] ∈ ^3 a vertex of the template mesh and q
[i] ∈ ^3 its corresponding point in the target point cloud. Putting point-to-plane distances and other optimizations aside, the ICP algorithm for estimating rigid transformation and anisotropic scale
minimizes the following quadratic error function while simultaneously establishing the correspondences between template meshand target point cloud:
Here, N is the number of vertices in the template mesh.
A straightforward optimization strategy for ICP is to linearize this error functional around the parameter estimate (s,θ,t) of the last iteration and solve a linear system to find a small change in
the parameters that decreases the error. Formally, this amounts to linearizing
in (∆s, ∆θ , ∆t). To this end, rotation can be approximated as
. where [p][x] is the cross product matrix
For a formal derivation, see, for example, [ WI95 ]. Expanding equation (2) and omitting higher order terms yields, after rearranging, the following linear system:
For meshes of practical interest, this system is overdetermined and can be solved in a least squares sense, for example by SVD. The parameters are updated as follows:
Note that rotation is maintained as a matrix and updated by multiplication rather than by adding up Euler angles in order to avoid gimbal lock. Finally, the updated transformation is applied and new
point correspondences are established for the next iteration.
Since laser-scans are often incomplete, some vertices of the reference mesh may be assigned to overly far points in the cloud. These vertices have a large effect on the optimization due to the use of
squared errors. Therefore, they should be treated as outliers by introducing a threshold on the distance between a template vertex and its target correspondence. Equations corresponding with outliers
can be eliminated from the linear system in equation (5) by multiplying both sides with a diagonal binary matrix.
Our geometric registration algorithm comprises three steps: First, landmark points are found on the laser-scan mesh using the ICP scheme. Second, a nonlinear transform is computed to match the
reference with the scan, using the landmark correspondences from step one as anchor points. Finally, the scan is resampled in the topology of the reference mesh and holes can be filled by
In order to localize a landmark, the template mesh is repeatedly registered with the target point cloud. In every iteration, the ICP error function is re-weighted in order to concentrate the
registration process on an increasingly smaller area around the landmark. The template mesh and the landmarks we use are shown in figure 1. The proposed localization strategy requires landmarks to
have a non-flat, geometrically distinctive vicinity. Currently, landmark locations in the template are determined manually. However, slippage analysis introduced by Gelfand and Guibas [ GG04 ] could
be an interesting approach to automatic search for suitable landmark locations.
For re-weighting, the ICP error of equation (1) is extended by a vertex specific factor w[i], i = 1 ⋯ N which yields
. This propagates to the linear system of equation (5) as a multiplication of the matrix with a diagonal weight matrix
The role of the weights is to reduce the influence of the vertices distant from the landmark in an iterative manner. Generally, the weighting scheme should be a function ofthe form
where p[i] is the vertex the weight of which is to be determined, L is the landmark that is currently processed, and α is a parameter that decreases with the iterations of the algorithm, controlling
the size of the neighborhood that is to be considered in the ICP error term. Also, the weighting scheme must satisfy the constraints
Hence, by setting α = ∞ in the first iteration, unweighted ICP is performed.
There are numerous possibilities to implement this strategy. The most simple weighting scheme probably is the binary function
where dist(p[i],L) is a measure of distance from a vertex P[i] to L. More complex weighting schemes replace the step function by a smooth falloff in L. Regarding the distance measure, there are again
multiple possibilities. In our implementation, we use approximate geodesic distances computed with the fast marching algorithm of [ KS98 ]. Note that the distance measures can be precomputed and
stored as long as the reference mesh stays constant.
In summary, the algorithm for finding all landmarks is the following:
An illustration of the weighting function in the inner loop is shown in figure 3 for the upper lip landmark and the function of equation (14). Examples of automatically placed landmarks in different
head scans are depicted in figure 2.
Depending on an individual head's geometry and on the quality of the laser-scan, the algorithm may converge to a bad localization of a landmark. Problem areas are the ears, where a good initial
position of the template is required for good localization results, as well as the mouth where the geometry is not very pronounced for some individuals. A typical failure case in several landmarks is
illustrated in figure 4. The algorithm also requires the scanned subjects to have the same facial expression as the template mesh. This is easiest to realize if both, subjects and template, display a
neutral expression. Landmark localization quality is hard to evaluate numerically on our database since the low resolution of the laser-scans and lack of textures make a precise localization
difficult for humans as well. However, the quality of the results is sufficient to build a morphable model that is capable of reproducing the facial characteristics of a wide range of individuals.
Figure 3. Landmark registration with an increasingly smaller area of influence for the error function: Example sequence of the influence area for the upper lip landmark with a binary weighting
Figure 4. Failure case of landmark registration: Ear, mouth and missing parts in the scan may lead to misregistration.
In the second step of the geometric algorithm, the template mesh is matched with the laser-scan point cloud in a global, nonlinear fashion using the established landmark correspondences. Denoting the
landmarks in the template mesh by p[i], i = 1 ⋯ K and their point cloud correspondences by q[i] , we seek a transformation Φ(⋅) such that Φ(p[i]) for all i and such that the points between the
landmarks are interpolated in a natural fashion. This is nicely realized by the Thin Plate Spline formalism [ Boo89 ], which yields a transformation minimizing a physically interpretable bending
energy. Similar non-linear deformation models have been used successfully in face processing before, e.g. in [ BKC08, BBA07 ].
From a Thin Plate Spline perspective we solve a three dimensional scattered data interpolation problem from ^3 to ^3 . The known values are the displacement vectors from the laser-scan landmark
points to the reference mesh landmarks,
The unknowns to be interpolated are the displacement vectors of all non-landmark points in the reference mesh. Dealing with a three dimensional problem, the radial basis function to use for the
spline is u(x) = |x| according to [ Boo89 ] and thus, with u[i,j] = || p[i] - p[j]||, we get a Thin Plate Spline matrix
Hence the weight matrix for the mesh transform is
and an arbitrary point v in the reference mesh transforms to
with u[i] = || v - p[i] ||.
Figure 5. Results of the geometric registration algorithm. Top row: Original laser-scans. Middle row: Result after registration and topology transfer. Bottom row: Final result after hole-filling by
offset interpolation.
The final step is to resample the laser-scan in the topology of the deformed reference mesh which now closely matches the scan. Therefore, for each vertex in the template mesh, its normal is computed
and all intersections of the laser-scan with a line through the vertex in normal direction are determined. If there are multiple intersections, the one closest to the vertex is used. Moreover, there
is a threshold on the distance between the vertex and the scan in order to exclude deficient matches at the scan's border. The chosen intersection point is taken as the correspondence to the template
mesh vertex.
To speed up the computation of the ray-mesh-intersection, the laser-scan mesh is represented as an axis aligned bounding box tree and the fast ray box intersection of Williams et al. [ WBMS05 ] is
used. Note that due to the size of a typical scan it is not feasible to build the tree down to the level of individual triangles. In our implementation with around 80 000 points in a scan, there are
100 triangles per leaf that have to be tested for each ray.
Due to scanning errors, the resampled scans have holes where no valid correspondence along the normal was found. The interpolation scheme described in section 4.2 can be used to fill these holes as
long as the missing geometry is smooth and not too complex; a model-based approach capable of filling in more complex geometry is described in section 5.3. Again, we can formulate the task as a
scattered data interpolation problem from ^3 to ^3 . Denote by h[i] the vertices in the template mesh that lack correspondence in the scan (i.e. the vertices that make up the "hole"). Denote by p[j]
vertices "surrounding the hole" that do have correspondences q[j] in the scan. The known values are the displacements at the surrounding vectors d[j] = p[j] - q[j] . To be interpolated are the
displacements h[i] is set to h[i]+
Final results of the registration algorithm before and after hole filling are shown in figure 5.
The model-driven algorithm is based on a morphable model of head shape, which was computed as described in section 2.2. The database required to build the model was generated with the geometric
registration algorithm reported above. For more details on the model-driven algorithm, including comparisons to other registration methods and issues of time complexity, see [ SE09 ].
In the following, be M the morphable shape model matrix and μ the model's mean vector. Hence the geometry of every mesh that can be represented by the model can be written as
is a single vector of all vertex coordinates and the parameter vector m is the location of the mesh in the model's "shape space". Note that our morphable model is a PCA mixture model comprising of
six subspace models for mutually overlapping regions. Therefore, the model matrix is not simply a matrix of principal components. Details on the construction of M for the mixture model are given in
section 5.1.
To use the morphable model for laser-scan registration, we again start from the ICP error functional of equation (1) and substitute equation (19) for the template mesh vertices p[i] :
Note that in the above equation M[i] and μ[i] refer to three-row blocks of M and μ that correspond with individual mesh vertices. Additional to the scale and rigid transform parameters, the error now
also depends on the shape parameter vector m. The addition of the shape model propagates to the matrix of the linear system of equation (5) as follows:
See [ SE09 ] for a more detailed derivation. In the ICP loop, the system is now also solved for a shape change ∆m. The shape is initialized by m ← 0 which boils down to the mean shape μ of the model
according to equation (19).
Once the model's shape and pose are fitted to the laser-scan, there are two ways to proceed: Either, the scan can be resampled by establishing correspondences along the normals of the template
vertices, as in the geometric algorithm (section 4.3). Alternatively, the fitted morphable model may itself be used as the registered version of the laser scan. In this case, the results are, of
course, confined to the space of shapes the model can represent. Which strategy is best depends on the application: Resampling the scan is computationally more expensive and gives noisier results but
it preserves idiosyncrasies of the scanned individual that may not be captured by the model.
Note that due to the shape parameters, there are significantly more degrees of freedom in the error function minimized by the model-driven method than in that minimized by rigid ICP. Therefore, the
result of the model-driven algorithm is more sensitive to the initial alignment of template and target. Best results are achieved when template and target are pre-aligned with respect to pose and
anisotropic scale by regular ICP.
For a simple morphable model with a single subspace, the model matrix M is simply the eigenvector matrix yielded by the PCA. If a PCA mixture model with multiple subspaces is used, M has to
constructed from the region's individual eigenvector matrices. See figure 6 for an illustration of the regions used in this work.
Figure 6. Partition of the head into regions for the PCA mixture model. Each part is described by its own subspace model.
Figure 7. Results of the model-driven registration algorithm. Original laser-scans in the bottom row, fitted morphable model in the top row. Six region models and 15 principal components per region
were used.
Assume the number of regions and hence the number of subspaces used is K and denote each region's eigenvector matrix by M^(j),j = 1 ⋯ K. Further be [1]⋯ [K] sets of vertex indices such that i ∈ [j]
if vertex p[i] belongs to region [j] .
If the regions do not overlap, i.e. [i] ∩ [j] = ∅ for i ≠ j, the matrix M has a block-diagonal structure:
The results from this type of model, however, show harsh discontinuities between the regions since each region is optimized independently from the others. Therefore, the region models have to be
coupled by introducing overlap at the region borders, i.e. vertices that belong to several regions. In consequence, the matrix M is no longer simply block diagonal.
If a vertex p[i] belongs to region j its location is determined by three consecutive rows of the region's eigenvector matrix M^(j) , multiplied with the shape parameters for that region. Denote these
three rows of M^(j) by M[i] ^(j) and define
where 0 is an all-zero matrix of appropriate size. Then, M can be defined block-wise as
where α[i] ^(j) are block-weights satisfying
Note that for a vertex p[i] that belongs just to one region, only one block M[i] ^(j) is non-zero. As regions are typically defined to overlap only at their borders, the majority of vertices will be
of this kind and M is fairly sparse.
The role of the block weights α[i] ^(j) is to normalize the contributions of different region models for vertices that belong to more than one region. The most simple block weight is α[i] ^(j) R[i]
is the number of regions the vertex belongs to. If the area of overlap is deeper than just one shared border vertex, more sophisticated weights may be considered, giving more influence to some
regions than to others. In the following we give an example of a weight that depends only on the mesh topology.
Be a border edge an edge with only one incident triangle and a border vertex a vertex that is incident to at least one border edge. Denote by [j] the set of border vertices of region [j] and by [j]
the index set of neighbors of the vertex with index i. Now consider an arbitrary vertex and define its "centricity" c[i] ^(j) with respect to region [j] recursively as
Now the blending weight α[i] ^(j) can be defined as
Equation (28) is equal to one for vertices that appear in only one part. For vertices appearing in multiple parts the weight is higher with respect to parts where the vertex is further away from the
part border. The sum of blending weights of one vertex is always one, thus satisfying the constraint in equation (26).
It is generally desirable to introduce as little overlap between the regions as possible: Large overlap increases the size of the linear system that has to be solved in every ICP iteration. Also,
from a modeling point of view, every sub-model should ideally describe it's own part of the domain (here: the face) and interfere with other parts as little as possible. The downside to small overlap
is the risk of discontinuities as shown in figure 8: The smaller the overlap, the more independent is the optimization of every region. As the optimization minimizes a sum of squared errors over all
vertices in the region, two regions may yield significantly different locations for their shared vertices.
