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After the explanation of van der Waals forces by Fritz London, several scientists soon realised that his definition could be extended from the interaction of two molecules with induced dipoles to
macro-scale objects by summing all of the forces between the molecules in each of the bodies involved. The theory is named after H. C. Hamaker, who derived the interaction between two spheres, a
sphere and a wall, and presented a general discussion in a heavily cited 1937 paper.^[1]
The interaction of two bodies is then treated as the pairwise interaction of a set of N molecules at positions: R[i] {i:1,2,... ...,N}. The distance between the molecules i and j is then:
${\displaystyle R_{ij}=|R_{i}-R_{j}|}$
The interaction energy of the system is taken to be:
${\displaystyle V_{\mathrm {int} }^{1,2,...N}={\frac {1}{2}}\sum _{i=0}^{\mathbb {N} }\sum _{j=0(eq i)}^{\mathbb {N} }V_{\mathrm {int} }^{ij}(R_{ij})}$
where ${\displaystyle V_{\mathrm {int} }^{ij}}$ is the interaction of molecules i and j in the absence of the influence of other molecules.
The theory is however only an approximation which assumes that the interactions can be treated independently, the theory must also be adjusted to take into account quantum perturbation theory.
1. ^ Hamaker, H. C. (1937) The London – van der Waals attraction between spherical particles. Physica 4(10), 1058–1072. doi:10.1016/S0031-8914(37)80203-7 | {"url":"https://www.knowpia.com/knowpedia/Hamaker_theory","timestamp":"2024-11-10T07:45:45Z","content_type":"text/html","content_length":"76404","record_id":"<urn:uuid:dd781a23-77ee-4365-8687-88594b7fc419>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00866.warc.gz"} |
How do you find the 1st and 2nd derivative of e^(x^2)? | HIX Tutor
How do you find the 1st and 2nd derivative of #e^(x^2)#?
Answer 1
$f ' \left(x\right) = 2 x {e}^{{x}^{2}} , f ' ' \left(x\right) = 2 {e}^{{x}^{2}} \left(2 {x}^{2} + 1\right)$
#d/dx(e^x)=e^x" and " d/dx(e^(g(x)))=e^(g(x)).g'(x)#
#color(blue)"First derivative"#
#color(blue)"Second derivative"#
Differentiate using the #color(red)"product rule"#
Differentiating #f'(x)=2xe^(x^2)#
now #g(x)=2xrArrg'(x)=2#
and #h(x)=e^(x^2)rArrh'(x)=2xe^(x^2)#
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Answer 2
To find the first derivative of ( e^{x^2} ), use the chain rule:
[ \frac{d}{dx} e^{x^2} = 2x \cdot e^{x^2} ]
To find the second derivative, differentiate the first derivative with respect to ( x ) again:
[ \frac{d^2}{dx^2} e^{x^2} = (2 + 4x^2) \cdot e^{x^2} ]
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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NEP 30 — Duck typing for NumPy arrays - implementation
NEP 30 — Duck typing for NumPy arrays - implementation#
Peter Andreas Entschev <pentschev@nvidia.com>
Stephan Hoyer <shoyer@google.com>
Standards Track
We propose the __duckarray__ protocol, following the high-level overview described in NEP 22, allowing downstream libraries to return arrays of their defined types, in contrast to np.asarray, that
coerces those array_like objects to NumPy arrays.
Detailed description#
NumPy’s API, including array definitions, is implemented and mimicked in countless other projects. By definition, many of those arrays are fairly similar in how they operate to the NumPy standard.
The introduction of __array_function__ allowed dispatching of functions implemented by several of these projects directly via NumPy’s API. This introduces a new requirement, returning the NumPy-like
array itself, rather than forcing a coercion into a pure NumPy array.
For the purpose above, NEP 22 introduced the concept of duck typing to NumPy arrays. The suggested solution described in the NEP allows libraries to avoid coercion of a NumPy-like array to a pure
NumPy array where necessary, while still allowing that NumPy-like array libraries that do not wish to implement the protocol to coerce arrays to a pure NumPy array via np.asarray.
Usage Guidance#
Code that uses np.duckarray is meant for supporting other ndarray-like objects that “follow the NumPy API”. That is an ill-defined concept at the moment – every known library implements the NumPy API
only partly, and many deviate intentionally in at least some minor ways. This cannot be easily remedied, so for users of np.duckarray we recommend the following strategy: check if the NumPy
functionality used by the code that follows your use of np.duckarray is present in Dask, CuPy and Sparse. If so, it’s reasonable to expect any duck array to work here. If not, we suggest you indicate
in your docstring what kinds of duck arrays are accepted, or what properties they need to have.
To exemplify the usage of duck arrays, suppose one wants to take the mean() of an array-like object arr. Using NumPy to achieve that, one could write np.asarray(arr).mean() to achieve the intended
result. If arr is not a NumPy array, this would create an actual NumPy array in order to call .mean(). However, if the array is an object that is compliant with the NumPy API (either in full or
partially) such as a CuPy, Sparse or a Dask array, then that copy would have been unnecessary. On the other hand, if one were to use the new __duckarray__ protocol: np.duckarray(arr).mean(), and arr
is an object compliant with the NumPy API, it would simply be returned rather than coerced into a pure NumPy array, avoiding unnecessary copies and potential loss of performance.
The implementation idea is fairly straightforward, requiring a new function duckarray to be introduced in NumPy, and a new method __duckarray__ in NumPy-like array classes. The new __duckarray__
method shall return the downstream array-like object itself, such as the self object, while the __array__ method raises TypeError. Alternatively, the __array__ method could create an actual NumPy
array and return that.
The new NumPy duckarray function can be implemented as follows:
def duckarray(array_like):
if hasattr(array_like, '__duckarray__'):
return array_like.__duckarray__()
return np.asarray(array_like)
Example for a project implementing NumPy-like arrays#
Now consider a library that implements a NumPy-compatible array class called NumPyLikeArray, this class shall implement the methods described above, and a complete implementation would look like the
class NumPyLikeArray:
def __duckarray__(self):
return self
def __array__(self):
raise TypeError("NumPyLikeArray can not be converted to a NumPy "
"array. You may want to use np.duckarray() instead.")
The implementation above exemplifies the simplest case, but the overall idea is that libraries will implement a __duckarray__ method that returns the original object, and an __array__ method that
either creates and returns an appropriate NumPy array, or raises a``TypeError`` to prevent unintentional use as an object in a NumPy array (if np.asarray is called on an arbitrary object that does
not implement __array__, it will create a NumPy array scalar).
In case of existing libraries that don’t already implement __array__ but would like to use duck array typing, it is advised that they introduce both __array__ and``__duckarray__`` methods.
An example of how the __duckarray__ protocol could be used to write a stack function based on concatenate, and its produced outcome, can be seen below. The example here was chosen not only to
demonstrate the usage of the duckarray function, but also to demonstrate its dependency on the NumPy API, demonstrated by checks on the array’s shape attribute. Note that the example is merely a
simplified version of NumPy’s actual implementation of stack working on the first axis, and it is assumed that Dask has implemented the __duckarray__ method.
def duckarray_stack(arrays):
arrays = [np.duckarray(arr) for arr in arrays]
shapes = {arr.shape for arr in arrays}
if len(shapes) != 1:
raise ValueError('all input arrays must have the same shape')
expanded_arrays = [arr[np.newaxis, ...] for arr in arrays]
return np.concatenate(expanded_arrays, axis=0)
dask_arr = dask.array.arange(10)
np_arr = np.arange(10)
np_like = list(range(10))
duckarray_stack((dask_arr, dask_arr)) # Returns dask.array
duckarray_stack((dask_arr, np_arr)) # Returns dask.array
duckarray_stack((dask_arr, np_like)) # Returns dask.array
In contrast, using only np.asarray (at the time of writing of this NEP, this is the usual method employed by library developers to ensure arrays are NumPy-like) has a different outcome:
def asarray_stack(arrays):
arrays = [np.asanyarray(arr) for arr in arrays]
# The remaining implementation is the same as that of
# ``duckarray_stack`` above
asarray_stack((dask_arr, dask_arr)) # Returns np.ndarray
asarray_stack((dask_arr, np_arr)) # Returns np.ndarray
asarray_stack((dask_arr, np_like)) # Returns np.ndarray
Backward compatibility#
This proposal does not raise any backward compatibility issues within NumPy, given that it only introduces a new function. However, downstream libraries that opt to introduce the __duckarray__
protocol may choose to remove the ability of coercing arrays back to a NumPy array via np.array or np.asarray functions, preventing unintended effects of coercion of such arrays back to a pure NumPy
array (as some libraries already do, such as CuPy and Sparse), but still leaving libraries not implementing the protocol with the choice of utilizing np.duckarray to promote array_like objects to
pure NumPy arrays.
Previous proposals and discussion#
The duck typing protocol proposed here was described in a high level in NEP 22.
Additionally, longer discussions about the protocol and related proposals took place in numpy/numpy #13831
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Available functions
Functions can be used with any field or function, as long as the expected types matches, which means that any args element can also be another object defining a function.
Arithmetic functions
Can contain any number of arguments and will be added together.
"function": "add",
"args": ["field_slug", 3, 5]
Can contain any number of arguments and will be subtracted from each other.
"function": "sub",
"args": ["field_slug", 4]
Can contain any number of arguments and will multiply all arguments.
"function": "mul",
"args": ["field_slug", 3]
Must contain two arguments that will be divided.
"function": "div",
"args": ["field_slug", 2]
Date functions
Take one datetime argument as input and truncates the time part to only return the date.
"function": "date",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-09-22".
Takes one date or datetime as argument and returns the year as an integer.
"function": "year",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be 2020.
Takes one date or datetime as argument and returns the month as an integer.
"function": "month",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be 9.
Takes one date or datetime as argument and returns a year-month string.
"function": "year_month",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-09".
First day of week
Takes one date or datetime as argument and returns a date string of the first of day of the corresponding week.
Can be used as dimension to aggregate data by week.
"function": "first_day_of_week",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-09-21".
First day of month
Takes one date or datetime as argument and returns a date string of the first of day of the corresponding month.
Can be used as dimension to aggregate data by month.
"function": "first_day_of_month",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-09-01".
Year and Week Number starting Monday
Takes one date or datetime as argument and returns a year-week_number string, The week number is follows the ISO week number definition.
"function": "year_week_number_starting_monday",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-39".
Year and Week Number starting Sunday
Takes one date or datetime as argument and returns a year-week_number string, The week number is follows the ISO week number definition but transposed to start on Sunday.
"function": "year_week_number_starting_sunday",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-39".
Truncate date at hour
Takes one date or datetime as argument and returns a datetime string where we truncate until the hours.
Useful to aggregate metrics by a datetime dimension hourly.
"function": "truncate_date_at_hour",
"args": ["date_crawled"]
If date_crawled is "2020-09-22 08:07:00Z", the result of this function will be "2020-09-22 08:00:00Z".
Aggregation functions
To use aggregation functions, see Metrics how to use them in a query.
Takes one argument and will add all aggregated entries.
"function": "sum",
"args": ["field_slug"]
Takes one argument and will average all aggregated entries.
"function": "avg",
"args": ["field_slug"]
Weighted Average
Takes two arguments and will do a weighted average all aggregated entries, depending on the corresponding weight field.
"function": "weighted_avg",
"args": ["field_slug", "weight_field_slug"]
Takes one argument and will return the smallest value among the aggregated entries.
"function": "min",
"args": ["field_slug"]
Takes one argument and will return the smallest value among the aggregated entries.
"function": "max",
"args": ["field_slug"]
Takes one argument and returns the number of aggregated entries.
"function": "count",
"args": ["field_slug"]
Count Distinct
Takes one argument and returns the number of different aggregated entries.
"function": "count_distinct",
"args": ["field_slug"]
Approximated Count Distinct
Takes one argument and returns the approximate number of different aggregated entries.
If the query takes some time, using an approximate can speed up the query.
"function": "count_distinct_approx",
"args": ["field_slug"]
Count True
Takes one argument and returns the number of true aggregated entries.
"function": "count_true",
"args": ["field_slug"]
Count False
Takes one argument and returns the number of false aggregated entries.
"function": "count_false",
"args": ["field_slug"]
Count Null
Takes one argument and returns the number of null aggregated entries.
"function": "count_null",
"args": ["field_slug"]
Count Equals
Takes two arguments and returns the number of aggregated entries where both values are equal.
"function": "count_eq",
"args": ["field_slug_1", "field_slug_2"]
Count Greater than
Takes two arguments and returns the number of aggregated entries where the first argument is greater than the second.
"function": "count_gt",
"args": ["field_slug_1", "field_slug_2"]
Count Greater than or equals
Takes two arguments and returns the number of aggregated entries where the first argument is greater than or equal to the second.
"function": "count_gte",
"args": ["field_slug_1", "field_slug_2"]
Count Lower than
Takes two arguments and returns the number of aggregated entries where the first argument is lower than the second.
"function": "count_lt",
"args": ["field_slug_1", "field_slug_2"]
Count Lower than or equals
Takes two arguments and returns the number of aggregated entries where the first argument is lower than or equal to the second.
"function": "count_lte",
"args": ["field_slug_1", "field_slug_2"]
Conditional functions
Takes three arguments as input. The first argument must be a boolean, and can either be a field or a function like eq that will return a boolean.
If the first argument is true, it returns the second argument. If the first argument is false, it returns the third argument.
"function": "if",
"args": [true, 1, 2]
Will always return 1. A more realistic example, if the number of crawls from Google equals the number of visits from Google, we return true, else we return null.
"function": "if",
"args": [
"function": "eq",
"args": [
If not exists
Takes two arguments as input. Return the first argument if it is not null, else it returns the second one.
"function": "if_not_exists",
"args": ["search_console.period_0.count_clicks", 0]
When joining collections, we might not have any clicks for a crawled URL, because it never appeared on a SERP. It can therefore be null, but we don't want to handle null and just replace it with 0.
Takes two arguments as input and returns true if they are equal.
"function": "eq",
"args": ["field_slug_1", "field_slug_2"]
Not equal
Takes two arguments as input and returns true if they are different.
"function": "ne",
"args": ["field_slug_1", "field_slug_2"]
Lower than
Takes two arguments as input and returns true if the first one is lower than the second.
"function": "lt",
"args": ["field_slug_1", "field_slug_2"]
Greater than
Takes two arguments as input and returns true if the first one is greater than the second.
"function": "gt",
"args": ["field_slug_1", "field_slug_2"]
Lower than or equal
Takes two arguments as input and returns true if the first one is lower than or equal to the second.
"function": "lte",
"args": ["field_slug_1", "field_slug_2"]
Greater than or equal
Takes two arguments as input and returns true if the first one is greater than or equal to the second.
"function": "gte",
"args": ["field_slug_1", "field_slug_2"]
Takes one argument as input and returns true if the argument is not null.
"function": "exists",
"args": ["field_slug"]
Not exists
Takes one argument as input and returns true if the argument is null.
"function": "not_exists",
"args": ["field_slug"]
Takes one argument as input and returns it's negated result.
"function": "not",
"args": ["field_slug"]
Takes any number of arguments as input and returns true if all arguments are true.
"function": "and",
"args": ["field_slug_1", "field_slug_2"]
Takes any number of arguments as input and returns true if at least one argument is true.
"function": "or",
"args": ["field_slug_1", "field_slug_2"]
Takes one argument as input and handle the input as a constant.
Can be used to hard-code strings instead of interpreting them as field slugs.
"function": "literal",
"args": ["constant_value"]
Will return "constant_value" for each entry.
List functions
These functions can only be applied of list type fields.
Takes one argument and returns the first element of the list as result.
"function": "first",
"args": ["field_slug"]
Takes one argument and returns the list with sorted elements.
"function": "sort",
"args": ["field_slug"]
Array to String
Takes two arguments and returns a string composed of each element of the list, with is the first argument, with a separator which is the second argument.
"function": "array_to_string",
"args": ["field_slug", "separator"]
HTTP code functions
HTTP code family
Takes one argument as input and returns the family of the HTTP code.
Families are:
• 2xx for HTTP codes between 200 and 299, and 304
• exx for HTTP codes below 0
• Yxx for all other HTTP codes, where Y is the first digit of the HTTP code
"function": "http_code_family",
"args": ["http_code"]
If the HTTP code is 429, this function returns 4xx.
HTTP code quality
Takes one argument as input and returns the quality of the HTTP code.
Quality values are:
• good for HTTP codes between 200 and 299, and 304
• bad for all others
"function": "http_code_quality",
"args": ["http_code"]
If the HTTP code is 429, this function returns bad.
Ranges function
The ranges function is explained in details in Dimensions.
It is called with two arguments. The first one if the field with continuous values that we want to aggregate on. The second argument is a list of objects containing one, or both, of the from and to
"function": "ranges",
"args": [
"to": 2
"from": 2,
"to": 5
"from": 5,
"to": 8
"from": 8
String functions
Takes any number of string arguments and concatenates them together.
"function": "concat",
"args": [
"function": "literal",
"args": ["&botify"]
Will concatenate the URL with a &botify suffix. (the example could be improved to handle if URL path doesn't contain any query string to use a ? instead. Leaving that as an exercise to the reader).
Matches content
Takes two string arguments as input and returns true if the first argument is contained by the second argument one, when both arguments are normalized.
Normlization means ignoring any non-ascii characters and spaces in the strings.
Used to verify if a keyword is contained the title of the URL for instance:
"function": "matches_content",
"args": [
Values list functions
Values list are mostly used for RealKeywords Keyword Groups feature, in order group together multiple keywords into one list.
Once you know your values list identifier, you will be able to query it through some functions.
Match lists
Takes as input one object with two keys: field and lists.
"function": "match_lists",
"args": [
"field": "keyword",
"lists": [
And it will return the value of the item it matches in one of the lists.
Lists are evaluated in the order they are passed to the function.
Special functions
By dimension
Allows breaking down a metric on another dimension than the ones specified in the BQL query.
Takes as input the metric and the dimension on which you want to break it down.
Used for sparklines in the Botify Application to break down the data by date, additionally to the selected dimensions:
"function": "by_dimension",
"args": [
To JSON string
Takes one argument as input and transforms it to a JSON string.
Useful to get rid of list fields and fit them into a single row.
"function": "to_json_string",
"args": ["query_string_keys"] | {"url":"https://developers.botify.com/docs/functions","timestamp":"2024-11-12T20:27:07Z","content_type":"text/html","content_length":"531625","record_id":"<urn:uuid:75d93435-ca33-45d2-9405-7a18ed3db07b>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00065.warc.gz"} |
LM 2_6 Graphs of velocity versus time Collection
2.6 Graphs of velocity versus time by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
2.6 Graphs of velocity versus time
Since changes in velocity play such a prominent role in physics, we need a better way to look at changes in velocity than by laboriously drawing tangent lines on `x`-versus-`t` graphs. A good method
is to draw a graph of velocity versus time. The examples on the left show the x-t and v-t graphs that might be produced by a car starting from a traffic light, speeding up, cruising for a while at
constant speed, and finally slowing down for a stop sign. If you have an air freshener hanging from your rear-view mirror, then you will see an effect on the air freshener during the beginning and
ending periods when the velocity is changing, but it will not be tilted during the period of constant velocity represented by the flat plateau in the middle of the v-t graph.
Students often mix up the things being represented on these two types of graphs. For instance, many students looking at the top graph say that the car is speeding up the whole time, since “the graph
is becoming greater.” What is getting
Similarly, many students would look at the bottom graph and think it showed the car backing up, because “it's going backwards at the end.” But what is decreasing at the end is `v`, not `x`. Having
both the x-t and v-t graphs in front of you like this is often convenient, because one graph may be easier to interpret than the other for a particular purpose. Stacking them like this means that
corresponding points on the two graphs' time axes are lined up with each other vertically. However, one thing that is a little counterintuitive about the arrangement is that in a situation like this
one involving a car, one is tempted to visualize the landscape stretching along the horizontal axis of one of the graphs. The horizontal axes, however, represent time, not position. The correct way
to visualize the landscape is by mentally rotating the horizon 90 degrees counterclockwise and imagining it stretching along the upright axis of the `x-t` graph, which is the only axis that
represents different positions in space.
2.6 Graphs of velocity versus time by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license. | {"url":"https://www.vcalc.com/wiki/vCollections/LM+2_6+Graphs+of+velocity+versus+time+Collection","timestamp":"2024-11-13T09:24:45Z","content_type":"text/html","content_length":"46347","record_id":"<urn:uuid:6a3c4682-9133-47d1-b22b-617115aeb8fd>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00824.warc.gz"} |
Understanding Mathematical Functions: How To Calculate Functions
Mathematical functions are a fundamental concept in mathematics that represent the relationship between input and output. They are essential in various fields, including engineering, physics, and
computer science. Understanding how to calculate functions is crucial for solving real-world problems and analyzing data. In this blog post, we will explore the definition of mathematical functions
and the importance of understanding how to calculate them.
Key Takeaways
• Mathematical functions represent the relationship between input and output and are essential in various fields.
• Understanding how to calculate functions is crucial for solving real-world problems and analyzing data.
• Functions can be defined and categorized into different types, such as linear, quadratic, and exponential.
• Components of a function include input, output, domain, and range, which are important for understanding its behavior.
• Techniques for calculating functions include substitution, graphical, and algebraic methods, and it's important to practice and use tools for complex functions.
Understanding Mathematical Functions: How to Calculate Functions
In mathematics, a function is a relation between a set of inputs and a set of possible outputs with the property that each input is related to exactly one output.
A. Definition of a function
A mathematical function can be thought of as a rule that assigns each input to exactly one output. It can be represented as f(x) = y, where f is the function, x is the input, and y is the output. The
input x is often referred to as the independent variable, while the output y is the dependent variable.
B. Examples of functions (linear, quadratic, exponential)
There are various types of mathematical functions, some of which include:
• Linear function: A linear function is a function that can be graphically represented as a straight line. Its general form is f(x) = mx + b, where m is the slope and b is the y-intercept.
• Quadratic function: A quadratic function is a function that can be graphically represented as a parabola. Its general form is f(x) = ax^2 + bx + c, where a, b, and c are constants and a ≠ 0.
• Exponential function: An exponential function is a function in which the variable appears in the exponent. Its general form is f(x) = a^x, where a is a positive constant.
Understanding the components of a function
When it comes to understanding mathematical functions, it's important to grasp the key components that make up a function. These components include the input and output, as well as the domain and
range of the function.
A. Input and output
The input of a function is the independent variable, which is typically denoted as 'x'. The output of a function is the dependent variable, usually denoted as 'f(x)'. This means that for every input
value, there is a corresponding output value. Understanding the relationship between the input and output is crucial in calculating functions.
B. Domain and range
The domain of a function refers to all the possible input values that the function can accept. It is important to identify the domain of a function to ensure that the input values are valid. The
range of a function, on the other hand, refers to all the possible output values that the function can produce. Determining the range of a function is essential in understanding the potential outputs
of the function.
Techniques for calculating functions
Understanding how to calculate mathematical functions is essential for solving problems in various fields such as engineering, physics, and economics. Several techniques are used to calculate
functions, including substitution method, graphical method, and algebraic method.
A. Substitution method
• Definition:
The substitution method involves replacing a variable in a function with a specific value, and then evaluating the resulting expression.
• Steps:
To calculate a function using the substitution method:
□ Replace the variable in the function with the given value.
□ Simplify the resulting expression by performing the necessary operations.
• Example:
Calculate the value of the function f(x) = 2x + 3 when x = 4 using the substitution method.
f(4) = 2(4) + 3 = 8 + 3 = 11
B. Graphical method
• Definition:
The graphical method involves plotting the function on a graph and determining the value of the function at a specific point.
• Steps:
To calculate a function using the graphical method:
□ Plot the function on a graph.
□ Locate the point corresponding to the given value of the variable on the graph.
□ Determine the value of the function at that point.
• Example:
Using the graph of the function f(x) = x^2, determine the value of f(3).
f(3) = 3^2 = 9
C. Algebraic method
• Definition:
The algebraic method involves manipulating the function algebraically to determine its value at a specific point.
• Steps:
To calculate a function using the algebraic method:
□ Replace the variable in the function with the given value.
□ Simplify the resulting expression using algebraic techniques such as factoring, expanding, or solving equations.
• Example:
Calculate the value of the function g(x) = 2x^2 - 5x + 3 when x = 2 using the algebraic method.
g(2) = 2(2)^2 - 5(2) + 3 = 8 - 10 + 3 = 1
Common functions and their calculations
Mathematical functions are essential tools for modeling and analyzing real-world situations. Understanding how to calculate different types of functions is fundamental for anyone working in fields
that require problem-solving and critical thinking. In this blog post, we will explore some of the most common mathematical functions and their calculations.
Linear functions
A linear function is a basic algebraic function that represents a straight line on a graph. The general form of a linear function is y = mx + b, where m is the slope of the line and b is the
• To calculate the value of y for a given x, simply plug the value of x into the equation and solve for y.
• Alternatively, if you have two points on the line, you can use the formula for the slope to calculate the value of m and then use the equation of the line to find the value of b.
Quadratic functions
A quadratic function is a type of function that can be represented by a parabola on a graph. The general form of a quadratic function is y = ax^2 + bx + c, where a, b, and c are constants.
• To calculate the value of y for a given x, simply plug the value of x into the equation and solve for y.
• If you are given the coordinates of the vertex of the parabola, you can use the formula for the vertex to find the value of y.
Exponential functions
An exponential function is a type of function that grows or decays at a constant rate. The general form of an exponential function is y = a * b^x, where a and b are constants, and b is the base of
the exponential.
• To calculate the value of y for a given x, simply plug the value of x into the equation and solve for y.
• If you are given the initial value of y and the growth or decay rate, you can use the formula for exponential growth/decay to calculate the value of y at a specific time or interval.
Tips for effectively calculating functions
Understanding mathematical functions is essential for various fields such as engineering, physics, and economics. Here are some tips for effectively calculating functions.
A. Practice solving various types of functions
• Understand the basics:
Before diving into complex functions, it is crucial to have a strong understanding of basic mathematical concepts such as addition, subtraction, multiplication, and division.
• Start with simple functions:
Practice solving simple linear and quadratic functions to build a foundation for solving more complex functions.
• Explore different types of functions:
Familiarize yourself with various types of functions such as exponential, logarithmic, trigonometric, and polynomial functions.
B. Use calculators and software for complex functions
• Utilize graphing calculators:
Graphing calculators can help visualize functions and identify key points such as intercepts, maxima, and minima.
• Explore mathematical software:
Mathematical software such as MATLAB, Mathematica, or Python can handle complex mathematical functions and provide accurate solutions.
• Take advantage of online resources:
There are various online tools and resources available for solving mathematical functions, which can be beneficial for learning and practicing.
In conclusion, it is essential to understand and calculate functions as they play a crucial role in various fields such as mathematics, science, engineering, and economics. Being proficient in
calculating functions allows for better problem-solving and decision-making. Continuous practice and learning are necessary for improving calculation skills and gaining a deeper understanding of
mathematical functions.
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Solutions of polymers
Tutorial: Solutions of polymers with SEB
Contributors: Lau Blom Grøndahl & Carsten Svaneborg (FKF SDU).
Illustration of a Gaussian chain model of a single flexible polymer molecule.
Before you start
Learning outcomes
In this tutorial you will learn about flexible polymers, and their form factor, in particular
• the conformational statistics of polymers,
• the Gaussian model for flexible polymers,
• how to derive the Debye form factor,
• and how to use SEB to plot the form factor of flexible polymer.
Polymers are made by chemically linking many identical monomers together to form long linear chains. Typical polymers are polyethylene (e.g. plastic bags), polycarbonate (e.g. automotive),
polyisoprene (e.g. rubbers), polystyrene (e.g. plastic cutlery), and polydimethylsiloxane (e.g. lubricants). There are also many biopolymers, such as DNA, collagen, fibrin, polypeptides,
polynucleotides, and polysaccharides which plays various roles in our cells and body.
Model of a single polymer
Here we will focus on a single synthetic polymer for instance polyethylene. The chemical bonds beween subsequent carbon atoms are 1.54Å long. The angle between subsequent carbon bonds is 109.5
degrees. These are set by fixed by chemistry. From a polymer physics pespective, what is interesting is the torsional angles defined by three subsequent bonds. The torsion angles can roughly be in
three different states (trans, gauge+, gauge-) where the energy difference of gauge+/gauge- and trans is comparable to room temperature. At absolute zero all torsion angles will be trans, and the
polymer will have a rod-like zig-zag conformation and have zero conformational entropy.
If we imagine increasing the temperature, then the polymer can increase its entropy by introducing additional gauge+/gauge- torsional states. This corresponds to introducing a number of kinks along
the polymer. The physics is analogous to the formation of vacancy-interstitial pairs in a crystal at finite temperature. Hence a single polymer is a very flexible and can adopt a vast number of
conformations depending on the random sequence of trans/gauge+/gauge- torsion angles along the polymer.
If we zoom out a bit from the detailed chemical structure, then the simplest polymer model, used in polymer physics, is that of a random-walk. The random walk consists of $N$ steps, where each step
has a constant length $b$ (aka the Kuhn length). All steps are in a random direction, and they are assumed to be statistically independent from each other. Then the total contour length of the random
walk is $L=Nb$, and its easy to show that the average spatial distance from one end to the other of the polymer is given by $\langle R^2\rangle = Nb^2$.
Then we can ask what is the probably distribution of the spatial distance between the ends or any internal monomers along the polymer? Let $n$ denotes the number of steps between the monomers (or the
ends in which case $n=N$). With a bit of mathematics, we obtain a Gaussian distribution:
$$ P(r;n) = \left(\frac{3}{2\pi b^2 n}\right)^{3/2} \exp\left(- \frac{3 r^2}{2 b^2 n} \right).$$
Mathematical note: here we have implicitly taken the limit $N\rightarrow\infty$,$b\rightarrow0$ keeping $\langle R^2\rangle$ constant. Thus here we regard $n$ as a real number $\in[0,N]$, and not
just an integer. This is known as the Gaussian polymer model.
Physically this corresponds to replacing each straight step with a more fine-detailed random walk. Thus turning the random walk model into a fractal. This approximation is ok, since polymer physics
is dominated by the large scale conformational properties. The short-scale chemical details rarely matters. However, for describing real polymers its clear that a Gaussian polymer models on length
scales below $b$. Later we will see how the polymer scattering is affected by this.
From real polymers to Gaussian polymer models
Different chemical polymer species corresponds to different values of the Kuhn length $b$. Different molecular weights of polymers corresponds to different number of random walk steps. See e.g. R.
Everaers et al. for the Kuhn length of other commodity polymers "Kremer−Grest Models for Commodity Polymer Melts: Linking Theory, Experiment, and Simulation at the Kuhn Scale" Macromolecules 53, 1917
(2020) or the polymer handbook.
Some examples:
• Polyethylene is relatively stiff and has a Kuhn length of $b=15.40Å$. One random walk step contains $6$ monomers or $12$ $CH_2$ groups.
• A much more flexible polymer is cis-polyisoprene (cis-PI) with $b=9.45Å$. One random walk step contains just $1.89$ monomers.
• We can also apply random walk models to DNA in which case $b=90nm$ and one step corresponds to $265$ base-pairs. Note these numbers depends on salt concentration and temperature.
• Actin is a very stiff biopolymer with $b=35.6\mu m$ that is an essential mechanical component in our muscles. However, in biology actin never gets so long as to be a polymer that is $N<1$.
Exercise 1
A polymer molecule has a contour length $L$ and a characteristic spatial size known as the radius of gyration $R_g=\sqrt{\langle R^2\rangle/6}$. Using the random walk model above
• 1a: calculate $N$, $R_g$ and $L$ for a polyethylene molecule with $1.000$ monomers.
• 1b: calculate $N$, $R_g$ and $L$ for a polyisoprene molecule with $1.000$ monomers.
• 1c: calculate $N$, $R_g$ and $L$ for a DNA molecule with $10^7$ base-pairs (approximately one chromosome).
The size of a human cell nucleus is $\sim 10\mu m$, how does that compare to the size of denatured chromosome DNA in solution?
Derivation of form factor of a single polymer
We think of a single polymer in a solution, where each monomer as a point scatterer, then the form factor can be stated as
$$F(q) = \left<\frac{\sin(q r)}{qr} \right>, $$
which is known as the Debye Formula. Since the polymer is free to rotate any way, the form factor only depends on the magnitute of the momentum transfer $q$, and the scattering pattern will be axis
symmetric around the direct beam.
The difficulty here is that polymers are flexible, so we have to average the scattering over the conformations they can adopt. In particular, we have to perform the average over 1) all pairs of
monomers along the polymer, 2) the distribution of spatial distances between a given pair of monomers.
Sketch of polymer showing the meaning of the symbols.
To average over pairs of monomers, we first imagine that each monomer is randomly chosen from an uniform distribution in the interval $[0,N]$. Thus $P(n_1)=1/N$ for $n_1\in [0,N]$ and zero elsewhere.
Hence the average over the first monomer position $n_1$ corresponds to performing the integral $\int_0^N dn_1 N^{-1} \cdots$ and similar for the second monomer position $n_2$. Then for two given
monomers $n_1,n_2$, the number of steps between them is $n=|n_1-n_2|$, and thus $P(r;n)$ above gives the probability of the distance between the monomers being $r$. Using spherical coordinates we can
integrate over the distribution of spatial distances: $\int_0^\infty dr 4\pi r^2 P(r;|n_1-n_2|) \cdots $. Thus we can express the three averages as:
$$ F(q) = \int_0^N \frac{dn_1}{N} \int_0^N \frac{dn_2}{N} \int_0^\infty dr 4\pi r^2 \left(\frac{3}{2\pi b^2 |n_1-n_2|}\right)^{3/2} \exp\left(- \frac{3 r^2}{2 b^2 |n_1-n_2|} \right) \frac{\sin(q r)}
after quite a bit of paper-and-pencil work (left as an exercise), the result is becomes the nice short expression $$ F(q) = \frac{2(\exp(-x)-1+x)}{x^2}, $$ with $x=q^2 R_g^2$ and $R_g^2=\langle R^2\
rangle/6=b^2 N/6$ is the radius of gyration of the polymer molecule. This describes the typical spatial extend of the molecule. Note that because all the three distributions that went into the
integrals were normalized, then the form factor is also normalized such that $F(0)=1$. This result was first derived by P. Debye (P. Debye, J. Phys. Coll. Chem 51, 18–32 (1947)).
Scattering Equation Builder (SEB)
Scattering Equation Builder (SEB) is a C++ library for analytical derivation of form factors of complex structures. The structures are build out of basic building blocks called sub-units. Polymers
and rods are two of the sub-units supported by SEB.
Before you can use SEB you need to install a working C++ compiler, the GiNaC, GSL and CLN libraries, and the SEB source code itself. See GitHub for the details of how to install SEB on various
operating system. Importantly, you need to remember the folder, where you put the SEB source code. It has a subfolder "work" where you can save and compile your own programs.
Exercise 2: Gaussian polymer with SEB
To calculate the form factor of a Gaussian polymer, cut'n'paste the following C++ program into an text editor (e.g. notepad). Save it as "Polymer.cpp" in the work folder under the SEB installation.
// Include SEB functionality
#include "SEB.hpp"
int main()
// Create world of sub-units
World w("World");
// Add a single polymer-subunit named "A"
GraphID p = w.Add(new GaussianPolymer(), "A");
// Wrap unit in a structure named Structure (this will make sense later)
w.Add(p, "Structure");
// Print out equation for the form factor
ex F=w.FormFactor("Structure");
cout << "Form Factor= " << F << "\n";
// To evaluate the equation, we need to define value of paramters
ParameterList params;
w.setParameter(params,"Rg_A",1); // Radius of gyration for "A" polymer
w.setParameter(params,"beta_A",1); // Scattering length
// Choose q values
DoubleVector qvec=w.logspace(0.01, 10.0, 1000 );
// Use Evaluate to save form factor data to a file
w.Evaluate( F, params, qvec, "formfactor_polymer.q", "Form factor of a polymer with beta=1 and Rg=1Å.");
Commandline prompt on Linux showing how to compile and run the program, also shown is some
of the content of the file "formfactor_polymer.q".
In a terminal, navigate to the folder where you installed SEB (source/SEB in my case above). In that folder run "make". The first time you run make, it will compile the whole SEB library and your own source file. Next time it will only compile the source files you have changed. In my example above, it only compiles "work/Polymer.cpp". The resulting executable is "work/Polymer" (work/Polymer.exe on Windows). Run the executable by typing "work/Polymer".
In the example above you can see it prints out the equation for the Debye form factor. After it has run, it has created a file "formfactor_polymer.q" in the current folder. The file starts with some comments, and then has two columns of numbers: q and F(q). Try plotting the file in log-log representation.