Figure 8. Effect of geometric regularization on a model with an overlap depth of one vertex. Left: A crease under the nose is clearly visible. Right: The same model fitted with geometric
This problem can be addressed even for the smallest possible overlap of one shared line of vertices along the region border by introducing a Tikhonov regularization term to the parameter
estimation system. The aim of the regularization is to force shared vertices into the same position. Taking into account the block structure of M as defined in equation (25), we can write the
location p[i] of a vertex given shape and pose parameters (s,θ,t,m) as
Note that m^(j) is the part of the shape vector that determines the shape of region [j]. Now assume that p[i] is shared by exactly two regions [S] and [T] (i.e. it is a vertex at the border between
two regions). Therefore, α[i] ^(j) = 0 for all values of j except for {S,T} and the summation in the above equation reduces to α[i] ^(S)M[i] ^(S)m^(S)+α[i] ^(T)M[i] ^(T)m^(T) . Clearly, the two
regions yield the same location for the vertex, if
This term can be expressed in matrix form and added to the linear system for every vertex that is a border vertex between two regions. For vertices shared by more than two regions, each unordered
pair of regions yields one regularization term. In practice, these vertices can be ignored without harm.
In section 4.3 we sketched an interpolation algorithm that can be used to fill in holes in the laser-scans after registration with the geometric algorithm. The data-driven methodology, too, lends
itself to a hole-filling in fact the use of a model allows much more radical " filling in", up to the point of generating a plausible head that closely matches a very small piece of given geometry.
This is sometimes referred to as shape completion in the literature.
Figure 9. Shape completion with the model-driven fitting algorithm. The fitting target is only a small piece of geometry (left). A plausible complete head (right) is fitted to the given data. Note
that there is no given data for the ear region.
Figure 10. In this shape completion example, the target is only the red profile curve, which was extracted from a photo. The fitted head closely matches the target profile (right).
In order to cope with holes in the scan, outlier handling should be added to the optimization as described in section 3. Once the model is fitted to the scan, no additional filling or interpolation
step is necessary: The shapes generated by the model are by construction always complete. Algebraically, this follows from the fact that the system of equation (22) loses rows when outliers occur
but not columns. Therefore and regardless of the outliers, a complete shape parameter vector m is always computed. A full mesh is simply obtained by multiplying m with the the original shape matrix
that contains all rows.
There is a caveat, however, if huge parts of the target geometry are missing, which is typical for the shape completion scenario. This can be seen by looking more closely at the effects of outlier
removal on the matrix M. When a vertex with index i is classified as outlier due to the lack of a nearby target point, a three row block
corresponding to the outlier vertex is removed from M. This is uncritical as long as there are sufficiently many vertices left in the regions to which the removed vertex belongs. What happens,
however, if all information in a region j is lost? It can be seen from the construction of M in equation (25), that in this case all M[i] ^(j) for which α[i] ^(j) ≠ 0 are lost and M contains a number
of zero-columns. In consequence, the elements of m that determine the shape of the "unknown" region are undetermined. If, however, geometric regularization is used, there is still sufficient
information in the Tikhonov matrix to determine the shape parameters of region [j] . Geometrically, this is the information that whatever the geometry of the region is, it must match the geometry of
its neighboring region in the shared vertices. Note that thisinformation also helps to prevent overfitting in the case of very few inlier points. Figures 9 and 10 show some extreme examples of shape
completion. Note that in both examples, the ears are completely undetermined by the given data.
We described two robust methods for registering laser-scans of human heads and transforming them to a new semantically consistent topology defined by a user-provided template mesh. The first
algorithm is based on finding landmark correspondences by iteratively registering the vicinity of a landmark with a re-weighted error function. Thin-plate spline interpolation is then used to deform
the template mesh and finally the scan is resampled in the topology of the deformed template. The second algorithm directly optimizes pose and shape of a morphable shape model which can be computed
using the results of the first algorithm. We outlined blending and regularization strategies for coping with PCA mixture models where the shape is split up into regions each described by an
individual subspace. For both algorithms, we addressed strategies for filling in missing geometry for incomplete laser-scans. While the interpolation-based approach can fill in small or smooth
regions, the model-driven strategy is capable of fitting a plausible complete head mesh to arbitrarily small geometry. We addressed the importance of regularization in the case of extreme shape
Regarding the geometric algorithm, future research will concentrate on different re-weighting schemes in order to improve precision of automatic landmark placement, especially for difficult cases
such as the ears and the corners of the mouth. For the model-driven method, an interesting open question is the choice of regions in the PCA mixture model. Currently, the regions are defined manually
and based on an intuitive segmentation of the head into nose, mouth, eyes, etc. It would be interesting to investigate data-driven strategies that yield a segmentation as the result of an
optimization process. The incorporation of classical ICP optimizations such as point-to-plane distances into both algorithms is another interesting direction of research.
License ¶
Any party may pass on this Work by electronic means and make it available for download under the terms and conditions of the Digital Peer Publishing License. The text of the license may be accessed
and retrieved at http://www.dipp.nrw.de/lizenzen/dppl/dppl/DPPL_v2_en_06-2004.html.
Recommended citation ¶
David C. Schneider, and Peter Eisert, Algorithms For Automatic And Robust Registration Of 3D Head Scans. JVRB - Journal of Virtual Reality and Broadcasting, 7(2010), no. 7. (urn:nbn:de:0009-6-26626)
Please provide the exact URL and date of your last visit when citing this article. | {"url":"https://www.jvrb.org/past-issues/7.2010/2662/view?set_language=en","timestamp":"2024-11-02T20:35:59Z","content_type":"application/xhtml+xml","content_length":"176788","record_id":"<urn:uuid:b4fae670-2923-4716-ba04-3a7c33c8ef4d>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00731.warc.gz"} |
What is the formula for P A and B?
Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). If the probability of one event doesn’t affect the other, you have an independent event.
Is P a B the same as P B A?
P(A|B) is the probability of A, given that B has already occurred. This is not the same as P(A)P(B.
What is PA intersection B?
Two events are mutually exclusive or disjoint if they cannot occur at the same time. The probability of the intersection of Events A and B is denoted by P(A ∩ B). If Events A and B are mutually
exclusive, P(A ∩ B) = 0. The probability that Events A or B occur is the probability of the union of A and B.
What is the formula of dependent events?
There are two types of events in probability which are often classified as dependent or independent events….Difference Between Independent and Dependent Events.
Dependent Events Independent Events
3. Formula can be written as: P(A and B) = P(A) × P(B | A) 3. Formula can be written as: P(A and B) = P(A) × P(B)
What is the value of P(A) and P(B) in the equation?
P (A) = 0.20, P (B) = 0.70, A and B are independent. The 0.14 is because the probability of A and B is the probability of A times the probability of B or 0.20 * 0.70 = 0.14.
What is the formula for conditional probability?
P(B|A) P(A and B) = P(A)P(B|A) or P(B|A) = P( A and B) P(A) Brent Clayton Homework #0 Summary of class notes 2/3/2011 The lecture notes for this particular day a centered on sections 2.4-2.5 in the
Devore book. Basically the idea of conditional probability is presented here.
What is the meaning of PP(B|a)?
P(B|A) This only applies when the events are independent of each other meaning event A has no effect on the probability of event B happening. The other case involes these two events when they are
What is the probability of A and B times the probability?
The 0.14 is because the probability of A and B is the probability of A times the probability of B or 0.20 * 0.70 = 0.14. | {"url":"https://profoundadvices.com/what-is-the-formula-for-p-a-and-b/","timestamp":"2024-11-03T03:24:45Z","content_type":"text/html","content_length":"53692","record_id":"<urn:uuid:c2772a4f-f381-48c5-86c6-5858dee46f7c>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00123.warc.gz"} |
Menlo Systems
A Terahertz imaging camera to aid landing helicopters
A group of researchers from the PSL University and Safran Electronics & Defense in France teamed up to investigate the potential of THz imaging through optically dense sand clouds using Menlo
Systems’ Terahertz time-domain spectrometer
During helicopter landing sand stirred up by strong air turbulence decreases visibility and thus presents a threat to the crew members. An imaging system operating in the THz range provides
sufficient resolution and penetration depth through such clouds. A team of researchers from the Institut Langevin at the PSL University and from Safran Electronics & Defense, both in France, used
Menlo Systems’ TERA K15 THz Time Domain Spectroscopy (TDS) system in their experimental setup to quantify the attenuation of THz waves under simulated brownout conditions. A simultaneous transmission
measurement using a laser diode at 650 nm allows direct comparison to the visible spectral range and serves as a means to estimate the scattering strength and density of the air suspended particles.
Compared to the size of sand grains, the wavelength of THz radiation is very long. Therefore THz waves are only weakly scattered and penetrate through a sand cloud nearly without disturbance. On the
other hand, the THz wavelength is short enough to provide sufficient resolution in an imaging system for the visualization of larger objects. For the assessment of such an active imaging system which
could aid helicopters during landing, its signal-to-noise-ratio (SNR) needs to be estimated. The calculation requires the input of the scattering properties of sand, which themselves depend on the
refractive index and on the size of the grains. Existing databases still lack of reliable information on the refractive index of sand in the THz range since they are based on measurements which do
not entirely account for the conditions in a realistic scenario, or where the parameters are not sufficiently specified. For a more accurate determination of the scattering properties, the team
proposed a method which is based on the measurement of the beam extinction through a suspension of particles in air. By simultaneously measuring in the THz and the visible (VIS) range, the visibility
can be directly compared, and the particles’ scattering strength and density is retrieved from the measurement in the VIS range. Together with the particle size distribution, the real part of the
refractive index and the particle density can be deduced.
In order to mimic the brownout conditions for the experiment as close as possible to reality, and to avoid disturbing effects such as coupling to a carrier object or position correlations of the
particles, the researchers chose a volume scattering sample of particles suspended in air. A fan fixed to the top of the chamber simulates the helicopter rotors stirring up particles of known grain
size distribution, the chamber being sealed by a thin layer of cellophane tape which has been previously characterized for THz attenuation. The chamber contained either sand or glass beads to compare
varying particle composition.
The group used Menlo Systems’ TERA K15 Terahertz-TDS system for the determination of the sample THz parameters and a 650 nm laser diode with for the visible domain, both propagating at a small angle
through the sample. For the THz field attenuation they performed the ratio of measurements in a suspension and in a reference without particles (fan switched off). Using a diaphragm before the
photodiode, the portion of forward scattered light at 650 nm was eliminated.
To characterize the optical properties of the scattering particles in the THz range, the absorption was neglected due to its minor contribution to the extinction. The group computed the THz
extinction due to brownout using the Beer-Lambert law, taking the ratio of transmitted intensities with and without particle suspension. From a Mie theory model they derived the refractive index of
the particles and their density. They monitored the particle density over the time by the measurement in the VIS range. The refractive indices were found to be 1.67 for sand and 2,54 for glass beads.
The measured value for sand containing a mixture of various molecules slightly deviates from the literature value of 1.96 for pure silica, while for glass beads is consistent with the refractive
index of bulk glasses. Due to less than 5 % variation the refractive index assumed to be constant over the measured THz range, what was confirmed by comparing experimentally determined and
theoretically calculated extinction efficiencies. For higher frequencies around 2 THz the Mie model does not perfectly match the experimental results. Further, since the refractive index highly
depends on the sand composition, it was suggested to characterize other types of sand for an evaluation closer to real conditions.
Author: Patrizia Krok
Original publication:
C. Prophète et al.: Terahertz and Visible Probing of Particles Suspended in Air; IEEE Transactions on Terahertz Science and Technology Vol. 9, p. 120 (2019)
DOI: https://doi.org/10.1109/TTHZ.2019.2891077 | {"url":"https://www.menlosystems.com/jp/events/application-news/view/2770","timestamp":"2024-11-02T08:33:37Z","content_type":"text/html","content_length":"32403","record_id":"<urn:uuid:79826328-81c9-4711-8a4f-0c60af214c73>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00180.warc.gz"} |
ITHEA International Scientific Society
: MATRIX POWER S-BOX ANALYSIS1
Abstract: Construction of symmetric cipher S-box based on matrix power function and dependant on key is analyzed. The matrix consisting of plain data bit strings is combined with three round key
matrices using arithmetical addition and exponent operations. The matrix power means the matrix powered by other matrix. This operation is linked with two sound one-way functions: the discrete
logarithm problem and decomposition problem. The latter is used in the infinite non-commutative group based public key cryptosystems. The mathematical description of proposed S-box in its nature
possesses a good “confusion and diffusion” properties and contains variables “of a complex type” as was formulated by Shannon. Core properties of matrix power operation are formulated and proven.
Some preliminary cryptographic characteristics of constructed S-box are calculated.
Keywords: Matrix power, symmetric encryption, S-box.
ACM Classification Keywords: E.3 Data Encryption, F.2.1 Numerical Algorithms and Problems.
Kestutis Luksys, Petras Nefas | {"url":"http://idr.ithea.org/tiki-read_article.php?articleId=441","timestamp":"2024-11-03T22:15:55Z","content_type":"text/html","content_length":"33824","record_id":"<urn:uuid:0a0bea4c-2cc8-476f-ac49-134b7bd34d94>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00335.warc.gz"} |
Variants of Chow's Lemma
Lemma 76.41.1. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a separated morphism of finite type. Then there exists a commutative
\[ \xymatrix{ X \ar[rd] & X' \ar[l] \ar[d] \ar[r] & \overline{X}' \ar[ld] \\ & Y } \]
where $X' \to X$ is proper surjective, $X' \to \overline{X}'$ is an open immersion, and $\overline{X}' \to Y$ is proper and representable morphism of algebraic spaces.
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Quantum Communication in Random Networks
Theorists at MPQ find surprising behaviours in quantum random networks.
Internet, networks of connections between Hollywood actors, etc, are examples of complex networks, whose properties have been intensively studied in recent times. The small-world property (that
everyone has a few-step connection to celebrities), for instance, is a prominent result derived in this field. A group of scientists around Professor Cirac, Director at the Max Planck Institute of
Quantum Optics (Garching near Munich) and Leader of the Theory Division, has now introduced complex networks in the microscopic, so called, quantum regime (Nature Physics, Advanced Online
Publication, DOI:10.1038/NPHYS1665). The scientists have proven that these quantum complex networks have surprising properties: even in a very weakly connected quantum network, performing some
measurements and other simple quantum operations allows to generate arbitrary graphs of connections that are otherwise impossible in their classical counterparts.
The behaviour of networks has been widely explored in the context of classical statistical mechanics. Periodic networks, by definition, have a regular structure, in which each node is connected to a
constant number of ‘geometrical’ neighbours. If one tries to enlarge these systems, their topology is not altered since the unit cell is just repeated ad aeternum. The construction of a random
network is completely different: each node has a small probability of being connected to any other node. Depending on the connection probability and in the limit of infinite size, such networks
exhibit some typical effects. For instance, if this probability is high enough, nearly all nodes will be part of one giant cluster; if it is too small only sparse groups of connected nodes will be
In a quantum network one link between neighbouring nodes is given by one pair of entangled qubits, for example atoms; in other words, one link in a quantum network represents the entanglement between
two qubits. Therefore, a node possesses exactly one qubit for each neighbour, and since it can act on these qubits it is called a ‘station’. This holds for any kind of quantum networks. However,
there are different ways of defining the entanglement between neighbouring qubits. Until now, quantum networks have been mostly modelled as periodically structured graphs, that is, lattices. In the
work described here the scientists set the amount of entanglement between two nodes to be equal to the connection probability of the classical random graphs.
In the classical case, some specific subgraphs appear suddenly if one lets the connection probability scale with the size of the network: for very low probabilities only trivial connections (simple
links) are present in the network, whereas for higher probabilities the subgraphs become more and more complex (e.g., triangles, squares, or stars). In quantum networks, on the other hand, a
qualitatively different behaviour emerges: even for the lowest non-trivial connection probability, i.e., if the entanglement between the nodes is, at first sight, just sufficient to get simple
connections, it is in fact possible to generate communication subgraphs of any complexity. This result mainly relies on the superposition principle and on the ability to coherently manipulate the
qubits at the stations.
“In our article we want to point out that networks with a disordered structure and not periodic lattices have to be studied in the context of quantum communication”, says Sébastien Perseguers, who
has worked on this topic in the frame of his doctoral thesis. “In fact, it is well known that real-world communication networks have a complex topology, and we may predict that this will also be the
case for quantum networks. Furthermore, we want to emphasize the fact that the best results are obtained if one ‘thinks quantumly’ not only at the connection scale, but also from a global network
perspective. In this respect, it is essential to deepen our knowledge of multipartite entanglement, that is, entanglement shared between more than two particles.” In the future the scientists are
going to extend their model to networks of a richer structure, the so-called complex networks which describe a wide variety of systems in nature and society, and they expect to find many new and
unexpected phenomena. Sébastien Perseguers/Olivia Meyer-Streng
Prof. Dr. J. Ignacio Cirac
Honorary Professor, Technische Universität München
Max Planck Institute of Quantum Optics
Hans-Kopfermann-Straße 1, 85748 Garching
Phone: +49 (0)89 32905 -705/736 / Fax: -336
E-mail: ignacio.cirac@mpq.mpg.de
Sébastien Perseguers
Max Planck Institute of Quantum Optics
Phone: +49 (0)89 32905 -345 / Fax: -336
E-mail: sebastien.perseguers@mpq.mpg.de
Dr. Olivia Meyer-Streng
Press & Public Relations
Max Planck Institute of Quantum Optics
Phone: +49 (0)89 32905 -213
E-mail: olivia.meyer-streng@mpq.mpg.de | {"url":"https://www.mpq.mpg.de/4857772/10_05_21","timestamp":"2024-11-09T14:09:58Z","content_type":"text/html","content_length":"184817","record_id":"<urn:uuid:a85764fe-fee6-469a-bcb8-ac4909f1fb72>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00609.warc.gz"} |
Thickness Gauging
The most simple application for EMATs is thickness gauging and they have been commercially used in such areas as steel boiler pipe and petrochemical pipe inspection (for more detailed information on
these areas see EMATs section).
We have developed a non-contact technique capable of measuring the thickness of moving or hot aluminium sheet that is more accurate than the measurement obtained using a digital micrometer in a
static test.
Aluminium sheet is used in a wide range of devices and products from sports cars to beverage cans. Accurate control of the thickness is not only of great manufacturing cost benefit but will also lead
to improved product safety, reduced wastage and increased manufacturing efficiency throughout the supply chain.
The technique that we have developed is an ultrasonic thickness gauging approach that employs a non-contact device called an Electromagnetic Acoustic Transducer (see EMAT section). The EMAT is used
to generate and receive broadband (very ‘sharp’ signals in the time domain) ultrasonic waves in metals and is particularly efficient when operating on aluminium. In order to achieve these results we
have used a combination of highly specialised hardware coupled together with mathematical analysis of the waveform using an algorithm called a Fast Fourier Transform (FFT). Using our approach it is
possible to measure the thickness of aluminium sheet to within sub-micron accuracy [1].
What sensors do we use and why do we use them?
Ultrasonic thickness gauging is typically performed using contact piezoelectric transducers for a wide range of materials and is a well-established technique [2]. Electromagnetic Acoustic Transducers
(EMATs) have been used sporadically in thickness gauging applications for over 25 years. To date, EMATs are not widely used as historically they have been large bulky devices only capable of
operating at low frequencies (below 1MHz). The advent of small, high field, rare-earth permanent magnets has meant that EMATs can now be manufactured without the need for an electromagnet and are now
comparable in size to conventional contact transducers.
Figure 1
Ultrasound is coupled through to the sample from the transducer, usually by a gel. The ultrasonic wave is reflected from the opposite face of the sample and some is coupled back through to the
Figure 2
Ultrasonic waves reverberate between the two outer faces of the sample and an exponentially decaying echo train is detected as shown above. The time between 2 consecutive signals corresponds to the
distance travelled in one ‘round trip’ i.e. two plate thicknesses.
In most metals the shear wave EMAT has the advantage that the shear wave travels at roughly half the longitudinal wave velocity and thus the first shear wave arrival has roughly double the transit
time of a longitudinal wave in the same sample. Higher accuracy can be obtained measuring the shear wave rather than the longitudinal wave as the shear wave has a much longer transit time and hence
the absolute temporal measurement error is roughly half for the shear wave.
Benefits of using EMATs
• Non-contact making online inspection more practical. (see figure 3)
• Only single sided access is required.
• EMATs are electromagnetically coupled and are thus insensitive to misalignment.
• Electromagnetic coupling allows a more rapid test (no delay through contacting couplant liquid)– important on moving samples.
• Broadband high frequency devices giving high measurement resolution.
Figure 3
Schematic diagram showing the set-up used in the EMAT thickness gauging technique.
What do we measure?
Consider a wave with amplitude 'E[1]' and angular frequency 'w' travelling through the aluminium sheet in the direction of increasing distance ‘z’. Such a wave can be described by the exponential
function for the displacement
w[1] = E[1] exp { i ( kz - wt ) } (equn. 1)
Where t is time and k is the wavenumber (2 p / wavelength). µ
And similarly the displacement ‘x[2]’ of a wave travelling in the opposite direction can be written as:-
w[2] = E[2] exp { -i ( kz +wt ) } (equn. 2)
Note that we have not defined the displacement as a shear of longitudinal wave displacement and the expressions are general and apply to either case.
Figure 4
Schematic diagram of waves in thin sheet.
The boundary conditions are that the aluminium-air boundary is stress free as the plate is unconstrained. The waves add linearly and we chose the first aluminium-air interface to be at z=0 and the
other to be at z=d, where d is the thickness of the sheet as shown below in figure 12. The stress at a general position z is given by a summation of the stresses due to each wave such that:-
Where r is the material density, stress (s) associated with x[1] is given by:-
s= r . (w/ k)^2 . dx[1]/ dz (equn. 3)
As the total stress (sum due to each wave) must be zero at each boundary (z=0 and z=d) for all values of t we obtain
E[1] = E[2] (from boundary condition at z=0) (equn. 4)
Thus at z=d we obtain:-
exp { ikd} = exp { -ikd} (equn. 5)
exp {2ikd} = 1 (equn. 6)
The solution to this final equation is w=npv/d (where n is an integer, v is the wave velocity) as before. Thus we have a more ‘formal’ explanation of why the resonant peaks occur at specific values
within a Fourier transform of the ultrasonic echo train.
Magnitude Fourier transforms of the ultrasonic waveforms in thin sheet
Prior to performing the Fast Fourier Transform (FFT) the data is windowed starting from a point one microsecond after the generation pulse to a point where the average peak amplitude falls to a value
below 10% of its maximum value. Ignoring data before one microsecond ensures that the ‘dead-time’ region is not transformed as it contains no useful information, and ignoring data below 10% of the
maximum amplitude value ensures that there is sufficiently high signal to noise in the data that is transformed. The particular region windowed (1-20?s) for the transform is arbitrary but it has been
used consistently throughout the data presented here. Prior to performing the FFT, a Hanning window is applied to the data and the data is ‘padded-out’ with null points. Padding out the data prior to
a transform increases the resolution of the resultant FFT.
Figure 5
Normalised magnitude FFT of the standard broadband shear wave EMAT waveform captured on the 500mm thick sample.
On closer examination it is evident that each of the peaks in the transform has a more complicated structure. Examining the fifth peak of the FFT for the 500mm thick sample (figure 5) it can clearly
be seen that there are actually 2 overlapping peaks centred on 15.5MHz. This is because the aluminium is mildly anisotropic and therefore acoustically birefringent due to microscopic changes effected
in the crystallographic structure of the metal during the rolling process. The net result is that for normal incidence shear waves the energy is guided into 2 discrete orthogonal polarisation
directions, each with a slightly different ultrasonic velocity. Thus the ‘repeat’ frequency for each shear wave pulse will be slightly different and they will be more clearly separated in the
transform at higher frequencies.
Figure 6
‘Zoom-in’ of the fifth peak in the FFT of the wider bandwidth broadband shear wave EMAT waveform captured on the 500mm thick sample.
In order to demonstrate the sub-micron accuracy it was necessary to measure samples with slightly different thicknesses but the same composition and rolling conditions. It was assumed that the shear
wave velocities (or texture) for each of these samples would be nominally identical. The samples were measured with a micrometer and the average thickness of each was found to be 279.3mm, 277.4mm and
277.2mm +/- 1mm in each case. The FFT for each ultrasonic waveform captured on these samples is shown in figure 7. The samples measured using a digital micrometer could not be sensibly differentiated
but are clearly separated by the ultrasonic measurement.
Figure 7
The first peaks in the FFT for the nominally 280mm thick samples. Note that the thickness values are calculated from averaged micrometer readings and that the ultrasonic measurements will actually
yield more reliable thickness values.
What's next?
The next stage of this work is to try to implement the technology on-line as it is in a 'ready-to-go' state for use in the aluminium industry. Single shot data at stand-offs of over 1mm can be used
to succesfully measure sheet thickness as shown below in figure 8.
The same technique can also be used to measure can wall thickness in formed cans, but as with any ultrasonic method the thickness accuracy obtained can only ever be as accurate as the callibration
velocity. Some further work is required to obtain the same level of performance on steels as they are generally less efficient for EMAT operation.
We are also currently looking to develop a method for measuring the thickness of foils using EMATs to generate the anti-symmetic Lamb wave mode (similar EMAT system described in Anisotropic Texture
section) and this approach looks very promising.
Figure 8
This FFT was obtained from a single waveform capture on the 250mm thick sample with an EMAT – sample separation of 1.5mm. A steel plate was positioned behind the sample to increase the magnetic
field. The inset shows the double peak structure at around 6.35MHz.
References for further reading
[1]. Dixon S et al, High accuracy non-contact ultrasonic thickness gauging of aluminium sheet using Electromagnetic Acoustic Transducers (EMATs), Ultrasonics, 39, pp445-453, 2001
[2]. Dixon S, Edwards C and Palmer SB, Recent developments using ElectroMagnetic Acoustic Transducers, Insight, 44, pp274-278, 2002 | {"url":"https://warwick.ac.uk/fac/sci/physics/research/ultra/research/thick/","timestamp":"2024-11-10T18:08:33Z","content_type":"text/html","content_length":"50987","record_id":"<urn:uuid:f25593df-9f20-43f0-9289-980dea0fa292>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00293.warc.gz"} |
Cloud WordNet Browser
Synonyms/Hypernyms (Ordered by Estimated Frequency) of noun polynomial
1 sense of polynomial
Sense 1
-- (a mathematical function that is the sum of a number of terms)
mathematical function
single-valued function
-- ((mathematics) a mathematical relation such that each element of a given set (the domain of the function) is associated with an element of another set (the range of the function))
Similarity of adj polynomial
1 sense of polynomial
Sense 1
-- (having the character of a polynomial; "a polynomial expression") | {"url":"https://cloudapps.herokuapp.com/wordnet/?q=polynomial","timestamp":"2024-11-02T09:23:40Z","content_type":"text/html","content_length":"17153","record_id":"<urn:uuid:81b40960-3771-4cc6-97d3-23300fae14f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00825.warc.gz"} |
How to Quickly Copy a Formula Down in Excel?How To Quickly Copy A Formula Down In Excel? (2024)
How to Quickly Copy a Formula Down in Excel?