Log-log plot of the form factor of a Gaussian polymer model with $Rg=1$ AU (red), $2$ AU (green), $0.5$ AU (black), where AU is an arbitrary length unit.
Your plot should look similar to the plot above. Since the form factor only depends on $x=q^2 R_g^2$ changing $R_g$ corresponds to a horizontal shift of the curve. Remember that log-log representation is good for reading off powerlaws. For $qR_g \ll 1$ corresponding to large scales the form factor is essentially flat thus $F(q) ~ q^0$, whereas for scales smaller than the polymer size $qR_g \gg 1$ the form factor follows a powerlaw $F(q)\sim q^{-2}$. These exponents are in fact directly related to the fractal dimension of the polymer. Seen at large scales any finite sized object is a point with $0$ fractal dimension, whereas a random-walk has fractal dimension $2$. This is a reflection of the $\langle R^2 \rangle \sim L$ relation between spatial distances and contour distances.
A note on units: SEB does not have any specific choice of unit built in. The reason is that
scattering expressions only depend on dimensionless combinations such as $qR_g$. Thus if choose AU=$m$, $\mu m$, $nm$, or $Å$ when you specify the radius of gyration. That is in the code you write $R_g=1$ but with some unit $AU$, then you have chosen AU$^{-1}=$$m^{-1}$, $\mu m^{-1}$, $nm^{-1}$, or $Å^{-1}$ as the unit for the $q$. In this case you would plot $q/AU^{-1}$ which is the same as $q AU$.
Exercise 3
For the three polymers in Exercise 1 calculate the following:
□ If $R_g$ is the characteristic spatial scale of a polymer, what is the corresponding characteristic reciprocal length $q$ value?
□ The step/Kuhn length $b$ sets a lower spatial length scale where Gaussian chain statistics applies. Calculate the corresponding $q_{max}$ up to which the Debye expression applies.
□ Use SEB to calculate the form factors of the three polymers in relevant range of $q$ values. Sketch what you expect the log-log plot to look like. Then make the plot. Did you get what you expected? | {"url":"https://sastutorials.org/SEB/SEB_Polymers/SEB_polymer.html","timestamp":"2024-11-08T18:05:00Z","content_type":"text/html","content_length":"17261","record_id":"<urn:uuid:6a96b674-be98-4d0f-9fb4-5acfe95d9e58>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00161.warc.gz"} |
Usotatami is a logic puzzle published by Nikoli. A rectangular or square grid contains numbers in some cells. The aim is to divide the grid into rectangular regions such that each region contains
exactly one number. Every region must be exactly one cell wide; the length of the other side is NOT equal to the number in this region. A grid dot must not be shared by the corners of four regions.
Cross+A can solve puzzles from 3 x 3 to 30 x 30. | {"url":"https://cross-plus-a.com/html/cros7uttm.htm","timestamp":"2024-11-06T23:36:21Z","content_type":"text/html","content_length":"1348","record_id":"<urn:uuid:4bcd7dd9-b166-4a78-a379-36cdbbdeaecb>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00384.warc.gz"} |
Can we use gravitational waves to rule out extra dimensions – and string theory with it?
Gravitational Waves,
Computer simulation.
Credits: Henze, NASA
Probably not.
Last week I learned from New Scientist that “
Gravitational waves could show hints of extra dimensions.
” The article is about a paper which recently appeared on the arxiv:
The claim in this paper is nothing but stunning. Authors Andriot and Gómez argue that if our universe has additional dimensions, no matter how small, then we could find out using gravitational waves
in the frequency regime accessible by LIGO.
While LIGO alone cannot do it because the measurement requires three independent detectors, soon upcoming experiments could either confirm or forever rule out extra dimensions – and kill string
theory along the way. That, ladies and gentlemen, would be the discovery of the millennium. And, almost equally stunning, you heard it first from New Scientist.
Additional dimensions are today primarily associated with string theory, but the idea is much older. In the context of general relativity, it dates back to the work of Kaluza and Klein the 1920s. I
came across their papers as an undergraduate and was fascinated. Kaluza and Klein showed that if you add a fourth space-like coordinate to our universe and curl it up to a tiny circle, you don’t get
back general relativity – you get back general relativity plus electrodynamics.
In the presently most widely used variants of string theory one has not one, but six additional dimensions and they can be curled up – or ‘compactified,’ as they say – to complicated shapes. But a
key feature of the original idea survives: Waves which extend into the extra dimension must have wavelengths in integer fractions of the extra dimension’s radius. This gives rise to an infinite
number of higher harmonics – the “Kaluza-Klein tower” – that appear like massive excitations of any particle that can travel into the extra dimensions.
The mass of these excitations is inversely proportional to the radius (in natural units). This means if the radius is small, one needs a lot of energy to create an excitation, and this explains why
he haven’t yet noticed the additional dimensions.
In the most commonly used model, one further assumes that the only particle that experiences the extra-dimensions is the graviton – the hypothetical quantum of the gravitational interaction. Since we
have not measured the gravitational interaction on short distances as precisely as the other interactions, such gravity-only extra-dimensions allow for larger radii than all-particle extra-dimensions
(known as “universal extra-dimensions”.) In the new paper, the authors deal with gravity-only extra-dimensions.
From the current lack of observation, one can then derive bounds on the size of the extra-dimension. These bounds depend on the number of extra-dimensions and on their intrinsic curvature. For the
simplest case – the flat extra-dimensions used in the paper – the bounds range from a few micrometers (for two extra-dimensions) to a few inverse MeV for six extra dimensions (natural units again).
Such extra-dimensions do more, however, than giving rise to a tower of massive graviton excitations. Gravitational waves have spin two regardless of the number of spacelike dimensions, but the number
of possible polarizations depends on the number of dimensions. More dimensions, more possible polarizations. And the number of polarizations, importantly, doesn’t depend on the size of the
extra-dimensions at all.
In the new paper, the authors point out that the additional polarization of the graviton affects the propagation even of the non-excited gravitational waves, ie the ones that we can measure. The
modified geometry of general relativity gives rise to a “breathing mode,” that is a gravitational wave which expands and contracts synchronously in the two (large) dimensions perpendicular to the
direction of the wave. Such a breathing mode does not exist in normal general relativity, but it is not specific to extra-dimensions; other modifications of general relativity also have a breathing
mode. Still, its non-observation would indicate no extra-dimensions.
But an old problem of Kaluza-Klein theories stands in the way of drawing this conclusion. The radii of the additional dimensions (also known as “moduli”) are unstable. You can assume that they have
particular initial values, but there is no reason for the radii to stay at these values. If you shake an extra-dimension, its radius tends to run away. That’s a problem because then it becomes very
difficult to explain why we haven’t yet noticed the extra-dimensions.
To deal with the unstable radius of an extra-dimension, theoretical physicists hence introduce a potential with a minimum at which the value of the radius is stuck. This isn’t optional – it’s
necessary to prevent conflict with observation. One can debate how well-motivated that is, but it’s certainly possible, and it removes the stability problem.
Fixing the radius of an extra-dimension, however, will also make it more difficult to wiggle it – after all, that’s exactly what the potential was made to do. Unfortunately, in the above mentioned
paper the authors don’t have stabilizing potentials.
I do not know for sure what stabilizing the extra-dimensions would do to their analysis. This would depend not only on the type and number of extra-dimension but also on the potential. Maybe there is
a range in parameter-space where the effect they speak of survives. But from the analysis provided so far it’s not clear, and I am – as always – skeptical.
In summary:
I don’t think we’ll rule out string theory any time soon.
[Updated to clarify breathing mode also appears in other modifications of general relativity.]
8 comments:
1. Hi Sabine,
very good post, thank you! I have two remarks on what you say they claim "While LIGO alone cannot do it because the measurement requires three independent detectors, soon upcoming experiments
could either confirm or forever rule out extra dimensions – and kill string theory along the way."
1/ Since one cannot predict the amplitude of the effects they describe, there is no way one can rule out extra-dimensions from a search of them : one can only put limits on their amplitudes. At
the same time, extra-polarization modes are generic in non-GR theories of gravitation. Therefore, observing them wouldn't be a confirmation of extra-dimension either. Although observing a
discrete set of high frequency signals in all polarization modes may be a smoking gun.
2/ It turns out that the LIGO detectors alone can detect extra-polarization modes. More, the two LIGO and the Virgo detectors can distinguish all the different extra-polarization modes. Not for
transient events indeed, but from the stochastic gravitational wave background: http://adsabs.harvard.edu/abs/2017arXiv170408373C (see figure 14 and 15).
2. Hi Olivier,
Thanks for pointing out, I have added a clarification on the first point. 2.) Interesting, I'll have a look at this!
3. Great post on a topic of much interest today, Dr. Bee. I don't read New Scientist any more, and I am far to skeptical of string theory to buy into it (but mainly because I can't imagine Nature
being so complicated).
Question: might the extra dimensions in string theory just be a bunch of parameters needed to make it work, kind of like curve fitting data to a polynomial?
4. "... extra dimensions add another way for gravitational waves to make space shape-shift, called a breathing mode. ...space expands and contracts as gravitational waves pass through, in addition
to stretching and squishing." Expect added degrees of freedom in gravitational radiation orbital decay and elsewhere, yes?
Pulsar binaries lack orbital decay anomalies. PSR B1913+16 Hulse–Taylor, pulsar-neutron star, 7.75-hr orbit. PSR J0348+0432, pulsar-white dwarf, 2.46-hr orbit. PSR J1903+0327, pulsar- solar star,
95.17-day orbit.
Hydrogen Lyman-alpha, 21-cm hyperfine transition, Lamb shift.
...H-like (91+) and He-like (90+) uranium ions
...Table 5.22; Section 5.9 Muonic atom decay
... Mössbauer spectroscopy. Fe-57, 14.4 KeV gamma-ray, 5×10^(-9) eV linewidth, 10^(-12) resolution.
...Casimir effect, 29.5 - 86 nm interaction range.
Dark, sterile, see-saw, selective extra dimensions?
5. Bill,
The number of dimensions is a parameter and a specific number is necessary to 'make it work' and I think I really don't understand the question.
6. With the exception that this is not KK but a warped compactification.
KK ansatz is a direct product (e.g. M₄ x S¹) for the vacuum but you always get moduli in the uncompactified directions.
To stabilize e.g. via flux compactification you have to use Warped ansatz (due to the flux you can't have a direct product anymore) where you have a warp factor in front of the line element of
the 4d metric that depends on the compactified coordinates.
But forget all these subtleties the big news is that String theory is falsifiable after all and It can finally get the legit scientific theory stump by the popperian bureaucrats!!
7. As the second author of the paper in question I fully agree with Olivier's comment (first comment here above), particularly with his first point. Moreover, I believe the whole point of vue of
your post is a little forced. Talking about falsifying string theory on the basis of our analysis is quite a stretch, to say the least, and I don't think this viewpoint conveys the message very
well or fits with the study at all. While ruling out string theory is a legitimate question to ask, I do not think one can answer it based on our simple study, which focuses on the propagation
(ignoring sources and emission) and furthermore is not directly related to string theory but more genetically to extra dimensions.
8. Gustavo,
Ruling out extra-dimensions is pretty much the only way to rule out string theory (at low energies). I agree with you of course that it wouldn't be as easy as my blogpost might have made it sound
because many other possible explanations for the absence of a signal would have to be ruled out.
COMMENTS ON THIS BLOG ARE PERMANENTLY CLOSED. You can join the discussion on Patreon.
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The Argument About Acceleration in Physics
We are going to deal with that later within this book. This is how it is with time. You may not have the response to that question!
Well, we could take thorough data and analyze it. Tell me, what do you need to do first!
What is Really Happening with Acceleration in Physics
The equations necessary to decide on the unknown are listed above. In physics, however, they don’t have the exact meaning and they’re distinct concepts. Physical quantities that are completely
specified by just giving out there magnitude are called scalars. All this usually means that the real” calculation is dependent on factors and assumptions that weren’t explicitly mentioned in the
It’s possible to so that it is possible to print them out to try on your own. The humble spring is among the three most important mechanical components to learn how to model. You’re able to calculate
velocity by utilizing a very simple formula which uses rate, distance, and time. Wasted energy takes the shape of heat and at times sound or light. Today we will likely look at volatile formulas.
Acceleration in Physics Fundamentals Explained
In your physics lab, you are requested to take a meter stick and set a mark on the ground of a lengthy hallway in 1 meter intervals for a total of 8 meters. The normal force points toward the middle,
therefore it ought to be given the positive price. Now locate the entire distance traveled. Now find the entire distance traveled.
Acceleration in Physics: the Ultimate Convenience!
That’s because lift offers upward push once it starts. In other words, the net force always points toward the middle of motion. There is going to be a place where the air resistance is big enough to
balance the gravitational force.
In the event http://saibabbarjewellers.com/2019/11/14/the-definitive-solution-for-what-is-reflection-in-physics-you-can-find-out-about-today/ the road curves, you will feel your body is attempting to
move toward the outside the curve. Such is true of orbiting satellites. Ariel Atom is among the fastest accelerating cars on the planet.
And fortunately, there’s a shortcut. The comprehensive comparison in tabular format is provided below. An object doesn’t always do precisely the same speed. Velocity is the speed at which an object
moves from one area to another.
What You Should Do to Find Out About Acceleration in Physics Before You’re Left Behind
The conceptual introduction is completed. By way of example, during a normal visit to school, there are lots of changes in speed. In each one of these activities, we’re altering the status of motion.
It is very important to try to remember this distinction. The common velocity formula describes the association between the length of your route and the time it can take to travel.
Furthermore, questions arose about the duration of the normal recovery curve for the majority of people. Now that would be a whole lot of work. The bulk of the predictions from these sorts of
theories are numerical. There are SEVEN unique types of word problems to choose from, that range from easy to advanced, and that means that you can create a wonderful number of worksheets.
By definition, acceleration is the very first derivative of velocity with regard to time. It’s complicated to memorize each and every arrangement of the 2 equations and I recommend that you practice
creating new combinations from the original equations. You are able to figure out the normal velocity of an object employing the formula for velocity. Yet they are quite fair units when you start to
think about the definition and equation for acceleration. Although this is definitely the most basic acceleration equation, there are different approaches to solve it as well.
The Ideal Approach to Acceleration in Physics
In Centripetal Force, we’ll think about the forces involved with circular motion. Motion along a curved path might be considered effectively one-dimensional if there is but one degree of freedom for
those objects involved. A real general statement would need to consider any initial velocity and the way the velocity was changing. Therefore a unit for force is really the kilogram-meter per second
squared. It’s vector, meaning it has both magnitude and direction.
Acceleration in Physics: No Longer a Mystery
This resembles when you press back on the gas pedal in a car on a straight portion of the freeway. Say you develop into a drag racer so as to analyze your acceleration farther down the dragway. When
you accelerate the auto, it usually means that you increase the speed of the vehicle. You might have to comprehend what your normal speed was since the speed varied during your travels. In that
situation, the speed of the vehicle can be measured but not its velocity. Locate the typical velocity of this vehicle.
Let’s look at a fast example to comprehend the difference between speed and acceleration. Information about one of the parameters can be employed to determine unknown information concerning the other
parameters. A velocity is constant if the next two conditions are satisfied.
It’s usually referred to as jerk. For our purposes within this lesson, we will concentrate on instantaneous measurements. It’s also essential to note that the brain is in danger for damage at
numerous points.
What Everybody Dislikes About Acceleration in Physics and Why
We can then find out the acceleration rate. This example illustrates acceleration as it is often understood, but acceleration in physics is considerably more than simply increasing speed. You know
the acceleration and the previous speed, and you would love to know about the full distance required to get to that speed. It’s the net force that is connected to acceleration.
To put it simply, when velocity changes, we’ve got acceleration. You are able to experience a speedy acceleration when you begin the vehicle from a zero relative velocity or from a standstill into
the greater speed. Over this kind of interval, the ordinary velocity becomes the instantaneous velocity or the velocity at a particular instant. Determine the normal speed and the typical velocity.
Acceleration in Physics – Is it a Scam?
So long as you’re consistent within an issue, it doesn’t matter. This is an easy problem, but nonetheless, it always helps to visualize it. | {"url":"https://www.lannakingdomelephantsanctuary.com/the-argument-about-acceleration-in-physics/","timestamp":"2024-11-01T20:39:48Z","content_type":"text/html","content_length":"26347","record_id":"<urn:uuid:76e63b28-6749-4e2e-aab6-de79de1aff01>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00000.warc.gz"} |
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Question Text Find the perimeter of trapezium whose parallel side are 13 cm and 20 cm and non parallel sides are 7 cm
Updated On Mar 17, 2024
Topic All topics
Subject Mathematics
Class Class 9
Answer Type Video solution: 3
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Lecture 008 - Rollercoasters
The problem: for $n$ people, figure out who is brave by sending them to the rollercoaster. If a person is brave, and being sent to rollercoaster, and the number of people on the rollercoaster, the
rollercoaster will start and tell you whether it is started.
• Mathematically: Given a function $f(B, A) : \{0, 1\}^n \times \{0, 1\}^n \to \{0, 1\}^n$ that is written as $f(B, A) = (A_1 \land B_1) \oplus ... \oplus (A_n \land B_n)$, given mystery bits $A_1,
..., A_n$, figure out the mystery bits by plug in $B_1, ..., B_n$ of your choice and observe the function's output.
• More simply, figure out the behavior of function $XOR_{B_1 ... B_n} (A_1, ..., A_n)$ (with constraints $f(B, A) = (A_1 \land B_1) \oplus ... \oplus (A_n \land B_n)$) by plug in $A_1, ..., A_n$ of
your choice.
XOR with bitmask: $XOR_{B_1 ... B_n} (A_1, ..., A_n) = (A_1 \land B_1) \oplus ... \oplus (A_n \land B_n)$
• observe since $\land$ is commutative, we can interchange the string $B_1 ... B_n$ with the string $A_1, ..., A_n$.
The subscript of XOR denotes whether the person is "brave".
Sign Computes XOR with Filter
We start by simulating the function $XOR_{1101}(A_1, ..., A_n): A_i \to \{0, 1\}$. On classical computer, it is identical to the following function: by the property of $XOR$:
XOR(x_1, x_2, ..., x_n) = \begin{cases} 1 & \text{if }|\{x | x = 1\}| \mod 2 = 1\\ 0 & \text{otherwise} \end{cases} = \sum_{i = 1}^n x_i
The effect of $XOR$ is simply counting how many input that is non-zero. $XOR(x_1, ..., x_n) = \sum_{i = 1}^n x_i$. The filtered $XOR_{B_1 ... B_n} (A_1, ..., A_n)$ can be seen as dot product of
vectors $\begin{bmatrix}A_1 & ... & A_n\end{bmatrix} \cdot \begin{bmatrix}B_1 & ... & B_n\end{bmatrix}$. Note that $(\sum_i a_i \cdot b_i) \mod 2 = \sum_i ((a_i \cdot b_i) \mod 2)$.
Problem: figure out vector $|A\rangle$ by giving you a blackbox $f(|B\rangle) = \langle A | B \rangle$, and you can only plug in $|B\rangle$.
def XOR_{1101}(A_1, ..., A_n):
CNOT A1 Ans
CNOT A2 Ans
(Skip A3 since the 3rd bit of mask is zero)
CNOT A4 Ans
return Ans
Observe that XOR is commutative and associative
From now we assume we have the secret strong 1101. Our method is to show that we can compute 1101 to the truth table of XOR_{1101}(A_1, ..., A_n). So if we can obtain the truth table, then 1101 can
be recovered. Note that obtaining the truth table itself do need 1101, but not necessary in a human-readable form, we only need a code that has a secret 1101 in it.
Load Filtered Input to Amplitude
If written in quantum code, $XOR_{1101}(A_1, ..., A_n)$ can be sign-compute as the following:
def If XOR_1101_Then_Minus(A1, A2, A3, A4):
If A1 Then Minus A1
If A2 Then Minus A2
(Skip A3 since the 3rd bit of mask is zero)
If A4 Then Minus A4
Intuition: We first prepare uniform superposition, and for every possible state, sign compute XOR_B on that state.
The circuit diagram can be written as follow
A1: H, Add_1, H
A2: H, Add_1, H
A3: H, Add_0, H
A4: H, Add_1, H
So we can re-write the code as follow using parallel computation
def If XOR_1101_Then_Minus(A1, A2, A3, A4):
// sign computes Add 1101 to A1, ..., An
H on A1, ..., An
Add 1101 to A1, ..., An
H on A1, ..., An
We are effectively load $XOR$ result into amplitude with $-1$ (un-normalized) or $0$ amplitudes, depending on the input.
Getting Truth Table Result to Amplitude Space
Now, if we initialize all bits to uniform superposition, and then do the above instruction, we can load the truth table of $XOR_{1101}$ into state.
@require A1, ..., An=0
def load_table():
// initialize to uniform superposition
H on A1, ..., An
// do sign compute
If XOR_1101_Then_Minus(A1, A2, A3, A4)
In above, code, we uniformly initialize $A_1, ..., A_n$ to equal amplitude, and negate the sign of the amplitude by If XOR_1101_Then_Minus(A1, A2, A3, A4).
Observe the above function is the same as: @require A1, ..., An=0 def load_table(): H on A1, ..., An // same as If XOR_1101_Then_Minus(A1, A2, A3, A4) below H on A1, ..., An Add 1101 to A1, ...,
An H on A1, ..., An Therefore can be simplified to @require A1, ..., An=0 def load_table(): // this two lines are canceled out Add 1101 to A1, ..., An // notice 1|1101> H on A1, ..., An
Therefore, we effectively sign computed If XOR_1101_Then_Minus(A1, A2, A3, A4)
Theorem: For any string $b \in \{0, 1\}^n$, if we initialize bits $B_1, ..., B_n$ to string $b$, and do a $H$ gate, then the resulting amplitude's sign encodes the truth table of $XOR_b$. That is:
Amplitude[a] = \begin{cases} +x & \text{if } XOR_b(a) = 0\\ -x & \text{if } XOR_b(a) = 1\\ \end{cases}
Corollary: if you start with truth table of $XOR_b$ as signed amplitude, then you could do the inverse of the above operation (H on A1, ..., An) and end up with amplitude 1 on string $b$. This is
because the second line of Add 1101 to A1, ..., An // notice 1|1101>.
Bernstein-Vazirani Problem (Cira 1991)
Input: classical code $C$ that computes function $F : \{0, 1\}^n \to \{0, 1\}$ where $F$ is guaranteed to be some $XOR_b$ function for some unknown $b$.
Output: deduce $b$
Classical Solution: run $C$ with $n \cdot |C|$ steps. (Where $|C|$ is the steps of the original code and n is the length of truth table, or how many people there are in rollarcoaster)
Quantum: do If XOR_1101_Then_Minus(A1, A2, A3, A4) with only $2|b|$ instruction (2 because we need garbage-free)
@require A1, ..., An = 0
def solve_bernstein_vazirani():
// make uniform superposition
Had on A1, ..., An
// load truth table
// assume we have blackbox XOR_b here
Sign Compute XOR_b(A1, ..., An)
H on A1, ..., An // by corollary, amplitude 1 on b
The above algorithm is $O(2n + 2|C|)$ instructions. The entire procedural is like: 1. sign compute on uniform superposition using the blackbox $XOR_b$ 2. try to reverse the blackbox, since we
know the general structure of the blackbox 3. but only reverse half of it, to the point when it adds $b$, so we only need to reverse one instruction $H$ 4. so after that we get some state that is
equivalent of blackbox's $H, H, CNOT b$ on $0$, but that is exactly $CNOT b$ on $0$ which is $b$
Summary: there are two ways to sign compute XOR_b function: one using blackbox, the other using $b$ which we don't know. We can sign compute using the first method, and reverse according to the
second method to get $b$. | {"url":"https://kokecacao.me/page/Course/F22/15-459/Lecture_008_-_Rollercoasters.md","timestamp":"2024-11-12T16:10:21Z","content_type":"text/html","content_length":"19871","record_id":"<urn:uuid:6ecc4c3c-8d37-4a1c-8ee5-dbbb7b8f548b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00736.warc.gz"} |
Talk:Maximal elements algorithms - electowikiMaximal elements algorithmsTalk:Maximal elements algorithms
As far as I know, Kosaraju's algorithm finds the "strongly connected components" of a directed graph. But the Schwartz set is not identical to the "strongly connected components", but to the
"communicating classes" of a directed graph.
A strongly connected component SCC is a set of knots with the following property:
1. If knot A is in SCC and knot B is in SCC \ {A}, then there is a directed path from knot A to knot B, that consists only of knots in SCC, and a directed path from knot B to knot A, that consists
only of knots in SCC.
On the other side, a communicating class CC is a set of knots with the following properties:
1. If knot A is in CC and knot B is in CC \ {A}, then there is a directed path from knot A to knot B and a directed path from knot B to knot A.
2. If knot A is in CC and there is a directed path from knot B to knot A, then also knot B is in CC.
Markus Schulze 04:44, 27 October 2006 (PDT)
Isn't the Beats and Beats-or-ties relation for Smith and Schwartz the wrong way around? The maximal element for Beat is the set whose members beat everybody outside of it -- that's Smith. And the
maximal element of Beat-or-tie is the set whose members beat or tie everybody outside of it -- that's Schwartz, if I'm not mistaken. Because Beat-or-tie is more lenient than Beat, the set can be
smaller (can exclude more candidates)... and the Schwartz set is a subset of the Smith set. Kristomun (talk) 23:27, 25 January 2020 (UTC)
I don't know for sure, but let me point out that Smith is associated with the beat-or-tie path, and Schwartz with the beatpath. Taking the Wikipedia example of 3 candidates (https://
en.wikipedia.org/wiki/Schwartz_set#Smith_set_comparison), with A>B, B>C, A=C, we can see that the only candidate with a beatpath to all others is A (A>B>C), since B doesn't beat anyone who beats
A, and C beats nobody. But a beat-or-tie path can be constructed from any of the 3 candidates to the others, since they all beat or tie someone who beats or ties someone else. This lines up with
A being the only member of the Schwartz set and all 3 being in the Smith set. Since this page mentions "The Schwartz set is associated with the beatpath order" and "The Smith set is associated
with the beat-or-tie order", I'm guessing this explains it. BetterVotingAdvocacy (talk) 20:00, 26 February 2020 (UTC) | {"url":"https://electowiki.org/wiki/Talk:Maximal_elements_algorithms","timestamp":"2024-11-13T04:50:30Z","content_type":"text/html","content_length":"41874","record_id":"<urn:uuid:5b020f4a-c8a6-43ef-abd7-0dd3c2203fc9>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00052.warc.gz"} |
On the Number of Solutions Generated by Dantzig’s Simplex Method for LP with Bounded Variables
We give an upper bound for the number of different basic feasible solutions generated by Dantzig’s simplex method (the simplex method with the most negative pivoting rule) for LP with bounded
variables by extending the result of Kitahara and Mizuno (2010). We refine the analysis by them and improve an upper bound for a standard form of LP. Then we utilize the improved bound for an LP with
bounded variables. We show some results when the bound is applied to the minimum cost flow problem and the maximum flow problem.
To appear in Pacific Journal of Optimization. | {"url":"https://optimization-online.org/2011/02/2914/","timestamp":"2024-11-12T15:49:09Z","content_type":"text/html","content_length":"82825","record_id":"<urn:uuid:d9433874-f5f5-47de-8acb-f12067e0520f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00155.warc.gz"} |
Process Control and Product Control - Statistical Quality Control (SQC) | Applied Statistics
Statistical Quality Control (SQC):
Process Control and Product Control
The main objective in any production process is to control and maintain a satisfactory quality level of the manufactured product. This is done by ‘Process Control’. In Process Control the proportion
of defective items in the production process is to be minimized and it is achieved through the technique of control charts. Product Control means that controlling the quality of the product by
critical examination through sampling inspection plans. Product Control aims at a certain quality level to be guaranteed to the customers. It attempts to ensure that the product sold does not contain
a large number of defective items. Thus it is concerned with classification of raw materials, semi-finished goods or finished goods into acceptable or rejectable products.
Control Charts
In an industry, there are two kinds of problems to be faced, namely
(i) To check whether the process is conforming to its standard level.
(ii) To improve the standard level and reduce the variability.
Shewhart’s control charts provide an answer to both. It is a simple technique used for detecting patterns of variations in the data. Control charts are simple to construct and easy to interpret. A
typical control charts consists of the following three lines.
(i) Centre Line (CL) indicates the desired standard level of the process.
(ii) Upper Control Limit (UCL) indicates the upper limit of tolerance.
(iii) Lower Control Limit (LCL) indicates the lower limit of tolerance.
If the data points fall within the control limits, then we can say that the process is in control, instead if one or more data points fall outside the control limits, then we can say that the process
is out of control.
For example, the following lines with the data points plotted, diagram shows all the three controlsince all the points falls within the control limits, we can say that the process is in control.
Control Charts for Variables
These charts may be applied to any quality characteristic that can be measured quantitatively. A quality characteristic which can be expressed in terms of a numerical value is called as a variable.
Many quality characteristics such as dimensions like length, width, temperature, tensile strength etc… of a product are measurable and are expressed in a specific unit of measurements. The variables
are of continuous type and are regarded to follow normal probability law. For quality control of such data, there are two types of control charts used. They are as follows :
(i) Charts for Mean ( )
(ii) Charts for Range (R) | {"url":"https://www.brainkart.com/article/Process-Control-and-Product-Control_39028/","timestamp":"2024-11-02T09:25:49Z","content_type":"text/html","content_length":"39337","record_id":"<urn:uuid:34b40f30-5307-4e8e-bea9-c2e398a0e421>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00097.warc.gz"} |
Miscalculation of percent
I am measuring weight with two load cells and hx711.
I am getting clean numbers of two weights (0,5 kg or 1 kg). I want to calculate how much percent of weight is on every cell, so 50% and 50% if I put 0,5 kg on every cell or 33% and 66% if I put 0,5
kg on one and 1 kg on another cell.
The funny thing is that I never get 100% as result. If I put only one weight on cell I can read load of 1 kg for example but percent is always 99% (should be 100%). Also, if I put 1kg on every cell I
don't get 50% and 50% but 49% and 50%. I have no idea why
I am using byte variable for percent and float for weight. I tried with int also but it didn't help.
I am calculating percent like this:
percent1 = ((weight1 * 100.0) / weightSum);
percent2 = ((weight2 * 100.0) / weightSum);
I just tried to multiply with 101 instead of 100 and now it works as it should, I get 50-50% or 100%. Anyway, I don't understand why.
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?
Can you please post a copy of your sketch, using code tags?
They are made with the </> icon in the reply Menu.
See section 7 http://forum.arduino.cc/index.php/topic,148850.0.html
So we can see you setup and how you are measuring and calculating your weights.
Thanks.. Tom...
Yes, that's the nature of using floating point math. You lose accuracy. You should scale up your values and use fixed point math.
Or you can round off the float before truncating it by adding 0.5 before you store it in an int.
Or you can round off the float before truncating it by adding 0.5 before you store it in an int.
Also, try defining percent1 and percent2 as floats. You should get very close to 100% now.
lg, couka | {"url":"https://forum.arduino.cc/t/miscalculation-of-percent/388656","timestamp":"2024-11-08T14:00:12Z","content_type":"text/html","content_length":"32521","record_id":"<urn:uuid:8a64a381-32db-434e-8f26-14e022a8d95b>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00548.warc.gz"} |
The high low method can be relatively accurate if the highest and lowest activity levels are representative of the overall cost behavior of the company. However, if the two extreme activity levels
are systematically different, then the high low method will produce inaccurate results. Due to the simplicity of using the high-low method to gain insight into the cost-activity relationship,
inherent risk vs control risk it does not consider small details such as variation in costs. The high-low method assumes that fixed and unit variable costs are constant, which is not the case in real
life. Because it uses only two data values in its calculation, variations in costs are not captured in the estimate. The average activity level and the average cost for the periods in the database
are then computed.
High low method uses the lowest production quantity and the highest production quantity and comparing the total cost at each production level. It uses only the lowest and highest production
activities to estimate the variable and fixed cost, by assuming the production quantity and cost increase in linear. It ignores the other points of productions, so it may be an error when the cost
does not increase in a linear graph. The two points are not representing the production cost at a normal level. Once we have arrived at variable costs, we can find the total variable cost for both
activities and subtract that value from the corresponding total cost to find a fixed cost.
1. The two main types of regression analysis are linear regression and multiple regression.
2. Given the dataset below, develop a cost model and predict the costs that will be incurred in September.
3. They include rent, the interest rate on loans, insurance charges, etc.
Furthermore, unless you have access to a computer, computations necessitated by the least squares approach are tedious and time-consuming. An example of a relevant cost is future cost and opportunity
cost, whereas irrelevant cost is sunk cost and committed cost. Cost accounting is used for several purposes, such as standard costing, activity-based costing, lean accounting, and marginal costing.
How to use the high-low method? – High-low method formula
In any business, three types of costs exist Fixed Cost, Variable Cost, and Mixed Cost (a combination of fixed and variable costs). ABC International produces 10,000 green widgets in June at a cost of
$50,000, and 5,000 green widgets in July at a cost of $35,000. There was an incremental change between the two periods of $15,000 and 5,000 units, so the variable cost per unit during July must be
$15,000 divided by 5,000 units, or $3 per unit. Since we have established that $15,000 of the costs incurred in July were variable, this means that the remaining $20,000 of costs were fixed.
High Low Method vs. Regression Analysis
Mixed cost is the combination of variable and fixed cost and it is also called “Semi Variable Cost”. The high-low accounting method estimates these costs for different production levels, mainly if
you have limited data to inform your decisions. This article describes the high-low method formula and how to use the high-low cost method calculator to estimate any business or production cost per
For example, the table below depicts the activity for a cake bakery for each of the 12 months of a given year.
The fixed cost is calculated by subtracting the variable cost for the average activity level from the total average cost. Multiply the variable cost per unit (step 2) by the number of units expected
to be produced in May to work out the total variable cost for the month. The high-low method is an accounting technique used to separate out fixed and variable costs in a limited set of data.
What are costs And cost behavior?
Used in the field of management accounting, which is an essential part of accounting. An easy split is to use Monday, Wednesday, Friday for high intensity, and Tuesday and Thursday for low intensity.
For this split, I recommend total-body lifting three days a week on high days. This is the best structure for sport, since sports are played with the total body. Furthermore, it allows you to hit
each body part three days a week, rather than the two days a week of an upper-lower split.
Once you have the variable cost per unit, you can calculate the fixed cost. High Low method will give us the estimation of fixed cost and variable cost, the result may be changed when the total unit
and cost of both point change. The method does not represent all the data provided since it relies on just two extreme activity levels. Those activity levels may not be representative of the costs
incurred, due to outlier costs that are higher or lower than what the organization incurs in other activity levels. Regression analysis helps forecast costs as well, by comparing the influence of one
predictive variable upon another value or criteria. However, regression analysis is only as good as the set of data points used, and the results suffer when the data set is incomplete.
The effect is represented on a straight line to approximate each of the data points. Another drawback of the high-low method is the ready availability of better cost estimation tools. For example,
the least-squares regression is a method https://intuit-payroll.org/ that takes into consideration all data points and creates an optimized cost estimate. This method has disadvantages in that it
fits a straight line to any set of cost data, regardless of how unpredictable the cost behavior pattern is.
Management accounting involves decision-making, planning, coordinating, controlling, communicating, and motivating. Similar to management accounting and financial accounting, there is cost accounting
to determine the cost of a product. No, there are other methods apart from the high-low method accounting formula. Some popular methods are the scatter plot method, accounting, and regression
analysis. But the high-low cost method provides a simple approach to achieve it. Variable costs are expenses that change depending on the quantity of production or number of units sold.
High-low method formula
However, to identify these costs, we need to observe the cost behaviors strongly. It can be calculated by subtracting the present realizable salvage value from the book value. For example, buying
2,000 shares of company A at $10 a share, for instance, represents a sunk cost of $20,000. Relevant/ Irrelevant costs – These are also known as avoidable and unavoidable costs. Avoidable costs are
the ones that are affected by the decision of a manager, whereas unavoidable costs are costs that are not affected by the decision of managers. Some common examples of these costs are supervision
costs and marketing costs.
Cost accounting also helps in minimizing product costs as it highlights the reports of profit. There are also other cost estimation tools that can provide more accurate results. The least-squares
regression method takes into consideration all data points and creates an optimized cost estimate. It can be easily and quickly used to yield significantly better estimates than the high-low method.
For an athlete needing both power and heavy conditioning, low days are great opportunities for tempos. The total volume of tempo runs depends on your sport, but these will help you recover from high
days while expanding the tank with some much needed aerobic work.