Do you need to apply the same formula to multiple rows in an Excel spreadsheet? There’s no need to manually type or copy-paste the formula into each individual cell. Excel provides several quick and
easy ways to auto-fill formulas down a column. In this article, we’ll show you how to quickly copy formulas in Excel using fill handle, Fill Down command, keyboard shortcuts, Ctrl-Enter, and absolute
cell references.
Using Fill Handle to Copy Formula Down
The fill handle is the small square at the bottom right corner of a selected cell or range. When you hover over the fill handle, the cursor changes to a black plus sign (+).
To use the fill handle to copy a formula down:
1. Select the cell with the formula you want to copy down
2. Move your cursor over the fill handle until it changes to a black (+)
3. Click and drag the fill handle down to extend the formula to adjacent cells below
4. Release the mouse to fill the formula down
The fill handle automatically adjusts the cell references in the formula as it copies down. For example, if the formula in cell B2 references cell A2, then when copied to B3 it will reference A3, in
B4 it references A4, and so on.
Fill Down Command
Another way to quickly copy a formula down in Excel is using the Fill Down command. To do this:
1. Select the cell with the formula and the range of cells you want to fill
2. Go to the Home tab on the ribbon
3. In the Editing group, click the Fill button and select Down
This copies the formula from the top cell down to the selected range below, adjusting the cell references accordingly. You can also access Fill Down from the right-click context menu.
Keyboard Shortcuts to Copy Formulas
Excel provides some handy keyboard shortcuts to speed up copying formulas down a column:
• To fill the formula down to adjacent cells, select the cell and press Ctrl + D
• To fill the formula up to adjacent cells, select the cell and press Ctrl + U
• To fill the formula down and across to adjacent cells, select the range and press Ctrl + R
Ctrl-Enter to Fill Formula Down
Here’s a quick way to copy a formula down a whole column in Excel:
1. Select the cell with the formula
2. Press Ctrl + Shift + Down Arrow to select all cells below in the column
3. Press Ctrl + Enter to fill the formula down to the selected cells
Ctrl-Enter fills the formula down and Enter moves the selection to the cell below.
Copy Formula Down with Absolute Cell References
By default, Excel uses relative cell references in formulas, where the column and row references change based on the relative position of the cell containing the formula.
Sometimes you may want a cell reference to remain constant when copying a formula down. To do this, use an absolute cell reference by placing a dollar sign ($) before the column and/or row reference,
like $A$1.
To quickly toggle between relative, mixed, and absolute cell references:
1. Double-click the cell to edit the formula
2. Select a cell reference in the formula
3. Press F4 to cycle through the reference types:
• A1 (relative)
• $A$1 (absolute)
• A$1 (mixed with absolute row)
• $A1 (mixed with absolute column)
Using absolute references anchors those cells so they don’t change when auto-filling the formula down.
Example: Copy SUM Formula Down
Let’s walk through an example of using these methods to quickly copy a SUM formula down in Excel.
Suppose you have a worksheet with sales data like this:
Sales Rep Jan Feb Mar
John 100 150 200
Lisa 250 300 350
Michael 400 450 500
To calculate the total sales for each rep:
1. In D2, enter the formula =SUM(B2:C2)
2. Double-click the fill handle to copy the formula down column D
3. The formula will auto-adjust to =SUM(B3:C3) in D3, =SUM(B4:C4) in D4, etc.
The completed table will look like:
Sales Rep Jan Feb Mar Total
John 100 150 200 450
Lisa 250 300 350 900
Michael 400 450 500 1,350
You can also use the Fill Down, Ctrl+D or Ctrl+Enter methods explained above to quickly copy the SUM formula down after entering it once in D2.
Copy Formula Down to Filtered Data
When you have a filtered dataset, Excel can auto-fill formulas down only to the visible cells.
To copy a formula down to filtered rows:
1. Enter the formula in the first visible row under the header row
2. Select the cell with the formula
3. Go to the Home tab and find the Editing group
4. Open the Fill drop-down menu and choose Down
This fills the formula down only to the visible rows after applying the filter, skipping the hidden ones.
Paste Only Formulas Down a Column
When you copy cells in Excel, it copies over the contents and the formatting by default. But sometimes you may want to paste only the formulas down without overwriting the existing data or
Here’s how to paste only the formulas down a column:
1. Select and copy the cells with the formulas you want to paste down
2. Right-click the first cell where you want to paste
3. From the context menu, go to Paste Special > Formulas
This pastes down only the formulas to the destination cells without changing their data or formatting.
Final Thoughts
Learning how to auto-fill formulas in Excel can help you work more efficiently when building spreadsheets. The fill handle, Fill Down command, keyboard shortcuts and Ctrl+Enter technique all provide
quick ways to copy a formula down to multiple rows.
For more precise control over formula copying behavior, you can use absolute and mixed cell references where needed. Excel also allows you to copy formulas down to only the visible cells in a
filtered list and paste only formulas to a column without overwriting existing data.
By applying these methods, you’ll be able to quickly populate worksheet columns with formulas and calculations, saving time and effort in Excel.
What is the quickest way to copy a formula down in Excel?
The quickest way to copy a formula down in Excel is to use the fill handle. Simply select the cell with the formula, hover over the fill handle (the small square at the bottom-right corner of the
cell) until the cursor changes to a black plus sign (+), and then click and drag the fill handle down to the desired range.
How do I copy a formula down using keyboard shortcuts in Excel?
To copy a formula down using keyboard shortcuts, select the cell with the formula and press Ctrl + D. This will copy the formula down to the adjacent cells below. To copy the formula up, use Ctrl + U
, and to copy the formula down and across, select the range and press Ctrl + R.
What is the difference between relative and absolute cell references in Excel formulas?
Relative cell references change based on the relative position of the cell containing the formula when copied, while absolute cell references remain constant. To create an absolute reference, place a
dollar sign ($) before the column and/or row reference, like $A$1. Use absolute references when you want a cell reference to stay the same when copying a formula down.
Can I copy a formula down to filtered rows only in Excel?
Yes, Excel can auto-fill formulas down to only the visible cells in a filtered dataset. To do this, enter the formula in the first visible row under the header row, select the cell with the formula,
go to the Home tab, open the Fill drop-down menu, and choose Down. This will fill the formula down only to the visible rows after applying the filter.
How can I paste only formulas down a column without overwriting existing data or formatting?
To paste only formulas down a column without changing the existing data or formatting, first copy the cells with the formulas. Then, right-click the first cell where you want to paste, and from the
context menu, go to Paste Special > Formulas. This will paste down only the formulas to the destination cells, preserving their original data and formatting.
Vaishvi Desai is the founder of Excelsamurai and a passionate Excel enthusiast with years of experience in data analysis and spreadsheet management. With a mission to help others harness the power of
Excel, Vaishvi shares her expertise through concise, easy-to-follow tutorials on shortcuts, formulas, Pivot Tables, and VBA.
Leave a Reply Cancel reply | {"url":"https://excelsamurai.com/how-to-quickly-copy-a-formula-down-in-excel/","timestamp":"2024-11-02T11:12:33Z","content_type":"text/html","content_length":"227123","record_id":"<urn:uuid:f55754d5-d48d-4450-aeb8-fec7778593eb>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00679.warc.gz"} |
[CIVIL] Idealized moment-curvature curves are not computed correctly.
Idealized moment-curvature curves are not computed correctly.
The moment-curvature curve is calculated using the section geometry, reinforcement status, and inelastic material properties, and the idealized moment-curvature curve is obtained using the
moment-curvature curve.
So, basically, the three items mentioned must be properly defined for the calculation to work.
First, for concrete materials, the Mander model simulates the effect of triaxial compression by cast reinforcement and transverse restraint bars to increase the strength of deep concrete. Therefore,
it may be an inappropriate application if the steel is not in closed reinforcement, or if the steel is embedded in the column core without reinforcement, such as in SRC sections.
The second thing to note is that you must enter the appropriate strain rate for concrete in the inelastic material properties.
Input mistakes, or using some strain input values as default values, can result in yield points being calculated at inappropriate locations or failure to draw an idealized curve, as shown in the
figure below.
If the idealized curve does not compute correctly, the input values of the inelastic material properties need to be rechecked.
[ Moment-curvature curves with incorrect yield strain values ]
Strain rates for concrete material properties can be determined by referring to the table below to determine extreme strain rates based on column cross-section geometry and rebar overlap.
[ Guidelines for Evaluating Seismic Performance of Existing Facilities (Bridges), <Table 4.2.1>. ]
Furthermore, the yield strain of concrete is suggested by the guidelines to be 70% of the strain at maximum strength.
In the Inelastic Material dialog, for example, you can see that the yield strain, ε_cy, is automatically calculated to be 0.0014 when the strain at maximum strength, ε_co, is 0.002.
[ Unconfined Concrete Dialog Box in Inelastic Material ]
However, the Constrained Concrete Input dialog box does not perform an automatic calculation, so you will need to manually enter the 70% strain by clicking the check box.
[ Confined Concrete Dialog Box in Inelastic Material ] | {"url":"https://support.midasuser.com/hc/en-us/articles/16135056338841--CIVIL-Idealized-moment-curvature-curves-are-not-computed-correctly","timestamp":"2024-11-09T14:23:35Z","content_type":"text/html","content_length":"37006","record_id":"<urn:uuid:d080d4f9-c561-4fc0-a3af-87b6f0f52470>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00149.warc.gz"} |
Resurgence in the 1 2 BPS sector
Gandote, Sonagnon Eunice Edwige
We study matrix models as a toy model for N = 4 Super Yang-Mills (SYM) theory which is a quantum eld theory. In particular we are interested in the gauge/gravity duality which conjectures an
equivalence between N = 4 SYM and IIB string theory on AdS5 S5. We discuss the planar 't Hooft limit where we x = g2Y MN while taking N ! 1. In this limit we nd 1=N2 in the matrix model is equivalent
to ~ of the string theory. When we study the N dependence of ribbon graphs, we nd that the 1 N expansion in the gauge theory can be interpreted as a sum over surfaces suggestive of the perturbation
expansion of a closed string theory. We then consider a non-planar but large N limit, allowing us to discuss the giant graviton. We nd that the group representation theory of the symmetric group and
unitary group organizes the physics of giant gravitons. We compute two, three and multi point functions of giant graviton operators. The large N expansion of giant graviton correlators is considered.
Giant gravitons are described using operators with a bare dimension of order N. In this case the usual 1=N expansion is not applicable and there are contributions to the correlator that are
non-perturbative in character. The machinery needed to determine the non-pertubative physics form the pertubative contributions is the origin of the term resurgence. By writing the (square of the)
correlators in terms of the hypergeometric function 2F1(a; b; c; 1), we clarify the structure of the 1=N expansion.
A dissertation submitted to the University of the Witwatersrand in ful lment of the requirements for candidacy for the degree of Master of Science. Johannesburg, 2019 | {"url":"https://wiredspace.wits.ac.za/items/2e5e3eba-2d98-4e2f-bd17-aacd9abd1f9e","timestamp":"2024-11-13T19:30:39Z","content_type":"text/html","content_length":"450645","record_id":"<urn:uuid:be004a19-5eea-4287-b8f5-c8575c952efc>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00117.warc.gz"} |
target-sum | Leetcode
Similar Problems
Similar Problems not available
Target Sum - Leetcode Solution
LeetCode: Target Sum Leetcode Solution
Difficulty: Medium
Topics: backtracking array dynamic-programming
Target Sum problem on LeetCode is a problem of finding out the number of possible combinations of target sum using given array of numbers. The detailed solution to Target Sum problem is given below:
Problem Statement: You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new
Find out how many ways to assign symbols to make sum of integers equal to target S.
Solution: The solution to this problem is based on dynamic programming approach where we maintain a 2D matrix to keep track of all possible sums for each index and each target value.
1. Initialize a 2D matrix dp of size (n+1)(2sum+1) with all values as 0. (where n is the length of array and sum is the total sum of array elements) Here, dp[i][j] will represent the number of ways
to achieve a sum of j using first i elements of the array.
2. Set dp[0][sum] to 1, since it will represent zero elements used to get the sum.
3. For each element in the array, we iterate over all possible values of j and update dp[i][j].
a. dp[i][j+nums[i-1]] += dp[i-1][j]
This means if we add nums[i-1] to include current element in the sum, then the required target becomes j+nums[i-1]
and we can add all possible ways we could have achieved a target of j without including current element by adding the
number of combinations from previous i-1 elements to the dp[i][j+nums[i-1]].
b. dp[i][j-nums[i-1]] += dp[i-1][j]
This means if we subtract nums[i-1] to include current element in the sum, then the required target becomes j-nums[i-1]
and we can add all possible ways we could have achieved a target of j without including current element by adding the
number of combinations from previous i-1 elements to the dp[i][j-nums[i-1]].
4. Return dp[n][sum+S] which will represent the number of ways to achieve the target sum S using all elements of the array.
class Solution { public: int findTargetSumWays(vector<int>& nums, int S) { int n = nums.size(); int sum = accumulate(nums.begin(), nums.end(), 0); if (sum < S || (sum + S) % 2 != 0) return 0; // if
sum is less than S or sum+S is odd, return 0 int target = (sum + S) / 2; // target value which we need to achieve vector<vector<int>> dp(n+1, vector<int>(2sum+1, 0)); dp[0][sum] = 1; for (int i = 1;
i <= n; i++) { for (int j = 0; j < 2sum+1; j++) { if (j+nums[i-1] < 2*sum+1) dp[i][j] += dp[i-1][j+nums[i-1]]; if (j-nums[i-1] >= 0) dp[i][j] += dp[i-1][j-nums[i-1]]; } } return dp[n][target]; } };
Time Complexity: O(nsum) where n is the length of array and sum is the total sum of array elements Space Complexity: O(nsum)
This algorithm can also be optimized to use space in O(sum) as it is only dependent on value of sum. | {"url":"https://prepfortech.io/leetcode-solutions/target-sum","timestamp":"2024-11-08T23:39:15Z","content_type":"text/html","content_length":"56715","record_id":"<urn:uuid:197d095b-4260-4bd0-b953-ecb4293ac031>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00191.warc.gz"} |
The Coin Problem Quiz
Questions and Answers
We all need to learn how to handle our money, and that comes with knowing the value of a certain coin to the dollar. We may not end up becoming financial geniuses one day but we need to know the
value of the very money we work so hard for. So, do you think you can solve our coin problem here? Take our quiz and find out.
• 1.
Josh has 4 more quarters than dimes. If he has a total of $1.70, how many quarters and dimes does he have?
Correct Answer
A. 2
Let's assume the number of dimes Josh has is x. Since he has 4 more quarters than dimes, the number of quarters he has is x + 4. The value of x dimes is 10x cents, and the value of (x + 4)
quarters is 25(x + 4) cents. The total value of his coins is $1.70, which is equal to 170 cents. Therefore, the equation can be formed as 10x + 25(x + 4) = 170. Solving this equation, we get x =
2. So, Josh has 2 dimes and 2 + 4 = 6 quarters.
• 2.
Chantel has $4.85 in coins. If she has 6 more nickels than dimes and twice as many quarters as dimes, how many coins of each type does she have?
Correct Answer
B. 7
Chantel has 7 coins in total. Let's assume she has x dimes. Since she has 6 more nickels than dimes, she has x + 6 nickels. Also, she has twice as many quarters as dimes, so she has 2x quarters.
The total value of the coins is $4.85. The value of the dimes is 0.10x, the value of the nickels is 0.05(x + 6), and the value of the quarters is 0.25(2x). Setting up the equation 0.10x + 0.05(x
+ 6) + 0.25(2x) = 4.85 and solving for x, we find that x = 7. Therefore, Chantel has 7 dimes, 13 nickels, and 14 quarters.
• 3.
Tarzan bought a pencil for Jane and received change for $3 and 20 coins, all nickels and quarters. How many of each kind are given?
□ A.
6 nickels and 10 quarters
□ B.
15 nickels and 10 quarters
□ C.
□ D.
10 nickels and 10 quarters
Correct Answer
D. 10 nickels and 10 quarters
Tarzan received change for $3 and 20 coins. Since all the coins are either nickels or quarters, we can set up a system of equations to represent the given information. Let's assume that Tarzan
received x nickels and y quarters. The value of x nickels is 5x cents, and the value of y quarters is 25y cents. According to the given information, the total value of the coins is $3, which is
equal to 300 cents. So, we can write the equation 5x + 25y = 300. Additionally, we know that the total number of coins is 20, so we can write another equation x + y = 20. By solving these two
equations simultaneously, we find that x = 10 and y = 10. Therefore, Tarzan received 10 nickels and 10 quarters.
• 4.
Magalia received a money order worth $13. She received 10 more dimes than nickels, and 22 more quarters than dimes. How many coins of each did she receive?
□ A.
5 nickels, 20 dimes, and 40 quarters.
□ B.
20 dimes, and 42 quarters.
□ C.
10 nickels, 20 dimes, and 42 quarters.
□ D.
10 nickels, 20 dimes, and 2 quarters.
Correct Answer
C. 10 nickels, 20 dimes, and 42 quarters.
Magalia received 10 nickels, 20 dimes, and 42 quarters. The question states that she received 10 more dimes than nickels, so the number of dimes is 10 more than the number of nickels.
Additionally, it states that she received 22 more quarters than dimes, so the number of quarters is 22 more than the number of dimes. Therefore, the number of nickels is 10, the number of dimes
is 20, and the number of quarters is 42.
• 5.
John has a total of 19 nickels and dimes worth $1.65. How many of each type of coin does he have?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 5 nickels and 14 dimes
John has a total of 19 coins, made up of nickels and dimes. Let's assume he has x nickels and y dimes.
The value of x nickels is 5x cents, and the value of y dimes is 10y cents.
According to the problem, the total value of all the coins is $1.65, which is equal to 165 cents.
So, we can write the equation: 5x + 10y = 165.
To find the values of x and y, we need to solve this equation.
By trial and error, we can see that when x = 5 and y = 14, the equation is satisfied.
Therefore, John has 5 nickels and 14 dimes.
• 6.
Paloma has quarters and nickels. She has twice as many quarters as nickels. If the value of the coins totals $4.40, how many quarters and nickels does Paloma have?
□ A.
8 nickels and 16 quarters
□ B.
4 nickels and 16 quarters
□ C.
8 nickels and 10 quarters
□ D.
18 nickels and 16 quarters
Correct Answer
A. 8 nickels and 16 quarters
Let's assume that Paloma has x nickels. Since she has twice as many quarters as nickels, she would have 2x quarters. The value of each nickel is $0.05 and the value of each quarter is $0.25. The
total value of the nickels would be 0.05x and the total value of the quarters would be 0.25(2x) = 0.50x. The total value of all the coins is $4.40, so we can set up the equation 0.05x + 0.50x =
4.40. Solving this equation, we find that x = 8. Therefore, Paloma has 8 nickels and 2(8) = 16 quarters.
• 7.
Mom had $11.85 in her purse. She only had nickels and quarters in her purse. The number of quarters is equal to one less than twice the number of nickels. How many of each kind of coin did she
□ A.
10 nickels and 43 quarters
□ B.
20 nickels and 43 quarters
□ C.
22 nickels and 43 quarters
□ D.
22 nickels and 40 quarters
Correct Answer
C. 22 nickels and 43 quarters
Let's assume the number of nickels is N and the number of quarters is Q. The value of N nickels is 0.05N and the value of Q quarters is 0.25Q. We can create two equations based on the given
information: Q = 2N - 1 (the number of quarters is equal to one less than twice the number of nickels) and 0.05N + 0.25Q = 11.85 (the total value of the coins is $11.85). By substituting the
value of Q from the first equation into the second equation, we get 0.05N + 0.25(2N - 1) = 11.85. Solving this equation, we find that N = 22 and Q = 43. Therefore, she had 22 nickels and 43
• 8.
Grandma saves all of her quarters and dimes in a jar. She collected 170 coins that are worth $26. How many of each coin does she have?
□ A.
60 quarters and 110 dimes
□ B.
40 quarters and 110 dimes
□ C.
60 quarters and 100 dimes
□ D.
Correct Answer
A. 60 quarters and 110 dimes
Grandma collected a total of 170 coins. Let's assume she has x quarters and y dimes.
Since a quarter is worth 25 cents and a dime is worth 10 cents, the total value of the quarters can be represented as 25x and the total value of the dimes can be represented as 10y.
According to the given information, the total value of all the coins is $26, so we can write the equation as 25x + 10y = 2600 (since 1 dollar = 100 cents).
To find the number of each coin, we need to solve this equation. The solution is x = 60 and y = 110, which means Grandma has 60 quarters and 110 dimes.
• 9.
Colin has $1.15 in his wallet, which contains 18 coins in nickels and dimes. Find how many of each kind coin he has in his wallet.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 13 nickels and 5 dimes
• 10.
In a collection of dimes and quarters there are 6 more dimes than quarters. If there is $29.65 overall, how many of each are there?
□ A.
183 quarters and 89 dimes
□ B.
□ C.
83 quarters and 189 dimes
□ D.
Correct Answer
B. 83 quarters and 89 dimes
Let's assume the number of quarters is x. Since there are 6 more dimes than quarters, the number of dimes would be x+6.
The total value of quarters would be 0.25x, and the total value of dimes would be 0.10(x+6).
Given that the overall amount is $29.65, we can set up the equation 0.25x + 0.10(x+6) = 29.65.
By solving this equation, we find that x = 83. Therefore, there are 83 quarters and 89 dimes. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=3dq-the-coin-problem-quiz","timestamp":"2024-11-04T04:09:44Z","content_type":"text/html","content_length":"435879","record_id":"<urn:uuid:23d9d311-e9d4-48ec-acc2-4a2f6723e4dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00422.warc.gz"} |
Calculating bond price on calculator with different spot rates
Question 16 from practice questions on the CFA website.
[question removed by admin]
This could be done manually, one cash flow at the time. But how could I solve this using functions on the BAII Plus? With the TVM functions it seems I can only discount with one interest rate.
No way to do this other than manually on the calculator as the CF function assumes an a-priori known constant discount rate.
However, once you have the result you could calculate the overall discount rate which leads to the calculated bond value. But this is pointless since you already have the result and the discount rate
would only be known on a posteriory basis.
In this casr the result would be 101.98 (Answer C): Using the calculator: PV= 101.98; N=5; PMT=-1; FV= -100 --> Solve for I/Yx2= 1.95%.
This would be discount rate (if known, which however is not the case here) which would make it possible to solve the question fully with the calculator. | {"url":"https://www.analystforum.com/t/calculating-bond-price-on-calculator-with-different-spot-rates/100816","timestamp":"2024-11-11T13:14:53Z","content_type":"text/html","content_length":"18763","record_id":"<urn:uuid:25735869-fdd6-49e1-bccc-96bc63986d35>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00268.warc.gz"} |
Growthclap Blog
Suppose you have been running an A/B test for a week now, and every day you are asked by your business stakeholders, “How long are we planning to run the test? Do we have a significance yet?”. This
is not an unusual situation. In fact all product managers run into this issue. Except that many a times we have no idea how long should we be running the test, so we look at the results in a hope
that we reach significance. The problem compounds if you are running a test but you expect no uplift — This could be either due to Aesthetic reasons or revenue upside. How long should you run it?
Tricky isn’t it?
We should ideally never start a test without knowing how many samples we are going to collect. Why? Otherwise, you will be looking at data and you will end up doing ‘Data Peeking’, which is stopping
the test as soon as you achieve significance. Here is an example — Suppose you have a coin and your hypothesis is that it is fair. How do you prove that? Simple — toss it 100 times. But what if you
tossed it 10 times and saw tails 10 times. It seems statistically significant to stop the test at this point in time and reject the Null hypothesis — that the coin is fair. What went wrong? You
stopped the test a little too soon. You had no idea to begin with how long you should have run the test. The other problem that you may run into if you have not calculated the sample size is that you
wont be able to say confidently how long you are going to run the test for.
So how do we approach this?
Follow the first rule of product management — Embrace the ambiguity but avoid the uncertainty.
This is how we can approach calculating the sample size: Suppose we are running an A/B test that where: Our current conversion rate for an event such as % of users signing up for email is 10% and we
expect a 10% uplift in conversion if the treatment wins. Then,
Baseline conversion: P1 = 20%
Uplift in conversion: 10% (This is what you estimated as the expected impact of your change). As part of growth team, we usually aim for 20% uplift but even 10% could be big depending on how matured
your product is. The higher the uplift the sooner you reach significance.
Expected conversion of the treatment group: P2 = 20%*(1+10%) = 22%
Significance level: This is the chance of a false positive i.e. at 5% significance level what is the chance that we will reject the null hypothesis when it was in reality (Which you would never know)
was true. Of course, we want to minimize this error so we choose 5%. If you have less traffic then you may want to increase this to 10% or even 20%.
False Positive: Type I error — Rejecting the null hypothesis when it is true
Statistical Power: This is the probability that you will get a false negative. Phew! Power (= 1 — Type II Error) is the probability of avoiding a Type II error or in other words Power is the
probability that the test will detect a deviation from the null hypothesis, should such a deviation exist. Typically we set it to 80%.
False Negative : Type II error — Failing to reject the null hypothesis when it is false
Now we have everything that we can actually go ahead and calculate the sample size needed. We can either use an online calculator, G power tool, or R. Depending upon which tool you are using you may
see slightly different numbers but that is okay.
Let us see each one of them one by one:
a) Online calculator such as this one here
b) Use G*Power tool: Download the tool from here. Go to Test family ‘Z tests’, Statistical tests as ‘Proportions: Difference between two independent proportions’ and add the P1, P2, Alpha
(Statistical significance), Power = 0.8.
Expected output:
c) R: The function that we are going to use is power.prop.test (man page).
power.prop.test(n = NULL, p1 = NULL, p2 = NULL, sig.level = 0.05, power = NULL, alternative = c(“two.sided”, “one.sided”), strict = FALSE)
Go to any online R compiler such as this one here and type the following command with n set to NULL.
power.prop.test(n = NULL, p1 = 0.2, p2 = 0.22, power = 0.8, alternative = 'two.sided', sig.level = 0.05)
This is the output that you will get in R
Two-sample comparison of proportions power calculation
n = 6509.467
p1 = 0.2
p2 = 0.22
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
This means that we would need about 6510 samples in each group. Which means we would need 13020 traffic.