In cost accounting, the high-low method is a way of attempting to separate out fixed and variable costs given a limited amount of data. The high-low method involves taking the highest level of
activity and the lowest level of activity and comparing the total costs at each level. The high-low method is a cost accounting technique that compares the total cost at the highest and lowest
production level of business activity. It uses this comparison to estimate the fixed cost, variable cost, and a cost function for finding the total cost of different production units.
You can us our labor cost calculator and VAT calculator to understand more on this topic. In contrast to the High Low Method, Regression analysis refers to a technique for estimating the relationship
between variables. It helps people understand how the value of a dependent variable changes when one independent variable is variable while another is held constant. The two main types of regression
analysis are linear regression and multiple regression. The high low method determines the fixed and variable components of a cost.
Is the high low method the only method for estimating fixed and variable costs?
Due to its unreliability, high low method should be carefully used, usually in cases where the data is simple and not too scattered. For complex scenarios, alternate methods should be considered such
as scatter-graph method and least-squares regression method. It’s also possible to draw incorrect conclusions by assuming that just because two sets of data correlate with each other, one must cause
changes in the other. Regression analysis is also best performed using a spreadsheet program or statistics program. We should be really careful when choosing the data for calculation with this tool,
as any small mistake can lead to an inaccurate result. Semi-Variable Cost – These expenses are not constant in total or per unit.
Management accounting refers to identifying, analyzing, and communicating financial information to a firm’s managers to achieve the company’s future goals. To understand the high-low method, first,
we need to understand management accounting. The high-low method is used in the field of management accounting, which is an essential part of accounting. | {"url":"https://web.setiaskyresidences.com/2023/01/27/high-low-method-definition-formulas-example/","timestamp":"2024-11-03T09:11:58Z","content_type":"text/html","content_length":"81517","record_id":"<urn:uuid:9280960f-3a9b-48ef-8b76-da502e92fc0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00249.warc.gz"} |
I have now completed a series of introductory lectures on Graph Theory, for 3rd/4th year students at the University of Bristol. A set of written notes plus (less usefully) my slides are mirrored
here. One student declared this to be “[the] most interesting course I have taken at Bristol Uni”, so take a look to …
Continue reading ‘‘Introduction to Graph Theory’ Lecture Materials’ »
Lecturing 2012/13
In the next academic year I’ll be one of the lecturers for Topics in Discrete Mathematics at the University of Bristol. This comes in two flavours- the level H/6 unit MATH30002 for third-year
students, and the level M/7 unit MATHM0009 for those in the fourth year (which adds a project component). This will be the …
Undergraduate Projects at the University of Bristol
I’m offering a couple of possible projects for third-year students at Bristol (i.e., for MATH32001 or MATH32200 depending on amount of time) in the upcoming academic year (2012/13). Other topics in
areas such as spectral graph theory or computational number theory could also be possibilities; students are welcome to get in touch (via my magdt@bristol …
Continue reading ‘Undergraduate Projects at the University of Bristol’ »
The Extended Euclidean Algorithm
Some notes on Euclid’s algorithm and its extension for solving linear Diophantine equations in two variables. | {"url":"https://maths.straylight.co.uk/archives/category/teaching","timestamp":"2024-11-03T03:10:50Z","content_type":"text/html","content_length":"34903","record_id":"<urn:uuid:e61b3c7e-4b37-4954-b662-636892d73bef>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00605.warc.gz"} |
480000 Milliseconds to Minutes
480000 ms to min conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in
the United Kingdom) and as a fraction (exact result). Every display form has its own advantages and in different situations particular form is more convenient than another. For example usage of
scientific notation when working with big numbers is recommended due to easier reading and comprehension. Usage of fractions is recommended when more precision is needed.
If we want to calculate how many Minutes are 480000 Milliseconds we have to multiply 480000 by 1 and divide the product by 60000. So for 480000 we have: (480000 × 1) ÷ 60000 = 480000 ÷ 60000 = 8
So finally 480000 ms = 8 min | {"url":"https://unitchefs.com/milliseconds/minutes/480000/","timestamp":"2024-11-03T23:37:01Z","content_type":"text/html","content_length":"22790","record_id":"<urn:uuid:f2996e31-0af5-4b21-8778-53ba05065b10>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00229.warc.gz"} |
Domain and Range Worksheets
This compilation of domain and range worksheet pdfs provides 8th grade and high school students with ample practice in determining the domain or the set of possible input values (x) and range, the
resultant or output values (y) using a variety of exercises with ordered pairs presented on graphs and in table format. Find the domain and range of relations from mapping diagrams, from finite and
infinite graphs and more. Get started with our free worksheets.
Write the Domain and Range | Relation - Ordered Pairs
State the domain and range of each relation represented as a set of ordered pairs in Part A and ordered pairs on a graph in Part B of these printable worksheets.
Write the Domain and Range | Relation - Mapping
Determine the domain and range in each of the relations presented in these relation mapping worksheets for grade 8 and high school students. Observe each relation and write the domain (x) and range
(y) values in set notation.
Write the Domain and Range | Relation - Table
This batch presents the ordered pairs in tables with input and output columns. Identify the domain and range and write them in ascending order for each of the tables featured in these domain and
range pdf worksheets.
Write the Domain and Range | Finite Graph
Observe the extent of the graph horizontally for the domain and the vertical extent of the graph for range and write the smallest and largest values of both in this set of identifying the domain and
range from finite graphs worksheets. Use apt brackets to show if the interval is open or closed.
Write the Domain and Range | Infinite Graph
Bolster skills in identifying the domain and range of functions with infinite graphs. Analyze each graph, write the minimum and maximum points for both domain and range. If there is no endpoint, then
it can be concluded it is infinite.
Write the Range | Function Rule - Level 1
In this set of pdf worksheets, the function rule is expressed as a linear function and the domain is also provided in each problem. Plug in the values of x in the function rule to determine the
Write the Range | Function Rule - Level 2
Substitute the input values or values of the domain in the given quadratic, polynomial, reciprocal or square root functions and determine the output values or range in this section of Level 2
Write the Domain and Range | Function - Mixed Review
Test skills acquired with this printable domain and range revision worksheets that provide a mix of absolute, square root, quadratic and reciprocal functions f(x). Determine the domain (x) and plug
in the possible x-values to find the range (y). | {"url":"https://www.mathworksheets4kids.com/domain-range.php","timestamp":"2024-11-03T12:32:05Z","content_type":"text/html","content_length":"40184","record_id":"<urn:uuid:f99bce0c-9fa4-4559-abd0-fdde2b11cdb7>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00653.warc.gz"} |
A Meshless Method for Magnetohydrodynamics and Applications to Protoplanetary Disks
2012 Theses Doctoral
A Meshless Method for Magnetohydrodynamics and Applications to Protoplanetary Disks
This thesis presents an algorithm for simulating the equations of ideal magnetohydrodynamics and other systems of differential equations on an unstructured set of points represented by sample
particles. Local, third-order, least-squares, polynomial interpolations (Moving Least Squares interpolations) are calculated from the field values of neighboring particles to obtain field values and
spatial derivatives at the particle position. Field values and particle positions are advanced in time with a second order predictor-corrector scheme. The particles move with the fluid, so the time
step is not limited by the Eulerian Courant-Friedrichs-Lewy condition. Full spatial adaptivity is implemented to ensure the particles fill the computational volume, which gives the algorithm
substantial flexibility and power. A target resolution is specified for each point in space, with particles being added and deleted as needed to meet this target. Particle addition and deletion is
based on a local void and clump detection algorithm. Dynamic artificial viscosity fields provide stability to the integration.
The resulting algorithm provides a robust solution for modeling flows that require Lagrangian or adaptive discretizations to resolve. The code has been parallelized by adapting the framework provided
by Gadget-2. A set of standard test problems, including one part in a million amplitude linear MHD waves, magnetized shock tubes, and Kelvin-Helmholtz instabilities are presented. Finally we
demonstrate good agreement with analytic predictions of linear growth rates for magnetorotational instability in a cylindrical geometry. We provide a rigorous methodology for verifying a numerical
method on two dimensional Kelvin-Helmholtz instability. The test problem was run in the Pencil Code, Athena, Enzo, NDSPHMHD, and Phurbas. A strict comparison, judgment, or ranking, between codes is
beyond the scope of this work, although this work provides the mathematical framework needed for such a study. Nonetheless, how the test is posed circumvents the issues raised by tests starting from
a sharp contact discontinuity yet it still shows the poor performance of Smoothed Particle Hydrodynamics.
We then comment on the connection between this behavior and the underlying lack of zeroth-order consistency in Smoothed Particle Hydrodynamics interpolation. In astrophysical magnetohydrodynamics
(MHD) and electrodynamics simulations, numerically enforcing the divergence free constraint on the magnetic field has been difficult. We observe that for point-based discretization, as used in
finite-difference type and pseudo-spectral methods, the divergence free constraint can be satisfied entirely by a choice of interpolation used to define the derivatives of the magnetic field. As an
example we demonstrate a new class of finite-difference type derivative operators on a regular grid which has the divergence free property. This principle clarifies the nature of magnetic monopole
The principles and techniques demonstrated in this chapter are particularly useful for the magnetic field, but can be applied to any vector field. Finally, we examine global zoom-in simulations of
turbulent magnetorotationally unstable flow. We extract and analyze the high-current regions produced in the turbulent flow. Basic parameters of these regions are abstracted, and we build one
dimensional models including non-ideal MHD, and radiative transfer. For sufficiently high temperatures, an instability resulting from the temperature dependence of the Ohmic resistivity is found.
This instability concentrates current sheets, resulting in the possibility of rapid heating from temperatures on the order of 600 Kelvin to 2000 Kelvin in magnetorotationally turbulent regions of
protoplanetary disks. This is a possible local mechanism for the melting of chondrules and the formation of other high-temperature materials in protoplanetary disks.
• McNally_columbia_0054D_10876.pdf application/pdf 10.9 MB Download File
More About This Work
Academic Units
Thesis Advisors
Mac Low, Mordecai-Mark
Ph.D., Columbia University
Published Here
August 17, 2012 | {"url":"https://academiccommons.columbia.edu/doi/10.7916/D8V4128Z","timestamp":"2024-11-10T03:42:45Z","content_type":"text/html","content_length":"27593","record_id":"<urn:uuid:c48be54b-68eb-4187-85ef-b232c52d7db4>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00839.warc.gz"} |
Electrostatics-10 Important basic MCQs (Quiz) Part 1
1.An isolated solid metallic sphere is given charge +Q. The charge will be distributed on the sphere
(A) Uniformly but only on surface
(B) Only on surface but non-uniformly
(C) Uniformly inside the volume
(D) Non-uniformly inside the volume
2. Four charges are arranged at the corners of a square ABCD, as shown in the adjoining figure. The force on the charge kept at the centre O is
(A) Zero
(B) Along the diagonal AC
(C) Along the diagonal BD
(D) Perpendicular to side AB
3.The ratio of the forces between two small spheres with constant charge in (a) air (b) in a medium of dielectric constant K is
(A) 1 : K
(B) K : 1
(C) $1: K^2$
(D) $K^2 : 1$
4. Three charges 4q , Q and q are in a straight line in the position of 0, $l/2$, and $l$ respectively. The resultant force on q will be zero, if Q=
(A) – q
(B) – 2q
(C) $-\frac q2$
(D) 4q
5.Two charges each of 1 coulomb are at a distance 1 km apart, the force between them is
(A) $9\;\times\;10^3\;Newton\;$
(B) $9\;\times\;10^{-3}\;Newton\;$
(C) $1.1\;\times\;10^{-4}\;Newton\;$
(D) $10^4\;Newton\;$
6. There are two metallic spheres of same radii but one is solid and the other is hollow, then
(A) Solid sphere can be given more charge
(B) Hollow sphere can be given more charge
(C) They can be charged equally (maximum)
(D) None of the above
7. Three equal charges are placed on the three corners of a square. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$ , the ratio of magnitudes $\frac{F_
{12}}{F_{13}}$ is
(A) $\frac12$
(B) $2$
(C) $\frac1{\sqrt2}\;$
(D) $\sqrt{2\;}$
8. Two small spheres each having the charge Q are suspended by insulating threads of length from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle
between the two suspensions and the tension in each will be
(A) $180^o,\;\frac1{4\mathrm\pi\;\in_{\mathrm o}}\frac{Q^2}{{(2L)}^2}$
(B)$90^o,\;\frac1{4\mathrm\pi\;\in_{\mathrm o}}\frac{Q^2}{{L}^2}$
(C) $180^o,\;\frac1{4\mathrm\pi\;\in_{\mathrm o}}\frac{Q^2}{2 L^2}$
(D) $180^o,\;\frac1{4\mathrm\pi\;\in_{\mathrm o}}\frac{Q^2}{ L^2}$
9. A soap bubble is given a negative charge, then its radius
(A) Decreases
(B) Increases
(C) Remains unchanged
(D) Nothing can be predicted as information is insufficient
10. With the rise in temperature, the dielectric constant of a liquid
(A) Remains unchanged
(B) Changes erratically
(C) Increases
(D) Decreases
1 thought on “Electrostatics-10 Important basic MCQs (Quiz) Part 1”
Leave a Comment | {"url":"https://pcmtutorials.in/electrostatics-10-important-basic-mcqs-quiz-part-1/","timestamp":"2024-11-06T14:58:42Z","content_type":"text/html","content_length":"148726","record_id":"<urn:uuid:ba31d654-aeca-4c3f-ac0e-21b027ea86e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00546.warc.gz"} |
This task is about using Cuisenaire rods to work out fractions.
Use the Cuisenaire rods to help you answer the following questions.
Hint: Place the Cuisenaire rods next to each other to find the fractions.
A diagram has been provided for the first question.
│ a) If orange is a whole then... │
│ i) What fraction is yellow? _____ │
│ │
│ ii) What fraction is white? _____ │
│ │
│ iii) What fraction is red? _____ │
│ │
b) If the dark green rod is a whole then:
i) What fraction is red? _____
ii) What fraction is light green? _____
iii) What fraction is yellow? _____
c) If the dark green rod is \(1 \over 2\) then:
i) What fraction is light green? _____
ii) What fraction is red? _____
iii) What fraction is blue? _____
d) If the light green is \(1 \over 2\) then:
i) What fraction is blue? _____
ii) What fraction is red? _____
iii) What fraction is dark green? _____ | {"url":"https://arbs.nzcer.org.nz/node/4472/search-preview","timestamp":"2024-11-11T06:28:24Z","content_type":"text/html","content_length":"24231","record_id":"<urn:uuid:163d72c4-da07-4dd5-8b54-772f92407f16>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00601.warc.gz"} |
GATE (TF) Textile 2009 Question Paper Solution | GATE/2009/TF/52 - Textile Triangle
GATE (TF) Textile 2009 Question Paper Solution | GATE/2009/TF/52
Common Data Questions
The feed rate of a card is 100 kg/hour and the delivery rate is 400 m/min. Licker-in droppings and flat strips are 3% and 1% respectively.
Question 52 (Textile Engineering & Fibre Science)
If the total draft in the card is decreased by 10%, the sliver linear density in ktex will be
[Show Answer]
Option C is correct
Given in the question-
Given in the question-
Feed rate of card=100 kg/hr
=100 x 1000/60 gm/min
=1666.66 gm/min
Delivery rate=400 m/min
Licker-in droppings=3%
Flat strips=1%
The count of card sliver(Ne)=?
Firstly we have to find gm/m
Now, tex is the weight of 1000 meter length
Tex after total waste removed-
Total waste=Licker-in dropping+Flat stripe
Total waste=3%+1%
Total waste=4%
Tex=4166.65 x (1-Total waste%)
Tex=4166.65 x (1-4%)
Tex=4166.65 x (1-0.04)
Tex=4166.65 x 0.96
Relationship between tex and Ne-
Delivered card sliver count(Ne)=0.15
Feed count(tex)=4.16665 x 1000
Feed count(tex)=4166.65
Total Draft=1.071
If total draft is decreased by 10%
Then, Draft=Total draft-Total draft x 10%
Draft=1.071-1.071 x 10%
Draft=1.071-1.071 x 0.1
New Draft=0.9639
Delivered count=4166.65/0.9639
Delivered count=4322.69 tex
Sliver linear density in ktex=4.32 (Ans)
Frequently Asked Questions | FAQs
GATE Textile Engineering and Fibre Science (TF) Question Papers | GATE Textile Question Answer | GATE Textile Solved Question Papers | GATE Textile Papers | GATE Textile Answer Key | {"url":"https://www.textiletriangle.com/gate-textile/question-paper/2009-2/52-2/","timestamp":"2024-11-02T01:52:48Z","content_type":"text/html","content_length":"144978","record_id":"<urn:uuid:bab36003-6031-4e12-a981-f0ec05949369>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00610.warc.gz"} |
MECHANIC MOTOR VEHICLE 1st year WEEK WISE NSQF SYLLABUS (SPILT UP)
MECHANIC MOTOR VEHICLE 1st year WEEK WISE NSQF SYLLABUS (SPILT UP)
WORKSHOP CALCULATION & SCIENCE (for 1st Year ) 40 Hrs.
1. Reference Learning Outcome :- Demonstrate basic mathematical concept and principles to perform practical operations. Understand and explain basic science in the field of study. (Mapped NOS: ASC/
2. Duration :- Professional Knowledge WCS – 40 Hrs (for above learning out come)
• Unit, Fractions
• Classification of unit system
• Fundamental and Derived units F.P.S, C.G.S, M.K.S and SI units
• Measurement units and conversion
• Factors, HCF, LCM and problems
• Fractions – Addition, substraction, multiplication &division
• Decimal fractions – Addition, subtraction, multiplication& division
• Solving problems by using calculator (4 hrs)
• Square root, Ratio and Proportions, Percentage
• Square and square root
• Simple problems using calculator
• Applications of Pythagoras theorem and related problems
• Ratio and proportion
• Ratio and proportion – Direct and indirect proportions
• Percentage Percentage – Changing percentage to decimal and fraction
• Material Science
• Types metals, types of ferrous and non ferrous metals Physical and mechanical properties of metals
• Introduction of iron and cast iron
• Difference between iron & steel, alloy steel and carbon steel
• Properties and uses of rubber, timber and insulating materials
• Mass, Weight, Volume and Density
• Mass, volume, density, weight and specific gravity
• Related problems for mass, volume, density, weight and specific gravity
• Speed and Velocity, Work, Power and Energy
• Speed and velocity – Rest, motion, speed, velocity, difference between speed and velocity, acceleration and retardation
• Speed and velocity – Related problems on speed & velocity
• Work, power, energy, HP, IHP, BHP and efficiency
• Potential energy, kinetic energy and related problems with assignment
• Heat & Temperature and Pressure
• Concept of heat and temperature, effects of heat, difference between heat and temperature, boiling point & melting point of different metals and non-metals
• Thermal conductivity and insulators
• Concept of pressure – Units of pressure, atmospheric pressure, absolute pressure, gauge pressure and gauges used for measuring pressure
• Basic Electricity
• Introduction and uses of electricity, electric current AC,DC their comparison, voltage, resistance and their units
• Conductor, insulator, types of connections – series and parallel
• Ohm’s law, relation between V.I.R & related problems
• Magnetic induction, self and mutual inductance and EMF generation
• Mensuration
• Surface area and volume of solids – cube, cuboid, cylinder, sphere and hollow cylinder
• Levers and Simple machines
• Lever & Simple machines – Lever and its types | {"url":"https://ititechguru.com/mechanic-motor-vehicle-1st-year-week-wise-nsqf-syllabus-spilt-up/10/","timestamp":"2024-11-03T22:55:50Z","content_type":"text/html","content_length":"171295","record_id":"<urn:uuid:4a2fbc36-d036-4b21-a13a-824cb54dfccf>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00544.warc.gz"} |
Meshing or discretization in Computational Fluid Dynamics (CFD)
Computational Fluid dynamic is also known as CFD is a well-known method in the engineering field to solve fluid mechanics and heat transfer complex problems numerically with the aid of a computer
which is tremendously advance nowadays. (Read Introduction to CFD here)
But, despite its advanced capabilities, the accuracy and validity of CFD compared to actual data is strongly dependent on its operator’s knowledge and skill, basically, it is just a calculator. The
physics of fluid dynamic itself is very complex in nature and may have different approximation methods depend on its flow regime (viscosity, speed, density, laminar, turbulence, etc.) as well as its
occurred phenomena (vorticity, adverse pressure gradient, etc.) hence, the parameters to be inputted will vary case by case. One important parameter to be considered is meshing or discretization.
Meshing will strongly affect the quality of the resulted simulation, wether its accuracy, resolution or even its convergence characteristic. In general, the CFD engineer spent their time 70-80% of
the whole simulation process to generate the right mesh. Before we go deeper into the selection of mesh type and more advanced topics, first we must understand what is meshing.
Meshing is a discretization process of continuous fluid domain converted into a discrete computational domain so that each part (also called an element) of its domain can be solved with the discrete
equation. Then, why we should use a discrete equation instead of an analytical equation? This is because the nature of fluid mechanic’s governing equations consists of a non-linear partial
differential equation, the Navier-stokes equation which is known as one of the unresolved yet mathematical problem.
To illustrate the discretization process, consider the area calculation below the curve above. If the curve is only a constant curve, we can simply calculate the area of the curve with length times
height of the rectangle, or if the curve is only a linear curve then resulted picture is a triangle, we may simply calculate the area using 1/2*length*height of the triangle. But, what if the curve
is a quadratic, cubic, polynomial, exponential or even an arbitrary curve? What we can do is to divide the area behind the curve into small section and calculates each section area then sum all of
those area to “approximate” the total area below the curve (maybe you will ask, why not we use integral calculus? Yes, basically this is the idea of integral calculus, but the section length is
infinitesimally small for integral calculus). The division process of the area is also known as meshing or discretization from the continuous curve into a discrete curve.
On the other hand, when we are about to calculate each section area, we use discrete area of each rectangle, the length times height equation, this is also known as equation discretization. In fluid
dynamics, we often use the Finite Volume Method to discretize the Navier-Stokes equation, and basically these discretization methods are the core of CFD itself.
We can observe the characteristics of discretization from the area calculation case above. (1) we could get the more accurate result if we make the discretization even smaller, (2) not just accurate
in term of its value, but also higher resolution graphically (smoother curve), (3) but, if the mesh is too small we also generate a lot of discrete equations to be solved, these equations need
computational effort and time, so, make the mesh smaller is not straightforward solution to generate the optimum mesh.
Low-resolution mesh:
Let see the curves above, suppose the blue curve is an actual curve and we want to fit the blue curve with the orange curve which it’s resolution is constrained with the mesh (black lines). The
orange curve not fits the blue curve well in the high-gradient region. This is the keyword, “high-gradient”, we should focus to refine the mesh in the high-gradient region so we can make a
well-fitted graph without sacrificing a lot of computer effort and time. Following is the refined mesh in the high gradient region.
Refined mesh around the high-gradient region:
In Computational Fluid Dynamic (CFD) post-processing (or data interpretation), the resolution of its color also governed by the mesh size, it is analogous with your television or monitor resolution,
higher the dot per inch, higher the resolution.
low-resolution mesh:
High-resolution mesh:
We will discuss more advanced topics about mesh in the next article, mesh types and mesh quality.
To read other articles, click here.
aeroengineering services is an online platform that provides engineering consulting with various solutions, from CAD drafting, animation, CFD, or FEA simulation which is the primary brand of CV.
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Calculus 3 - mathXplain
5 topics, 51 short and super clear episodes
This Calculus 3 course includes 51 short and super clear episodes that take you through 5 topics and help you navigate the bumpy roads of Calculus 3. The casual style makes you feel like you are
discussing some simple issue, such as cooking scrambled eggs.
Table of contents:
The course consists of 5 sections: Matrices and vectors, Determinants, eigenvectors and eigenvalues, Functions of two variables, Double integrals, Differential equations
• Matrices - Matrices are really harmless creatures in mathematics. An nXk matrix is simply a rectangular array of numbers, arranged in n rows and k columns.
• Matrix operations - Scalar multiplication, addition and multiplication.
• Square and diagonal matrices - It is a square-shaped matrix with the same number of rows and columns.The diagonal matrix is a square matrix where all elements outside the main diagonal are zero.
• Transpose - The transpose matrix is created by swapping the rows and the columns of the matrix
• Normal region - The region is from a to b on the x axis, and from c to d on the y axis.
• Double integrals - Double integrals can be used to compute volumes under various surfaces.
• Polar coordinates - The idea of polar coordinates is that we replace the x and y coordinates with new ones. | {"url":"http://www.mathxplain.com:8080/calculus-3","timestamp":"2024-11-11T05:05:34Z","content_type":"text/html","content_length":"90076","record_id":"<urn:uuid:d7ed0a21-3e52-444d-908f-bce1e3b00653>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00326.warc.gz"} |
8 Week SQL Challenge: Case Study #2 Pizza Runner
Danny was scrolling through his Instagram feed when something really caught his eye — “80s Retro Styling 🎸 and Pizza 🍕 Is The Future!”
Danny was sold on the idea, but he knew that pizza alone was not going to help him get seed funding to expand his new Pizza Empire — so he had one more genius idea to combine with it — he was going
to Uberize it — and so Pizza Runner was launched!
Danny started by recruiting “runners” to deliver fresh pizza from Pizza Runner Headquarters (otherwise known as Danny’s house) and also maxed out his credit card to pay freelance developers to build
a mobile app to accept orders from customers.
Table Relationship
• customer_orders — Customers’ pizza orders with 1 row each for individual pizza with topping exclusions and extras, and order time.
• runner_orders — Orders assigned to runners documenting the pickup time, distance and duration from Pizza Runner HQ to customer, and cancellation remark. runners — Runner IDs and registration date
• pizza_names — Pizza IDs and name
• pizza_recipes — Pizza IDs and topping names
• pizza_toppings — Topping IDs and name
Case Study Questions
This case study has LOTS of questions — they are broken up by area of focus including:
• A. Pizza Metrics
• B. Runner and Customer Experience
• C. Ingredient Optimisation
• D. Pricing and Ratings
• E. Bonus DML Challenges (DML = Data Manipulation Language)
Data Cleaning and Transformation
Before I start with the solutions, I investigate the data and found that there are some cleaning and transformation to do, specifically on the
• null values and data types in the customer_orders table
• null values and data types in the runner_orders table
• Alter data type in pizza_names table
Firstly, to clean up exclusions and extras in the customer_orders — we create TEMP TABLE #customer_orders and use CASE WHEN.
SELECT order_id, customer_id, pizza_id,
WHEN exclusions IS null OR exclusions LIKE 'null' THEN ' '
ELSE exclusions
END AS exclusions,
WHEN extras IS NULL or extras LIKE 'null' THEN ' '
ELSE extras
END AS extras,
INTO #customer_orders -- create TEMP TABLE
FROM customer_orders;
Then, we clean the runner_orders table with CASE WHEN and TRIM and create TEMP TABLE #runner_orders.
In summary,
• pickup_time — Remove nulls and replace with ‘ ‘
• distance — Remove ‘km’ and nulls
• duration — Remove ‘minutes’ and nulls
• cancellation — Remove NULL and null and replace with ‘ ‘
SELECT order_id, runner_id,
WHEN pickup_time LIKE 'null' THEN ' '
ELSE pickup_time
END AS pickup_time,
WHEN distance LIKE 'null' THEN ' '
WHEN distance LIKE '%km' THEN TRIM('km' from distance)
ELSE distance END AS distance,
WHEN duration LIKE 'null' THEN ' '
WHEN duration LIKE '%mins' THEN TRIM('mins' from duration)
WHEN duration LIKE '%minute' THEN TRIM('minute' from duration)
WHEN duration LIKE '%minutes' THEN TRIM('minutes' from duration)
ELSE duration END AS duration,
WHEN cancellation IS NULL or cancellation LIKE 'null' THEN ''
ELSE cancellation END AS cancellation
INTO #runner_orders
FROM runner_orders;
Then, we alter the date according to its correct data type.
• pickup_time to DATETIME type
• distance to FLOAT type
• duration to INT type
ALTER TABLE #runner_orders
ALTER COLUMN pickup_time DATETIME,
ALTER COLUMN distance FLOAT,
ALTER COLUMN duration INT;
Now that the data has been cleaned and transformed, let’s move on solving the questions! 😉
A. Pizza Metrics
How many pizzas were ordered?
SELECT COUNT(*) AS pizza_order_count
FROM #customer_orders;
• Total pizzas ordered are 14.
• How many unique customer orders were made?
SELECT COUNT(DISTINCT order_id) AS unique_order_count
FROM #customer_orders;
• There are 10 unique customer orders made.
1. How many successful orders were delivered by each runner?
SELECT runner_id, COUNT(order_id) AS successful_orders
FROM #runner_orders
WHERE distance != 0
GROUP BY runner_id;
• Runner 1 has 4 successful delivered orders.
• Runner 2 has 3 successful delivered orders.
• Runner 3 has 1 successful delivered order.
1. How many of each type of pizza was delivered?
SELECT p.pizza_name, COUNT(c.pizza_id) AS delivered_pizza_count
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
JOIN pizza_names AS p
ON c.pizza_id = p.pizza_id
WHERE r.distance != 0
GROUP BY p.pizza_name;
• There are 9 delivered Meatlovers pizzas.
• There are 3 delivered Vegetarian pizzas.
1. How many Vegetarian and Meatlovers were ordered by each customer?
SELECT c.customer_id, p.pizza_name, COUNT(p.pizza_name) AS order_count
FROM #customer_orders AS c
JOIN pizza_names AS p
ON c.pizza_id= p.pizza_id
GROUP BY c.customer_id, p.pizza_name
ORDER BY c.customer_id;
• Customer 101 ordered 2 Meatlovers pizzas and 1 Vegetarian pizza.
• Customer 102 ordered 2 Meatlovers pizzas and 2 Vegetarian pizzas.
• Customer 103 ordered 3 Meatlovers pizzas and 1 Vegetarian pizza.
• Customer 104 ordered 1 Meatlovers pizza.
• Customer 105 ordered 1 Vegetarian pizza.
1. What was the maximum number of pizzas delivered in a single order?
WITH pizza_count_cte AS
SELECT c.order_id, COUNT(c.pizza_id) AS pizza_per_order
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.distance != 0
GROUP BY c.order_id
SELECT MAX(pizza_per_order) AS pizza_count
FROM pizza_count_cte;
• Maximum number of pizza delivered in a single order is 3 pizzas.
1. For each customer, how many delivered pizzas had at least 1 change and how many had no changes?
SELECT c.customer_id,
WHEN c.exclusions <> ' ' OR c.extras <> ' ' THEN 1
ELSE 0
END) AS at_least_1_change,
WHEN c.exclusions = ' ' AND c.extras = ' ' THEN 1
ELSE 0
END) AS no_change
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.distance != 0
GROUP BY c.customer_id
ORDER BY c.customer_id;
• Customer 101 and 102 likes his/her pizzas per the original recipe.
• Customer 103, 104 and 105 have their own preference for pizza topping and requested at least 1 change (extra or exclusion topping) on their pizza.
1. How many pizzas were delivered that had both exclusions and extras?
WHEN exclusions IS NOT NULL AND extras IS NOT NULL THEN 1
ELSE 0
END) AS pizza_count_w_exclusions_extras
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.distance >= 1
AND exclusions <> ' '
AND extras <> ' ';
• Only 1 pizza delivered that had both extra and exclusion topping. That’s one fussy customer!
1. What was the total volume of pizzas ordered for each hour of the day?
SELECT DATEPART(HOUR, [order_time]) AS hour_of_day,
COUNT(order_id) AS pizza_count
FROM #customer_orders
GROUP BY DATEPART(HOUR, [order_time]);
• Highest volume of pizza ordered is at 13 (1:00 pm), 18 (6:00 pm) and 21 (9:00 pm).
• Lowest volume of pizza ordered is at 11 (11:00 am), 19 (7:00 pm) and 23 (11:00 pm).
• What was the volume of orders for each day of the week?
SELECT FORMAT(DATEADD(DAY, 2, order_time),'dddd') AS day_of_week,
-- add 2 to adjust 1st day of the week as Monday
COUNT(order_id) AS total_pizzas_ordered
FROM #customer_orders
GROUP BY FORMAT(DATEADD(DAY, 2, order_time),'dddd');
• There are 5 pizzas ordered on Friday and Monday.
• There are 3 pizzas ordered on Saturday.
• There is 1 pizza ordered on Sunday.
B. Runner and Customer Experience
How many runners signed up for each 1 week period? (i.e. week starts 2021-01-01)
SELECT DATEPART(WEEK, registration_date) AS registration_week,
COUNT(runner_id) AS runner_signup
FROM runners
GROUP BY DATEPART(WEEK, registration_date);
• On Week 1 of Jan 2021, 2 new runners signed up.
• On Week 2 and 3 of Jan 2021, 1 new runner signed up.
1. What was the average time in minutes it took for each runner to arrive at the Pizza Runner HQ to pickup the order?
WITH time_taken_cte AS
SELECT c.order_id, c.order_time, r.pickup_time,
DATEDIFF(MINUTE, c.order_time, r.pickup_time) AS pickup_minutes
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.distance != 0
GROUP BY c.order_id, c.order_time, r.pickup_time
SELECT AVG(pickup_minutes) AS avg_pickup_minutes
FROM time_taken_cte
WHERE pickup_minutes > 1;
• The average time taken in minutes by runners to arrive at Pizza Runner HQ to pick up the order is 15 minutes.
1. Is there any relationship between the number of pizzas and how long the order takes to prepare?
WITH prep_time_cte AS
SELECT c.order_id, COUNT(c.order_id) AS pizza_order,
c.order_time, r.pickup_time,
DATEDIFF(MINUTE, c.order_time, r.pickup_time) AS prep_time_minutes
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.distance != 0
GROUP BY c.order_id, c.order_time, r.pickup_time
SELECT pizza_order, AVG(prep_time_minutes) AS avg_prep_time_minutes
FROM prep_time_cte
WHERE prep_time_minutes > 1
GROUP BY pizza_order;
• On average, a single pizza order takes 12 minutes to prepare.
• An order with 3 pizzas takes 30 minutes at an average of 10 minutes per pizza.
• It takes 16 minutes to prepare an order with 2 pizzas which is 8 minutes per pizza — making 2 pizzas in a single order the ultimate efficiency rate.
1. What was the average distance travelled for each customer?
SELECT c.customer_id, AVG(r.distance) AS avg_distance
FROM #customer_orders AS c
JOIN #runner_orders AS r
ON c.order_id = r.order_id
WHERE r.duration != 0
GROUP BY c.customer_id;
(Assuming that distance is calculated from Pizza Runner HQ to customer’s place)
• Customer 104 stays the nearest to Pizza Runner HQ at average distance of 10km, whereas Customer 105 stays the furthest at 25km.
1. What was the difference between the longest and shortest delivery times for all orders?
Firstly, let’s see all the durations for the orders.
SELECT order_id, duration
FROM #runner_orders
WHERE duration not like ' ';
Then, we find the difference by deducting the shortest (MIN) from the longest (MAX) delivery times.
MAX(duration::NUMERIC) - MIN(duration::NUMERIC) AS delivery_time_difference
FROM #runner_orders
WHERE duration not like '% %'
• The difference between longest (40 minutes) and shortest (10 minutes) delivery time for all orders is 30 minutes.
1. What was the average speed for each runner for each delivery and do you notice any trend for these values?
SELECT r.runner_id, c.customer_id, c.order_id,
COUNT(c.order_id) AS pizza_count,
r.distance, (r.duration / 60) AS duration_hr ,
ROUND((r.distance/r.duration * 60), 2) AS avg_speed
FROM #runner_orders AS r
JOIN #customer_orders AS c
ON r.order_id = c.order_id
WHERE distance != 0
GROUP BY r.runner_id, c.customer_id, c.order_id, r.distance, r.duration
ORDER BY c.order_id;
(Average speed = Distance in km / Duration in hour)
• Runner 1’s average speed runs from 37.5km/h to 60km/h.
• Runner 2’s average speed runs from 35.1km/h to 93.6km/h. Danny should investigate Runner 2 as the average speed has a 300% fluctuation rate!
• Runner 3’s average speed is 40km/h
1. What is the successful delivery percentage for each runner?
SELECT runner_id,
ROUND(100 * SUM
(CASE WHEN distance = 0 THEN 0
ELSE 1
END) / COUNT(*), 0) AS success_perc
FROM #runner_orders
GROUP BY runner_id;
• Runner 1 has 100% successful delivery.
• Runner 2 has 75% successful delivery.
• Runner 3 has 50% successful delivery
(It’s not right to attribute successful delivery to runners as order cancellations are out of the runner’s control.)
• What are the standard ingredients for each pizza?
• What was the most commonly added extra?
• What was the most common exclusion?
• Generate an order item for each record in the customers_orders table in the format of one of the following:
• Meat Lovers
• Meat Lovers - Exclude Beef
• Meat Lovers - Extra Bacon
• Meat Lovers - Exclude Cheese, Bacon - Extra Mushroom, Peppers
1. Generate an alphabetically ordered comma separated ingredient list for each pizza order from the customer_orders table and add a 2x in front of any relevant ingredients
For example: "Meat Lovers: 2xBacon, Beef, ... , Salami"
1. What is the total quantity of each ingredient used in all delivered pizzas sorted by most frequent first?
D. Pricing and Ratings
If a Meat Lovers pizza costs $12 and Vegetarian costs $10 and there were no charges for changes — how much money has Pizza Runner made so far if there are no delivery fees?
What if there was an additional $1 charge for any pizza extras?
1. The Pizza Runner team now wants to add an additional ratings system that allows customers to rate their runner, how would you design an additional table for this new dataset — generate a schema
for this new table and insert your own data for ratings for each successful customer order between 1 to 5.
2. Using your newly generated table — can you join all of the information together to form a table which has the following information for successful deliveries?
• customer_id
• order_id
• runner_id -rating -order_time
• pickup_time
• Time between order and pickup
• Delivery duration
• Average speed
• Total number of pizzas
1. If a Meat Lovers pizza was $12 and Vegetarian $10 fixed prices with no cost for extras and each runner is paid $0.30 per kilometre travelled — how much money does Pizza Runner have left over
after these deliveries?
E. Bonus Questions
If Danny wants to expand his range of pizzas — how would this impact the existing data design? Write an INSERT statement to demonstrate what would happen if a new Supreme pizza with all the toppings
was added to the Pizza Runner menu?
Top comments (2)
Aaron Reese •
I've not read the whole article (too long!) But your first two code blocks could have been achieved using ISNULL(), REPLACE() and CAST() and avoided the CASE statements and ALTER column types.
Cleaner code and less steps.
yaswanthteja • • Edited on • Edited
Hi Aaron thanks for your suggestion, i'm just started MySql .
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/yaswanthteja/8-week-sql-challenge-case-study-2-pizza-runner-e0o","timestamp":"2024-11-06T03:03:52Z","content_type":"text/html","content_length":"130363","record_id":"<urn:uuid:ffe13d63-6950-4eae-8555-ca8e884952a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00657.warc.gz"} |
Basic arithmetic
The basic arithmetic operations (also called basic arithmetic operations or simply arithmetic operations ) are the four mathematical operations of addition , subtraction , multiplication and division
. The mastery of basic arithmetic is one of the basic skills of reading , writing and arithmetic that students have to acquire during their school days .
Of the four basic arithmetic operations, addition and multiplication are regarded as basic operations and subtraction and division as derived operations. A number of calculation rules apply to the
two basic operations , such as the commutative laws , the associative laws and the distributive laws . In algebra , these concepts are then abstracted so that they can be transferred to other
mathematical objects .
The four basic arithmetic operations
Example of an addition
${\ displaystyle 1 + 2 = 3}$
The addition is the process of co-counting two (or more) numbers. The operator for the addition is the plus sign +, the operands are called summands , the term sum and the result is called the sum
value / value of the sum:
Summand + summand = sum value
The result of adding natural numbers is again a natural number. By memorizing and using elementary arithmetic techniques , small numbers can be added up in your head. The addition of large numbers
can be done by hand with the help of written addition .
Example of a subtraction
${\ displaystyle 5-1 = 4}$
Subtraction is the act of subtracting one number from another number. The operator for the subtraction is the minus sign -, the two operands are called the minuend and subtrahend , the term
difference and the result is called the difference value / value of the difference.
Minuend - Subtrahend = difference value
However, the result of subtracting two natural numbers is only a natural number again if the minuend is greater than the subtrahend. If the minuend and subtrahend are the same, the result is the
number zero , which is often also counted among the natural numbers. If the subtrahend is greater than the minuend, the result is a negative number . In order to be able to carry out the subtraction
without restrictions, the number range is therefore extended to the whole numbers . The subtraction of large numbers can be done by hand with the help of written subtraction .
Example of a multiplication
${\ displaystyle 3 \ cdot 5 = 15}$
Multiplication is the act of taking two (or more) numbers. The operator for multiplication is the mark of · (or x), the operands are multiplier called and multiplicand, the term product and the
result is product value / value of the product:
Multiplicand · Multiplier = product value
If there is no need to distinguish between multiplier and multiplicand, both are often referred to collectively as factors .
If the factors are natural or whole numbers, the result of the multiplication is again a natural or whole number. By memorizing the multiplication tables , small numbers can be multiplied in your
head. The multiplication of large numbers can be done by hand with the help of written multiplication .
Example of a division
${\ displaystyle 12: 3 = 4}$
Division is the process of dividing one number by another number. The operator for the division is the divided sign : (or /), the two operands are called dividend and divisor , the term quotient and
the result is called the quotient value / value of the quotient:
Dividend: Divisor = quotient value
However, the result of dividing two natural or whole numbers is only a natural or whole number again if the dividend is a multiple of the divisor. Otherwise you get a fraction . In order to be able
to carry out the division without restrictions, the number range is therefore expanded to include the rational numbers . The division by zero , however, can not be meaningfully defined. The division
of large numbers can be done by hand using the written division .
Basic arithmetic in class
The basic arithmetic operations are dealt with in mathematics lessons during the first years of school . In elementary school ( primary level ) arithmetic is first taught with small natural numbers
and later expanded to include larger numbers. Lessons also include the multiplication tables, division by remainder , solving simple equations and the rule of three . There are mental arithmetic ,
written arithmetic, rollover calculations and applications in the form of word problems practiced. Simple calculation laws are used for advantageous calculation. In the first years of a secondary
school ( secondary level I ) negative numbers are also considered, fractions and thus rational numbers are introduced, and the laws relating to the connection of the four basic arithmetic operations
are dealt with.
Calculation rules
The following are , and numbers from the underlying number range. The commutative laws apply to addition and multiplication${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle c}$
${\ displaystyle a + b = b + a}$ and ,${\ displaystyle a \ cdot b = b \ cdot a}$
that is, the result of a sum or a product is independent of the order of the summands or factors. Next apply associative laws
${\ displaystyle (a + b) + c = a + (b + c)}$ and .${\ displaystyle (a \ cdot b) \ cdot c = a \ cdot (b \ cdot c)}$
When adding or multiplying several numbers, it does not matter in which order the partial sums or partial products are formed. Therefore the brackets can be omitted from totals and products. In
addition, the distributive laws apply
${\ displaystyle a \ cdot (b + c) = a \ cdot b + a \ cdot c}$ and ,${\ displaystyle (a + b) \ cdot c = a \ cdot c + b \ cdot c}$
with which a product can be converted into a sum by multiplying and vice versa by excluding a sum into a product. Furthermore, the number behaves neutrally with regard to addition and the number
neutrally with regard to multiplication, that is ${\ displaystyle 0}$ ${\ displaystyle 1}$
${\ displaystyle a + 0 = 0 + a = a}$ and .${\ displaystyle a \ cdot 1 = 1 \ cdot a = a}$
These laws do not apply to subtraction and division, or only to a limited extent. Further calculation rules, such as dot before line , the rules in brackets and the laws of fractions , can be found
in the arithmetic formula collection .
Basic operations and derived operations
Subtraction as addition
${\ displaystyle 5-1 = 5 + ({-} 1)}$
Division as multiplication
${\ displaystyle 12: 3 = 12 \ cdot {\ tfrac {1} {3}}}$
In arithmetic , addition and multiplication are considered basic operations. The addition of natural numbers is seen as repeated determination of the successor of a summand and the multiplication of
natural numbers as repeated addition of a factor to itself. This view is then transferred to other number ranges, such as whole or rational numbers.
Subtraction and division are introduced as derived mathematical operations of the basic operations. One arrives at subtraction and division by asking about the solution of elementary equations of the
${\ displaystyle a + x = b}$ or ,${\ displaystyle a \ cdot x = b}$
where and are given numbers from the underlying number range and the number is sought. To solve these equations, an inverse operation of addition is required, namely subtraction, and also an inverse
operation of multiplication, namely division: ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle x}$
${\ displaystyle x = ba}$ or .${\ displaystyle x = b \, / \, a}$
The subtraction of a number is now defined as addition with the opposite number and division by a number as multiplication with the reciprocal value : ${\ displaystyle a}$ ${\ displaystyle -a}$${\
displaystyle a}$ ${\ displaystyle {\ tfrac {1} {a}}}$
${\ displaystyle x = b + (- a)}$ or .${\ displaystyle x = b \ cdot {\ tfrac {1} {a}}}$
The opposite number and the reciprocal of a number are called the inverse numbers with respect to addition and multiplication. In this way, the calculation rules for addition and multiplication can
also be applied to subtraction and division.
Algebraic structures
In algebra , these concepts that were initially created for arithmetic are abstracted in order to be able to transfer them to other mathematical objects . An algebraic structure then consists of a
set of carriers (here a set of numbers) and one or more links on this set (here the arithmetic operations) that do not lead out of it . The various algebraic structures then differ only in terms of
the properties of the links (the rules of calculation), which are defined as axioms , but not in terms of the concrete elements of the set of carriers. The following algebraic structures are obtained
for the basic operations:
• The set of natural numbers forms a commutative semigroup with the addition , in which the associative law and the commutative law apply to the link.${\ displaystyle (\ mathbb {N}, +)}$
• The set of natural numbers also forms a commutative semigroup with multiplication .${\ displaystyle (\ mathbb {N}, \ cdot)}$
• Together with the addition, the set of whole numbers forms a commutative group in which there is also a neutral element and an inverse element for each element.${\ displaystyle (\ mathbb {Z}, +)}
• The set of whole numbers forms a commutative ring with addition and multiplication , in which the distributive laws also apply to the links.${\ displaystyle (\ mathbb {Z}, +, \ cdot)}$
• With addition and multiplication, the set of rational numbers forms a field in which every element apart from zero has an inverse element with regard to multiplication.${\ displaystyle (\ mathbb
{Q}, +, \ cdot)}$
According to the principle of permanence , all calculation rules of a basic structure (here a simple number range with the basic operations) also apply in a correspondingly more specific structure
(here an extended number range with the same operations). This structuring and axiomatization now makes it possible to transfer knowledge gained from numbers to other mathematical objects. For
example, corresponding operations at are vectors the vector addition and matrices , the matrix addition . Special structures arise when considering finite sets , for example remainder class rings as
a mathematical abstraction of a division with remainder .
All four basic arithmetic operations were already known in ancient Egyptian mathematics and in Babylonian mathematics . However, multiplication and division were not arithmetic operations in their
own right. The multiplication of natural numbers was traced back to the continued doubling ( duplication ) of a factor and subsequent addition of the partial results. In the case of non-integer
quotients, the division was carried out approximately by means of continued halving ( mediation ). Multiplication and division can only be found as independent operations in ancient Greek mathematics
, for example in Euclid and Pappos .
Which arithmetic operations are included in the basic arithmetic operations has changed significantly over time. With Heron and Diophantos , squaring and taking square roots were added as further
basic arithmetic operations to the four known arithmetic operations . In Indian mathematics , these operations have been replaced by the more general exponentiation and extraction of the roots , and
in more recent times the logarithm has been added as the seventh basic arithmetic function. In Islamic mathematics , starting with Al-Chwarizmi , the duplatio and the mediatio were also viewed as
separate arithmetic operations.
In the arithmetic books of the Middle Ages there were further additions to the basic arithmetic operations, which were called "species" there. Around 1225, Johannes de Sacrobosco found a total of
nine of these species: Numeratio , Additio , Subtractio , Duplatio , Multiplicatio , Mediatio , Divisio , Progressio and Radicum extractio . The Numeratio dealt with counting, reading and writing
numbers, the Progressio was the summation of successive natural numbers and the Extractio only included the taking of square roots. It was only in 1494 rejected Luca Pacioli the Duplatio and Mediatio
as special cases of multiplication and division again. This was followed by further reductions until Gemma Frisius in 1540 was one of the first authors to limit the basic arithmetic operations to the
four known.
Individual evidence
Web links | {"url":"https://de.zxc.wiki/wiki/Grundrechenart","timestamp":"2024-11-04T15:07:52Z","content_type":"text/html","content_length":"84841","record_id":"<urn:uuid:1cba4aae-925d-42bc-8b3b-cae4363d919d>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00060.warc.gz"} |
6 Times Table Test - Free Printable
6 Times Table Test
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Help!! programming hp 48G
hey! I hope someone could help me!
I have a set of equations that I would like to solve with a program:
A = bd - zd^2
P = b + 2d v(1 + z^2)
R = A/P
V = 1/n * R^(2/3) *(1/2)
Q = A * V
I have already done it with the Multiple Equiation Solver, but I'd like to know if there's another way of doing it. This is what I did:
{'A = bd - zd^2'
'P = b + 2d v(1 + z^2)'
'R = A/P'
'V = 1/n * R^(2/3) *(1/2)'
'Q = A * V'}
'EQ' STO
And then I created another program to erase the variables that it stored:
{A b d z P R V n S EQ Mpar Q}
Please help me, I think this is not the best way to do it!
03-17-2007, 02:55 PM
Hi Marco,
This is what I do: I put each formulae in a variable F1, F2, F3, F4.
Then I add a description in the variable TITLE and next I complete the sequence of the params in a list and store it in LVARI.
Be shure the list is complete.
The main program then looks clearly structured:
<< F1 F2 F3 F4 4 \-> LIST STEQ TITLE LVARI MINIT MITM MSOLVR>>
Using STEQ shortens the program.
Hope this helps.
Kind regards,
Benny Vanrutten
03-17-2007, 05:20 PM
Hey Benny thanks for your answer!
I have sone doubts in what you told me: how can I add a title in the global variable?, and what does LVARI do? thanks for your answer!, but I have another doubt, When I don't give a lot of values it
tells me that it has TOO MANY UNKNOWS, but with the same values that I give it I can solve the problem mannually but it takes some time, I don't know if I can apply in the program an iteration methor
or something, what would you suggest me?
03-18-2007, 06:43 PM
Hi Marco,
In the variable LVARI you put ALL your parameters in the sequence you like them to appear on the display above the F-keys: e.g.
{ A b d z P R V ...}
In TITLE you put the title you want to appear on top of the display e.g. "MY EQUATIONS"
TO MANY UNKNOWNS: that is normaly what you get when you did not define enough variables. Perhaps the program needs startvalues.
You can find more in the hp50G users guide on pages 7-10 ... 7-21 (hp50gug.pdf) which you can download from www.hpcalc.org (november 24 2006).
If you like to Purge you can call LVARI and issue PURGE. That saves time. The best way is to run the program in a separate directory.
Benny Vanrutten
03-23-2007, 02:09 PM
Try my multiple equation solver. Scroll down to multiple equation solver for the HP28C. It works on the 48G too. | {"url":"https://archived.hpcalc.org/museumforum/thread-110374-post-110805.html#pid110805","timestamp":"2024-11-03T00:54:22Z","content_type":"application/xhtml+xml","content_length":"41723","record_id":"<urn:uuid:655c0b1d-c984-4264-be53-467f6bffca9f>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00046.warc.gz"} |
Second-order Linear ODE's with Constant Coefficients | JustToThePointSecond-order Linear ODE's with Constant Coefficients
And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous
Differential equations
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
1. Dependent and independent variables. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent
2. Constants. Fixed numerical values that do not change.
3. Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables
themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
• Dependent variables: Variables that depend on one or more other variables (y).
• Independent variables: Variables upon which the dependent variables depend (x).
• Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
• The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x[0], y[0]) and
• Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x[0], y[0]).
Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x[0], y[0]) .
A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) =
0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.
The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$
Second-order Linear Homogeneous ODE’s with Constant Coefficients
A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:
• y is the dependent variable (a function of the independent variable t),
• y′ and y′′ are the first and second derivatives of y with respect to t,
• t is the independent variable,
• A and B are constants.
This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.
General Solution
To solve this ODE, we seek two linearly independent solutions y[1](t) and y[2](t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c[1] and c[2] are two
arbitrary constants determined by initial conditions.
Physical Example: Mass on a Spring
Consider a mass-spring system where a mass m hangs on a spring. The position (displacement) of the mass at any time t is described by x(t), measured along the x-axis.
x-axis is vertical (it is not so in the diagram, sorry). When the mass is in equilibrium (denoted by 0), the spring is stretched. The origin of the x-axis is chosen so that the mass’s equilibrium
position corresponds to x = 0. This means the spring is exerting an upward force to balance the downward gravitational force on the mass.
The movement of the mass is governed by Newton’s second law, which can be written as: ma = mx’’ = F[total] = -kx (Hooke’s law states that the force needed to extend or compress a spring by some
distance x scales linearly with respect to that distance, F[spring] = -kx where k is a constant factor characteristic of the spring, the spring’s stiffness) -cx’ (Damping Force: F[damping] = -cx’
where c is the damping constant and x’ is the velocity) where m is the mass and a is the acceleration (Refer to Figure 1 for a visual representation and aid in understanding it)
Damping forces are a special type of force that are used to slow down or stop a motion. A dashpot is a device for cushioning or damping a movement (as of a mechanical part) to avoid shock.
This gives the equation of motion: mx’’ + cx’ + kx = 0 ↭[Dividing through by m, we get the simplified form:] $x’’ + \frac{c}{m}x’+ \frac{k}{m}x = 0$.
This is a second-order linear homogeneous ODE with constant coefficients that models the oscillatory motion of the spring system. To solve this equation, we need to find two independent solutions.
General Solution Method
To solve the second-order linear ODE: y’’ + Ay’ + By = 0, we assume the solution is of the form $y = e^{rt}$ where r is a constant to be determined.
Substituting or plugging $y = e^{rt}$ into the differential equation gives:
$r^2e^{rt} + Are^{rt} + Be^{rt} = 0$ ⇒[Cancelling the common factor $e^{rt}$ (which is never zero), we obtain the characteristic equation:] $r^2 + Ar + B = 0$.
The form of the general solution to the ODE depends on the nature of the roots of this quadratic equation.
Cases Based on the Roots
Case 1: Real and Distinct Roots (Overdamped System)
If the characteristic equation has two distinct real roots, say r[1] and r[2], then the general solution is: y = $c_1e^{r_1t} + c_2e^{r_2t}$.
In mechanical or electrical systems, this is akin to a system where the damping is so strong that the system returns to equilibrium without oscillating. The resistance to motion is so high that the
mass slowly creeps back to its equilibrium position (more slowly than if there were no damping) without oscillating. The terms $e^{r_1t}$ and $e^{r_2t}$ represent two exponentially decaying
Example y’’ + 4y’ + 3y = 0
Consider the equation: y’’ + 4y’ + 3y = 0.
The characteristic equation for the differential equation is: r^2 + 4y + 3 = 0 ↭[Factoring gives] (r + 3)(r + 1) = 0. Thus, the roots are r[1] = -3 and r[2] = -1.
The general solution for the differential equation is: y = $c_1e^{-3t} + c_2e^{-t}$
Suppose we are given the initial condition y(0) = 1, y’(0) = 0.
To determine c[1] and c[2], we first need to calculate the derivate of y: $y’ =-3c_1e^{-3t}- c_2e^{-t} $
Substituting the initial condition t = 0, y(0) = 1: 1 = c[1] + c[2].
Similarly, substituting y’(0) = 0, 0 = -3c[1] -c[2]. Adding both 1 = -2c[1], so c[1] = ^-1⁄[2], c[2] = -3c[1] = ^3⁄[2]
The final solution to the differential equation is: y = $\frac{-1}{2}e^{-3t} + \frac{3}{2}e^{-t}$ (Refer to Figure 2 for a visual representation and aid in understanding it)
Case 2. Complex roots (Underdamped System)
If the characteristic equation has complex roots, say r = a ± bi, the general solution involves sine and cosine terms.
We get two complex solutions, one of them is $y=e^{(a+bi)t} = e^{at}e^{bti} = e^{at}(cos(bt)+isin(bt))$⇒ Its real part is $e^{at}cos(bt)$ and its imaginary part is $e^{at}sin(bt)$ ⇒[Theorem, the real
and imaginary parts of a complex solution are themselves solutions] $e^{at}cos(bt)$ and $e^{at}sin(bt)$ are two independent real solutions ⇒ Thus, the general solution is: y = $e^{at}(c_1cos(bt)
+c_2sin(bt)) =$[Using the trigonometry identity a·cos(θ)+b·sin(θ) = c·cos(θ - Φ)] $e^{at}(c·cos(bt- Φ))$ where:
• $c = \sqrt{c_1^2+c_2^2}$
• $Φ = arctan(\frac{c_2}{c_1})$.
This corresponds to an underdamped system, where the system oscillates with decreasing amplitude.
Theorem (the real and imaginary parts of a complex solution are themselves solutions) If y(t) = u(t) + iv(t) is a complex solution to a differential equation y'' + Ay' + By = 0 with real coefficients
A and B, then both u(t) and v(t) satisfy the differential equation (they are both real solutions).
By assumption u(t) + iv(t) is a solution to a differential equation y’’ + Ay’ + By = 0, hence (u + iv)’’ + A(u + iv)’ + B(u + iv) = 0.
Compute derivatives and group real and imaginary parts taking into consideration that A and B are real numbers:
(u’’ + Au’ + Bu) + (v’’ + Av’ + Bv)i = 0 ⇒ u’’ + Au’ + Bu = 0, v’’ + Av’ + Bv = 0 ⇒Thus, u(t) and v(t) are real solutions.
Example y’’ + 4y’ + 5y = 0
Consider the equation y’’ + 4y’ + 5y = 0.
The characteristic equation for the differential equation is: r^2 + 4r + 5 = 0 ↭[Using the quadratic formula:] $r = \frac{-4±\sqrt{-4}}{2} = -2±i$. Thus, the roots are: r = -2 ± i.
The complex roots indicate that the general solution can be expressed in terms of exponential and trigonometric functions. One complex solution is: y = $e^{-(2+i)t} = e^{-2t}[cos(t)+i·sin(t)]$ ⇒
[Theorem. Its real and imaginary parts are two independent real solutions] $e^{-2t}cos(t), \text{ and } e^{-2t}sin(t)$ are two independent real solutions, and the real solution is finally y = $e^
{-2t}(c_1cos(t) + c_2sin(t))$.
Applying the initial conditions y(0) = 1, y’(0) = 0.
y(0) = 1 = $e^{0}(c_1cos(0) + c_2sin(0)) = c_1,$ so c[1] = 1.
Next, we we differentiate y(t): y’(t): y’(t) = $e^{-2t}[-2(c_1cos(t)+c_2sin(t)) + (-c_1sin(t)+c_2cos(t))]$
y’(0) = 0 ⇒ 0 = -2c[1] + c[2] = -2 + c[2] ⇒ c[2] = 2.
Therefore, the solution is: $e^{-2t}(cos(t) + 2sin(t)) = \sqrt{5}e^{-2t}cos(t- Φ)$ where $c = \sqrt{c_1^2+c_2^2} = \sqrt{1^2+2^2} = \sqrt{5}$ and Φ is a phase angle given by $Φ = arctan(\frac{c_2}
{c_1}) = tan^{-1}(\frac{2}{1})≈ 63.43° [\text{or (degrees × π) ÷ 180 = radians}] π/3$ radians (Refer to Figure 3 for a visual representation and aid in understanding it)
In this case, y = $e^{at}(c_1cos(bt)+c_2sin(bt)) = e^{at}(ccos(bt- Φ))$, the system oscillates while gradually returning to equilibrium. The oscillations are damped over time, meaning they decrease
in amplitude (the oscillations die out at a rate governed by $e^{at}$), but the system does not return to equilibrium immediately.
Two equal roots (Critically damped system)
In the study of second-order linear homogeneous ordinary differential equations (ODEs) with constant coefficients y′’ +Ay′ +By = 0, we encounter different cases based on the nature of the roots of
the characteristic equation. One such case is when the characteristic equation has two equal roots, which leads to a critically damped system.
The characteristic equation for this ODE is derived from assuming a solution of the form y = e^rt where r is a constant to be determined. Substituting this assumed solution into the ODE gives the
characteristic equation: r^2 + Ar + B = 0.
When this quadratic equation has two equal roots, it means that the discriminant is zero, Δ = A^2-4B = 0. This implies that the two roots of the characteristic equation are both r = −a, where a = $\
frac{A}{2}.$ So the characteristic equation is: r^2 + 2ar + a^2 = 0.
This quadratic can be factored as (r+a)^2 = 0. Thus, the characteristic equation has a repeated root r = -a.
Challenge of Finding Two Independent Solutions
In general, for a second-order linear ODE, we need two independent solutions to form the general solution. However, when the characteristic equation has a repeated root, we only have one solution of
the form: y[1](t) = e^-at. The challenge is that we need another linearly independent solution. To find the second solution, we use a special method.
Theorem (Reduction of Order). If y[1] is a known solution to the differential equation y’’ + py’ + qy = 0, then there exists another linearly independent solution y[2](t) of the form y[2](t) = y[1]u
(t), where u(t) is a function to be determined.
This is called the reduction of order method.
We know that y[1](t) is one solution. We assume that the second solution y[2](t) takes the form: y[2](t) = y[1](t)·u(t) where u(t) is an unknown function that we need to find.
We need to calculate the first and second derivatives of y[2](t).
• y’[2](t) =[Using the product rule] y[1]’(t)u(t)+y[1](t)u’(t)
• y’’[2](t) =[Again, applying the product rule to y’[2](t)] y’’[1](t)u(t) + 2y’[1](t)u’(t) + y[1](t)u’’(t)
Now, substitute into the original ODE y’’ + p(t)y’ + q(t)y = 0:
y’’[1](t)u(t) + 2y’[1](t)u’(t) + y[1](t)u’’(t) + p(t)(y[1]’(t)u(t)+y[1](t)u’(t)) + q(t)y[1](t)·u(t) = 0
Next, group terms involving u(t), u’(t) and u’’(t): (y’’[1](t) + p(t)y[1]’(t)+q(t)y[1](t))·u(t) +(2y’[1](t) + p(t)y[1](t))·u’(t) + y[1](t)u’’(t) = 0 ⇒[Simplify, y[1]’’(t) + p(t)y[1]’(t) + q(t)y[1](t)
= 0 because y[1](t) is a known solution of the ODE. Therefore, the term involving u(t) vanishes:] 0·u(t) +(2y’[1](t) + p(t)y[1](t))u’(t) + y[1](t)u’’(t) = 0 ⇒(2y’[1](t) + p(t)y[1](t))u’(t) + y[1](t)
u’’(t) = 0
Assuming y[1](t)≠ 0, divide the entire equation by y[1](t):
$u’’(t) + \frac{2y_1’(t)+p(t)y_1(t)}{y_1(t)}u’(t) = 0↭[\text{Let’s simplify the coefficients of u’(t)}] u’’(t) + (2\frac{y_1’(t)}{y_1(t)}+p(t))u’(t) = 0$
The equation becomes a first-order linear ODE in u’(t). Let’s introduce the substitution v(t) = u’(t) to reduce the order:
$v’(t) + (2\frac{y_1’(t)}{y_1(t)}+p(t))v(t) = 0$
This is a first-order linear ODE for v(t), and it can be solved using an integrating factor. The integrating factor is: μ(t) = $e^{\int (2\frac{y_1’(t)}{y_1(t)}+p(t))dt}$
We can split the integral into two parts: μ(t) = $e^{\int (2\frac{y_1’(t)}{y_1(t)})dt}e^{\int p(t)dt} = e^{2ln|y_1(t)|}e^{\int p(t)dt} = e^{ln|y_1(t)|^2}e^{\int p(t)dt} = y_1(t)^2e^{\int p(t)dt}$
Now, we multiply the original ODE by the integrating factor μ(t):
$y_1(t)^2e^{\int p(t)dt}v’(t)+y_1(t)^2e^{\int p(t)dt}(2\frac{y_1’(t)}{y_1(t)}+p(t))v(t) = 0$
Simplify the left-hand side into the derivative of the product μ(t)v(t): $\frac{d}{dt}(μ(t)v(t)) = 0$
To solve for v(t), integrate both sides with respect to t: μ(t)v(t) = C[1] where C[1] is a constant of integration.
Now, solve for v(t): $v(t)=\frac{C_1}{μ(t)} = \frac{C_1}{y_1(t)^2e^{\int p(t)dt}}$, so this is the solution for v(t). Once v(t) is found, recall that v(t)=u′(t). Integrate v(t) to find u(t), u(t) = $
\int v(t)dt + C_2$ where C[2] is another constant of integration. Finally, the second solution to the ODE is: y[2](t) = y[1](t)·u(t).
The general solution to the original second-order ODE is the linear combination of the two independent solutions y[1](t) and y[2](t): y(t) = C[1]y[1](t) + C[2]y[2](t) where C[1] and C[2] are
arbitrary constants.∎
Application to the Critically Damped Case
In our particular case, we already know that y[1](t) = e^-at is a solution. So, we can assume that the second solution is of the form y[2](t) = e^-at·u(t).
We now plug y[2](t) into the original ODE. To do this, we need to compute the first and second derivatives of y[2](t):
y’[2](t) =[Using the product rule] -ae^-atu + e^-atu’, y[2](t)’’ = a^2e^-atu -2ae^-atu’ + e^-atu''
Now, substitute y[2], y[2]’ and y[2]’’ into the original equation y’’ + 2ay’ + a^2y = 0, a^2e^-atu -2ae^-atu’ + e^-atu’’ -2a^2e^-atu +2ae^-atu’ + a^2e^-atu = 0 [Simplifying the equation] e^-atu’’ +
(-2ae^-at+2ae^-at)u’ + (a^2e^-at-2a^2e^-at+ a^2e^-at)u = 0⇒e^-atu’’⇒[Since e^-at ≠ 0, we are left with the differential equation] u’’(t) = 0
u’’(t) = 0 ⇒[If u′′(t) = 0, then u′(t) must be a constant because the derivative of any constant function is zero] u’(t) = c[1] where c[1] is a constant ⇒[Integrating again] u(t) = c[1]t + c[2] where
c[2] is another constant that comes from the integration process.
Thus, the solution to u’’(t) = 0 is a linear function: u(t) = c[1]t + c[2].
Thus, the second solution y[2](t) is y[2](t) = e^-at·u = e^-at·(c[1]t + c[2]). Since c[2]e^-at is just a multiple of the first solution y[1](t) = e^-at, we discard it for the sake of linear
independence (so y[2] is not a merely a scalar multiple of y[1]) Therefore, the second independent solution is y[2](t) = te^-at (For simplicity and without loss of generality, we can set c^1 = 1).
The general solution to the critically damped ODE is the linear combination of the two independent solutions: y(t) = c[1]y[1](t) + c[2]y[2](t) = c[1]e^-at + c[2]te^-at where c[1] and c[2] are
arbitrary constants determined by initial conditions.
In physical systems (e.g., a mass-spring-damper system), critical damping occurs when the damping is just sufficient to bring the system back to equilibrium as quickly as possible without
oscillating. The system returns to equilibrium without oscillating, but more slowly than in the overdamped case. The solution describes this behaviour:
This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare. | {"url":"https://justtothepoint.com/calculus/secondorderode/","timestamp":"2024-11-03T15:37:45Z","content_type":"text/html","content_length":"36206","record_id":"<urn:uuid:cc338cc2-7e7a-41e0-a33c-11123dd415d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00754.warc.gz"} |
What Inverter Size Do I Need to Run a Hair Dryer? - portablesolarexpert.com
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When we think of high powered appliances, the fridge and AC quickly come to mind. But hair dryers use a lot of electricity too, which is why a lot of people ask, what inverter size do I need to run a
hair dryer?
The average hair dryer uses 1500 to 1800 watts at the highest setting, so a 2000 watt inverter is ideal. Smaller hair dryers will consume 800 watts so a 1000 watt inverter will be sufficient.
Calculate Hair Dryer Inverter Size Requirements
Hair dryers come in different styles, designs and functionality. Some use more power than others, so we need to take a closer look at the numbers to find out what inverter size you need.
Most hair dryers have a power consumption ranging from 800 to 1800 watts. This electricity usage however is on a per hour basis. Unless you blow dry your hair for an hour, the usage will be lower.
The following is a power consumption guide for a 1500 watt hair dryer. These are only estimates and the watts usage might be different with yours.
• 60 minutes: 1500 watts
• 30 minutes: 750 watts
• 15 minutes: 375 watts
• 5 minutes: 160 watts
A person spends 15 minutes a day using a hair dryer on average. Assuming it is a 1500 watt model at maximum setting, power consumption will be 375 to 400 watts.
Again this figure is based off a 1500 watt inverter. High powered blow dryers might use 2200 watts or more. At the other end of the spectrum are low powered hair dryers that max out at 800 watts.
With this in mind, we can draw the following conclusions:
• A 1500 watt hair dryer is not going to use 1500 watts of inverter power, not unless you use it for an hour.
• If you blow dry your hair for 15 minutes a day (the average for most people), inverter power consumption will be 375 to 400 watts.
• A more powerful hair dryer will use more watts. But it might dry your hair faster so runtime will be shorter.
• The opposite might be true with low powered hair blowers. An 800W hair dryer might take more time to dry your hair so you end up using it longer than a 1500 watt model.
In other words, the power consumption might eventually come out the same. But it will depend on the product design and efficiency.
Technically you might get by with a 500 watt inverter. But we recommend a 1500 watt inverter in case of a power surge or extended use. There are plenty to choose from, and one of our favorites is the
POTEK 1500W Power Inverter as it is easy to use and durable.
How Many Batteries Do I Need to Run a Hair Dryer?
If you are living off the grid, you need a battery bank to run appliances and other electronics off an inverter. This is particularly true for a hair dryer as it needs a steady stream of power.
A 125ah deep cycle battery can run a 1500 watt hair dryer for an hour before it is fully discharged. Hair blowers that use 2000 watts or more require a minimum 200ah battery bank.
A Renogy 200ah 12V AGM battery will do nicely here. Once plugged into the inverter it will run your hair dryer as if it is on 120V.
There are a few things you need to know before running a hair dryer – or any appliance – off a battery bank. Namely, the depth of discharge and the appliance’s actual usage.
A hair dryer, just like a solar powered microwave , is usually not run for hours on end. So you have to calculate its watt usage by the minutes it operates.
The guide provided earlier illustrates how many watts a 1500 watt hair dryer might use over a specific period of time. We need to take the same approach with the batteries.
If you run a 1500 watt hair dryer for an hour, it consumes 1500 watts. For this you need a 12V 125ah battery (watts divided by volts = amps).
If you use the hair blower for 30 minutes, you would need 62.5ah. Use the blow dryer for 15 minutes and it would require 31.25ah and so on.
Depth of Discharge
So if you don’t use the hair dryer for an hour, you don’t need a 125ah battery. If you just use it for 15 minutes a day, a 31.25ah battery is sufficient, right?
Not quite, because of depth of discharge or DOD. The DOD determines when the battery should be recharged, and with lead acid batteries that is 50%.
So while you can fully discharge a 31.25ah battery, it is not a good idea. This will shorten the battery life cycle and might lead to long term damage.
The solution is to double the capacity to 62.5ah. Even if you use the hair dryer for 15 minutes there would still be 50% capacity left, just right for the DOD recharge.
There is no 62.5ah battery available though, so get the next largest available. That would be 75ah to 100 ah. 100ah is the standard size used in RVs and homes, so might as well go for that.
Will a 1000W Inverter Run a Hair Dryer?
A 1000 watt inverter can power a hair dryer provided there is enough energy in the battery bank. This also assumes the hair dryer uses less than 1000 watts when it runs.
In our examples above we have been using a 1500 watt hair dryer. But if you have one that consumes less than 1000 watts, a smaller inverter will work.
A small inverter should have no problems with an 800 watt hair dryer. As long as the installation and cables are the right size , the device will function properly.
For the hair dryer to run effectively, it must be the only appliance loaded on the inverter. Yes there is about 200 watts power in reserve, but it is actually less than that due to inefficiency.
Inverters lose energy when DC is converted to AC. The losses range from 15% to 5%.
This is expressed in efficiency ratings like 85%, 95% and so on. Due to this, an inverter consumes more power than its load.
If you have an 800 watt hair dryer loaded onto an 85% efficient 1000 watt inverter, the system uses 15% more watts:
800 x 115% = 920
While the hair dryer uses 800 watts, the inverter ends up using another 120 watts due to inefficiency. At 920 watts it is pretty close to the system capacity.