Now suppose you know historically that your website traffic is 2000 visitors then you know you have to run your hypothesis testing for 6.51 days or 7 days.
Bonus point: It is always a good idea to cover all days of the week as most of the businesses have ‘weeklikality’ in their demand pattern.
Now next time you are about to run the A/B test, pre-calculate the sample size needed so that you can set the right expectations with your business stakeholders.
Just in case you found the sample size much larger that you don’t think you will get to significance given the traffic that your website has, don’t worry, in another post I will share some cool
tricks on how to run A/B test when you do not have enough traffic. Until then, happy A/B testing.
Read our other articles on Product Leadership, Product Growth, Pricing & Monetization strategy, and AI/ML here. | {"url":"https://www.growthclap.com/blogs/a-b-testing-how-to-calculate-sample-size-before-launching-your-test","timestamp":"2024-11-13T12:54:25Z","content_type":"text/html","content_length":"24124","record_id":"<urn:uuid:a6a36558-6832-479d-bf10-5ab499352381>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00425.warc.gz"} |
Radiolab - Zeroworld
SPEAKER_03: Listen to Support It, WNYC Studios. This week on The New Yorker Radio Hour, staff writer Dexter Filkins traveled to the southern border this year looking for answers to what seems like an
impossible dilemma. That's The New Yorker Radio Hour, wherever you listen to podcasts. Wait, you're listening? Okay. All right. Okay. Okay. All right. You're listening to Radio Lab. Radio Lab. From
WNYC. SPEAKER_10: See? Yeah. Hey. Hello. I have to find your window. Hi. Are you tired? Uh, no. No, I'm all right. Oh. How are you? Oh, good. I'm good. I'm excited for this random little thing. Hey,
I'm Latif Nasser. I'm Lulu Miller. This is Radio Lab. A mystery guest is going to appear momentarily. SPEAKER_04: Oh, we see you. Hey. I can hear you both. Perfect. Hi. SPEAKER_13: Well, okay. So,
Kareem Latif. Hi. It's very nice to meet you. My pleasure. SPEAKER_05: Where are you? I'm in Alexandria, outside of DC. SPEAKER_06: Okay. So I guess the best way to set you up is that Kareem is here
because he has broken one of SPEAKER_03: the most forbidden rules of the universe. SPEAKER_03: Are you a cannibal? SPEAKER_12: Is that what I'm about to learn? SPEAKER_03: I haven't broken anything.
It's the question of whether to break it and how to break it. SPEAKER_03: What are the consequences of breaking it? SPEAKER_10: Could you break it? SPEAKER_03: Should I? Should I try to? Should I?
SPEAKER_13: Whoa. You seem like you're on a precipice. Brother. SPEAKER_03: Okay. So what is this rule that- SPEAKER_13: The rule- Yeah, what's the rule? In mathematics, you're allowed to do
everything for the most part. SPEAKER_03: You can multiply, you can divide. SPEAKER_13: But as you may recall from school, there's one thing in mathematics you're not allowed SPEAKER_03: to do. Do
you remember? This is dividing by zero? Dividing by zero. We have this entire structure of mathematics that is incredibly useful. SPEAKER_13: It's incredibly powerful. And it all kind of hinges upon
our agreeing to not go through this one door that has on SPEAKER_03: it- There is a sign on this door that says, division by zero, don't open this door because SPEAKER_03: what's on the other side of
this door is all sorts of craziness. SPEAKER_13: An infinite loop. Where everything is the same. Don't divide by zero. When you make the two into one, and when you make the inner like the outer, then
you will SPEAKER_13: enter the kingdom. It's like the sign you hang on the elevator that's not working. It's like out of order. Like please do not go through here. Well, here's the thing though. It
isn't that the elevator is out of order. It's that the elevator goes to a dimension that is so problematic to our way of thinking SPEAKER_13: in this dimension that as long as you agree to not go
into that kind of elevator shaft SPEAKER_13: wormhole, we're good. You can have your airplanes, you can have your computers. So today we've got a story about a paper that Karim Ani wrote almost 20
years ago about SPEAKER_02: dividing by zero. I happened across it about 10 years ago and it really tickled something in me. Over the years I would think about it. SPEAKER_11: I'd wonder whatever
happened to this guy who wanted to divide by zero. I'd wonder if there were consequences for math. I'd wonder if there were real consequences for reality, for my reality, for his reality. SPEAKER_13:
I didn't know. But I thought that as we ourselves are rounding the clock of the calendar year, passing through zero to start a new, I thought now might be a nice time to call him up and try to
understand. So my friends, leave your calculators at the door because we are going to try to enter SPEAKER_03: a new kind of math. Here we go. Well, I think what a mathematician would say is- Yeah.
Are you a mathematician? Yeah, sorry. We should say who you are by the way. Who are you, Karim? What do you do? Like what do you do? So I am the founder of Citizen Math and what we do is we write
lessons for middle and high school classrooms around real issues. So students are using mathematics to discuss, should the federal government increase the minimum wage? Or why do airlines oversell
their flights? Using mathematics as a tool for discussing and analyzing the world around us. Love it. But coming back to this idea of division by zero. Yeah. Okay. So what if we just start with, why
can't you divide by zero? Like why is that such a hard and fast rule? SPEAKER_13: Okay. Well, so one reason is because it violates a mathematical principle that every operation SPEAKER_03: needs to
be undoable. SPEAKER_03: Anything you do, you need to be able to undo. Okay. SPEAKER_13: Let's say you start with 10 and you divide by five. And so now you're at- Two. Now you need to be able to get
back to 10 though. And so you can go from two and multiply it by five to get back to 10. 10 divided by five gets you to two, two times five gets you back to 10. But if you now try that with zero. Ten
divided by zero is some number. SPEAKER_03: Well, to go backwards, now that some number times zero, how can that get you back to 10? So it violates- Because zero times anything is zero. SPEAKER_13:
It kind of sucked back into the black hole of zero-ness. Right. Right. That's the, it violates this custom, let's say, or a law- Because there'd be no thing you can multiply by zero to get to the
number 10. Exactly. Once you do- Right. Right. So mathematicians created this rule, this kind of barricaded door that basically says, do not try to divide by zero because the answer is undefined.
SPEAKER_13: There is no answer. You can't do it. There have been people who have gone through that door. Hi there. Hi, Steve. Lulu, I don't think we've ever done this together. I know. Isn't that
wild? I have never actually gotten to meet you. SPEAKER_03: This is so nice. Well, it's nice. Yeah. SPEAKER_13: Hi. Okay. This is Steve. Steve Strogatz, and I'm a mathematician and a math professor
at Cornell University. And Steve has not walked through the door of dividing by zero, but he says that these SPEAKER_03: SPEAKER_13: sorts of rules, these sorts of barricades, in math, it's always
been important to break SPEAKER_03: them. Exactly. That's actually some of the most fruitful parts of math, that when you try to do something that seems impossible, it often leads to the creation of
whole new universes. SPEAKER_13: So for example, Steve was like, okay, let's think about square roots. SPEAKER_13: So if you take a number like the number three- SPEAKER_08: Okay, so three times
three, that'd be in the jargon three squared. SPEAKER_03: Three times three, three squared is nine. SPEAKER_08: So the undoing of that is that the square root of nine is three. But now let's say you
wanted to take the square root of negative nine. SPEAKER_03: Your first thought would be negative three maybe is the square root of negative nine, but it doesn't work. If you do negative three times
negative three, you get positive nine, not negative nine. SPEAKER_08: Because in math, and we're not going to go into why, if you multiply two negative numbers, you get a, boom, positive number. So
you can't do it. You can't take the square root of negative nine. There is no number that will work. SPEAKER_03: So a long, long time ago, mathematicians were like, okay, there is a rule, no square
roots of negative numbers. But then in like the late 1500s, a bunch of new rambunctious upcoming disobedient mathematicians SPEAKER_08: said, well, what if we just broke that rule? And to make the
math work, we just invented a whole universe of new numbers. SPEAKER_08: That is so bizarre that mathematicians called these imaginary numbers. Numbers that are not technically negative and they're
not technically positive. They were sometimes called fictitious numbers. SPEAKER_03: But they allow the math to work in such a way that you can start doing square roots of negative numbers.
SPEAKER_08: Because you just wish it to be so. Just invent some new numbers? Yeah, it's invention. SPEAKER_03: Exactly. It's invention in the artistic sense. You can invent something that didn't
previously exist. But was anyone like, no, we have a rule. You can't take a square root of a negative number. Yes, absolutely. It's like anything else that human beings do. They're always
reactionaries. There are always people who say you're muddying the waters. You're messing up the pristine and beautiful world of math with your ugly ideas. SPEAKER_08: Because these ideas have a lot
at stake intellectually and there's always resistance. SPEAKER_03: But that's where the breakthroughs happen. You take something that earlier generations say was impossible and you say what if.
SPEAKER_03: And then you try it and you figure out a way to do it. SPEAKER_03: And that's where the progress happens. SPEAKER_08: But like what is like an imaginary number give us? That gives us the
modern world. Like concrete stuff. I'm going to tell you. OK. I mean imaginary numbers. OK. So if we fast forward to the 20th century, this is not why imaginary numbers are invented. They're invented
much earlier than that. But in the 20th century when the theory of the atom starts to be worked out, we learn how to describe what's going on with hydrogen atoms and helium and how light works. In
other words, we invent, we, the collective of scientists in the 1920s invent quantum mechanics. SPEAKER_08: So it's our most accurate physical theory there is. It gives us today everything.
SPEAKER_03: It gives us what we're doing right now, talking over the internet. It gives us lasers. SPEAKER_08: It gives us transistors, chips. SPEAKER_08: Everything in the modern world has an
underpinning in quantum theory and the electronic revolution SPEAKER_08: that it made possible. The math of quantum theory is built on imaginary numbers. You can't do quantum mechanics without
comfort with imaginary numbers. And it's crazy in that what was thought to be imaginary a few decades or really more like a few centuries later, turns out to be the mathematics of reality. And to
Steve, this is sort of the beauty and the artistry of math. I mean that in math we have creative freedom. We can do anything we want as long as it's logical. Science in many ways is a chronicle of
humans understanding of reality and logic. Kind of a chronicle of how we think. SPEAKER_06: Like it began with early humans coming up with the idea of what we call natural numbers, SPEAKER_08: one,
two, three, and so on. Then the Sumerians and Mesopotamia and the Mayans each independently came up with the idea of zero, which blows its way around the globe. SPEAKER_03: And then a few thousand
years later, the third century in China, negative numbers show up and they too spread across the world and math gets more and more complicated. SPEAKER_08: And so we start to come up with rules and
then we try to break those rules. And in the wake of that breakage, we often invent new numbers like imaginary numbers SPEAKER_13: or rational numbers or real numbers or complex numbers. We come up
with all these different tools that we've invented by pushing at the rules, SPEAKER_03: pushing at the boundaries of math that then help us to better understand the world around us. But this is where
division by zero is different, categorically different, because it's so beyond the, like it leads to these results that would undermine all of mathematics. And that would break math as we know it.
And this is where, for me, this becomes actually quite existential. When we come back, we are stepping through the door. SPEAKER_13: This week on The New Yorker Radio Hour, with immigration policy
front and center in Washington, staff writer Dexter Filkins traveled along the southern border looking for answers. I think it's difficult to appreciate the scale and the magnitude of what's
happening there unless you see it by yourself up close. SPEAKER_03: The dilemmas at the border. That's The New Yorker Radio Hour from WNYC Studios. Listen wherever you get your podcasts. SPEAKER_10:
Grab our calculators and watch what happens when we do. Because there are actually all these videos on YouTube. SPEAKER_00: Where sweet nerdy men will take these old mechanical calculators, punch in
some number, divide it by zero, we hit equals, so here we go. And what happens is the numbers on these calculators just keep rolling over and over and over. SPEAKER_03: And what happens is that it
gets into an infinite loop. SPEAKER_03: And over. And it will never stop and I guess it heats up so eventually it would catch fire. SPEAKER_03: Like the mechanisms driving that calculator just get
stuck. And it is right here for Karim. SPEAKER_07: Where this becomes actually quite existential. SPEAKER_07: Because he explains to understand what's driving that looping, you have to think about
the SPEAKER_09: math going on. He said, you know, take for example the number 10. SPEAKER_09: If you take 10 and divide it by 10, you get one. SPEAKER_03: 10 divided by five is two. 10 divided by
half is 20. SPEAKER_09: The smaller the number and the bottom, the number that you're dividing by, the larger SPEAKER_09: the result. And so by that reasoning. SPEAKER_03: If you divide by zero, the
smallest nothingness number we can conceive of, then your answer would be infinity. Why isn't it infinity? Infinity feels like a great answer. SPEAKER_07: Because infinity in mathematics isn't
actually a number. SPEAKER_07: It's a direction. It's a direction that we can move towards, but it isn't a destination that we can get SPEAKER_07: to. And the reason is because if you allow for
infinity, then you get really weird results. SPEAKER_03: SPEAKER_13: For instance, infinity plus zero is infinity. SPEAKER_03: Infinity plus one is infinity. Infinity plus two is infinity. Infinity
plus three is infinity. And what that would suggest is zero is equal to one is equal to two is equal to three is SPEAKER_13: equal to four. And that would break math as we know it. Again Steve
Strogatz. Because then as your friend says, all numbers would become the same number. SPEAKER_03: Which you know, for math, the whole vast interconnected web of it would be a problem. The world of
fluid dynamics, calculus, geometry, physics, all this stuff depends on numbers SPEAKER_03: being individual, discrete things. SPEAKER_13: But if you allow for division by zero, that all goes away.