This is why you need to have reserve power available. Extra power capacity is also essential in case of a sudden power spike. Grounding solar wires is also required for these reasons. The bottom
line, he more battery power available, the longer you can run the hair dryer and other appliances.
Do I Need Solar Panels to Run a Hair Dryer?
You do not need solar panels to use a hair dryer. It can run off the inverter battery bank. But if you are off the grid you probably have a solar system installed already.
While you do not need solar panels, the PV modules are necessary to recharge the batteries. Solar panels charge the battery bank so you can use it to power the inverter and your hair dryer.
If you want to use solar panels to run a hair dryer, it will take a 5 x 300W solar array. This will be enough to power an 800 to 1500W model for at least 5 hours.
This solar array can produce up to 1500 watts an hour. Due to changing weather patterns, panel efficiency, set up etc. the output will probably be lower. Even in ideal condition, the array may reach
peak (close or reach 300 watts) for a couple of hours only.
But even so, the output should be more than 800 watts, enough to get the hair dryer going. And if you are only using it for 15 minutes a day, the runtime will not be a problem.
But most likely you will use the panels to power up the batteries. An 800 watt hair dryer needs a 12V 100ah battery. A 3 x 300W solar array can recharge it in an hour or so. For a 1500 watt hair
blower you will need a 5 to 6 x 300W solar panels.
If you are going to use a hair dryer in an off grid system, be prepared to reserve a bit of power. You can use a battery bank as mentioned here, or shore power or even a generator.
I am an advocate of solar power. Through portablesolarexpert.com I want to share with all of you what I have learned and cotinue to learn about renewable energy. | {"url":"https://www.portablesolarexpert.com/what-inverter-size-do-i-need-to-run-a-hair-dryer/","timestamp":"2024-11-09T23:57:11Z","content_type":"text/html","content_length":"124611","record_id":"<urn:uuid:cd45ad38-b5e5-4108-a270-71a0d4716198>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00067.warc.gz"} |
Differential Topology, Infinite-Dimensional Lie Algebras, and Applications: D. B. Fuchs’ 60th Anniversary Collectionsearch
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Differential Topology, Infinite-Dimensional Lie Algebras, and Applications: D. B. Fuchs’ 60th Anniversary Collection
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Product Code: TRANS2/194
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Differential Topology, Infinite-Dimensional Lie Algebras, and Applications: D. B. Fuchs’ 60th Anniversary Collection
Hardcover ISBN: 978-0-8218-2032-2
Product Code: TRANS2/194
List Price: $175.00
MAA Member Price: $157.50
AMS Member Price: $140.00
eBook ISBN: 978-1-4704-3405-2
Product Code: TRANS2/194.E
List Price: $165.00
MAA Member Price: $148.50
AMS Member Price: $132.00
Hardcover ISBN: 978-0-8218-2032-2
eBook ISBN: 978-1-4704-3405-2
Product Code: TRANS2/194.B
List Price: $340.00 $257.50
MAA Member Price: $306.00 $231.75
AMS Member Price: $272.00 $206.00
• American Mathematical Society Translations - Series 2
Advances in the Mathematical Sciences
Volume: 194; 1999; 313 pp
MSC: Primary 14; 17; 58
This volume presents contributions by leading experts in the field. The articles are dedicated to D. B. Fuchs on the occasion of his 60th birthday. Contributors to the book were directly
influenced by Professor Fuchs and include his students, friends, and professional colleagues. In addition to their research, they offer personal reminicences about Professor Fuchs, giving insight
into the history of Russian mathematics.
The main topics addressed in this unique work are infinite-dimensional Lie algebras with applications (vertex operator algebras, conformal field theory, quantum integrable systems, etc.) and
differential topology. The volume provides an excellent introduction to current research in the field.
Graduate students, research mathematicians, and physicists interested in algebraic geometry; theoretical physicists.
□ Chapters
□ V. I. Arnold — First steps of local symplectic algebra
□ Pavel Etingof — Whittaker functions on quantum groups and $q$-deformed Toda operators
□ Boris Feigin and Edward Frenkel — Integrable hierarchies and Wakimoto modules
□ B. Feigin and S. Loktev — On generalized Kostka polynomials and the quantum Verlinde rule
□ Michael Finkelberg and Ivan Mirković — Semi-infinite flags. I. Case of global curve $\mathbb {P}^1 $
□ Boris Feigin, Michael Finkelberg, Alexander Kuznetsov and Ivan Mirković — Semi-infinite flags. II. Local and global intersection cohomology of quasimaps’ spaces
□ Fyodor Malikov and Vadim Schechtman — Chiral de Rham complex. II
□ E. Mukhin and A. Varchenko — On algebraic equations satisfied by hypergeometric solutions of the qKZ equation
□ V. Ovsienko and C. Roger — Deforming the Lie algebra of vector fields on $S^1$ inside the Lie algebra of pseudodifferential symbols on $S^1$
□ Alexander Postnikov, Boris Shapiro and Mikhail Shapiro — Algebras of curvature forms on homogeneous manifolds
□ Vladimir Retakh, Christophe Reutenauer and Arkady Vaintrob — Noncommutative rational functions and Farber’s invariants of boundary links
□ Serge Tabachnikov — Remarks on the geometry of exact transverse line fields
□ B. Tsygan — Formality conjectures for chains
□ V. A. Vassiliev — On finite order invariants of triple point free plane curves
□ Albert Schwarz, Alexei Sossinsky, Claude Roger, Boris Feigin, Sergei Tabachnikov and Alexander Astashkevich — Appendix. Personal notes
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Requests
Advances in the Mathematical Sciences
Volume: 194; 1999; 313 pp
MSC: Primary 14; 17; 58
This volume presents contributions by leading experts in the field. The articles are dedicated to D. B. Fuchs on the occasion of his 60th birthday. Contributors to the book were directly influenced
by Professor Fuchs and include his students, friends, and professional colleagues. In addition to their research, they offer personal reminicences about Professor Fuchs, giving insight into the
history of Russian mathematics.
The main topics addressed in this unique work are infinite-dimensional Lie algebras with applications (vertex operator algebras, conformal field theory, quantum integrable systems, etc.) and
differential topology. The volume provides an excellent introduction to current research in the field.
Graduate students, research mathematicians, and physicists interested in algebraic geometry; theoretical physicists.
• Chapters
• V. I. Arnold — First steps of local symplectic algebra
• Pavel Etingof — Whittaker functions on quantum groups and $q$-deformed Toda operators
• Boris Feigin and Edward Frenkel — Integrable hierarchies and Wakimoto modules
• B. Feigin and S. Loktev — On generalized Kostka polynomials and the quantum Verlinde rule
• Michael Finkelberg and Ivan Mirković — Semi-infinite flags. I. Case of global curve $\mathbb {P}^1 $
• Boris Feigin, Michael Finkelberg, Alexander Kuznetsov and Ivan Mirković — Semi-infinite flags. II. Local and global intersection cohomology of quasimaps’ spaces
• Fyodor Malikov and Vadim Schechtman — Chiral de Rham complex. II
• E. Mukhin and A. Varchenko — On algebraic equations satisfied by hypergeometric solutions of the qKZ equation
• V. Ovsienko and C. Roger — Deforming the Lie algebra of vector fields on $S^1$ inside the Lie algebra of pseudodifferential symbols on $S^1$
• Alexander Postnikov, Boris Shapiro and Mikhail Shapiro — Algebras of curvature forms on homogeneous manifolds
• Vladimir Retakh, Christophe Reutenauer and Arkady Vaintrob — Noncommutative rational functions and Farber’s invariants of boundary links
• Serge Tabachnikov — Remarks on the geometry of exact transverse line fields
• B. Tsygan — Formality conjectures for chains
• V. A. Vassiliev — On finite order invariants of triple point free plane curves
• Albert Schwarz, Alexei Sossinsky, Claude Roger, Boris Feigin, Sergei Tabachnikov and Alexander Astashkevich — Appendix. Personal notes
Permission – for use of book, eBook, or Journal content
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Let me make it clear about how pages that are many 1000 terms?
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Tobias Ekholm, Triangle Topology Seminar - Department of Mathematics
Tobias Ekholm, Triangle Topology Seminar
November 29, 2016 @ 4:30 pm - 5:00 pm
Title: Wrapped Floer cohomology and Legendrian surgery
Abstract: We first review the relation between wrapped Floer cohomology of co-core disks after Lagrangian handle attachment and the Legendrian DGA of the corresponding attaching spheres. Then we
discuss a generalization of this result to the partially wrapped setting where the Legendrian dga should be enriched with loop space coefficients, and describe several cases when explicit
calculations are possible via parallel copies or local coefficient systems. We also discuss applications of these ideas to the topology of Lagrangian fillings of Legendrian submanifolds. The talk
reports on joint work with Y. Lekili. | {"url":"https://math.unc.edu/event/triangle-topology-seminar-3/","timestamp":"2024-11-09T07:07:07Z","content_type":"text/html","content_length":"109826","record_id":"<urn:uuid:38ac938a-d623-48da-9124-567c01309cbb>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00209.warc.gz"} |
Many-Body Localization and its Discontents | Stanford Institute for Theoretical Physics
Main content start
Many-Body Localization and its Discontents
Mon March 7th 2022, 2:00pm
University of Virginia
Event Sponsor
Stanford Institute for Theoretical Physics
Abstract: A quantum system is said to be many-body localized (MBL) if it remains close to its initial state, i.e., it fails to thermalize. In 2016, I published a proof that certain one-dimensional
spin chains have an MBL phase (the proof depended on a certain assumption on level statistics). Some recent numerical studies have raised questions about whether there is a true MBL phase. I will
attempt to summarize the issues raised, but the fact remains that the mechanisms for the breakdown of MBL phase are well understood theoretically. In recent work with Morningstar and Huse (PRB,
2020), we develop specific RG flow equations. These are similar to the Kosterlitz-Thouless (KT) flow as previously shown, but there are important differences that place the MBL transition in a new
universality class. | {"url":"https://sitp.stanford.edu/events/many-body-localization-and-its-discontents","timestamp":"2024-11-11T16:07:01Z","content_type":"text/html","content_length":"34538","record_id":"<urn:uuid:055c1634-d165-40c3-8a4a-3f812835569e>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00094.warc.gz"} |
The variational approach to concentrationThe variational approach to concentration
The variational approach to concentration
June 10, 2024
\[\newcommand{\norm}[1]{\|#1\|} \newcommand{\Re}{\mathbb{R}} \renewcommand{\Mspace}[1]{\mathcal{M}({#1})} \newcommand{\E}{\mathbb{E}} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \renewcommand
{\Pr}{\mathbb{P}} \newcommand{\sd}{\mathbb{S}^{d-1}} \newcommand{\Tr}{\text{Tr}} \newcommand{\kl}{D_{\text{KL}}} \newcommand{\ind}{\mathbf{1}} \newcommand{\d}{\text{d}}\]
Let \((S_n)\) be some stochastic process in, say, \(\Re^d\). For instance, \(S_n = \sum_{i=1}^n X_i\) for multivariate observations \(X_i\in\Re^d\). We are aiming to generate a high probability bound
\[\norm{S_n} = \sup_{v:\norm{v}=1} \la v,S_n\ra.\]
There are several well-known ways to attempt this. Among them are covering, chaining, and Doob decomposition (done, e.g., here). Here we’ll explore a separate (and relatively new) approach.
The idea is to use the variational inequality which is at the core of the PAC-Bayesian methodology (so we could equivalently call this the PAC-Bayes approach to concentration). This lets us
simultaneously bound \(\la v,S_n\ra\) in each direction \(v\). Recall that a PAC-Bayes bound has the form
\[$$\label{eq:pb-basic-bound} \Pr( \forall \rho \in \Mspace{\Theta}: \text{Something holds}) \geq 1-\delta, \tag{1}$$\]
where \(\Theta\) is some parameter space and \(\Mspace{\Theta}\) is the set of all probability measures over that space. A PAC-Bayes bound provides a high probability bound simultaneously over all
posteriors. The variational approach to concentration translates this into a high probability bound simultaneously over all directions.
It’s worth taking a moment to understand why this simultaneity property is valuable. A natural thought is to treat \(\la v,S_n\ra\) as a scalar-valued process and apply well understood concentration
results for real-valued functions to it. Doing this in the naive way would give a separate bound for each \(v\in\sd\). So we’d have a result of the form:
\[\forall v\in\sd, \Pr(\la v,S_n\ra \geq B_n) \leq \delta.\]
But now we’re stuck. This is not a bound on \(\sup_{v\in\sd}\la v,S_n\ra\). One’s first thought is to take a union bound over \(v\) in order to move the “\(\forall v\in\sd\)” inside the probability
statement, which would give us the result. But there are uncountably many such vectors. The approach explored here solves this problem by translating the “\(\forall \rho\in\Mspace{\Theta}\)” in \
eqref{eq:pb-basic-bound} into “\(\forall v\in\sd\)”.
This variational approach was pioneered by Catoni and Giulini (here and here), and has now been used by a few authors to prove bounds in a variety of settings:
A general variational inequality
Different authors use seemingly different PAC-Bayesian inequalities to achieve their results. However, we recently showed that all of these inequalities are specific instantiations of the following
more general result.
Let \(\Theta\) be some measurable parameter space and let \(N(\theta)\) be nonnegative and have expected value at most 1 (it’s an e-value, if you like) for all \(\theta\in\Theta\). Then
\[\Pr(\forall\rho\in\Mspace{\Theta}: \E_\rho \log N(\theta) \leq \kl(\rho\|\nu) + \log(1/\delta))\geq 1-\delta,\]
where, as above, \(\Mspace{\Theta}\) is the set of all probability measures on \(\Theta\). The variational approach to concentration involves (i) finding some appropriate family of random variables \
(N(\theta)\), (ii) choosing \(\nu\) and a family of distributions \(\{\rho_\theta:\theta\in\Theta\}\) such that \(\sup_\theta \kl(\rho_\theta \|\nu)\) is small, and (iii) applying this “master
Example 1: Sub-Gaussian random vectors
This comes from our paper on time-uniform confidence spheres. Consider \(n\) iid copies \(X_1,\dots,X_n\) of a \(\Sigma\)-sub-Gaussian random vector \(X\in\Re^d\). That is,
\[\E\exp(\lambda \la \theta, X\ra) \leq \exp\left(\frac{\lambda^2}{2}\la\theta,\Sigma\theta\ra\right),\]
for all \(\lambda\in\Re\) and \(\theta\in\Re^d\). This implies that
\[N(\theta) = \exp\left\{\lambda\sum_{i\leq n}\la \theta, X_i\ra - \frac{n\lambda^2}{2}\la\theta,\Sigma\theta\ra\right\},\]
has expectation at most 1. Let \(\nu\) be a Gaussian with mean 0 and covariance \(\beta^{-1}I\) for some \(\beta>0\). Consider the family of distributions \(\{\rho_u:\norm{u}=1\}\) where \(\rho_u\)
is a Gaussian with mean \(u\) and covariance \(\beta^{-1}I\). Then the KL divergence between \(\rho_u\) and \(\nu\) is \(\kl(\rho_u\|\nu) = \beta/2\). Using the master theorem above, we obtain that,
with probability \(1-\delta\), simultaneously for all distributions \(\rho\),
\[\lambda\sum_{i\leq n} \E_\rho \la\theta, X_i\ra \leq \frac{n\lambda^2}{2}\E_\rho \la \theta, \Sigma\theta\ra + \frac{\beta}{2} + \log(1/\delta).\]
Now, for \(\rho=\rho_u\), \(\E_\rho \la \theta, X_i\ra = \la u,X_i\ra\) and
\[\E_\rho \la \theta, \Sigma\theta\ra = \la u,\Sigma u\ra + \beta^{-1}\Tr(\Sigma) \leq \norm{\Sigma} + \beta^{-1}\Tr(\Sigma),\]
using basic properties of the expectation of quadratic forms under Gaussian distributions (see e.g. here), and definition of the operator norm as \(\norm{A} = \sup_{u,v:\norm{u}=\norm{v}=1}\la u,\
Sigma v \ra\). Since this holds simultaneously for all \(\rho_u\), we obtain that, with probability \(1-\delta\),
\[\sup_u \lambda \sum_{i\leq n} \la u,X_i\ra \leq \frac{n\lambda^2}{2}(\norm{\Sigma} + \beta^{-1}\Tr(\Sigma)) + \frac{\beta}{2} + \log(1/\delta).\]
The left hand side is equal to \(\lambda \norm{\sum_{i\leq n}X_i}\), which gives us our concentration result. One can then optimize \(\lambda\) using some calculus. This gives us state-of-the-art
concentration up to an additive factor of \((\Tr(\Sigma^2)/n)^{1/4}\).
Example 2: Random matrices with finite Orlicz-norm
This example is adapted from Zhivotovskiy (2024). Let \(M_1,\dots,M_n\) be iid copies of a random matrix \(M\) with finite sub-exponential Orlicz norm, in the sense that, for some \(C>0\),
\[\norm{\la \theta, M\phi\ra}_{\psi_1} \leq C \la\theta, \Sigma\phi\ra,\]
for all \(\theta, \phi\in\Re^d\) where \(\Sigma = \E M\). Here
\[\norm{Y}_{\psi_1} = \inf\left\{u>0: \E\exp(|Y|/u)\leq 2\right\}.\]
We take our parameter space in the master theorem above to be \(\Theta = \Re^d\times \Re^d\). Let \(\nu\) again be Gaussian with mean 0 and covariance \(\beta^{-1}\Sigma\) and let \(\mu_u\) be a
truncated Gaussian mean \(u\), covariance \(\beta^{-1}\Sigma\) and radius \(r\). Being slightly loose with notation and writing \(\d\mu\) for the density of \(\mu\), the density of the truncated
Gaussian can be written as
\[\d\mu_{u}(x) = \frac{\ind\{\norm{x - u}\leq r\}}{Z}\d \rho_u,\]
where \(Z\) is some normalizing constant and \(\rho_u\) is a non-truncated Gaussian. For a vector \(u\in \Sigma^{1/2}\mathbb{S}^{d-1}\), the KL-divergence between a truncated normal \(\mu_u\) and \(\
nu\) is therefore
\[\begin{align*} \kl(\mu_{u} \| \nu) &= \int\log\left(\frac{1}{Z}\frac{\d \rho_u}{\d\nu}(\theta)\right) {\mu}_{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{1}{2}\int (\la \theta-u,\beta\
Sigma^{-1}(\theta-u)\ra + \la \theta, \beta\Sigma^{-1}\theta\ra )\mu_{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{\beta}{2}\int (2\la \theta,\Sigma^{-1}u\ra - \la u, \Sigma^{-1}u\ra )\mu_
{u}(\d\theta) \\ &= \log\left(\frac{1}{Z}\right) + \frac{\beta\la u,\Sigma^{-1} u\ra}{2} \leq \log\left(\frac{1}{Z}\right) + \frac{\beta}{2}. \end{align*}\]
Here \(Z = \Pr(\norm{\theta - u}\leq r)\) where \(\theta\sim \rho_{u}\). Equivalently, \(Z=\Pr(\norm{Y}\leq r)\) where \(Y\) is a normal with mean \(0\) and covariance \(\beta^{-1}\Sigma\). Hence \(1
- Z = \Pr(\norm{Y}>r)\leq \E\norm{Y}^2/r^2 = \beta^{-1}\Tr(\Sigma)/r^2\). Thus, taking \(r = \sqrt{2 \beta^{-1}\Tr(\Sigma)}\) yields \(Z\geq 1/2\), and we obtain
\[ \kl(\mu_u\|\nu) \leq \log(2) + \frac{\beta}{2}. \]
Therefore, since the KL divergence is additive on product measures, we have that
\[\kl(\rho_u\times\rho_v\|\nu\times\nu) \leq 2\log(2) + \beta.\]
Now it remains to construct a relevant quantity to use in the PAC-Bayes theorem. Consider
\[N(\theta,\phi) = \exp\left\{\lambda\sum_{i\leq n}\la\theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra - n\log\E\exp(\lambda\la \theta, \Sigma^{-1/2}M\Sigma^{-1/2}\phi\ra)\right\},\]
where the expectation is over \(M\). It’s easy to see that this has expectation at most 1 (it can be written as the product of terms each with expectation exactly one). We apply the master theorem
with the product distribution \(\mu_u\times \mu_v\) for \(u,v\in\Sigma^{1/2}\sd\) where \(\sd = \{x:\norm{x}=1\}\) is the unit sphere. Therefore, \(u = \Sigma^{1/2}u'\) for \(v = \Sigma^{1/2}v'\) for
some \(u',v'\in\sd\). We obtain that with probability \(1-\delta\), for all \(u,v\),
\[ &\lambda \sum_{i\leq n}\E_{\mu_u\times \mu_v} \la \theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra \\ &\leq n \E_{\mu_u\times\mu_v} \log \E\exp(\lambda\la\theta, \Sigma^{-1/2}M\Sigma^{1/2}\phi\ra) + \
frac{\beta}{2} + \log(2/\delta).\label{eq:matrix_bound1}\tag{2} \]
The truncated Gaussian is symmetric about its mean, so the left hand side above becomes
\[\E_{\mu_u\times\mu_v}\la \theta, \Sigma^{-1/2}M_i\Sigma^{-1/2}\phi\ra = \la \Sigma^{-1/2}u,M_i\Sigma^{-1/2}v\ra = \la u', M_i v'\ra,\]
where, as before, \(u',v'\in\mathbb{S}^{d-1}\). It remains to bound the right hand side of \eqref{eq:matrix_bound1}. For this we appeal to a result which bounds the MGF of a random variable in terms
of its \(\psi_1\)-norm. In particular, we appeal to an exponential inequality, which states that for a random variable \(Y\),
\[$$\label{eq:exp_ineq} \E[\exp(\lambda (Y - \E Y))]\leq \exp(4\lambda^2orm{Y-\E Y}_{\psi_1}^2),\quad \forall |\lambda| \leq \frac{1}{2orm{Y-\E Y}_{\psi_1}}.\tag{3}$$\]
Applying this with \(Y = \la \theta,\Sigma^{-1/2} M\Sigma^{-1/2}\phi\ra\) and noting that \(\E Y = \la \theta, \phi\ra\), we have
\[ & \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra - \la \theta, \Sigma^{-1/2}\E M \Sigma^{1/2} \phi\ra}_{\psi_1} \\ &\leq \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra}_{\psi_1} +
\norm{\la \theta, \Sigma^{-1/2}\E M \Sigma^{-1/2} \phi\ra}_{\psi_1} \\ &\leq \norm{\la \theta, \Sigma^{-1/2}M \Sigma^{-1/2} \phi\ra}_{\psi_1} + \E\norm{\la \theta, \Sigma^{-1/2} \Sigma^{-1/2} \phi\
ra}_{\psi_1} \\ &= 2 \norm{\la \Sigma^{-1/2}\theta, M \Sigma^{-1/2} \phi\ra}_{\psi_1} \\ &\leq 2C \la \Sigma^{-1/2}\theta, \Sigma\Sigma^{-1/2}\phi\ra = 2C\la \theta, \phi\ra \leq C(\norm{\theta}^2 +
\norm{\phi}^2). \]
Therefore, \eqref{eq:exp_ineq} yields
\[ \E\exp(\lambda \la\theta,\Sigma^{-1/2}M\Sigma^{-1/2}\phi\ra) &\leq \exp(\lambda \la \theta, \phi\ra + 4C^2\lambda^2(\norm{\theta}^2 + \norm{\phi}^2)^2). \label{eq:Eexp} \tag{4} \]
Note that \(\norm{\theta - u}\leq r\) and \(\norm{u} = \norm{\Sigma^{1/2}u'} \leq \norm{\Sigma^{1/2}} \leq \sqrt{\norm{\Sigma}}\), so
\[\norm{\theta}^2 \leq (r + \norm{u})^2 \leq \left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\right)^2.\]
The same bound holds for \(\norm{\phi}^2\). Therefore, \eqref{eq:Eexp} gives
\[ &\E_{\rho_u\times\rho_v} \log \E \exp(\lambda\la\theta, \Sigma^{-1/2} M\Sigma^{-1/2}\phi\ra ) \\ &\leq \lambda \la u,v\ra + 8C^2\lambda^2\left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\
right)^4 \\ &= \lambda \la u',\Sigma v'\ra + 8C^2\lambda^2\left(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}}\right)^4, \]
assuming that
\[|\lambda| \leq \frac{1}{4C(\sqrt{2\beta^{-1}\Tr(\Sigma)} + \sqrt{\norm{\Sigma}})^2}.\]
Choosing \(\beta = 2\Tr(\Sigma)/\norm{\Sigma}\) and putting everything together, \eqref{eq:matrix_bound1} gives that with probability \(1-\delta,\)
\[\sup_{u',v'\in\sd}\lambda\sum_{i\leq n} \la u', (M_i-\Sigma) v'\ra \lesssim C^2\lambda^2 \norm{\Sigma}^2 + \frac{\Tr(\Sigma)}{\norm{\Sigma}} + \log(1/\delta).\]
Dividing by \(n\) and optimizing over \(\lambda\) gives us a final bound that matches state-of-the-art concentration bounds for random matrices.
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What are some nonfiction books about specific numbers? - The Handy Math Answer Book
Mathematical Resources
Popular Resources
What are some nonfiction books about specific numbers?
Although most people wouldn’t think that a book about a number would be interesting, the following shows that’s not always true:
The Golden Ratio: The Story of Phi, the World’s Most Astonishing Number by Mario Livio (Broadway Books, 2003; ISBN: 0767908163)—A history of the number phi (1.6180339887), also known as the golden
ratio or divine proportion. There are examples from nature, as well as phi’s use in architecture and art throughout human history.
Pi: A Biography of the World’s Most Mysterious Number by Alfred S. Posamentier and Ingmar Lehmann (Prometheus Books, 2004; ISBN: 1591022002)—The story of the number pi throughout history, from the
Old Testament to modern politics. An epilogue has pi expressed to 100,000 decimal places.
e: The Story of a Number by Eli Major (Princeton University Press, 2009; ISBN: 9780691141343). In this new edition of his book, Major traces “e” from the 16th century to the present, winding his
story around the properties of this well-known number.
Zero: The Biography of a Dangerous Idea by Charles Seife and Matt Zimet (Penguin Books, 2000; ISBN: 0140296476)—An entertaining story about (literally) nothing. The development and use of nothing, or
zero, is covered in detail from ancient times to the present.
An Imaginary Tale: The Story of [the Square Root of Minus One] by Paul Nahin (Princeton University Press, 2010; ISBN: 9780691146003). This instructive book will take the reader through not only the
history of complex numbers, but why such imaginary numbers are important to mathematics.
And a book about the importance of all numbers:
Cosmic Numbers: The Numbers that Define Our Universe by James D. Stein (Basic Books, 2011; ISBN: 0465021980). This book traces the power of numbers, noting the “discovery, evolution, and
interrelationship of figures that define our world.” In other words, it gives the reader a good reason why numbers are so important to everyone. | {"url":"https://www.papertrell.com/apps/preview/The-Handy-Math-Answer-Book/Handy%20Answer%20book/What-are-some-nonfiction-books-about-specific-numbers/001137022/content/SC/52cb02a782fad14abfa5c2e0_default.html","timestamp":"2024-11-14T15:35:20Z","content_type":"text/html","content_length":"12582","record_id":"<urn:uuid:73df8a31-306e-428e-82e4-5d8dd9e77b9c>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00491.warc.gz"} |
Derivative of 2 to the x - Formula, Proof, Examples | Derivative of 2^x (2024)
The derivative of 2 to the x is equal to 2^x ln 2. We can calculate this derivative using various methods of differentiation such as the first principle of derivatives, and the formula for the
derivative of the exponential function, and using the natural logarithmic function followed by implicit differentiation. Mathematically, we can write the formula for the derivative of 2 to the x as d
(2^x)/dx = 2^x ln 2. The formula for the derivative of the function f(x) = a^x is given by a^x ln a. Using this formula, the derivative of 2 to the x is given by, (2^x)' = 2^x ln 2.
Further, in this article, we will explore the derivative of 2 to the x and its formula using different methods of evaluating derivatives. We will also solve various examples related to the derivative
of 2 to the x and other functions for a better understanding of the concept.
1. What is the Derivative of 2 to the x?
2. Derivative of 2 to the x Using First Principle
3. Derivative of 2 to the x Using Logarithmic Differentiation
4. Derivative of 2 to the x Using Chain Rule
5. FAQs on Derivative of 2 to the x
What is the Derivative of 2 to the x?
The derivative of 2 to the x is 2^x ln 2. We can write this as d/dx (2^x) = 2^x ln 2 (or) (2^x)' = 2^x ln 2. Since "ln" is nothing but natural logarithm (log with base 'e'), we can write this formula
as d/dx (2^x) = 2^x logₑ 2. i.e.,
2 to the x is mathematically written as 2^x and it is an exponential function (but NOT a power function). Because its base (2) is a constant and its exponent (x) is a variable. So we use the formula
d/dx(a^x) = a^x ln a to find the derivative of 2 to the x but we are not supposed to use the power rule d/dx (x^n) = n x^n-1 here as 2^x is NOT a power function.
To prove the derivative of 2 to the x, the straightforward method is using the derivative of exponential function a^x formula which says,
d/dx(a^x) = a^x ln a
Substitute a = 2 on both sides, we get
d/dx(2^x) = 2^x ln 2
Hence the formula is proved.
Derivative of 2 to the x Formula
As observed above, the formula for the derivative of 2 to the x is given by d(2^x)/dx = 2^x ln 2 (or) (2^x)' = 2^x ln 2. There are various other ways to prove the formula of the derivative of 2 to
the x. Here are a few of them.
• Using the first principle
• Using logarithmic differentiation
• Using chain rule
Let us prove the formula in each of these cases.
Derivative of 2 to the x Using First Principle
The limit definition of the derivative, which is also known as the first principle, says that the derivative of a function y = f(x) is found by using the limit:
f'(x) = lim[h→0][f(x + h) - f(x)] / h --- (1)
Since f(x) = 2^x, we have f(x + h) = 2^x + h.
Substituting these values in (1):
f '(x) = lim[h→0] [2^x + h - 2^x] / h
Using one of the properties of exponents, a^m + n = a^m · a^n. Using this, we have
f'(x) = lim[h→0] [2^x · 2^h - 2^x] / h
= lim[h→0] 2^x [ 2^h - 1] / h
= lim[h→0] 2^x · limₕ→ ₀ [ 2^h - 1] / h
= 2^x · lim[h→0] [ 2^h - 1] / h
Using one of the limit formulas, lim[h→0] [a^h - 1] / h = ln a.
f'(x) = 2^x ln 2
Hence the derivative of 2 to the x formula is proved.
Derivative of 2 to the x Using Logarithmic Differentiation
We use logarithmic differentiation to find the derivative of a function that has a variable in the exponent. In this process, we apply "log" (or) "ln" on both sides and then differentiate on both
sides. Let us assume the function to be differentiated to be y = 2^x. Taking "ln" on both sides,
ln y = ln 2^x
Using the properties of logarithms, ln a^m = m ln a. Using this,
ln y = x ln 2
Differentiating both sides with respect to x,
d/dx (ln y) = d/dx (x ln 2)
Using the constant multiplication rule of derivatives,
d/dx (ln y) = ln 2 d/dx (x)
Using the derivative of ln x rule, d/dx (ln x) = 1/x and also the chain rule on the left side,
(1/y) dy/dx = ln 2 (1)
Multiplying both sides by y,
dy/dx = y ln 2
Substituting y = 2^x here,
d/dx (2^x) = 2^x ln 2
Hence, we proved the derivative of 2 to the x to be 2^x ln 2. You can try deriving the same formula by applying "log" on both sides.
Derivative of 2 to the x Using Chain Rule
Using one of the properties of natural logarithms, e^ln a = a, for any 'a'. By this, we have
e^ln 2 = 2 (or) 2 = e^ln 2
Raising the exponent both sides by x,
2^x = (e^ln 2) ^x
We have (a^m) ^n = a^mn. By using this in the above step,
2^x = e^x ln 2
Differentiating both sides with respect to x,
d/dx (2^x) = d/dx (e^x ln 2)
We know that the derivative of ex is e^x and also applying the chain rule on the right side,
d/dx (2^x) = e^x ln 2 · d/dx (x ln 2)
= e^x ln 2 · (ln 2)
= e^ln 2^x · (ln 2)
Using the same property e^ln a = a again,
d/dx (2^x) = 2^x ln 2
Hence, the derivative of 2 to the x formula is derived.
Important Points on Derivative of 2 to the x:
• The derivative of 2 to the x power is, d/dx (2^x) = 2^x ln 2 (or) 2^x logₑ 2.
• Note that 2^x is an exponential function but NOT a power function.
• Use the derivative of a^x formula but NOT the derivative of x^n formula to find the derivative of 2 to the x.
☛Related Topics:
• Derivative Rules
• Inverse Trig Derivatives
• Implicit Differentiation
FAQs on Derivative of 2 to the x
What is the Derivative of 2 to the Power of x?
The derivative of 2 to the power of x has two formulas:
• d/dx (2^x) = 2^x ln 2
• d/dx (2^x) = 2^x logₑ 2
How to Find the Derivative of 2 to the x?
To find the derivative of 2 to the x, just apply the formula d/dx (a^x) = a^x ln a and substitute a = 2 in this formula. Then we get d/dx (2^x) = 2^x ln 2. We can also find the derivative of 2 to the
x using the first principle of derivatives, chain rule and implicit differentiation.
Is the Derivative of 2^x Equal to 2^x Itself?
No, the derivative of 2^x is NOT itself, the derivative of 2^x is 2^x ln 2. This comes from the formula d/dx (a^x) = a^x ln a.
What is the n^th Derivative of 2 to the x?
We know that d/dx (2^x) = 2^x ln 2. Let us differentiate it multiple times to identify the pattern.
• The 1^st derivative of 2^x is 2^x ln 2.
• The 2^nd derivative of 2^x is 2^x (ln 2)^2.
• The 3^rd derivative of 2^x is 2^x (ln 2)^3.
• ...
• The n^th derivative of 2^x is 2^x (ln 2)^n.
What is the Derivative of 2 to the x in Terms of Ln?
The derivative of an exponential function is, (a^x) ' = a^x ln a. By substituting a = 2 i this, (2^x)' = 2^x ln 2.
What is the Derivative of 2 to the x in Terms of Log?
The derivative of 2^x is usually expressed in terms of "ln" to be d/dx (2^x) = 2^x ln 2. But we know that ln = logₑ and hence the same formula can be written alternatively to be d/dx (2^x) = 2^x logₑ
What is the Second Derivative of 2 to the x?
The second derivative of 2 to the power x is given by 2^x (ln 2)^2 and its formula can be written as d^2/dx^2 (2^x) = 2^x (ln 2)^2. | {"url":"https://modatakip.net/article/derivative-of-2-to-the-x-formula-proof-examples-derivative-of-2-x","timestamp":"2024-11-11T22:01:16Z","content_type":"text/html","content_length":"73039","record_id":"<urn:uuid:54272ad6-1481-4add-bfca-e5cf4bc6898e>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00402.warc.gz"} |
How do I get rid of Na in Vlookup?
How do I get rid of Na in Vlookup?
Use IFERROR with VLOOKUP to Get Rid of #N/A Errors
1. =IFERROR(value, value_if_error)
2. Use IFERROR when you want to treat all kinds of errors.
3. Use IFNA when you want to treat only #N/A errors, which are more likely to be caused by VLOOKUP formula not being able to find the lookup value.
How do you write a copyright for a book?
Create Your Copyright Page
1. The copyright notice.
2. The year of publication of the book.
3. The name of the owner of the works, which is usually the author or publishing house name.
4. Ordering information.
5. Reservation of rights.
6. Copyright notice.
7. Book editions.
8. ISBN Number.
What is the copyright page called in a book?
edition notice
Why do I get NA in Vlookup?
The most common cause of the #N/A error is with VLOOKUP, HLOOKUP, LOOKUP, or MATCH functions if a formula can’t find a referenced value. For example, your lookup value doesn’t exist in the source
data. In this case there is no “Banana” listed in the lookup table, so VLOOKUP returns a #N/A error.
Is Vlookup an error?
When VLOOKUP can’t find a value in a lookup table, it returns the #N/A error. The IFERROR function allows you to catch errors and return your own custom value when there is an error. If VLOOKUP
returns the #N/A error, IFERROR takes over and returns the value you supply.
How do you write a legal disclaimer?
In your disclaimer, cover any and all liabilities for the product or service that you provide. You should warn consumers of any dangers or hazards posed by your product. You should list specific
risks while at the same time acknowledging that the list is not exhaustive. For example, you could write, “NOTICE OF RISK.
Is index better than Vlookup?
When deciding between which vertical lookup formula to use, the majority of Excel experts agree that INDEX MATCH is a better formula than VLOOKUP. However, many people still resort to using VLOOKUP
because it’s a simpler formula.
What is the difference between Vlookup and index match?
VLOOKUP uses the static data reference while looking up the values. INDEX MATCH uses the dynamic data ranges while looking up the values. Inserting or Deleting a column affects the VLOOKUP result.
Inserting or deleting a column does not affect the INDEX MATCH result.
How do I do a Vlookup with two criteria?
VLOOKUP with Multiple Criteria – Using a Helper Column
1. Insert a Helper Column between column B and C.
2. Use the following formula in the helper column:=A2&”|”&B2.
3. Use the following formula in G3 =VLOOKUP($F3&”|”&G$2,$C$2:$D$19,2,0)
4. Copy for all the cells.
What is a disclaimer in a book?
A disclaimer is a statement meant to protect you, as an author, from legal action against you for something contained in your book/ebook. Essentially, a disclaimer says that you may not be held
liable, or responsible, for anything based on what you’ve written.
What is a book index example?
Examples are an index in the back matter of a book and an index that serves as a library catalog. In a traditional back-of-the-book index, the headings will include names of people, places, events,
and concepts selected by the indexer as being relevant and of interest to a possible reader of the book.
How do you use index?
#1 How to Use the INDEX Formula
1. Type “=INDEX(” and select the area of the table then add a comma.
2. Type the row number for Kevin, which is “4” and add a comma.
3. Type the column number for Height, which is “2” and close the bracket.
4. The result is “5.8”
What should be included in a book index?
The Rules of Index Entries
• Use nouns the reader is likely to look for. Whenever possible, index entries should begin with nouns or noun phrases.
• Use lowercase letters.
• Use subentries to make things easier to find.
• Set image references in bold or italics.
• Use cross-references as needed.
• You don’t need to include everything.
What is an example of an index?
The definition of an index is a guide, list or sign, or a number used to measure change. An example of an index is a list of employee names, addresses and phone numbers. An example of an index is a
stock market index which is based on a standard set at a particular time.
What is the alternative for Vlookup?
Index Match
How do I solve Na error in Vlookup?
Problem: The lookup column is not sorted in the ascending order
1. Change the VLOOKUP function to look for an exact match. To do that, set the range_lookup argument to FALSE. No sorting is necessary for FALSE.
2. Use the INDEX/MATCH function to look up a value in an unsorted table.
How do I use index and match instead of Vlookup?
VLOOKUP requires a static column reference whereas INDEX MATCH requires a dynamic column reference. With VLOOKUP you need to manually enter a number referencing the column you want to return the
value from. As you are using a static reference, adding a new column to the table breaks the VLOOKUP formula.
What is difference between index and contents?
Table of Contents implies an organized list containing the chapter-wise headings and sub-headings along with page numbers. Index refers to a page which acts as a pointer to find out the keywords and
key terms, which the book contains.
What is index of a book?
A back-of-the-book index is a list of words with corresponding page references that point readers to the locations of various topics within a book. Indexes are generally an alphabetical list of
topics with subheadings appearing below multi-faceted topics that appear numerous times throughout a book.
How do you read a book index?
Indexing helpful hints
1. Read the proofs or manuscript.
2. Make a list of terms to appear.
3. Separate these terms into main entries and subentries.
4. Add the page numbers for every meaningful reference to a selected term.
5. Alphabetize all main entries and main words of subentries.
Can you put a Vlookup in an if statement?
Did you know that you can use Excel IF statements along with VLOOKUPs? For example, if you wanted to enter a value from a table into a cell, based on what was in another cell, you could start with an
IF statement and then enter the VLOOKUP in the “value if true” part of the IF statement.
What is called index?
An index is an indicator or measure of something. In finance, it typically refers to a statistical measure of change in a securities market. In the case of financial markets, stock and bond market
indexes consist of a hypothetical portfolio of securities representing a particular market or a segment of it.
Where do you put the disclaimer in a book?
You can put it at the start of the book, at the end, or on a Post-It note at your desk. It is common to have it at the back of the title page, and that’s where people will go look for it, but it
doesn’t have to.
What are the two main causes of errors for Vlookup?
VLOOKUP Troubleshooting Part 2 Another common cause for VLOOKUP errors is extra characters in one of the cells – usually extra space characters. Using the LEN function, I checked the length of the
string in each cell. The TRIM function will remove leading, trailing, and duplicate spaces.
What is another word for index?
What is another word for index?
indication guide
indicator mark
sign clue
evidence signal
token hint
Why is index and match better than Vlookup?
With unsorted data, VLOOKUP and INDEX-MATCH have about the same calculation times. With sorted data and an approximate match, INDEX-MATCH is about 30% faster than VLOOKUP. With sorted data and a fast
technique to find an exact match, INDEX-MATCH is about 13% faster than VLOOKUP.
Why is Vlookup bad?
It can not lookup and return a value which is to the left of the lookup value. It works only with data which is arranged vertically. VLOOKUP would give a wrong result if you add/delete a new column
in your data (as the column number value now refers to the wrong column). | {"url":"https://www.meatandsupplyco.com/how-do-i-get-rid-of-na-in-vlookup/","timestamp":"2024-11-06T16:40:39Z","content_type":"text/html","content_length":"57907","record_id":"<urn:uuid:be340991-e881-4120-9b4a-6b91c2805fa3>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00215.warc.gz"} |
4.6: Zeros of Polynomial Functions
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Learning Objectives
• Evaluate a polynomial using the Remainder Theorem.
• Use the Factor Theorem to solve a polynomial equation.
• Use the Rational Zero Theorem to find rational zeros.
• Find zeros of a polynomial function.
• Use the Linear Factorization Theorem to find polynomials with given zeros.
• Use Descartes’ Rule of Signs.
• Solve real-world applications of polynomial equations
A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of
a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the
cake pan be?
This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions
and solving polynomial equations.
Evaluating a Polynomial Using the Remainder Theorem
In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by \(x–k\), the
remainder may be found quickly by evaluating the polynomial function at \(k\), that is, \(f(k)\). Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend \(f(x)\) and a non-zero polynomial divisor \(d(x)\) where the degree of \(d(x)\) is less than or equal to the degree of \(f
(x)\),there exist unique polynomials \(q(x)\) and \(r(x)\) such that
\[f(x)=d(x)q(x)+r(x) \nonumber\]
If the divisor, \(d(x)\), is \(x−k\), this takes the form
\[f(x)=(x−k)q(x)+r \nonumber\]
Since the divisor \(x−k\)
is linear, the remainder will be a constant, \(r\). And, if we evaluate this for \(x=k\), we have
\[\begin{align*} f(k)&=(k−k)q(k)+r \\[4pt] &=0{\cdot}q(k)+r \\[4pt] &=r \end{align*}\]
In other words, \(f(k)\) is the remainder obtained by dividing \(f(x)\)by \(x−k\).
The Remainder Theorem
If a polynomial \(f(x)\) is divided by \(x−k\),then the remainder is the value \(f(k)\).
Given a polynomial function \(f\), evaluate \(f(x)\) at \(x=k\) using the Remainder Theorem.
1. Use synthetic division to divide the polynomial by \(x−k\).
2. The remainder is the value \(f(k)\).
Example \(\PageIndex{1}\): Using the Remainder Theorem to Evaluate a Polynomial
Use the Remainder Theorem to evaluate \(f(x)=6x^4−x^3−15x^2+2x−7\) at \(x=2\).
To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by \(x−2\).
\[ 2 \begin{array}{|ccccc} \; 6 & −1 & −15 & 2 & −7 \\ \text{} & 12 & 22 & 14 & 32 \\ \hline \end{array} \\ \begin{array}{ccccc} 6 & 11 & \; 7 & \;\;16 & \;\; 25 \end{array} \]
The remainder is 25. Therefore, \(f(2)=25\).
We can check our answer by evaluating \(f(2)\).
\[\begin{align*} f(x)&=6x^4−x^3−15x^2+2x−7 \\ f(2)&=6(2)^4−(2)^3−15(2)^2+2(2)−7 \\ &=25 \end{align*}\]
Exercise \(\PageIndex{1}\)
se the Remainder Theorem to evaluate \(f(x)=2x^5−3x^4−9x^3+8x^2+2\) at \(x=−3\).
Using the Factor Theorem to Solve a Polynomial Equation
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm.
If \(k\) is a zero, then the remainder \(r\) is \(f(k)=0\) and \(f (x)=(x−k)q(x)+0\) or \(f(x)=(x−k)q(x)\).
Notice, written in this form, \(x−k\) is a factor of \(f(x)\). We can conclude if \(k\) is a zero of \(f(x)\), then \(x−k\) is a factor of \(f(x)\).
Similarly, if \(x−k\) is a factor of \(f(x)\), then the remainder of the Division Algorithm \(f(x)=(x−k)q(x)+r\) is \(0\). This tells us that \(k\) is a zero.
This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree \(n\) in the complex number system will have \(n\) zeros. We can use the Factor Theorem to completely
factor a polynomial into the product of \(n\) factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.
According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\).
How to: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial
1. Use synthetic division to divide the polynomial by \((x−k)\).
2. Confirm that the remainder is \(0\).
3. Write the polynomial as the product of \((x−k)\) and the quadratic quotient.
4. If possible, factor the quadratic.
5. Write the polynomial as the product of factors.
Example \(\PageIndex{2}\): Using the Factor Theorem to Solve a Polynomial Equation
Show that \((x+2)\) is a factor of \(x^3−6x^2−x+30\). Find the remaining factors. Use the factors to determine the zeros of the polynomial.
We can use synthetic division to show that \((x+2)\) is a factor of the polynomial.
\[ -2 \begin{array}{|cccc} \; 1 & −6 & −1 & 30 \\ \text{} & -2 & 16 & -30 \\ \hline \end{array} \\ \begin{array}{cccc} 1 & -8 & \; 15 & \;\;0 \end{array} \]
The remainder is zero, so \((x+2)\) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:
We can factor the quadratic factor to write the polynomial as
By the Factor Theorem, the zeros of \(x^3−6x^2−x+30\) are –2, 3, and 5.
Exercise \(\PageIndex{2}\)
Use the Factor Theorem to find the zeros of \(f(x)=x^3+4x^2−4x−16\) given that \((x−2)\) is a factor of the polynomial.
The zeros are 2, –2, and –4.
Using the Rational Zero Theorem to Find Rational Zeros
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us
to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial
Consider a quadratic function with two zeros, \(x=\frac{2}{5}\) and \(x=\frac{3}{4}\). By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and
then construct the original quadratic function absent its stretching factor.
Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two
denominators from the original rational roots: 5 and 4.
We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational
Zero Theorem; it is a means to give us a pool of possible rational zeros.
The Rational Zero Theorem states that, if the polynomial \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) has integer coefficients, then every rational zero of \(f(x)\) has the form \(\frac{p}{q}\) where
\(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
How to: Given a polynomial function \(f(x)\), use the Rational Zero Theorem to find rational zeros.
1. Determine all factors of the constant term and all factors of the leading coefficient.
2. Determine all possible values of \(\dfrac{p}{q}\), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. Be sure to include both positive and negative
3. Determine which possible zeros are actual zeros by evaluating each case of \(f(\frac{p}{q})\).
Example \(\PageIndex{3}\): Listing All Possible Rational Zeros
List all possible rational zeros of \(f(x)=2x^4−5x^3+x^2−4\).
The only possible rational zeros of \(f(x)\) are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.
The constant term is –4; the factors of –4 are \(p=±1,±2,±4\).
The leading coefficient is 2; the factors of 2 are \(q=±1,±2\).
If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.
\[\dfrac{p}{q}=±\dfrac{1}{1},±\dfrac{1}{2} \; \; \; \; \; \; \frac{p}{q}=±\dfrac{2}{1},±\dfrac{2}{2} \; \; \; \; \; \; \dfrac{p}{q}=±\dfrac{4}{1},±\dfrac{4}{2} \nonumber\]
Note that \(\frac{2}{2}=1\) and \(\frac{4}{2}=2\), which have already been listed. So we can shorten our list.
\[\dfrac{p}{q} = \dfrac{\text{Factors of the last}}{\text{Factors of the first}}=±1,±2,±4,±\dfrac{1}{2}\nonumber \]
Example \(\PageIndex{4}\): Using the Rational Zero Theorem to Find Rational Zeros
Use the Rational Zero Theorem to find the rational zeros of \(f(x)=2x^3+x^2−4x+1\).
The Rational Zero Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of 1 and \(q\) is a factor of 2.
\[ \begin{align*} \dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] &=\dfrac{factor\space of\space 1}{factor\space of\space 2} \
The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for \(\frac{p}{q}\) are ±1 and \(±\frac{1}{2}\). These are the possible rational zeros for the function. We can
determine which of the possible zeros are actual zeros by substituting these values for \(x\) in \(f(x)\).
Of those, \(−1\),\(−\dfrac{1}{2}\), and \(\dfrac{1}{2}\) are not zeros of \(f(x)\). 1 is the only rational zero of \(f(x)\).
Exercise \(\PageIndex{3}\)
Use the Rational Zero Theorem to find the rational zeros of \(f(x)=x^3−5x^2+2x+1\).
There are no rational zeros.
Finding the Zeros of Polynomial Functions
The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of
the zeros of a polynomial function.
How to: Given a polynomial function \(f\), use synthetic division to find its zeros.
1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero,
discard the candidate.
3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.
Example \(\PageIndex{5}\): Finding the Zeros of a Polynomial Function with Repeated Real Zeros
Find the zeros of \(f(x)=4x^3−3x−1\).
The Rational Zero Theorem tells us that if \(\dfrac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of –1 and \(q\) is a factor of 4.
\[\begin{align*}\dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] =\dfrac{factor\space of\space -1}{factor\space of\space 4} \end
The factors of –1 are ±1 and the factors of 4 are ±1,±2, and ±4. The possible values for \(\dfrac{p}{q}\) are \(±1\),\(±\dfrac{1}{2}\), and \(±\dfrac{1}{4}\). These are the possible rational zeros
for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.
Dividing by \((x−1)\) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as
\[(x−1)(4x^2+4x+1) \nonumber\]
The quadratic is a perfect square. \(f(x)\) can be written as
We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.
\[ \begin{align*} 2x+1=0 \\[4pt] x &=−\dfrac{1}{2} \end{align*}\]
The zeros of the function are 1 and \(−\frac{1}{2}\) with multiplicity 2.
Look at the graph of the function \(f\) in Figure \(\PageIndex{1}\). Notice, at \(x =−0.5\), the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero −0.5. At \(x=1\),
the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero \(x=1\).
Figure \(\PageIndex{1}\).
Using the Fundamental Theorem of Algebra
Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells
us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.
Suppose \(f\) is a polynomial function of degree four, and \(f (x)=0\). The Fundamental Theorem of Algebra states that there is at least one complex solution, call it \(c_1\). By the Factor Theorem,
we can write \(f(x)\) as a product of \(x−c_1\) and a polynomial quotient. Since \(x−c_1\) is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra
to the third-degree polynomial quotient. It will have at least one complex zero, call it \(c_2\). So we can write the polynomial quotient as a product of \(x−c_2\) and a new polynomial quotient of
degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of \(f(x)\).
The Fundamental Theorem of Algebra states that, if \(f(x)\) is a polynomial of degree \(n > 0\), then \(f(x)\) has at least one complex zero.
We can use this theorem to argue that, if \(f(x)\) is a polynomial of degree \(n >0\), and a is a non-zero real number, then \(f(x)\) has exactly \(n\) linear factors
where \(c_1,c_2\),...,\(c_n\) are complex numbers. Therefore, \(f(x)\) has \(n\) roots if we allow for multiplicities.
Q&A: Does every polynomial have at least one imaginary zero?
No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers.
Example \(\PageIndex{6}\): Finding the Zeros of a Polynomial Function with Complex Zeros
Find the zeros of \(f(x)=3x^3+9x^2+x+3\).
The Rational Zero Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of 3 and \(q\) is a factor of 3.
\[ \begin{align*} \dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] &=\dfrac{factor\space of\space 3}{factor\space of\space 3} \
The factors of 3 are ±1 and ±3. The possible values for \(\dfrac{p}{q}\), and therefore the possible rational zeros for the function, are ±3,±1, and \(±\dfrac{1}{3}\). We will use synthetic division
to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3.
Dividing by \((x+3)\) gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as
\[(x+3)(3x^2+1) \nonumber\]
We can then set the quadratic equal to 0 and solve to find the other zeros of the function.
\[ \begin{align*} 3x^2+1=0 \\[4pt] x^2 &=−\dfrac{1}{3} \\[4pt] x&=±−\sqrt{\dfrac{1}{3}} \\[4pt] &=±\dfrac{i\sqrt{3}}{3} \end{align*}\]
The zeros of \(f(x)\) are \(–3\) and \(±\dfrac{i\sqrt{3}}{3}\).
Look at the graph of the function \(f\) in Figure \(\PageIndex{2}\). Notice that, at \(x =−3\), the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero \(x=–3\). Also note the
presence of the two turning points. This means that, since there is a \(3^{rd}\) degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without
bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of \(x=−3\) is 1 and there are two complex
solutions, which is what we found, or the multiplicity at \(x =−3\) is three. Either way, our result is correct.
Figure \(\PageIndex{2}\).
Find the zeros of \(f(x)=2x^3+5x^2−11x+4\).
The zeros are \(–4\), \(\frac{1}{2}\), and \(1\).
Using the Linear Factorization Theorem to Find Polynomials with Given Zeros
A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have \(n\) zeros in the set of complex numbers, if we allow for
multiplicities. This means that we can factor the polynomial function into \(n\) factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as
its degree, and that each factor will be in the form \((x−c)\), where c is a complex number.
Let \(f\) be a polynomial function with real coefficients, and suppose \(a +bi\), \(b≠0\), is a zero of \(f(x)\). Then, by the Factor Theorem, \(x−(a+bi)\) is a factor of \(f(x)\). For \(f\) to have
real coefficients, \(x−(a−bi)\) must also be a factor of \(f(x)\). This is true because any factor other than \(x−(a−bi)\), when multiplied by \(x−(a+bi)\), will leave imaginary components in the
product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function \(f\) with real coefficients has a
complex zero \(a +bi\), then the complex conjugate \(a−bi\) must also be a zero of \(f(x)\). This is called the Complex Conjugate Theorem.
According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form \((x−c)\), where \(c\) is a complex
If the polynomial function \(f\) has real coefficients and a complex zero in the form \(a+bi\), then the complex conjugate of the zero, \(a−bi\), is also a zero.
How to
Given the zeros of a polynomial function \(f\) and a point \((c, f(c))\) on the graph of \(f\), use the Linear Factorization Theorem to find the polynomial function.
1. Use the zeros to construct the linear factors of the polynomial.
2. Multiply the linear factors to expand the polynomial.
3. Substitute \((c,f(c))\) into the function to determine the leading coefficient.
4. Simplify.
Example \(\PageIndex{7}\): Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros
Find a fourth degree polynomial with real coefficients that has zeros of \(–3\), \(2\), \(i\), such that \(f(−2)=100\).
Because \(x =i\) is a zero, by the Complex Conjugate Theorem \(x =–i\) is also a zero. The polynomial must have factors of \((x+3),(x−2),(x−i)\), and \((x+i)\). Since we are looking for a degree 4
polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.
\[ f(x) & =a(x+3)(x−2)(x−i)(x+i) \\ f(x) & =a(x^2+x−6)(x^2+1) \\ f(x) & =a(x^4+x^3−5x^2+x−6) \]
We need to find \(a\) to ensure \(f(–2)=100\). Substitute \(x=–2\) and \(f (-2)=100\) into \(f (x)\).
\[ 100=a({(−2)}^4+{(−2)}^3−5{(−2)}^2+(−2)−6) \\ 100=a(−20) \\ −5=a \]
So the polynomial function is
\[f(x)=−5x^4−5x^3+25x^2−5x+30\] Analysis
We found that both \(i\) and \(−i\) were zeros, but only one of these zeros needed to be given. If \(i\) is a zero of a polynomial with real coefficients, then \(−i\) must also be a zero of the
polynomial because \(−i\) is the complex conjugate of \(i\).
If \(2+3i\) were given as a zero of a polynomial with real coefficients, would \(2−3i\) also need to be a zero?
Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.
Find a third degree polynomial with real coefficients that has zeros of \(5\) and \(−2i\) such that \(f (1)=10\).
Using Descartes’ Rule of Signs
There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of
Signs tells us of a relationship between the number of sign changes in \(f(x)\) and the number of positive real zeros. For example, the polynomial function below has one sign change.
This tells us that the function must have 1 positive real zero.
There is a similar relationship between the number of sign changes in \(f(−x)\) and the number of negative real zeros.
In this case, \(f(−x)\) has 3 sign changes. This tells us that \(f(x)\) could have 3 or 1 negative real zeros.
According to Descartes’ Rule of Signs, if we let \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) be a polynomial function with real coefficients:
• The number of positive real zeros is either equal to the number of sign changes of \(f(x)\) or is less than the number of sign changes by an even integer.
• The number of negative real zeros is either equal to the number of sign changes of \(f(−x)\) or is less than the number of sign changes by an even integer.
Example \(\PageIndex{8}\): Using Descartes’ Rule of Signs
Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for \(f(x)=−x^4−3x^3+6x^2−4x−12\).
Begin by determining the number of sign changes.
There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine \(f(−x)\) to determine the number of negative real roots.
\[ f(−x) & =−{(−x)}^4−3{(−x)}^3+6{(−x)}^2−4(−x)−12 \\ f(−x) & =−x^4+3x^3+6x^2+4x−12 \]
Again, there are two sign changes, so there are either 2 or 0 negative real roots.
There are four possibilities, as we can see in Table \(\PageIndex{1}\).
Table \(\PageIndex{1}\)
Positive Real Zeros Negative Real Zeros Complex Zeros Total Zeros
We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure \(\PageIndex{3}\). We can see from the graph that the function has 0 positive real
roots and 2 negative real roots.
Figure \(\PageIndex{3}\).
Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for \(f(x)=2x^4−10x^3+11x^2−15x+12\). Use a graph to verify the numbers of positive and
negative real zeros for the function.
There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.
Solving Real-World Applications
We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.
Example \(\PageIndex{9}\)
Solving Polynomial Equations
A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of
a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the
cake pan be?
Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by \(V=lwh\). We were given that the length must be four inches longer than the width, so we can
express the length of the cake as \(l=w+4\). We were given that the height of the cake is one-third of the width, so we can express the height of the cake as \(h=\dfrac{1}{3}w\). Let’s write the
volume of the cake in terms of width of the cake.
\[V=(w+4)(w)(\dfrac{1}{3}w)\] \[V=\dfrac{1}{3}w^3+\dfrac{4}{3}w^2\]
Substitute the given volume into this equation.
Substitute 351 for V.
\(1053=w^3+4w^2\) Multiply both sides by 3. \(0=w^3+7w^2−1053\)
Subtract 1053 from both sides.
Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are \(\pm 1,±3,±9,±13,±27,±39,±81,±117,±351,\) and \(±1053\). We
can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make
the most sense as dimensions for a small sheet cake. Use synthetic division to check \(x=1\).
Since 1 is not a solution, we will check \(x=3\).
Since 3 is not a solution either, we will test \(x=9\).
Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet
cake pan.
\(l=w+4=9+4=13\) and \(h=\dfrac{1}{3}w=\dfrac{1}{3}(9)=3\)
The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches
A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be
one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?
3 meters by 4 meters by 7 meters
Access these online resources for additional instruction and practice with zeros of polynomial functions.
Key Concepts
• To find \(f(k)\), determine the remainder of the polynomial \(f(x)\) when it is divided by \(x−k\). This is known as the Remainder Theorem. See Example \(\PageIndex{1}\).
• According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\). See Example\(\PageIndex{2}\).
• According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading
coefficient. See Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\).
• When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
• Synthetic division can be used to find the zeros of a polynomial function. See Example \(\PageIndex{5}\).
• According to the Fundamental Theorem, every polynomial function with degree greater than 0 has at least one complex zero. See Example \(\PageIndex{6}\).
• Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form \((x−c)\), where \(c\) is a complex number. See Example \(\
• The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.
• The number of negative real zeros of a polynomial function is either the number of sign changes of \(f(−x)\) or less than the number of sign changes by an even integer. See Example \(\PageIndex
• Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example \(\PageIndex{9}\).
Descartes’ Rule of Signs
a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of \(f(x)\) and \(f(−x)\)
Factor Theorem
\(k\) is a zero of polynomial function \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\)
Fundamental Theorem of Algebra
a polynomial function with degree greater than 0 has at least one complex zero
Linear Factorization Theorem
allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form \((x−c)\), where \(c\) is a complex number
Rational Zero Theorem
the possible rational zeros of a polynomial function have the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.
Remainder Theorem
if a polynomial \(f(x)\) is divided by \(x−k\),then the remainder is equal to the value \(f(k)\) | {"url":"https://math.libretexts.org/Courses/Mission_College/Math_1%3A_College_Algebra_(Carr)/04%3A_Polynomial_and_Rational_Functions/4.06%3A_Zeros_of_Polynomial_Functions","timestamp":"2024-11-08T14:07:02Z","content_type":"text/html","content_length":"177841","record_id":"<urn:uuid:3b40ec0f-89a4-429d-bc5a-9230425a4c00>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00654.warc.gz"} |
Integer programming and game theory
1. Define Integer Programming Problem (IPP)? (DEC ’07)
A linear programming problem in which some or all of the variables in the optimal solution are restricted to assume non-negative integer values is called an Integer Programming Problem (IPP) or
Integer Linear Programming
2. Explain the importance of Integer programming problem?
In LPP the values for the variables are real in the optimal solution. However in certain problems this assumption is unrealistic. For example if a problem has a solution of 81/2 cars to
be produced in a manufacturing company is meaningless. These types of problems require integer values for the decision variables. Therefore IPP is necessary to round off the fractional values.
3. List out some of the applications of IPP? (MAY ’09) (DEC ’07) (MAY ’07)
IPP occur quite frequently in business and industry.
All transportation, assignment and traveling salesman problems are IPP, since the decision variables are either Zero or one.
All sequencing and routing decisions are IPP as it requires the integer values of the decision variables.
Capital budgeting and production scheduling problem are PP. In fact, any situation involving decisions of the type either to do a job or not to do can be treated as an IPP.
All allocation problems involving the allocation of goods, men, machines, give rise to IPP since such commodities can be assigned only integer and not fractional values.
4. List the various types of integer programming? (MAY ’07)
Mixed IPP
Pure IPP
5. What is pure IPP?
In a linear programming problem, if all the variables in the optimal solution are restricted to assume non-negative integer values, then it is called the pure (all) IPP.
6. What is Mixed IPP?
In a linear programming problem, if only some of the variables in the optimal solution are restricted to assume non-negative integer values, while the remaining variables are free to take any
non-negative values, then it is called A Mixed IPP.
7. What is Zero-one problem?
If all the variables in the optimum solution are allowed to take values either 0 or 1 as in ‘do’ or ‘not to do’ type decisions, then the problem is called Zero-one problem or standard discrete
programming problem.
8. What is the difference between pure integer programming & mixed integer programming?
When an optimization problem, if all the decision variables are restricted to take integer values, then it is referred as pure integer programming. If some of the variables are allowed to take
integer values, then it is referred as mixed integer programming.
9. Explain the importance of Integer Programming?
In linear programming problem, all the decision variables allowed to take any non-negative real values, as it is quite possible and appropriate to have fractional values in many situations. However
in many situations, especially in business and industry, these decision variables make sense only if they have integer values in the optimal solution. Hence a new procedure has been developed in this
direction for the case of LPP subjected to additional restriction that the decision variables must have integer values.
10. Why not round off the optimum values instead of resorting to IP? (MAY ’08)
There is no guarantee that the integer valued solution (obtained by simplex method) will satisfy the constraints. i.e. ., it may not satisfy one or more constraints and as such the new solution may
not feasible. So there is a need for developing a systematic and efficient algorithm for obtaining the exact optimum integer solution to an IPP.
11. What are methods for IPP? (MAY ’08)
Integer programming can be categorized as
(i) Cutting methods
(ii) Search Methods.
12. What is cutting method?
A systematic procedure for solving pure IPP was first developed by R.E.Gomory in 1958. Later on, he extended the procedure to solve mixed IPP, named as cutting plane algorithm; the method
consists in first solving the IPP as ordinary LPP. By ignoring the integrity restriction and then introducing additional constraints one after the other to cut certain part of the solution space
until an integral solution is obtained.
13. What is search method?
It is an enumeration method in which all feasible integer points are enumerated. The widely used search method is the Branch and Bound Technique. It also starts with the continuous optimum, but
systematically partitions the solution space into sub problems that eliminate parts that contain no feasible integer solution. It was originally developed by A.H.Land and A.G.Doig.
14. Explain the concept of Branch and Bound Technique?
The widely used search method is the Branch and Bound Technique. It starts with the continuous optimum, but systematically partitions the solution space into sub problems that eliminate parts that
contain no feasible integer solution. It was originally developed by A.H.Land and A.G.Doig.
15. Give the general format of IPP?
The general IPP is given by
Maximize Z = CX
Subject to the constraints
AX ≤ b,
X ≥ 0 and some or all variables are integer.
16. Write an algorithm for Gomory’s Fractional Cut algorithm?
1. Convert the minimization IPP into an equivalent maximization IPP and all the
coefficients and constraints should be integers.
2. Find the optimum solution of the resulting maximization LPP by using simplex
3. Test the integrity of the optimum solution.
4. Rewrite each X[Bi]
5. Express each of the negative fractions if any, in the k^th row of the optimum simplex
table as the sum of a negative integer and a non-negative fraction.
6. Find the fractional cut constraint
7. Add the fractional cut constraint at the bottom of optimum simplex table obtained in
step 2.
8. Go to step 3 and repeat the procedure until an optimum integer solution is obtained.
17. What is the purpose of Fractional cut constraints?
In the cutting plane method, the fractional cut constraints cut the unuseful area of the feasible region in the graphical solution of the problem. i.e. cut that area which has no integer-valued
feasible solution. Thus these constraints eliminate all the non-integral solutions without loosing any integer-valued solution.
18.A manufacturer of baby dolls makes two types of dolls, doll X and doll Y. Processing of these dolls is done on two machines A and B. Doll X requires 2 hours on machine A and 6 hours on Machine B.
Doll Y requires 5 hours on machine A and 5 hours on Machine B. There are 16 hours of time per day available on machine A and 30 hours on machine B. The profit is gained on both the dolls is same.
Format this as IPP?
Let the manufacturer decide to manufacture x[1] the number of doll X and x[2] number of doll Y so as to maximize the profit. The complete formulation of the IPP is given by
Maximize Z = x[1]+x[2]
Subject to 2 x[1] + 5 x[2] ≤16
6 x[1]+ 5 x[2] ≤30
and ≥0 and are integers.
19. Explain Gomory’s Mixed Integer Method?
The problem is first solved by continuous LPP by ignoring the integrity condition. If the values of the integer constrained variables are integers, then the current solution is an optimal solution to
the given mixed IPP. Else select the source row which corresponds to the largest fractional part among these basic variables which are constrained to be integers. Then construct the Gomarian
constraint from the source row. Add this secondary constraint at the bottom of the optimum simplex table and use dual simplex method to obtain the new feasible optimal solution. Repeat this procedure
until the values of the integer restricted variables are integers in the optimum solution obtained.
20. What is the geometrical meaning of portioned or branched the original problem?
Geometrically it means that the branching process eliminates portion of the feasible region that contains no feasible-integer solution. Each of the sub-problems solved separately as a LPP.
21. What is standard discrete programming problem?
If all the variables in the optimum solution are allowed to take values either 0 or 1 as in ‘do’ or ‘not to do’ type decisions, then the problem is called standard discrete programming problem.
22. What is the disadvantage of branched or portioned method?
It requires the optimum solution of each sub problem. In large problems this could be very tedious job.
23. How can you improve the efficiency of portioned method?
The computational efficiency of portioned method is increased by using the concept of bounding. By this concept whenever the continuous optimum solution of a sub problem yields a value of the
objective function lower than that of the best available integer solution it is useless to explore the problem any further consideration. Thus once a feasible integer solution is obtained, its
associative objective function can be taken as a lower bound to delete inferior sub-problems. Hence efficiency of a branch and bound method depends upon how soon the successive sub-problems are
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Day 6 Project: Fizz Buzz
Welcome to the day 6 project for the 30 Days of Python series. Today's project is actually a very common interview question, which revolves around a childhood counting game called Fizz Buzz.
In case you're not familiar with the game, it goes like this:
• One player starts by saying the number 1.
• Each player then takes it in turns to say the next number, counting one at a time.
• If the number is divisible by 3, instead of saying the number, the player should say, "Fizz".
• If the number is divisible by 5, instead of saying the number, the player should say, "Buzz".
• If the number is divisible by 3and5, instead of saying the number, the player should say, "Fizz Buzz".
• If you make a mistake, you're usually eliminated from the game, and the game continues until there's only a single player remaining.
If there are no mistakes, the first 15 rounds of Fizz Buzz should look like this:
Fizz Buzz
Below you'll find a brief explaining what to do for our version, and you'll also find a model solution with an accompanying explanation. I'd really recommend you try to do this on your own before
checking out our version.
Just like with the day 3 project, there's nothing wrong with looking back at the content for the last 6 days, or referencing your notes. You also shouldn't be worried if your solution is a little
different to ours, as there are many, many ways to tackle this particular problem.
The brief
For our version, we're only going to have a single player, the computer, and it's going to play the first 100 rounds of Fizz Buzz all by itself. In other words, we need to print out the first 100
items in the sequence, starting from 1.
In order to complete this exercise, you're going to need to use loops, and you can generate your list of numbers using range. You're also going to need conditionals, and you're going to need be able
to check if something is divisible by 3 or 5.
For this last part, you can use an operator called modulo, which uses the percent symbol (%). Modulo will give you the remainder of a division, so if a number is divisible by 3, the value of number %
3 will be 0.
If you want to learn about modulo in more detail, we have a post that you can check out.
An alternative you can make use of is the is_integer method. We can call this on a float to check if it's an integral (whole) number. For example, we can write something like this:
(2.0).is_integer() # True
(3.7).is_integer() # False
And we can test the result of a division like this:
(12 / 4).is_integer() # True
(12 / 5).is_integer() # False
Good luck!
Solution walkthrough
Just like we did for the day 3 project, we're going to break the project up into smaller chunks. This is going to make it much easier to catch our mistakes early, and it's also going to make the
project a little less daunting.
If you'd prefer to watch a video walkthrough for this project, see here.
I think a good first step is just getting our range of numbers. We can this with the range function, and we're going to need to pass in two arguments: a start value, and a stop value. We need a start
value because range will start at 0 by default.
Don't forget that the stop value for range is not inclusive, so we need to specify a range from 1 to 101:
We can check that we have everything we need by converting it to a list a printing it. Remember that we can't print range directly, because range is lazy, and doesn't calculate its values until we
ask for them.
numbers = list(range(1, 101))
Assuming we don't have any problems, I think the next logical step is to print the numbers using a loop. We can do away with our list and numbers variable at this point, and just put our range in the
loop directly:
for number in range(1, 101):
This should give us 1 to 100 printed to the console, with each number on a different line.
Now we need to start filtering out the numbers we don't want to print from this output. Let's start by accounting for Fizz numbers: those divisible by 3.
Using the modulo approach, we can do something like this:
for number in range(1, 101):
if number % 3 == 0:
Here we check to see if a number is divisible by 3. If it is, we print "Fizz"; otherwise, we print the number itself. We've seen conditionals a few times now, so this is hopefully relatively
straightforward at this point.
If we wanted to use the is_integer method, the solution looks very similar:
for number in range(1, 101):
if (number / 3).is_integer():
Okay, now we that we have Fizz numbers taken care of, let's expand our conditions to account for Buzz.
The process is very similar, we just have to add an elif clause checking for a second condition. That condition is whether or not the number is divisible by 5.
for number in range(1, 101):
if number % 3 == 0:
elif number % 5 == 0:
This step is a place where people often trip up. If we don't use an elif clause, and instead use a new if statement, we end up in a situation where we have two lots of output for many of the numbers.
Let's look at an incorrect version and think about what's going on:
for number in range(1, 101):
if number % 3 == 0:
if number % 5 == 0:
If a number is divisible by 3, we trigger this first conditional. We then check the second condition. If the number is divisible by 5, we end up also printing "Buzz" on the next line. While we want
this to happen, we want it all on the same line, and we're going to use another step for this. If the number isn't divisible by 5, we end up printing "Fizz"and the number. Make sure you don't fall
into this trap.