And you get into all of these strange consequences like one equaling zero equaling two equaling infinity equaling four. And so in order to protect math and all the things we use it for, like making
computers and planes and all modern technology, mathematicians said that when you try to divide by zero, the answer is undefined. It's undefined. There's no sensible definition. And that's why they
put up that barricaded door. Because what's beyond the door is it just seems impossible. SPEAKER_08: It seems very difficult to get our heads around. Because effectively what we're saying is
everything is one thing. Now Karim says, when I first started thinking about this 10 years ago, or however long that was, it was something fun to think about. SPEAKER_03: It was something fun to
write a grad school paper about. But he says more recently, he's had this feeling that's grown and grown. SPEAKER_13: Of this isn't complete. There's something else here. Now maybe this is something
you have felt at some point in your life. SPEAKER_03: Maybe you're even feeling it right now that the daily stuff of it isn't all there is. But there's something else out there. And for Karim, he's
like, look, I'm not religious. SPEAKER_13: He's devoted basically his whole life to math. SPEAKER_08: And mathematics is kind of a representative of one way of thinking about not just the SPEAKER_03:
world, but one way of thinking about reality. SPEAKER_13: And so to Karim, it perplexes him, it sort of tugs at him to see math itself saying, when you actually follow out the operation of dividing
by zero, you end up in a completely SPEAKER_03: different realm. SPEAKER_13: Where one equals two equals three equals infinity. That all of these numbers are one and the same, that everything is
effectively one thing. SPEAKER_03: Everything is equal to everything else. And this world of division, I don't mean political division, but that too, this world of duality, SPEAKER_03: of
differences, of things being discrete from one another, that all goes away. And Karim can't help but to notice that's the sort of stuff you hear from. Jesus said to them, people like Jesus, when you
make the two into one and Buddha, or people SPEAKER_03: who follow Taoism or people who have done intense meditation or intense hallucinogenics. SPEAKER_13: Oftentimes those people come back and the
thing that they say is, I felt like I was SPEAKER_03: one with everything. So you see in these like religious texts, you see literally like the collapse of the integer system. I'm seeing math being a
way of thinking about reality and thinking about the nature of nature. SPEAKER_03: And to Karim, because the math itself leads to this undefined place where numbers work SPEAKER_13: really
differently. Where all of these numbers are one and the same. To him? That suggests that there is something else. And I'm not saying that's God or whatever it is. It's just there's something else
here. I can't, by definition, I cannot on this side of the door, articulate what that reality SPEAKER_03: would look like. But I'm middle aged. SPEAKER_02: Now that Karim is rolling into his mid 40s.
SPEAKER_03: I don't have children, a spouse. He finds himself unable to stop wondering about what that something else could really SPEAKER_13: look like. I look at my life and I think, well, after 44
years, you're still not content with SPEAKER_03: this. That must be a sign that either you're doomed to be discontented or that's a sign that SPEAKER_13: like you're not going to find it here.
SPEAKER_03: You need to go through the door because honestly, what's your alternative? But how do you actually do it? SPEAKER_13: Like how do you, I don't get how you, how do you actually divide by
zero and go through SPEAKER_13: the door? I don't know. I have no idea what it would mean practically to divide by zero. But he says he does know it would have to start with some pretty major
changes. Like he would definitely need to quit his job. He would need to leave behind his house in the DC burbs. SPEAKER_03: Look, I'm Arab. I feel this weird like attraction to the desert.
SPEAKER_03: Like I would probably go take camping gear and go find a desert and sit in the desert. SPEAKER_03: And then, well, he's not entirely sure. All he knows is that he would need to connect
with that mathy part of his brain he has been SPEAKER_13: using for decades, thinking about numbers as these discrete and different things and then try to turn it off. That is the thing that I will
need to put down. SPEAKER_13: And then maybe if he listened really close, he could begin to hear or feel the something SPEAKER_03: else behind all of this. SPEAKER_13: Now, okay, so what's my
personal reaction to that? SPEAKER_03: By the way, there's a guy named Steve Strogatz. We talked to him about you. We were behind your back and we talked to him about you. And we told him about how
you were thinking about trying to access a world where there SPEAKER_13: are no differences in numbers. I would say you can do that. SPEAKER_03: If you want to do that, you can do it. You can make a
universe in your mind where all numbers are the same number. Let me describe that universe. There's a universe I'm going to call zero world. Welcome to zero world. Where in fact there's only one
number. SPEAKER_13: And here are the properties of the mathematical zero world. SPEAKER_03: Zero plus zero. Equals zero. And that's true no matter how many times you add zero. You can't get any new
numbers in this world because there are no additional numbers. SPEAKER_13: There's only zero. SPEAKER_08: Zero plus zero plus zero plus zero. As far as the eye can see. SPEAKER_03: And that's it.
That's your universe. It's the universe of zero. All numbers are the same because they're all zero. And are you happy now? SPEAKER_03: He keeps going. He keeps going. That's like such a solipsistic,
pathetic little universe. SPEAKER_08: That is the ultimate in navel gazing. That does nothing for anybody. But it's self-consistent. You can live in that universe if you want to pretend there's
nothing but zero. SPEAKER_08: Oh, see, okay. And let me respond to that then. Because Steven Stromgast is a really smart dude. SPEAKER_08: But the question, that first question of are you happy now?
I would say, well, Steven, if you live in one world or where every number is distinct from one another, like if you're happy in that world, great. I'm not. Because I have this question in the back of
my mind. This question of what is actually on the other side of that door? To me, it is zero world and it's a very, I just find it incredibly stultifying. SPEAKER_08: It's a very impoverished little
self-contained logical place. Stultifying but mathematically sound? I think it is. It's defensible. You can have it. There's nothing wrong with it. It's just as minimal as a thing can be. It has no
potential for anything beyond itself. But it's just a fine little solipsist looking at its own belly button. But the inside your belly button is everyone and everything. It's like, I don't know, I'm
just trying to defend him because he's not here. SPEAKER_13: I don't know if I want to go there. You can try. I'm not buying it. No, but it's like division. He kept saying division goes away.
SPEAKER_13: Political division, spiritual division, duality goes away. SPEAKER_13: Let me try to make the case for it. The case for it, I guess, is this is a noble impulse to see the unity. It's also
a productive impulse. Scientifically looking for unified theories has historically been the way to great progress SPEAKER_03: in physics. SPEAKER_08: So to recognize that electricity and magnetism
are actually two sides of the same coin that we now call electromagnetism. That was a great invention, a great breakthrough of the middle 1800s that gave us modern things SPEAKER_08: like wireless
and telegraphs and telephone. And then Einstein unifying space and time, matter and energy. This is a trend. We've been doing this unification program in physics for the past 150 years and it's
SPEAKER_03: very, very successful. And it reveals these underlying deep commonalities among things that are superficially different. SPEAKER_08: So the idea that there's great insight to be had by
realizing that things that look SPEAKER_03: different are actually deep down the same, that's a good move. SPEAKER_08: That is historically a very good move much of the time. But there's also the
move that along with the unifying impulse, you also have to have the diversifying impulse. You have to realize that not all things are the same, that there is great abundance in the world, all kinds
of diversity, whether of people or biological species or phenomena. And there are two kinds of scientists or more than two, but I mean there are unifiers and diversifiers and there's a need for both.
And I guess I want to argue for the happy middle that if you're all about diversity, you won't see patterns. And if you're all about unity, you won't see richness. And I think both are blinkered
visions of the world. I just don't believe in either extreme. SPEAKER_08: And in some ways, talking to Steve and talking to Karim, I think the question we were really kicking around is, does your
experience of the world feel fulfilling and complete, even true? And for Steve, there is a deep pleasure and joy and a benefit, like a real tangible benefit to accepting math exactly as it is and
reveling in how it describes reality. And for Karim... Every day I sit at my computer. There isn't. Kind of rewriting our lessons to tighten things up. The one I was working on yesterday was about
concert tickets and about all the fees and like our secondary ticket brokers discourage or are they actually like correcting kind of a market failure. That sounds interesting. Oh, yeah. All of our
lessons are interesting. I think. But that is so based on math. SPEAKER_08: And it sounds like your every day you're staring at these things that you believe are confining SPEAKER_03: you, these
numbers, and you're literally not just staring at them. You're like working with them even more intimately than most people because you're trying to like fit them around the universe and explain that
back to kids. Like you're playing with these tools that sound like they have you feel like are failing you or maybe not failing you, but they aren't all that's there. I sort of feel like I'm spinning
my wheels needlessly. I feel like I'm ready for something. I feel like I'm ready for whatever is the next thing. But what's crazy to me is like, but to do that because of the nature of what you do
SPEAKER_13: and what your passion has been, you have to turn your back on math. It sort of sounds like. I mean, I think, look, I think we live our lives in phases and that isn't I'm not going
SPEAKER_13: to put it down and then stomp all over it. Yeah. Right. SPEAKER_03: It's like it's a gentle putting down. It's not throwing it on the ground, but I feel like I've sucked all the juice out
of that orange for me. Okay. One last question. When you think about the world, when you think about zero world mathematically, where one equals two equals zero equals infinity, everything gets
sucked into the black hole of zero. Yeah. This place that you, it sounds like you yearn for that you want to go experience and understand and feel, right? I mean, is that okay? SPEAKER_13: What does,
has it, has thinking about it and spending time there theoretically, has it SPEAKER_03: changed your understanding of numbers or math at all? Has it expanded math for you at all? I respect math more
by virtue of it writing the sign. Writing the sign? SPEAKER_13: Yeah. What does that mean? Mathematics saying, mathematics saying there's something we can't account for. I admire that. SPEAKER_04:
Why? SPEAKER_13: I admire that. Why? Why? Because everybody, I am Christian. This is the truth. There is no truth but for this. I am Muslim. This is the truth. SPEAKER_03: There is no truth but for
this. Mathematics is an incredibly powerful tool. And for the institution or for mathematics personified to say, I'm an exceptionally powerful tool. SPEAKER_13: If you master me and if you use me,
you're going to be able to do so much, but I'm not SPEAKER_03: complete. There is something I can't account for. I think that humility, I really, I think that is enviable. When I first wrote that
paper about Division by Zero, I was like, I'm really going to stick it to math. SPEAKER_13: And now it's more like, what a wonderful gift for this powerful tool that we use to do so SPEAKER_13: much
to say, but if you want to go further, you need to put me down now. SPEAKER_13: SPEAKER_13: This episode was produced by Matthew Kielty with help from Laquetti Foster Keys and Alyssa Jean Perry.
Mixing help from Arianne Wack, fact checking by Diane Kelly. It was edited by Pat Walters. Steve Strogatz, by the way, also hosts a podcast all about math where he zips and zazzles through
SPEAKER_03: different puzzles and questions with all kinds of fun guests. It is called The Joy of Why, W-H-Y, The Joy of Why. And Kareem wrote a book all about how to get kids talking about how math
interplays with real world puzzles. It's called Dear Citizen Math. And you can check out citizenmath.com to see all sorts of neat lessons he and his team have dreamed up over the years for middle
school and high school classrooms. That'll do it for today. That'll do it for this year. Thank you so much for listening to Radiolab. I hope you all get a little bit of zero world over the break,
like where nothing is happening. Just low stress, low thought. Rest? Dare we say rest? Bye. Welcome back to zero world. Oh, well, there are no phones. Yes, your precious little phone is gone. Oh, no.
Oh, no. Oh, God. Going somewhere? I don't think so. There are no cars. There's no planes, motorcycles, bicycles. None of it. No money. No money. Oh, how good, freedom. No money. You can't even count
here. There's nothing. Nothing but zero. As far as the eye can see. Are you happy now? Hi, I'm Hazel and I'm from Silver Spring. SPEAKER_05: Radiolab was created by Chad Bohmak and is edited by Soren
Wheeler. Lulu Miller and Latif Nasser are our co-hosts. Dylan Keith is our director of sound design. Our staff includes Simon Adler, Jeremy Bloom, Becca Bressler, Eketty Foster-Kees, W. Harry
Bortuna, David Gabel, Maria Paz-Gutitis, Sinju Naines-Sump-Badan, Matt Kielty, Annie Nacoon, Alex Neeson, Sara Khari, Alyssa Jung-Perry, Sarah Sandback, Arian Wack, Pat Walters, and Molly Webster.
Our fact checkers are Diane Kelly, Emily Krueger, and Natalie Middleton. SPEAKER_04: Thank you. Hi, I'm Ram from India. Leadership support for Radiolab's science programming is provided by the Gordon
and Betty Moore Foundation, Science Sandbox, Assignment Foundation Initiative, and the John Templeton SPEAKER_08: Foundation. Foundation support for Radiolab was provided by the Alford P. Sloan
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Lomax Distribution (Pareto II)
What is the Lomax Distribution?
The Lomax distribution is a heavy tailed distribution originally proposed by Lomax (1987), who used it in his analysis of business failure lifetime data. The distribution, which is basically a
shifted Pareto distribution, is widely used in survival analysis, and has many applications in actuarial science, economics, and business.
Two parameters define the distribution: scale parameter λ and shape parameter κ (sometimes denoted as α). The shorthand X ∼ Lomax(λ,κ) indicates the random variable X has a Lomax distribution with
those two parameters.