Now that we've done "Fizz" and "Buzz", we need to account for "Fizz Buzz". This is another place people often trip up, because the order of the conditions matters. You'll see what I mean in a second.
First, let's look at how we can do this using nested conditions:
for number in range(1, 101):
if number % 3 == 0:
if number % 5 == 0:
print("Fizz Buzz")
elif number % 5 == 0:
This solution works because for any numbers divisible by 3, we perform a second check. If the number is divisible by 5 as well, we know that the conditions have been met for "Fizz Buzz", so we can
print that to the console. If this second condition wasn't met, we know the number is only divisible by 3, so we can print "Fizz" instead.
That solution works great, but what about a solution that doesn't use these nested conditions? We have two major options, but we'll save the second one for the bonus content at the end.
The option we can use here is checking if something is divisible by 15. That's because any number divisible by both 3 and 5 is also divisible by 15. As I mentioned already, where we put this
condition is really important. For example, if we try the following, we're not going to get the output we want:
for number in range(1, 101):
if number % 3 == 0:
elif number % 5 == 0:
elif number % 15 == 0:
print("Fizz Buzz")
This is because Python is only going to check conditions until it finds one that's true. Any number divisible by 15 is also divisible by 3, so this first condition catches these numbers before we hit
this third branch. We therefore get "Fizz" printed where we expect "Fizz Buzz".
To correct this, we need to put the more specific conditions first. In this instance, something being divisible by 3 is a broader condition than being divisible by 15, because the numbers divisible
by 15 are a smaller subset of the numbers divisible by 3. The condition checking for divisibility by 15 therefore needs to come first.
for number in range(1, 101):
if number % 15 == 0:
print("Fizz Buzz")
elif number % 3 == 0:
elif number % 5 == 0:
With that, both our solutions work, and we're done!
Bonus material
Dividing by 15 is a neat trick to keep the solution down to a single conditional block, but maybe it's not super clear to some people what's going on. It would maybe be better if we could be direct
about what we're checking here.
We can actually evaluate multiple expressions using a pair of special Boolean operators called and and or. Their use in this case is relatively straightforward, and very easy to read:
for number in range(1, 101):
if number % 3 == 0 and number % 5 == 0:
print("Fizz Buzz")
elif number % 3 == 0:
elif number % 5 == 0:
However, there are some important details about how and and or works, so if you're interested in using these operators, you should read our post on this topic.
We also have some additional solutions to Fizz Buzz in another post. If you're interested in seeing some of these alternative approaches, you can find them here. | {"url":"https://teclado.com/30-days-of-python/python-30-day-6-project/","timestamp":"2024-11-09T11:14:42Z","content_type":"text/html","content_length":"170863","record_id":"<urn:uuid:e98e43fa-1349-4707-b406-d258b4ceb92c>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00582.warc.gz"} |
The “Bridge” of Comprehension | The Healing Garden
About Lesson
Understanding that Math evolved and grew out of Logic, which grew out of Linguistics is vital to understanding the need for Number, Value, Measurement, and Word.
This is a Language and it must be treated like a Language for many people to understand it. Throwing non-number people into Mathematics without the Language will result in more confusion than
A mark of the Oxymoron “Bad Teacher.”
In Math, we are Defining and then we are Measuring.
In Math, we are measuring 3 things:
Upon measuring these three things, Balance must always be maintained. No matter what. The Balance is “weighing” or Matching the “Copy” against the “Original” (Mother Nature).
This is all about Precision.
In Linguistics, we learn how words Define and “Give Shape” to Feeling and Comprehension.
In Logic, we learn how the Comprehension moves and Grows and changes.
In Mathematics, we measure the space, value, definition, and change of that Growth.
In Physics, we study the Laws of that Change.
All of Math begins with Words and Definitions.
For this, you will require the Mathematical Notation:
This is the Language of Math as Defined prior to beginning. | {"url":"https://www.annashealinggarden.org/courses/mathematics-demystified/lesson/the-bridge-of-comprehension/?v=eb65bcceaa5f","timestamp":"2024-11-10T02:44:45Z","content_type":"text/html","content_length":"167510","record_id":"<urn:uuid:ba178960-23bd-4088-a160-4b35750b46c4>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00520.warc.gz"} |
centroid and centre of gravity
"Difference Between Centre of Gravity and Centroid." Therefore, being perfectly balanced the bar should remain in any … The term is peculiar to the English language. The point where the gravitational
force acts upon the body with any density is called Centre of gravity. In that case the centre of gravity is equivalent with the centroid of the body, Centroid is the average position of all the
points of an object. Centre of gravity is denoted by the symbol or letter ‘g’ or “cg”. We equivalently represent the system of forces by single force acting at a specific point. While balancing, the
centre of the object is where the gravitational force is concentrated. PDF. CENTROID CENTRE OF GRAVITY Centre of gravity : It of a body is the point at which the whole weight of the body may be
assumed to be concentrated. It is represented or denoted by C.G. Centre of gravity or centre of mass is the point where the whole mass of the body is concentrated. The term "centroid" is of recent
coinage (1814). Centre of Gravity refers to a physical characteristic of a body whereas Centroid refers to a geometric characteristic of an object. They are Centre of Gravity and Centroid. ….. Thus
centre of gravity, centre of mass and the centroid, all are physically the same but theoretically different. Centroid is the geometric centre of the object. There are different ways to calculate
these two things. in fact we replace that all of these small forces by a single equivalent force (the resultant force). If the object does not have uniform density, then its focus point is called
Centre of Gravity. If the body is homogeneous (having constant density), then its center of gravity is … Centre of gravity is the point where total weight of the object acts. There is no need to
resubmit your comment. The total area of the membership function distribution used to represent the combined control action is divided into a number of sub-areas. In case of irregular geometric
bodies, the centre of gravity is located in the intersection of the gravity lines. Free PDF. 4. The centre of gravity of the triangle is in the cross-section of the angle bisections and the centre of
gravity of the cube in the cross section of its diagonals. Difference between the Centre of Gravity and Centroid los difference between the terms Centre of Gravity and Centroid is that the former
refers to the point where the total weight of the object is focused at. Ask Any Difference is a website that is owned and operated by Indragni Solutions. Fig. Centre of Gravity(cg) can be calculated
using the equation W=S x dw. Centre of Gravity is applicable to objects with any density. This direction is called the centre of gravity of the object. • Categorized under Mathematics &
Statistics,Physics,Science | Difference Between Centre of Gravity and Centroid. Which means the object has its weight distributed equally across all parts of the body. PDF | On Jun 7, 2018, Christo
Ananth published Engineering Mechanics - Centroids and Center of Gravity | Find, read and cite all the research you need on ResearchGate 9.3.2 Relationship between the centroid, center of mass, and
center of gravity of a body Centroid is a geometric property of a body; the center of mass entails the distribution of the mass of the body in addition to the geometry. Difference Between
Polyurethane and Lacquer (With Table), Difference Between Kayak and Canoe (With Table), “The purpose of Ask Any Difference is to help people know the difference between the two terms of interest.
whereas the centre of a gravity is the point through which the weight of a body acts. It is the point at which a cutout of the shape could be perfectly balanced. The main difference between Centre of
Gravity and Centroid is that the former refers to the point where the total weight of the body is concentrated whereas the latter the refers to the geometric centre of an object. Download Full PDF
Package. Centre of Gravity denotes the total weight of an object with any density. A short summary of this paper. This is where the gravitational force (weight) of the body acts for any orientation
of the body. If it is tilted slightly, the gravitational pull at the centre creates a new point where the weight is concentrated. The Centre of Gravity of any object plays an important role while
trying to balance that object. Centroid can be calculated by using the plumb line method or by taking the mean of median, in case of a triangle. In physics, centroid of a body is defined as the focus
point of the vectors’ collection of the gravitational acceleration of all the material points of the same object. Here is an example to make it clear: Consider a can placed on a flat surface. The
position of the centre of the mass of the rigid body in the Cartesian coordinate system is determined by the radius vector rS = (∫rρdV) / M, where r is the unit vector, ρ is the body’s density, V
volume, and M is the mass of the body. that the body is homogeneous. DifferenceBetween.net. Centroid and center of gravity are two important concepts in statics. This will lead to the can being
pulled into a stable position. 2.1 The centre of gravity of a body is the point at which all the mass of … The centroid is also represented by C.G. Both these terms involve the concepts of density,
weight and balance of any object. This paper. In each of the following figures G represents the centroid, and if each area was suspended from this point it would balance. Centre of Gravity or cg can
be calculated by the equation mentioned above; W = S x dw where. In order to find out the Centroid of an object, you can use the plumb line method proposed by Archimedes. Calculating centre of
gravity is not a simple procedure because the mass (and weight) may not be uniformly distributed throughout the object. or. centroid is a geometrical centre of a plane figure or an object, it divides
the shape into regions of equal moments. Figure 5(a) also shows that the turning moments acting on each arm of the bar, due to force of gravity, are equal and opposite. Both Centre of Gravity and
Centroid are concepts that are important, to find out the central focus point of objects where the gravitational force acts on the body. Notify me of followup comments via e-mail, Written by :
Emilija Angelovska. Centre of gravity is the point where the total weight of the body acts while centroid is the geometric centre of the object. Download PDF Package. The position of the centre of
gravity for the particle system in the Cartesian coordinate system is determined by the radius vector rS = Σmiri / Σmi, where mi are the masses of the particles, and ri are the radius vectors of the
particles. In general when a rigid body lies in a field of force acts on each particle of the body. In each of the following figures 'G' represents the centroid, and if each area was suspended from
this point it would balance. Whereas the latter refers to the geometric centre of an object. Like the centre of gravity, the centroid of an area is also denoted by the letter G. Some examples of
centroids are shown in Fig. and updated on February 14, 2020, Difference Between Similar Terms and Objects, Difference Between Centre of Gravity and Centroid, Difference Between Thermodynamics and
Kinetics, Difference Between Additive Colors and Subtractive Colors, Difference Between Horizontal and Vertical Asymptote, Difference Between Leading and Lagging Power Factor, Difference Between
Commutative and Associative, Difference Between Systematic Error and Random Error, Difference Between Quantum Mechanics and General Relativity, Difference Between Horizontal and Vertical Axis Wind
Turbine, Difference Between Genome Sequencing and Genome Mapping, Difference Between Covid-19 and Allergies, Difference Between Virulence and Infectivity, Difference Between Vitamin D and Vitamin D3,
Difference Between LCD and LED Televisions, Difference Between Mark Zuckerberg and Bill Gates, Difference Between Civil War and Revolution, Centre of mass of a geometric object with any density,
Centre of mass of a geometric object of uniform density, Point where weight of a body or system may be considered to act, The gravitational forces of the elementary parts of which the body is
composed can be replaced with the action of a resultant force with, The centre of gravity is located in the intersection of the gravitational lines, and in the correct geometric bodies is determined
geometrically. This term is used to denote the centre of a body with uniform density. 24 Full PDFs related to this paper . The centroid of an object in -dimensional space is the intersection of all
hyperplanes that divide into two parts of equal moment about the hyperplane. The following is a list of centroids of various two-dimensional and three-dimensional objects. The Center of Gravity is
the same as the centroid when the density is the same throughout. Centroid can be found with methods such as the plumb line method discussed above. CENTROIDS AND CENTRE OF GRAVITY. Centre of Gravity
and Centroids are related to planes. Entire mass is supposed to be concentrated at this point. Centroid is the centre of gravity for objects of uniform density. Centre of gravity is referred to with
the use of the letter ‘g’. Hang plumb on the same nail. : AJIT MORE.In this video, basic difference between centroid and centre of gravity is exaplined. of a uniform rod lies at its middle point.
Whereas if it is of uniform density, then the Centre of Gravity is equal to its Centroid. Centroid is the central point of objects with uniform density. Centroid is often denoted by the symbol or
letter “c”. If a particle or body system moves under the influence of an external force, the point at which the centre of gravity is located moves as if it contains all the mass of the system or
body. Is there a difference between the centroid and the center of gravity of an object? Emilija Angelovska. The words centre and gravity are derived from the Latin (or Greek) words “centrum” and
“gravitatio”. Centre of gravity of a body is defined as the point through which the whole weight of the body acts. PDF. Equilibrium Equilibrium was introduced in previoussection. But Centroid is the
centre of an object with uniform density. Centroid and center of gravity will be at same point and we must note it here that centroid of a body will also be indicated by C.G or G. Ever since then,
we've been tearing up the trails and immersing ourselves in this wonderful hobby of writing about the differences and comparisons. It is used as a substitute for the older terms " center of gravity,"
and " center of mass ", when the purely geometrical aspects of that point are to be emphasized. Michael Corleone. To calculate the center of gravity, a person can use the formula CG W = S x dw.
Center of Mass and Centroids Center of Mass: Following equations independent of g (Vector representation) Unique point [= f(ρ)] :: Centre of Mass (CM) CM coincides with CG as long as gravity field is
treated as uniform and parallel CG or CM may lie outside the body m zdm z m ydm y m xdm x ³ m ³ dm r r ³ ³ ³ ³ ³ ³ dV z dV z ydV y x x U U That is, the centre of any object where the density is
evenly distributed over its body is called the Centroid. It is represented by CG. The centre of mass can also be located outside the body’s mass limits, which depends on its shape. 1. Centroid and
center of gravity are two important concepts in statics. When in context of triangles, the centroid of a triangle is the point of intersection of its three medians. The centroid of an area is
situated at its geometrical centre. the term centre of gravity applies to the bodies with mass and weight while the term centroid applies to plane areas. The Centroid is the average position of all
the points of an object. of a rectangle (or a parallelogram or a square) lies at the point where its … We write on the topics: Food, Technology, Business, Pets, Travel, Finance, and Science”. Not a
simple procedure because the mass of the body where the is. Central point of intersection of the body can be found with methods centroid and centre of gravity. Business, Pets, Travel, Finance, and
Science ” centre creates a new point where the gravitational at. Centroid '' is of uniform density homogeneous density, weight and balance of any object plays an important while. Find out the
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Module 2: Introduction to Culturally Responsive Math Modeling
Learn about Culturally Responsive Math Modeling
This is the second module in the 6 part series. Some of the modules around the tasks can be used in different sequences but we feel that Modules 1-2 Intro to Culturally Responsive Math Modeling
(CRMM) is foundational to build on any of the other work.
Please refer to the professional learning cycle by starting with the Introduce-Prepare-Enact-Reflect phases for the Core Practices for Culturally Responsive Math Modeling (CRMM) .
Thank you for your interest in this work and feel free to reach to the project team at eqstemm@gmail.com
Jenn, Julia, Erin, Mary Alice, Elizabeth
For Full Module for 2a: Intro to Math Modeling -Click here
Module 2a: How does Math Modeling support Culturally Responsive Teaching?
Learn about Math Modeling and a routine to get your students mathematizing the world.
Collaboratively watch a lesson with the CRMT lens.
Teach and collect work for formative assessment
Debrief and Reflect on the Enactment
Module 2b: Deeper Dive into Culturally Responsive Teaching through Lesson Enactments
Honoring Student Thinking (CRMT) | {"url":"https://eqstemm.org/about/module-2a/","timestamp":"2024-11-08T08:04:40Z","content_type":"text/html","content_length":"71941","record_id":"<urn:uuid:319b1ee8-a602-4d8c-8465-1d63f1f3eebb>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00566.warc.gz"} |
Digital Waveform Generation: Approximate a Sine Wave
This example shows how to design and evaluate a sine wave data table for use in digital waveform synthesis applications in embedded systems and arbitrary waveform generation instruments.
Even small systems use real-time direct digital synthesis of analog waveforms using embedded processors and digital signal processors (DSPs) connected to digital-to-analog converters (DACs). Using
MATLAB® and Simulink®, you can develop and analyze the waveform generation algorithm and its associated data before implementing it with Simulink® Coder™ on target hardware.
The most accurate way to digitally synthesize a sine wave is to compute the full-precision sin function directly for each time step, folding omega*t into the interval 0 to 2*pi. In real-time systems,
the computational burden is typically too large for this approach. Alternatively, you can use a table of values to approximate the behavior of the sin function, either from 0 to 2*pi, half-wave, or
quarter-wave data.
Tradeoffs between the two approaches include algorithm efficiency, data ROM size required, and the accuracy and spectral purity of the implementation. Similar analysis is needed when performing your
own waveform designs. The table data and look-up algorithm alone do not determine performance in the field. Additional considerations, such as the accuracy and stability of the real-time clock and
digital-to-analog conversion, are required to assess overall performance. The Signal Processing Toolbox™ and the DSP System Toolbox™ complement the capabilities of MATLAB and Simulink for work in
this area.
Another method to approximate the behavior of the sine wave is to use the COordinate Rotation DIgital Computer (CORDIC) approximation. The Givens rotation-based CORDIC algorithm is one of the most
hardware-efficient algorithms because it requires only shift-add iterative operations.
Create Table in Double Precision Floating Point
These commands make a 256-point sine wave and measure its total harmonic distortion when sampled first on the points and then by jumping with a delta of 2.5 points per step using linear
interpolation. Similar computations are done by replacing the sine values with CORDIC sine approximation. For frequency-based applications, spectral purity can be more important than absolute error
in the table.
In this example, the file ssinthd.m calculates the total harmonic distortion (THD) for digital sine wave generation with or without interpolation. This THD algorithm proceeds over an integral number
of waves to achieve accurate results. The number of wave cycles used is A. The step size 'delta' is A/B. Traversing A waves hit all points in the table at least one time, which is needed to
accurately find the average THD across a full cycle.
The relationship used to calculate THD is
where ET = total energy and EF = fundamental energy.
The energy difference between ET and EF is spurious energy.
N = 256;
angle = 2*pi*(0:(N-1))/N;
s = sin(angle)';
thd_ref_1 = ssinthd(s,1,N,1,'direct' );
thd_ref_2p5 = ssinthd(s,5/2,2*N,5,'linear' );
cs = cordicsin(angle,50)';
thd_ref_1c = ssinthd(cs,1,N,1,'direct' );
thd_ref_2p5c = ssinthd(cs,5/2,2*N,5,'linear' );
Use Sine Wave Approximations in a Model
You can put the sine wave you design into a Simulink model and see how it works as a direct lookup with linear interpolation and with CORDIC approximation. This model compares the output of the
floating-point tables to the sin function. As expected from the THD calculations, the linear interpolation has a lower error than the direct table lookup in comparison to the sin function. The CORDIC
approximation shows a lower error margin when compared to the linear interpolation method. This margin depends on the number of iterations when computing the CORDIC sin approximation. You can
typically achieve greater accuracy by increasing the number of iterations. The CORDIC approximation eliminates the need for explicit multipliers. This is used when multipliers are less efficient or
non-existent in hardware.
Open and simulate the sldemo_tonegen model.
set_param('sldemo_tonegen', 'StopFcn','');
out = sim('sldemo_tonegen');
Plot the simulation output.
currentFig = figure('Color',[1,1,1]);
subplot(3,1,1), plot(out.tonegenOut.time, out.tonegenOut.signals(1).values); grid
title({'Difference between direct lookup', 'and reference signal'});
subplot(3,1,2), plot(out.tonegenOut.time, out.tonegenOut.signals(2).values); grid
title({'Difference between interpolated lookup', 'and reference signal'});
subplot(3,1,3), plot(out.tonegenOut.time, out.tonegenOut.signals(3).values); grid
title({'Difference between CORDIC sine', 'and reference signal'});
Examine Waveform Accuracy
When you examine signals, you can see different characteristics of the different algorithms. For example, zoom in on the signals between 4.5 and 5.2 seconds of simulation time.
ax = get(currentFig,'Children');
set(ax(3),'xlim',[4.8, 5.2]);
set(ax(2),'xlim',[4.8, 5.2]);
set(ax(1),'xlim',[4.8, 5.2]);
Implement Table in Fixed Point
You can convert the floating-point table into a 24-bit fractional number using nearest rounding. Test the new table for total harmonic distortion in direct lookup mode at 1, 2, and 3 points per step
and then with fixed-point linear interpolation.
bits = 24;
is = num2fixpt(s,sfrac(bits),[],'Nearest','on');
thd_direct1 = ssinthd(is,1,N,1,'direct');
thd_direct2 = ssinthd(is,2,N,2,'direct');
thd_direct3 = ssinthd(is,3,N,3,'direct');
thd_linterp_2p5 = ssinthd(is,5/2,2*N,5,'fixptlinear');
Compare Results
Choosing a table step rate of 8.25 points per step (33/4), jump through the double-precision and fixed-point tables in direct and linear modes and compare distortion results.
thd_double_direct = ssinthd(s,33/4,4*N,33,'direct');
thd_sfrac24_direct= ssinthd(is,33/4,4*N,33,'direct');
thd_double_linear = ssinthd( s,33/4,4*N,33,'linear');
thd_sfrac24_linear = ssinthd(is,33/4,4*N,33,'fixptlinear');
Use Preconfigured Sine Wave Blocks
Simulink also includes a Sine Wave block with continuous and discrete modes, plus fixed-point Sine and Cosine function blocks that implement the function approximation with a linearly interpolated
lookup table that exploits the quarter wave symmetry of sine and cosine. The model sldemo_tonegen_fixpt uses a sampled sine wave source as the reference signal and compares it with a lookup table
with or without interpolation, and with CORDIC sine approximation in fixed-point data types.
Open the sldemo_tonegen_fixpt model.
set_param('sldemo_tonegen_fixpt', 'StopFcn','');
out = sim('sldemo_tonegen_fixpt');
Plot the simulation output.
subplot(3,1,1), plot(out.tonegenOut.time, out.tonegenOut.signals(1).values); grid
title({'Difference between direct lookup', 'and reference signal'});
subplot(3,1,2), plot(out.tonegenOut.time, out.tonegenOut.signals(2).values); grid
title({'Difference between interpolated lookup', 'and reference signal'});
subplot(3,1,3), plot(out.tonegenOut.time, out.tonegenOut.signals(3).values); grid
title({'Difference between CORDIC sine', 'and reference signal'});
Use Sine Function with Clock Input
The model also compares the sine wave source reference with the sin function whose input angle in radians is time-based, or computed using a clock. You can test the assumption that the clock input
would return repeatable results from the sin function for period 2*pi. This plot shows that the sin function accumulates error when its input is time-based. The plot also shows that a sampled sine
wave source is more accurate to use as a waveform generator.
subplot(1,1,1), plot(out.tonegenOut.time, out.tonegenOut.signals(4).values); grid
title({'Difference between time-based sin function', 'and reference signal'});
Survey of Behavior for Direct Lookup and Linear Interpolation
To perform a full-frequency sweep of the fixed-point tables and gain insight to the behavior of this design, run the file sldemo_sweeptable_thd.m. Total harmonic distortion of the 24-bit fractional
fixed-point table is measured at each step size, and moves through D points at a time, where D is a number from 1 to N/2, incremented by 0.25 points. N is 256 points in this example. Frequency is
discrete and therefore a function of the sample rate.
Notice the modes of the distortion behavior in the plot. When retrieving from the table precisely at a point, the error is smallest. Linear interpolation has a smaller error than direct lookup
between points. However, the error is relatively constant for each of the modes up to the Nyquist frequency.
tic, sldemo_sweeptable_thd(24, 256), toc;
Elapsed time is 1.052861 seconds.
Next Steps
Using CORDIC approximation, you can run this example using different numbers of iterations to see the effects on accuracy and computation time. Try different implementation options for waveform
synthesis algorithms using automatic code generation available in Simulink Coder and production code generation using Embedded Coder™. Embedded target products offer direct connections to a variety
of real-time processors and DSPs, including connection back to the Simulink model while the target is running in real-time. The Signal Processing Toolbox and DSP System Toolbox offer capabilities for
designing and implementing a wide variety of sample-based and frame-based signal processing systems with MATLAB and Simulink.
[1] Chrysafis, Andrea. "Digital Sine-Wave Synthesis Using the DSP56001/2." Motorola, 1988.
See Also
Sine, Cosine | Sine Wave | Sine Wave Function
Related Topics | {"url":"https://fr.mathworks.com/help/simulink/slref/digital-waveform-generation-approximating-a-sine-wave.html","timestamp":"2024-11-05T03:11:15Z","content_type":"text/html","content_length":"85709","record_id":"<urn:uuid:133442e7-e4a4-41eb-889d-370e3168ce9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00532.warc.gz"} |
RYG formula in response to a numbered cell
I am having a bit of trouble figuring out how to write a formula to automate RYG symbols. Basically I want the rules to be as follows:
1. If value is less than 4, RED
2. If value is greater than 4 but less than 6, YELLOW
3. If value is greater than 6, GREEN
So far I have figured out how to do a single formula for 1 & 3 (written in the format of: =IF([13-Mar]1 > 6, "Green"), but not combined in the same cell and I have not at all figured out how to
write the formula for 2.
I hope that all makes sense... I can clarify further if needed.
Any guidance will be extremely appreciated!
• Try this, I got a variation in my sheet to work
=IF([13-Mar]1 <4, "Red", IF([13-Mar]1 =5, "Yellow", IF([13-Mar]1 >6, "Green")))
• Hi!
What your are looking for is Nested If. In your case, this would look like :
"= IF([13-Mar]1 > 6, "Green", IF([13-Mar]1 > 4, "Yellow", "Red") ) "
Your should read this like : If [13-Mar]1 > 6 then Green, ElseIf [13-Mar]1 > 4 then Yellow, Else Red.
Hope this helps!
• That was exactly what I needed!
Thank you so much! You're a lifesaver!
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/22061/ryg-formula-in-response-to-a-numbered-cell","timestamp":"2024-11-06T05:09:34Z","content_type":"text/html","content_length":"397146","record_id":"<urn:uuid:956c2d52-022d-4da6-aa6b-64e6a94a3b74>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00607.warc.gz"} |
Reinventing Math: turning negatives to positives
Remember those ridiculous word problems in math class that were laughably convoluted and unrealistic? What if we were to inject a bit of reality combined with some consciousness-raising into word
problems. We could get kids motivated to solve real world problems and oh, by the way, learn a little math in the process. Here are some possible word problem examples (which of course involve
biochar!) based on a recent New York Times article about air pollution in India.
India produces 34 M tons of crop residues, mostly rice and wheat straw. Although illegal, most farmers burn the straw as this is the quickest way to clear fields for the next crop. Instead of
burning, which contributes significantly to air pollution, farmers could carbonize these residues, creating a valuable nutrient carrier, biochar, which can also sequester carbon in the soil. Solve
the following problems:
1. 800 liters of straw will yield 200 liters of biochar. What is the yield as a percent?
2. If all farmers in India produced biochar instead of burning crop residues, how many tons of biochar could be produced?
3. If straw char contains 36% carbon, how many tons of carbon could India create from crop residues per year?
4. A typical car emits 4.7 metric tons of greenhouse gases per year. What is the equivalent in the number of cars, to the amount of carbon which India could create if all crop residues were
5. A small farmer with 1 hectare is fined USD$38 for burning his crops. Crop residues per hectare vary from .4 tons – 3.0 tons depending on many factors. Assume Farmer A has 2 tons of crop
residues and that one laborer who is paid $8 per day can harvest and carbonize 800 kg of straw in one day. Will the farmer be better off burning his crops or carbonizing them?
6. How much biochar will Farmer A produce?
7. For every kilo of biochar produced, assume the farmer can reduce purchases of lime for his fields. (Lime is needed for fields that are acidic.) Lime costs $30 per ton. How much will the farmer
save assuming a 1:1 ratio for biochar:lime.
8. What is the effective cost of producing biochar when labor and savings in lime are included?
The list of word problems could go on and on just for this single scenario. The questions could even get more complicated for older students, but you get the general idea. Math word problems could
be customized for different regions or for different problems which resonate with different cultures around the world.
Maybe, just maybe, if we create a math curriculum focused on climate change solving math word problems, we could start turning negatives into positives….in more ways than one! Kids could educate
their parents on this new math, offering new solutions which will not only reduce air pollution and rebalance carbon levels but could also improve farmer livelihoods too! | {"url":"https://fingerlakesbiochar.com/reinventing-math-turning-negatives-to-positives/","timestamp":"2024-11-13T12:31:30Z","content_type":"text/html","content_length":"46660","record_id":"<urn:uuid:70b02b16-c890-4060-ac9a-75a9d1b699c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00084.warc.gz"} |
Kriging Toolkit for Python.
The code supports 2D and 3D ordinary and universal kriging. Standard variogram models (linear, power, spherical, gaussian, exponential) are built in, but custom variogram models can also be used. The
2D universal kriging code currently supports regional-linear, point-logarithmic, and external drift terms, while the 3D universal kriging code supports a regional-linear drift term in all three
spatial dimensions. Both universal kriging classes also support generic ‘specified’ and ‘functional’ drift capabilities. With the ‘specified’ drift capability, the user may manually specify the
values of the drift(s) at each data point and all grid points. With the ‘functional’ drift capability, the user may provide callable function(s) of the spatial coordinates that define the drift(s).
The package includes a module that contains functions that should be useful in working with ASCII grid files (\*.asc).
See the documentation at http://pykrige.readthedocs.io/ for more details and examples.
PyKrige requires Python 3.5+ as well as numpy, scipy. It can be installed from PyPi with,
scikit-learn is an optional dependency needed for parameter tuning and regression kriging. matplotlib is an optional dependency needed for plotting.
If you use conda, PyKrige can be installed from the conda-forge channel with,
conda install -c conda-forge pykrige
Kriging algorithms
• OrdinaryKriging: 2D ordinary kriging with estimated mean
• UniversalKriging: 2D universal kriging providing drift terms
• OrdinaryKriging3D: 3D ordinary kriging
• UniversalKriging3D: 3D universal kriging
• RegressionKriging: An implementation of Regression-Kriging
• ClassificationKriging: An implementation of Simplicial Indicator Kriging
• rk.Krige: A scikit-learn wrapper class for Ordinary and Universal Kriging
• kriging_tools.write_asc_grid: Writes gridded data to ASCII grid file (\*.asc)
• kriging_tools.read_asc_grid: Reads ASCII grid file (\*.asc)
• kriging_tools.write_zmap_grid: Writes gridded data to zmap file (\*.zmap)
• kriging_tools.read_zmap_grid: Reads zmap file (\*.zmap)
Regression Kriging
Regression kriging can be performed with pykrige.rk.RegressionKriging. This class takes as parameters a scikit-learn regression model, and details of either the OrdinaryKriging or the
UniversalKriging class, and performs a correction step on the ML regression prediction.
A demonstration of the regression kriging is provided in the corresponding example.
PyKrige uses the BSD 3-Clause License. | {"url":"https://geostat-framework.readthedocs.io/projects/pykrige/en/stable/","timestamp":"2024-11-13T13:00:56Z","content_type":"text/html","content_length":"19428","record_id":"<urn:uuid:cb11f110-c87b-4ebf-9fd4-8d9f2c45528e>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00030.warc.gz"} |
(PDF) Seek and Decode: Random Multiple Access with Multiuser Detection and Physical-Layer Network Coding
... On the PHY layer, previous investigations focused on applying Compressive Sensing (CS) in multi-user detection contexts by exploiting the capabilities of sparse signal processing [2], [3].
Furthermore, joint MAC-and PHY-layer protocol design has been identified as one important enabler to support MMC [4], [5], [9] and [4] has shown that the combination of Coded Random Access (CRA) and
Compressed Sensing Multi-User Detection (CS-MUD) is a promising joint protocol to enhance system performance. Therefore, CS-MUD not only is known as an efficient PHY layer technique for sparse
Multi-User Detection (MUD) problem [1]- [3] but also serves as promising PHY layer enabler in joint MAC-and PHY-layer protocol design [4], [5]. ...
... With (7) updated by (9), the probability term is bounded by 1/β due to γ > 0. In the case, the domain of γ has changed by the probability bounding term, which yields the effective noise variance
σ 2 eff and optimal γ eff for the LASSO algorithm considering a discrete BPSK alphabet. ... | {"url":"https://www.researchgate.net/publication/262225905_Seek_and_Decode_Random_Multiple_Access_with_Multiuser_Detection_and_Physical-Layer_Network_Coding","timestamp":"2024-11-05T07:43:05Z","content_type":"text/html","content_length":"750099","record_id":"<urn:uuid:5ffd6262-0822-416f-90d8-f121de306b8a>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00389.warc.gz"} |
Determining Which of Given Shapes Is Not Congruent to the Others
Question Video: Determining Which of Given Shapes Is Not Congruent to the Others Mathematics
Which shape is not congruent to the others?
Video Transcript
Which shape is not congruent to the others?
This problem is a sort of odd-one-out question. We’re given some shapes and we need to find the one that’s different. This is because when shapes are congruent, we know they’re exactly the same size;
they’re exactly the same shape. So, four out of our five shapes are exactly the same size and shape. But one of them is not congruent. We’re looking then for the shape that’s different.
Now, as we’ve just said, there are two parts, two shapes, being congruent. They must be the same size, and they must be exactly the same shape. It doesn’t matter what color they are; it doesn’t
matter what position we put them in. But those two factors must be true. Now, if we look at our five shapes, we can see that they’re all in different positions. As we’ve just said, this doesn’t mean
they can’t be congruent. We know that two shapes can be exactly the same, but just in a different position. But it does make them harder to spot.
One way that we could check which of these shapes are the same even though they’re in different positions would be to put a little piece of tracing paper or some sort of paper we can see through over
one of the shapes then trace it and then move it to see if it fits exactly on top of any of the other shapes. For example, we’ve just proved that shapes (a) and (d) are congruent.
But you know, perhaps there’s a quicker way to find the answer here, because not only the shapes need to be exactly the same shape to be congruent, they need to be exactly the same size. And one of
these shapes is not the same size as any of the others. We don’t need to get a ruler or anything like that. We can see just by looking with our eyes that shape (b) is larger than any of the other
shapes. So, although we could go from (a) to (e) and check whether they’re exactly the same shape, perhaps using tracing paper, we’ve identified what we could call the odd one out because it’s not
the same size as any of the others. The shape that is not congruent is shape (b). | {"url":"https://www.nagwa.com/en/videos/767105306505/","timestamp":"2024-11-06T17:07:56Z","content_type":"text/html","content_length":"243130","record_id":"<urn:uuid:1931dae6-abad-4f70-8fed-fa50a8a80eae>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00810.warc.gz"} |
Re: [Icehouse] [Zendo] Rule clarification [Mondo]
Sorry if this has been dealt with somewhere (I did try searching for it)
In PwP, there's a paragraph in the advanced rules section that advises not to call Mondo if you're not sure (or something... Sorry - I don't have it in front of me), because if you do, you'll have to
reveal your guess.
Is that a mistake? Because my understanding of Mondo is that you just reveal a black or white stone and you're either right or wrong (and get a guessing stone or not).
Why is it not a good strategy to call mondo on every turn - you've got a chance of getting a guessing stone each time and no penalty. (Unless i'm wrong and that's not a mistake...) | {"url":"http://archive.looneylabs.com/mailing-lists/icehouse/msg04965.html","timestamp":"2024-11-03T06:01:47Z","content_type":"text/html","content_length":"7707","record_id":"<urn:uuid:a6dd1b21-460d-41d7-a650-891be3817b64>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00147.warc.gz"} |
[Solved] A soil has a bulk density of 135 lb/ft 3 | SolutionInn
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A soil has a bulk density of 135 lb/ft 3 and a dry density of 120 lb/ft 3 , and the specific gravity of the
A soil has a bulk density of 135 lb/ft^3 and a dry density of 120 lb/ft^3, and the specific gravity of the soil particles is 2.75. Determine (a) moisture content, (b) degree of saturation, (c) void
ratio, and (d) porosity.
There are 3 Steps involved in it
Step: 1
Solution a The bulk density and dry density d as following d 1w Where w is the moisture content ...
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Recommended Textbook for
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View Answer in SolutionInn App | {"url":"https://www.solutioninn.com/study-help/questions/a-soil-has-a-bulk-density-of-135-lbft-3-413687","timestamp":"2024-11-11T00:41:34Z","content_type":"text/html","content_length":"109711","record_id":"<urn:uuid:7276c6c2-29e0-4632-9023-92a52a941a52>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00762.warc.gz"} |
Hydraulic Ram Pump
Hydraulic Ram Pump
The Hydraulic Ram Pump is a water pump that requires no electricity or fuel to operate. As long as you have a waterway with falling water you can install and use this pump.
If you would like to buy a pre-assembled Ram Pump then click here.
Check out this video on how to build, install, and operate the pump. Below the video is a link to the Free eBook showing you the parts list and all the steps you need to build the pump and get it
Here is the free eBook:
Click to Download the Ebook:
This Free “how to” eBook will guide you through the steps needed to make the pump.