The probability density function for the Lomax distribution is:
For all x, λ and κ greater than zero.
Lomax distribution showing two different sets of parameter variations.
Variants of the Lomax may have more than the basic two parameters. For example, the beta exponentiated Lomax is a five-parameter continuous model (Mead 2016 ).
Many other variants of the Lomax distribution exist, including:
• Exponential Lomax
• Exponentiated Lomax
• McDonald Lomax
• Poisson Lomax
• Power Lomax.
• Transmuted Lomax
• Weibull Lomax
• Weighted Lomax
These are just a few: there are many more. New variants are being proposed all of the time. For example, see Oguntunde et al’s 2017 proposed new generalization.
A. Hassan and A. Al-Ghamdi, “Optimum step stress accelerated life testing for Lomax distribution,” Journal of Applied Sciences Research, vol. 5, pp. 2153–2164, 2009.
Kotz, S.; et al., eds. (2006), Encyclopedia of Statistical Sciences, Wiley.
Everitt, B. S.; Skrondal, A. (2010), The Cambridge Dictionary of Statistics, Cambridge University Press.
K. Lomax. Business failures: another example of the analysis of failure data. J Am Stat Assoc. 1987;49:847–852
M.E. Mead. On Five-Parameter Lomax Distribution:Properties and Applications. Pakistan Journal of Statistics and Operation Research. Vol 12. No.1. 2016.
E. Oguntunde et al. “A New Generalization of the Lomax Distribution with Increasing, Decreasing, and Constant Failure Rate.” Modelling and Simulation in Engineering Volume 2017. | {"url":"https://www.statisticshowto.com/lomax-distribution/","timestamp":"2024-11-13T14:20:35Z","content_type":"text/html","content_length":"68519","record_id":"<urn:uuid:abd20ccf-08bf-4b0b-9096-3a4bafbe3f21>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00476.warc.gz"} |
Voting Theory has a HOLE — EA Forum
[[epistemic status: I am sure of the facts I present regarding algorithms' functionality, as well as the shared limitation inherent in Gibbard's Theorem, et al. Yet, these small certainties serve my
ultimate conclusion: "There is an unexplored domain, for which we have not made any definitive conclusions." I then hypothesize how we might gain some certainty regarding the enlarged class of voting
algorithms, though I am likely wrong!]]
TL;DR - Voting theory hit a wall in '73; Gibbard's Theorem proved that no voting method can avoid strategic voting (unless it is dictatorial or two-choices only). Yet, Gibbard's premise rests upon a
particular DATA-TYPE - a ranked list. In contrast, we know for a fact from machine-learning research that other data-types can succeed even when a ranked list fails. So, the 'high-dimensional
latent-space vectors' of artificial neural networks are NOT inherently limited as Gibbard's rank lists were. Either Gibbard does apply to that data-type, as well (which is a new research paper
waiting to happen) OR Gibbard does not apply, in which case we might find a strategy-less voting method. That seems worth looking into!
Gibbard's Theorem
I am not taking issue with any of Gibbard's steps; I am pointing to a gap in his premise. He restricted his analysis to all "strategies which can be expressed as a preference n-tuple"... a ranked
list. And, the assessment presumed as a result of this by Voting Theorists seems to be "because ranked-lists fail, then ALL data-types must also fail." That claim is factually incorrect, and a
failure of logic.
Disproof: In machine learning, a Variational Auto-Encoder converts an input into a vector in a high-dimensional latent-space. Because those vectors maintain important relational data, then the VAE is
able to accurately reconstruct the inputs. Yet, if you converted those latent-space vectors into scalars, using ANY distance-metric, then you have LOST all that information. Now, with only a
ranked-list to compute upon, the Variational Auto-Encoder will fail. The same is true for Transformers, the most successful form of deep neural networks (Transformers use a dot-product to compare
vector similarity; any reduction to scalars becomes meaningless).
This is proof by existence that "An algorithm CAN function properly when fed latent-space vectors, DESPITE that algorithm failing when given ONLY a ranked-list." So, the claim of Voting Theorists
that "if ranked-lists fail, then everything must fail" is categorically false. There is an entire domain left unexplored, abandoned. We can't make sure claims upon "the impossibility of strategy-less
voting algorithms" when given latent-space vectors. No one has looked there, yet.
Some Hope?
There are reasons to suspect that we might find a strategy-less voting algorithm in that latent-space. Fundamentally, when neural networks optimize, they are 'guaranteed convex' - eventually, they
roll to the true bottom. So, if one was to 'minimize regret' for that neural network, we should expect it to reach the true minimum. (eventually!) However, the more important reason for hope comes
from looking at how latent-spaces cluster their voters, and what that would do to a strategic ballot.
In a latent-space, each self-similar group forms a distinct cluster. So, a naïve voting algorithm could seek some 'mode' of the clusters, trimming-away the most aberrant ballots first. Wait! Those
'aberrant' ballots are the ones who didn't fit in a cluster - and while a few weirdoes might be among them (myself included), that set of ballots which are outside the clusters will ALSO contain ALL
the strategic ballots! Thus, a naïve 'modal' algorithm would ignore strategic ballots first, because those ballots are idiosyncratic.
If we can find a strategy-less voting algorithm, that would be a far superior future. I don't know where to begin to evaluate the gains from better decision-making. Though, considering that
governments' budgets consume a large fraction of the global economy, there are likely trillions of dollars which could be better-allocated. That's the value-statement, justifying an earnest effort to
find such a voting algorithm OR prove that none can exist in this case, as well.
I hope to hear your thoughts, and if anyone would like to sit down to flesh-out a paper, I'm game. :)
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Guy Raveh 3
I'm having some trouble understanding your post, and I think it could be useful to:
1. State the exact problem setting you are addressing,
2. State Gibbard's theorem, and
3. Show how exactly machine learning has solutions for that problem.
As far as I understand Gibbard's theorem, it does not only apply to ranked lists (as opposed to Arrow's theorem). Rather, it applies for any situation where each participant has to choose from some
set of personal actions, and there's some mechanism that translates every possible combination of actions to a social choice (of one result from a set). In this context either the mechanism limits
the choice to only two results; or there's a dictatorship, or there's an option for strategic voting.
This isn't my area so I'm taking a risk here that I might say something stupid, but: The existence of strategic voting is a problem, since it disincentivises truthfulness; But intuitively that
doesn't mean it's the end of the world - all voting mechanisms we've had so far were very obviously open to strategic voting, and we still have functioning governments. Also choices that are limited
to two options are often important and interesting, e.g. whether we improve the world or harm it, whether we expand into space or not, or whether the world should act to limit population growth. | {"url":"https://forum.effectivealtruism.org/posts/zyXtnrokGyR7yhqsg/voting-theory-has-a-hole","timestamp":"2024-11-03T10:11:33Z","content_type":"text/html","content_length":"204780","record_id":"<urn:uuid:4de2ed45-1976-4115-9162-2cff1b4c1be5>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00672.warc.gz"} |
Review of Magnetic Resonance Imaging (MRI)
Review of Magnetic Resonance Imaging (MRI)
The hydrogen atom has a nucleus consisting of only a single proton. Because of this it exhibits a relatively large magnetic moment. Hydrogen is also plentiful throughout the body due to its presence
in every water molecule (
Where k is the Boltzmann constant, T is the temperature, and ΔE = ħγ
Here, γ is the gyromagnetic ratio and radiofrequency (RF) pulse. The pulse, tuned to the Larmor frequency, produces a perpendicular magnetic field known as the B1 field.
The presence of the
In order to generate spatial localization of the transverse magnetization additional spatially varying external magnetic fields are applied in the X, Y, and Z directions. These are called gradients
and their magnitudes are represented by the symbols GX, GY and GZ respectively. As the name implies, these fields are not uniform but instead add or subtract from the main field in order to produce a
linear variation of the field strength in each direction. This spatially encodes each spin in the imaging region with a unique field strength dependent on its location.
With these ideas in place the process of acquiring an image by playing out a pulse sequence can be described. Firstly, a gradient in the Z-direction, GZ, is applied. Next, an RF sync pulse is
transmitted during application of Gz whose bandwidth selects out a rectangular slab of spins along the longitudinal direction and tips them into the transverse plane (Figure 1). Following this, a
phase encode gradient is applied to spatially encode the spins along the Y-axis. Next, a frequency encode gradient is applied along the X-axis. During the readout period, the voltage induced in the
receiver coils by the rotating transverse magnetization is digitally sampled. A basic pulse sequence is depicted in Figure 2.
The entire signal acquisition process can be described by the signal equation:
The full derivation of (1) can be found in. The key observation is that the exponential term matches that of the two-dimensional Fourier transform. That is, s(t) is the Fourier transform of the
magnetization distribution, m. If m lies in image space and is described by spatial location variables <x, y>, then its Fourier transform, M(kx, ky) lies in k-space and <kx, ky> represent spatial
frequency. This form reveals that the position in k-space is dictated by the time-integral of the gradients applied. Thus the gradients can be thought of as tracing out a trajectory through k-space.
In this case, the Gx (or readout) gradient sweeps from left to right across k-space, and each successive Gy gradient (or phase encode) “blips” the trajectory up another line. As samples of s(t) are
acquired over time, k-space is filled in by placing each sample into its appropriate location as given by (2) & (3). An image is then recovered by simple application of the 2D inverse Fourier
transform. Most MRI data is acquired using a Cartesian sampling trajectory. That is, k-space is traversed in a raster scan and samples are acquired at regular intervals, <Δkx, Δky>. Samples are
acquired along the readout direction with a temporal spacing of Δt, or, a sampling rate of Fs = 1/Δt. The total distance travelled in each direction is given by kx,max = NxΔkx and ky,max = NyΔky,
where Nx, Ny are the number of samples acquired in each direction. The reconstructed image then possesses a field of view FOVx = 1/Δkx and FOVy=1/Δky. The distance travelled in k-space gives the
resolution: Δx = 1/2kx,max, Δy=1/2ky,max. The scan time for a single frequency encode is the repetition time, given by TR = NxΔt; and therefore for the total image, Timg = TR*Ny. K-space need not be
filled in by only a raster trajectory; in principle any arbitrary traversal may be used.
Lastly the source of contrast in an MR image will be described. Immediately following the application of the RF pulse all spins precess at the same frequency and in-phase. However inhomogeneity in
the local magnetic field as seen by each spin will cause them to precess at different frequencies. This results in a loss of phase coherence and causes the transverse magnetization to decay over
time. Broadly speaking there are two sources of field inhomogeneity. Random interactions of the protons in the form of dipole (magnetic) interactions, thermal rotation, and displacement occur at the
microscopic level. Collectively these are referred to as T2 effects and are tissue-dependent.
A second source arises from macroscopic spatial variations that occur in the main field due to imperfect magnet construction, imperfect shimming, and changes in susceptibility at tissue boundaries.
These effects are static (or slowly varying) in time and are known collectively as T2′ effects:
ΔB is the variation in the field arising from the aforementioned sources. As these effects are external, larger voxels may encompass larger regions of variability, and therefore exhibit greater
apparent dephasing. The contribution of both of these effects is known as the apparent T2, or T2* and is shorter than either effect:
The overall decay of the transverse magnetization is commonly described by an exponential equation:
Where M0 is the magnitude of the transverse magnetization immediately following the RF excitation. It is noted that this is only an approximation and the actual decay function is the Fourier
transform of the frequency distribution of spins within the voxel. At the same time, the longitudinal magnetization slowly returns to its original equilibrium value (M0) as excited protons lose
energy into the surrounding tissues and realign with the main field. This is known as T1 recovery and is also characterized by an exponential function:
Different tissues possess different T1 and T2 (and T2*) values. By adjusting the echo time (TE) and repetition time (TR) of a scan, the resulting image can be weighted towards T1 or T2 contrast.
Table 1 shows the effects of different TR and TE values on two tissues with different T1 and T2 values.
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An algorithm for approximate tandem repeats
A perfect single tandem repeat is defined as a nonempty string that can be divided into two identical substrings, e.g., abcabc. An approximate single tandem repeat is one in which the substrings are
similar, but not identical, e.g., abcdaacd. In this paper we consider two criterions of similarity: the Hamming distance (k mismatches) and the edit distance (k differences). For a string S of length
n and an integer k our algorithm reports all locally optimal approximate repeats, r = ūû, for which the Hamming distance of ū and û is at most k, in O (nk log(n/ k)) time, or all those for which the
edit distance of ū and û is at most k, in O (nk log k log(n/ k)) time. This paper concentrates on a more general type of repeat called multiple tandem repeats. A multiple tandem repeat in a sequence
S is a (periodic) substring r of S of the form r = u^au′, where u is a prefix of r and u′ is a prefix of u. An approximate multiple tandem repeat is a multiple repeat with errors; the repeated
subsequences are similar but not identical. We precisely define approximate multiple repeats, and present an algorithm that finds all repeats that concur with our definition. The time complexity of
the algorithm, when searching for repeats with up to k errors in a string S of length n, is O (nka log(n/ k)) where a is the maximum number of periods in any reported repeat. We present some
experimental results concerning the performance and sensitivity of our algorithm. The problem of finding repeats within a string is a computational problem with important applications in the field of
molecular biology. Both exact and inexact repeats occur frequently in the genome, and certain repeats occurring in the genome are known to be related to diseases in the human.
ASJC Scopus subject areas
• Modeling and Simulation
• Molecular Biology
• Genetics
• Computational Mathematics
• Computational Theory and Mathematics
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