-New in Version 1.2: Pictures, Standpipe info.
-New in version 1.1: Standpipes, Snifter Valves, Drive Pipe Ratios, Pressure Tank Size.
File is 19mb
Example Page
Old version:
File size : 16mb
Hydraulic Ram Pump v1
Hydraulic Ram Pump Results:
My Brother In-Law and I worked with the Ram Pump and found some data that you might be interested in. We use two methods to find the delivery height. We determine the water Flow Rate at the height
that is discovered. And lastly we show you a method to find the feet of head that your pump is working with.
Check out this video for the results:
How far, Horizontally, can the Ram Pump carry water?
With 6′ of head falling into the pump I test out the distance the water will travel at approximately 15 feet above the pump.
I am using garden hose as the delivery pipe because it is the easiest tube to acquire and does not cost as much as other options.
After 415′ of hose set out horizontally I am still getting water out of the hose at the rate of around 1 gpm
185 thoughts on “Hydraulic Ram Pump”
1. Pingback: Ram Pump Flooding | Land To House
2. Suppose your water sourc is other than a creek and you don’t want to lose 80 or 90 percent of your source water is there some kewl way to retreve that would be lost water without using
1. As far as I know there is not another system that will pump water with out electricity. There has to be some form of energy (moving water) to run the pump. Now you might be able to get a
solar panel that will charge batteries to run a small electric pump. That way you would not lose any water and you would be essentially electricity free because the energy is renewable.
1. hi Seth, generally, I get a total different idea to pump water to 10 times higher place than yours same without electricity. my draft design is to erect a n shape pipe in creek, let’s say
the n shape get height around 10m; fill water in pipe and it will keep flowing by siphon theory; now put a Venturi on top of n shape to mix air in water, accumulating out air into a
container and utilize air pressure to pump water to higher place. several air containers connecting in a row but sitting at different height, each container pump water to a certain place,
finally, water out of pipe at much higher place than hydraulic ram pump. what do you think. Cheers
1. Very interesting. I have not every worked with the Venturi effect and it sounds passible. It sounds like you will need a few more parts than are found in the hydraulic ram but if you
are able to pump without as much water loss then you would be saving in the end. Please do some testing and keep me updated with some video.
I hope that it works well
2. I know of a way. It can be done.
3. It’s possible to put a pipe to conduit the water flowing from the pressure valve back to the water source tank in the case of the ramp pump would be used in aquaponic system to move the
water between the fish tank and the growing beds. This backpressure how could affect the system performance? I’m thinking about reducing the water usage, wich is one of the greatest
advantages of aquaponic systems. The point of discharge of this back flow must be allocated under the pressure tank cap or this level isn’t related to the discharge energy in the pressure
Or the ramp pump could be partially allocated in the fish tank?, so only after de check valve prior the pressure tank wiil be outside of the fish tank and the water that flows from
thepressure valve doesn’t waste.
I wish had been clear in my questions.
Thanx in advance for your response
1. Hello. From what I have seen it is best to use a 12v solar pump with a bell syphon instead of the ram pump to exchange water between auquaponics beds. This is because the loss from
the waste valve is so great. When you send water back to the source from a ram pump you will only be getting a small amount of the water back. The ram pump is very lossy pump but free
so it works out.
It is possible to submerge the pump in water and it will still run. So you can place the pump in the fish tank and still have an output.
It might be worth trying the ram pump for your setup but I have not seen a successful ram pump with aquaponics.
4. how about if its night time
1. You might need to have a charge controller and battery bank if you need it to run over night.
3. I have a Water Ram I will be installing I have made a reservoir and have constructed a drive pipe of 2″ pvc with rubber connectors. the Drive pipe is 130″ long with 10 feet of head to the pump. I
also have a 90Deg. bend in the drive pipe. at about 55 ft. at the stream bend. I tested the pump but could not get it to come to prime . I will be hooking up ball valves on intake side and supply
side. I also had many air leaks and my supply inlet pipe was out of the water.
when the pump did work it shot water out the supply line at high pressure. and I did not have the 3/4″ supply line hooked up. I need to pump 70ft up hill. Will I need a stand pipe for this to
work. I will try testing later in the week when I have more time after I fix all these little problems .
Your site is very useful thank you for the information and I love your videos ./and E book to Thanks John F.
1. Installing pumps is so fun! Thank you for the good comment!
It sounds like you have a nice setup. From everything that I have read and heard you will need a stand pipe for anything over 100′. (my pump is only 75′ long)Try it without first and see what
happens. If it does not work then you will need to go up from the pump around the 100′ mark and install your stand pipe. it will have to be a tall one to reach the height of the source tank.
If the pipe is not heigh enough then the water will just pour out of it from the source.
Now you bring up an important point that I did not talk about in my Ebook. The air in the pipe during priming will stop it every time. when you start up the pump hold down your check valve
and let lots of water run out to make sure that you have all the air out of the pipe.
adjusting ball valves will let it prime much faster. I would install those for sure.
So I would suggest that you stop those air leaks and make sure that all the air is out and it should start working every time without issues.
4. Hi Seth , I worked on my Ram pump the last few hours after work this week on two separate days.
I fixed all the leaks in the drive pipe and located the drive pipe under the water level of my reservoir.
I added a 2″ ball Valve on the intake side and a 3/4″ Ball valve on the delivery pipe to the delivery line. I had it pumping for 3-4 minutes it was cycling at 1 pump every 3 seconds which I think
is slow?
It stopped pumping!. The pressure and water volume is really excellent. I noticed some leaves in the pump cylinder. My pump has a glass ball inside that opens and closed in a rubber Lined steel
I started the 2″ ball valve at 80-90% and keep the delivery at only 20-30% open .
I noticed when I turned the delivery ball valve open all the way to 100% the system lost prime and shut down faster.
what is the theory of the stand pipe and do you think I need one .
I have my delivery pipe at 2″
I used pvc with rubber connecters
I have 8-9 feet of head drop to the pump
Drive line length is 135 feet long,{I have a 90 deg bend at 55′ } Lost wave speed?
my distance to water source will be 75 feet rise
I have a 3/4″ delivery line I will have to travel 650Ft to source
I have 5-6 gallon per minute water flow to pump
1. Your pump setup seems like an extensive one but fun. Those ball valves will be important for getting the pressure tank primed the first time and they make it much easier to clean later.
because your delivery pipe is a lot smaller than your drive pipe you might need to adjust the ball valve on the drive pipe side so that the ratio is correct. (not totally sure about that one)
A cycle every 3 seconds is slow and that is from the time it takes the wave to get to the end of the drive pipe and back to the pump. If you have an 8-9 foot of head and a 135 foot drive pipe
you could place a stand pipe at the 40 foot mark or less from the pump that was 1′ above the source tank.
so basically take a ” T ” pvc joint and place it on some section of the drive pipe. and add a pipe that stands at least one foot above the source tank. this will move the source water to the
place the stand pipe is. just place it below that 55′ bend because it is slowing the wave a lot.
Now I am not sure that the pump is going to get the hight that you are looking for. the 2″ pipe might make a difference. what I have understood is that you will have a 1 to 7 ratio. so your
8′ of head * 7 = 56′. then you will likely lose a little on the distance of the delivery pipe from friction.
when you had it pumping how high did you get the water to flow? Thank you for the good comments.
1. Hi Seth, I have pumped water uphill about 13-15 ft. at a distance of 300ft of black plastic 3/4″ pipe. I need to go up about 350’ftmore and uphill 65-70′ ft. I had good pressure when it
was pumping . But I had to almost close the ball valve on the delivery side to make the pump stay primed and pumping but this also failed.
the water was coming out much slower then when the delivery pipe was fully open.
should I place the Stand pipe at 100′ ft. after the 90deg bend which is about at 55’ft o w the stand pipe would be 45’ft farther down the Drive pipe. And should I use a 3″ or 4″ wide pipe
for the stand pipe. thanks for your help and the conversations John.
I really have to measure my feet in head it could be higher than 8 feet I will have to figure out. thanks John.
1. Just so that we are on the same page I made this little picture. let me know if it is right.
I have not ever used a stand pipe so I cant really say what the best size is.
You might want to check out this video showing some info on drive pipe with stand pipe.
basically if you are using a 2″ drive pipe the supply line to the stand pipe should be 4″ and the stand pipe even larger 6″.
before you go into all the work of installing a stand pipe. did you try to bleed out all the air in the drive pipe? by holding open the pump and letting water flow out into the creek?
2. I have not tested but have a feeling the pulser pump is better. It just needs more testing. With the pulser pump the air getting in is desirable that is how it works and it is quiet
too. The vibrations of the ram pump could harm wildlife in the water. The pulser can probably have leaves pass thru no problem and run longer without issues.
5. Thanks for the video it was very helpful a lot to reconsider. the guy I bought this from through a lot of miss information and the size of piping I could use. 4″ supply will cost me a lot more
and 3″ will too.
I,m thinking of changing out to 3″ before the standpipe for supply from my reservoir and making the stand pipe 4″ and then 2″ into 11/2 ” to the pump .
the reason 11/2 ” the supply line must be half of what the intake is .that is what the guy was explaining in the video. I could buy 1″ flex tubing and use the 3/4 inch for another project. what
do you think I should do Seth now I’m totally confused in what I should do. thank you John.
1. Well you already have a lot of 2″ pipe. What I would try is start at the ram pump and walk about 25′ to 35′ up the drive pipe and install a 3″ standpipe there. That would move the source to
that point. it would bypass the 90deg bend and it would shorten the drive pipe a lot. use the 2″ pipe that is already there for the supply line. then use 3″ for the stand pipe. then go down
to 1″ to the ram (or 1-1/4″)
this would save you some money. you would only need to get the 1″ or 1-1/4″ pipe. and the stand pipe.
When I made my installation I was not worried about ratios. I just want out there and started testing. the 1-1/4″ drive pipe and the 3/4″ delivery worked very well. but 6′ of head only gave
me 35′ of pump lift then it stopped pumping. So I am still worried that your 8 to 10 feet of head will stop lifting before getting to the desired destination.
6. Thank you Seth for the help. I have decided to start next spring some time in late march or April of 2014 and pickup where I started. I have run out of money for this project for the year . I
will probably go for the 4″ pipe for the drive line and go with the 6″ stand pipe . I can use some of the 2″ later for moving water on other part of my land. I will probably change my mind an do
it this Fall before the freeze. I live in Northern New York State in the Adirondack MT’s.
I will keep you posted on my updates and I will make a video when it is working and pumping.
1. Thank you for the correspondence! I would love to see your pump in a video when you get it working. Your setup is much more complicated than my “set in the creek and turn it on” pump. Seeing
the creative things that you do to get it just right will be good.
Wow I bet it gets cold up there ! and stays cold.
1. Hay, Seth, I got the system working yesterday, I set it aside for 5 days and just let it run without the delivery pipe installed. and started it up and has been running now for over 15
hours. I measured the gallons per minute at 3/4 per minute. the run is 300 ft. to my pond and 15ft up hill. pressure seems good!.
I also reconstructed my dam uphill and lowered the intake drive pipe and extra 6″ . this made a real difference in performance . the pipe was sticking up out of the water I had to place a
cinder block and a few large rocks on top to get it under the water. I just did not like this setup one bit.
My son and I will test the system some more next week to see how far it will pump up hill.
I will still install a stand pipe 3″ on this and tweak out the drive pipe to send more pressure to the drive pipe and to the pump.
thanks John I will keep you updated on my progress.
1. very nice! 300 foot and 15 hours is a good start. getting 3/4 gallons per min would give you over 500 gallons a day. Not half bad at all.
Adding an extra 6″ should give you 3 or 4 feet more lift.
making sure that the pump is flat is important if you are using the swing check valves. if the pump is leaning to one side or the other the swing will not be working as well as it
Once you get the pump working for the first time it seems to get easier.
Keep me updated .
2. Did you ever get this working?
I am going through the same problem.
7. Hello Seth have you considered doing a design on a linear hydraulic ram pump. The linear pum seems like a cheaper pump to build.
1. I have seen those! I might get into them at some point but for now I am testing out variations to the one that I have in the videos. Thank you for your interest in these amazing pumps.
8. I was watching another video where it said you were supposed to drill a small hole somewhere near I think the horizontal check valve or near the pressure tank. i cannot remember. have you heard
of this? if so why, and is it really necessary? By the way. i am looking forward to making one of these and your information so far has been awesome. Thanks.
1. Hello Dale. That small hole is called a snifter air valve. It allows air to escape a piston in a hydraulic cylinder. I have seen people use them in the Ram Pump but I have made several pumps
that have been working for a long time without the snifter valve. basically water can flood the pressure tank that is on the top of the system and that will stop the pump from working.
I dont see the need for it in the pump that you can build from my little ebook. some people will also insert a pool noodle or a bike tube to keep the pressure tank from filling with water.
If you go with my design then you can simply unscrew the tank and push a bike tube in then fill it with air and that will do the same as the snifter valve if you ever find that the tank fills
with water.
I hope that this helps answer your question. Go build one! they are lots of fun.
9. I have read in your notes about the delivery pipe being the same size of the 1st check valve being very important. My problems is i have a huge pile of 2.5 used pipe laying around I wanted to use
as my delivery pipe..My delivery pipe is going to be almost 250 foot long (going with a stand pipe) But if i reduced my drive pipe down from 2.5 to 1.25 for the 1st check valve, maybe 10′ above
the pump, think it would work? Want to build this soon, have a stocked fish pond and the fish are growing awesome, but the water level isn’t. Or should I stick to the rules..I saw some 2.5 check
valves on the interweb, and maybe go with a 4″ pressure vessel? Any advice be greatly appreciated!
1. Thanks for the awesome question!
I should say that the important thing about the valve and drive pipe is that you cannot use a large check valve and small drive pipe. But in your case you are using a large drive pipe and
wanting to use a small check valve. That should be just fine.
I would place your stand pipe someplace between the pump and the 75′ mark. just make sure that you have at least 1′ of pipe above the water level of the source. Something that I did not
mention in my ebook was the ratios of using a stand pipe. You really should have a 3,2,1 ratio. so the drive pipe from the source should be the largest size then the stand pipe should be the
middle size and lastly the drive pipe from the stand pipe to the pump should be the smaller of the three. so if your main drive pipe is 2.5″ then you would use a 2″ stand pipe and then go
down to the 1.25 at the end. this ensures that no loss occurs in the system due to pipe size.
For the pressure vessel, if you are using a 1.25 valve, I would go with a 3″ pipe and make sure that it is at least 3′ tall.
From what I have found the ram pump is robust and works even under less than ideal conditions so you should be just fine in your setup with a large drive pipe and smaller check valves. Those
check valves are expensive and I see why you want to use the smaller one.
Hope this helps. Let me know about your progress.
10. Hi Seth, Great detailed videos, Followed your plans to build a rampump, just set it up yesterday in fact. Am fortunate enough to have a good flow of water from stream with plenty of fall!, I used
a different check valve to the one you used,it not working too good .
Only starts pumping when I keep my finger on it ! ,and without applied pressure to it, pumps way too slow for the amount of water coming into it. Question being, Is the check valve the problem do
you think! ?
Cheers Brian.
1. I am happy to hear you have a pump made! When you get them working they are amazing.
Is your first check valve held open by gravity?
Its been my experience that a slow pump is caused by a few things. the drive pipe is to long or the check valves are to small to name a couple.
If your drive pipe is more than 100′ your pump will be slow. Add a stand pipe closer to the pump. If your drive pipe is bigger than your check valves it will cause problems.
If the pump won’t start it could be air in the drive pipe or a water logged pressure tank or you don’t have enough back pressure on the delivery pipe to keep your output open fully.
I hope these are enough to get you started. Let me know.
11. Seth,
How would I use this ram pump on my water well?
The well is at the bottom of the hill and I want to pump the well water up the hill to my house.
Thank you for your help.
I am wanting to purchase your ram pump, but I need to know how to apply it to my well.
1. Hi Susan. Thank you for checking out my pumps!
The Ram Pump works by way of a pressure wave and has to have falling water. Does water pour out of the well like a spring? If that is the case then you can use a ram pump. If you have to use
a well pump to get the water out of the well then you are better off pushing the water up the hill with that pump. The Ram Pump will have lots of loss and you will be using more water than
you should. Basically the ram pump works best in a creek or pond.
I hope this answers your question. 🙂
12. Hi- would i be able to use your Ram Pump to pump water from the ocean in a marina or harbor area?
1. Hello. The ram pump works by water falling into the pump and creating a pressure wave. If you have water falling you can use the pump in salt water for sure. The ram pump won’t work on flat
13. Hello Seth, Thank you for delivering all this information. I want to ask is it possible or will it work if i attach my feed pipe to a tank in my attic or what happens when the tank fills up. I
would have a ball cock to stop it overflowing, but what about the ongoing supply from ram. I would appreciate your reply Eileen
1. Hello Eileen. Thank you for checking out my site!
The output of a ram pump is fairly consistent but the pressure is going to be low. You can hold your finger over the delivery pipe of a ram pump and stop the water most of the time. The pump
itself will continue to operate even if the valve on the delivery pipe has been closed. I say this to show that you can use a ballcock valve to stop the flow of the ram pump delivery pipe
when the tank is full and it will not effect the pump. So when the tank is lowered the delivery pipe will be opened again and allow water to flow.
It is completely possible to do this just remember that the pump will pump 24/7 so make sure that the ballcock valve is working or you will flood the house.
Thank you for the good question!
14. Hi Seth,
I have been watching your videos and think this application will work for me. If I understand head, it is the distance the water falls to the pump. I could conceivably have 20′ of fall to the
pump. the drawback is it will have to pump uphill that 20′ plus 50′ more to where I placed the storage tank, approximately 350′ distance. I measured my flow at around 4.5 gallons per minute at
its slowest time of year.
I do not have a clue what size pump I will need given these parameters. I saw where you put some check valves in the output side at the point in elevation where water stopped flowing.
I have a culvert that runs under the drive and thought about building a collection area to force all the water into the drive pipe or running a 4″ pipe and reducing it just before it enters the
pump. Any ideas will help and let me know which pump to order.
1. Hello. Thank you for watching my videos. You are right! Feet of head is the amount of water fall into the pump from the source. Your setup is very plausible. In my testing I have achieved 70′
vertical lift with only 12′ of head. If you have 20′ available then you will be able to pump to that hight. The horizontal distance of 350′ will not make a difference. (Some pumps have gone
over a mile) The water flow that you have is what will determine the size pump you will need. The 3/4″ ram pump needs about 3gpm to run at full but I have managed to get it to work on 1gpm.
The ratio of input to output is the same on all the pumps the difference is the flow rate. smaller pump = less flow. Using the one way check valves inline with your delivery will make a big
difference in vertical potential. So if the flow is small you can add the valves. 🙂
The intake is a little tricky because creeks often have a lot of silt in them that will stop up the pipe. So if you can make a little pond as the intake or use a 4″ pipe as you say to collect
water it will be a lot better as far as maintenance. I hope this answers your questions well enough. Please ask more if you need to. …… You would need the 3/4″ pump.
1. Thank you for your reply. I also noticed that you use 1.25″ galvanized nipples in the construction. Why not 1.25 schedule 40 PVC? Seems like that would be more cost effective.
I have to share this story. Yesterday when I was walking down through the woods to get to where I was going to measure water flow, I heard a slight noise. I usually play close attention
while walking but this time, I was more concerned about spider webs so I was waving a stick in front of me to break the webs. Well, the noise made me look and, less than four feet from
me, there on the ground was a copper head coiled and ready to strike. I gently backed up. It pays to watch everything. This could have put everything on the back burner.
1. Oh you are watching my old video on how to make the Ram pump. The models I sell on the site are all pvc other than the brass check valve. You are right. The pvc is the way to go.
Gosh that would be a day spoiler! I have a story like that. I was in Mississippi walking around a lake in the woods and almost stepped on a copper head. Those things are so bad.
15. Hello Seth Johnson.i need ur help.i tried to built a 1 inch ram pump at home.the drive pipe is 10 m long with same dimensions as the swing check valve.when i opened the inlet valve the first
swing check valve closes.the pressure is good enough.i used a water bottle for the pressure chamber.no doubt water gets up in the water bottle but the hammering effect doesnot occur.Plz help…
1. hello. I am happy to hear that you have made a ram pump for yourself. Normally when the first check valve snaps closed and won’t start the hammer effect, it is because there is air in the
drive pipe. Make sure the pipe is 100% free of air. just hold the valve open until it stops bubbling. If that does not fix the problem sometimes if a rock is in the second check valve,
holding it open, the pump will not start.
try these things and let me know if it works. 🙂
1. Thanks Johnson for the help.I worked on ur advice and made the drive pipe free of air and it worked beautifully.Tried for 5 times before but the pump was not working.Really grateful to u.
Will the pump perform better if the first swing check valve is larger in dimension than the second check valve? Does the height of pressure chamber effects the performance of pumping?What
about those pumps where the second swing check valve is perpendicular to the first or in line with the pressure chamber.which will pump better?
1. Oh good! That is normally what the problem is. It can be annoying to get all the air out of the pipe but it has to happen.
It has been my understanding that the two check valves need to be the same size but I have not tested making the second smaller. that is an interesting idea though. The hight of the
pressure chamber does not make a difference but the volume of the tank does. if the tank is too small the potential is reduced. If the tank is to large it just takes a while to get
the pressure up.
I have seen the pumps that have the vertical second valve and I dont like them. It just seems that the inline pumps work better. I like the idea of having the pressure wave inline
with the input and output and then having the pressure tank on the top.
Happy your setup is working!
16. Seth,
Great article on ram pumps! Thanks. I built a linear pump a few years ago and tested it with water supply from the house. We’ve had a fairly long drought here in Central Texas, so our creek has
only been running in short cycles until the water table builds back up. My ram pump has 1 1/4″ inlet and 3/4″ drive pipes. I have found lots of information and tables on
flow/drive/lift, but none on how far (linear distance) the pump will transfer water. I read somewhere that it is better to use a rigid pipe on the delivery side versus a garden hose.
I was planning to use garden hose to start as I can cobble together a bunch around the ranch for a test.
How far do you think a ram pump will push water? Would there be much difference between your 300-400 ft and approx.1,000 ft due to friction losses?
My goal is to pump water from the creek bed approx 35′ in elevation over a distance of about 1200 to 1500 feet to a stock pond. If I have to, I can explore a solar transfer pump to move the water
whatever the last part of the distance will be that the ram pump can’t make. It would be awesome if the ram pump will make the distance!
Thank you,
1. Hello
Thank you for checking out my stuff. My test of 300+ feet resulted in No loss basically. I have heard of people shipping water over 1 mile from a ram pump. These pumps are so forgiving that I
think you will have no problem with your distance. One thing that you might do is place an inline one way valve in the delivery pipe close to the pump. This will reduce the pressure of the
water on the pump itself.
I have another test where I use 350′ of garden hose with a 70′ lift and I ran out of hose before I was out of water. I am sure that using a ridged pipe for both drive and delivery pipes would
be the most ideal but you will be just fine with a garden hose. especially if you already have them.
17. hello seth … after watching a several video about your hydraulic ram pump i think i should have one!! .. and now im already make my own hydraulic ram pump and the size is 3/4 … my problem is the
check valve sometimes operate not constantly and it will operate with low velocity of water only … for your information i only test my pump with the water coming from garden tap only …. …. what
should i do ?? …
what is the recommended size of pressure tank for my 3/4 hydraulic ram pump ?? …
can i use the same type for both check valve ?? …. for example, can i use spring check valve only for both check valve or i need to use 1 swing check valve and 1 spring check valve??
do you have any suggestion if i don’t want to waste the water that coming from waste valve??? … my plan is to attached the garden hose at the top of waste valve and supply the water to the
delivery pipe??… does it possible??..
glad to hear back from you!! … thank you…
1. Hello
Thank you for watching my videos.
The ram pump needs to operate by a pressure wave so connecting to a tap actually will prevent the pressure wave from working correctly. If you connect the pump to a bucket or install it in a
creek then you can have the pressure wave.
For the 3/4″ pump size I would use a 2″ pipe that is 15″ tall.
The first check valve in my design needs to have a swing valve. It much hang open by gravity. Now the second one can be a spring valve.
The pump is going to have a 50 to 90% loss out of the waste valve. You can use that water for all kinds of things. Typically my pumps are installed into creeks so I just let the water run
back in the creek. If you connect the waste valve to the delivery pipe the pump will stop because the waste valve is low/no pressure and the delivery pipe is high pressure. Now you can use
the waste pipe as a flow down hill.
1. hye seth…. thanks for your comment … im following the every step that u make in your videos … and im facing another problem … please help me.
when i open the ball valve at drive pipe …. my first check valve will slam shut… and i push the valve to get rid of air inside the drive pipe…. and looks like the pump start working and
start to build a presure inside the chamber… my problem is when i open my second ball valve that connected witth delivery pipe ,… my first check valve slam shot and not pumping the water
… what should i do ??
for your information… my chamber is 2 inch and the height is 12 inch ,,, my both check valve is 1 inch and my drive pipe also 1 inch …
1. This is a common issue that you have. when you open the second ball valve you are releasing the pressure in the tank. I suggest opening the valve slowly until water has reached the
end of the delivery pipe. then you can open the valve more to allow more flow out. If you do not have your delivery pipe high enough then the pressure in the tank will quickly shoot
out water from the delivery pipe and the pump will stop.
Basically what I am saying is this: make sure the delivery pipe is uphill from the pump. slowly open the ball valve until the pipe is full of water. then open the valve all the way.
If the pump still stops then you dont have your delivery pipe high enough to support the pressure in the tank. you will have to only open the ball valve half way.
1. hye seth… thanks for your advise… now my pump start working …. but , im already attached a garden hose for my delivery pipe and i put the end of my delivery pipe at my rooftop …
but … whenever i open the second ball valve all the way… the pump will not working ….
is the related to my air chamber … im only use 2 inch diameter pvc pipe and the height is 12 inch ….
fyi… my length for delivery pipe is 10 m …. please help me thank you
2. That is a small pressure tank. I would use one that is at least two feet tall of that size but three inch pipe would be better. When you open the ball valve only a little does the
pump continue to work? If the pump works with the valve open only some but when you open it all the way it stops then the issue is one of two things. 1) The pressure tank is too
small. 2) the pump has too much head falling into the pump and when you open the valve the pressure is all released and the pump stops. you can reduce the feet of head entering
the pump or you can just open that second valve partially.
If you open the valve only some does the water reach the roof? It should reach but be a slow flow.
3. hye seth… thanks for replying my message… im very appreciate that..
the pump continue to work when i open a second ball valve a little… the water does reach the roof but in a slow flow… my head falling into the pump is about 2 feet and the drive
pipe is about 6 feet. for this im only test my pump using a water storage tank as a source of water.
4. OK so you are getting results. That 2′ of head is enough to get you between 14 and 18′ of lift so you are likely opening the ball valve and a surge of water flows out then the
pressure tank is empty and the pump stops. You can add hight to the delivery and that is good news. The bad part is that you cant open the ball valve all the way without losing
this pressure.
Also 6′ is a very short drive pipe. I typically dont go less than 15′.
5. hye seth… thank for replying …. now im going to fix the problem that i had… im going to change the size of air chamber as u recommended… hope i can get the better result with that
im watch your videos about attached a one way check valve at the delivery pipe to increase the efficiency of the pump …. how does it works?? … does it will increase the efficiency
of pumping the water??? … im really into it … glad if u can share some information with me about that … thanks sir …
6. The inline valve in the delivery pipe will reduce the weight of the water resting on the pump. So by adding the valve in the delivery pipe you will allow the pump to work with
less effort. Your setup has no issues with weight on the pump because your delivery pipe is not very high in the air. If you try to pump another 10-15′ you would benefit from the
valve in the delivery pipe but for your current 10′ hight on your roof you will not see any change.
18. Hi Jhonson,i have followed your instructions and assembled one.I got good results.I’m thinking of providing another set of 1 1/4 swing check valves ahead of the actual set as in your video.Is it
possible to increase more pressure by providing what i said? Or are there any modifications to increase pressure to pump more height,i mean to increase efficiency of ram pump.
1. Hello. I am happy that you have made a Ram Pump! Yes you can add an inline check valve to the delivery pipe to increase the potential of the pump. I have had the best success with inline
spring valves.
To increase the efficiency of the pump you can make sure the Drive pipe is ridged (does not have flex such as Steel or hard PVC). In my videos I use a black flex pipe and this is not ideal.
Also you can make sure that the pump is standing upright and the first check valve is also upright. Tilting the first check valve will reduce the efficiency.
19. i want to a file on hydraulic ram pump. can u mail??
1. I dont understand your question. Are you referring to the PDF download of the Free Ram Pump book?
20. Ho Seth. Is there a way to increase the delivery height with your design by increasing the diameter of the pump fittings, drive pipe, etc? I have very limited head of water available, but there
is plenty of flow. I would need to increase the 1:7 ratio of head:delivery height, substantially.
1. Hello
There are a few things that can improve the ratio. First is install a check valve in the delivery pipe a little way from the pump. This will relieve the pressure on the pump and allow it to
work better. Also you can use ridged pipe for the drive pipe. Still you will not get much better than 1:9.
21. Hello Seth Johnson,
I Was Hoping You Could Help Me With Some Information please Sir. I Have a Waterfall Producing 30-50 gpm and Flows Almost 150 Ft Downhill Into A Creek. The Proposed House Site Sits 75ft aprox.
Above The Pool base Of the Waterfall and 400ft Away. So Im thinking I Have 100ft Or Better Of Fall(head) From Pool To Creek. Is It Possible With Your Largest Pump To Get Water 150-200 ft If It
Runs The Side Of Thr Cliff Horizontal And Vertical Gradually,and 400ft Away?
1. Hello and thank you for checking out my stuff. That is a nice waterfall. So the horizontal distance of 400 ft is no problem at all. I have pushed water 500 ft with no change in flow. The
vertical is what will get you. 70 ft is the highest I have tested my large pump but some of my customers have reported over 100 ft. One issue that you will run into with my design is the
limitations of the brass swing valve. It is light weight and all that pressure can keep it closed if you add to much head. That said you should be able to get 100 ft from the pump just fine.
You will only need 15 to 20 ft of head to get a good flow at the top. If you want the water at 75 ft you will be really good but I assume you want a little pressure fall into the house spot?
2. Hi Seth
I have not built any pumps, but with a drop like that I would imagine that a
“pulse pump” is a way better deal. It has no moving parts whatsoever so there is nothing to wear out.
And there is also not as much shock waves going on.
The theory is simple.
-You have a large u shaped pipe with the outlet a fair bit lower than the inlet for a good flow rate.
-At the inlet you introduce a whole heap of air straws that angle down inside the pipe. These will suck air because the water flowing down over them creates a vacuum.
– The air never gets to the other side of the u however because you put a tank way down at the very bottom of the u which catches the air as it tries to go past, allowing the air and water to
separate. Only most of the the water at the bottom of this tank continues up the other side. The trick is that the water head of the lighter water-and-air mix going down will exceed the water
head of the denser plain water coming up the other side, so the u must be a fair bit lower on the exit side to account for the higher density.
– Now for the pipe that you will be interested in. You add a third small pipe that comes from near the top of the tank that you installed. This pipe will gulp all of the air and a little bit
of the water when all the air has been gulped. The density of this mix, having ALL the air, but only some of the water is the least dense of all. It stands to reason, that the height that
this will flow to is much greater than any of the pipes so far, as it will take much more height to create the same pressure head. There you have it. No valves required and nothing likely to
wear out in a hurry! This pump will happily pulse away delivering air, water, air, water…
The same principle can be used to make an air compressor, if you take less fluid off with the third pipe and allow the air to fill the chamber so that no water goes up your third pipe. In
that case you add a fourth pipe lower down in the water, but not as low as the main large u pipe. Every now and then (rarely if you set a valve installed on the third pipe right), the fourth
pipe will hit air and vent a bit of air and water thus keeping the air level right. The reason for having the fourth pipe is so that air never goes up the main U. This ensures that the
densities stay as you planned. If the density was to drop in the U pipe, then the air pressure might drop too.
1. Just as another note on the pulser pump (sorry I had the name wrong – it is not a pulse pump but a pulser pump)
There is a good link that will give you a good start if I am allowed to put links
It is http://www.appropedia.org/Pulser_pump
I really think that given a very large drop it is a better way to go.
1. I have actually seen one of these on a news story! It was used as a filter pump for a river that had lots of pollution. Because it has such potential to work without moving parts it
is good for a long long time. My brother in law was talking about such a pump used in very large fish setups because they can be used to get air into the water. He did say that the
pipes must be very carfully placed and sized or the system wont work. Someday I should try to make one. .. Thank you so much for sharing!
22. Hi
Seth, I’ve been trying to test the unit out but it seems the check valve doesn’t close or doesn’t open. I’ve being testing it direct from a hose line. My trouble is priming the unit. If there is
any slight leaks in the pipe thread connections will that be ta problem. And my Pvc one way check valve I am using are they pressure rated ?
1. Hello
The pump actually wont work on a garden hose. This is because the pump creates a pressure wave that travels up the drive pipe and then back down to the pump. With the 60psi of a garden hose
you are unable to get this pressure wave.
Make sure the drive pipe inner diameter is the same size as the the brass check valve. This is a must if you are to get the pump to work.
Small leaks will cause the pump to stop working over time but should not stop the pump from priming and starting for a while.
23. Visited your site, and I believe I watched all of your videos about the pump; as well as others before installing our own. Thank you for publishing all of this great info. We finally got to
install our pump and worked great. Very little pressure at the top of the hill but we filled a tank and used that. It worked for us for about two months? We only had to reprime once or twice. Not
Then the other day the tank started getting lower and lower. Checked on the pump. It was still going. We checked the black pex pipe for delivery to ensure it was still conmected. It was no holes
no problem. Checked for a clog in the lines. We think there may have been one but unsure. Either way when we pour water in at the top of the hill it makes it to the bottom. Found out after that
that the delivery had slightly come disconnected from the pump. ‘re connected ‘re primed and started the pump again. Still nothing. What are we missing. Beside the fact that a Hunter or other
tresspasser lodged a large stick into the check valve at some point surfing our scratching out heads. It appears the pump and valves are all fine and in tact.
What we have:
20 foot long 1 1/2 pvc drive and check
300 ft 3/4 in delivery pex
15 to 25 ft lift (I say closer to 15 hubby says 25 or more. We didn’t measure)
Guessing 4 -6 foot drop.
25-28 gpm (don’t remember exactly)
We do not have a pressure gauge.
Thanks for your help. We sure don’t know what else to check.
1. Hello and thank you for watching my videos. It sounds like you have air in the pressure tank. The pump will still work or look like it is working but in fact there is not enough pressure in
the tank to get the water up the hill. What I do on my pumps is turn off the delivery pipe and drive pipe. Then disconnect the delivery pipe to let the water in the tank flow out. Some pumps
have a snifter valve that allows a small gulp of air with every action of the pump but the design that I use only has an inner tube in the tank.
SO the first thing that I would try is purge the tank of water and then start it again and see if your lift returns. If that does not fix the issue we will explore other options. Hope this
1. Hi Seth
Great site, information and attentiveness to it. Thank you.
I have a problem that I think is similar to Mary’s: We inherited a pump (commercial but cannot find another site as informative as yours) and have had it running for about 13 years now.
However I’ve been battling with it over the last 2 years – complete loss of pump height (delivery pressure) with pump still going strong! Disconnecting it all and letting the water flow
out solved it for a while but now it cannot get the pressure up at all, and it stops after an hour or two anyway. When I try and restart the pump air gurgles out of the waste valve. No
obvious leaks in the drive pipe, delivery pipe or connections. So I think air is stopping it but cannot find a leak. Given your experience what is most likely the cause (and as per your
above answer shouldn’t there be air in the pressure tank)? Any help greatly appreciated.
1. Hi
Thank you! I have enjoyed getting to use these great pumps.
Since you have a commercial pump I am going to assume it has a snifter valve to gulp air with each cycle. First I would make sure that valve spits water every time the pump clicks.
(if it has a snifter) This will make sure the tank is getting air.
Since the pump is stopping after only a short time I would say this is a small bit of air in the drive pipe. Often times you can hear a small slosh sound in the pipe. Another common
reason for the pump to stop after a while is a leak in the delivery pipe. This might also be why you have lost pressure.
SO… Yes the pressure tank must have air in it. The drive pipe must be air free. also the delivery pipe must not have a leak in it. make sure the snifter valve is working if there is
Those are the main points to look at first. I do hope that helps. If not please email me a picture of your pump.
24. Pingback: Free Water Pumping | JackCollier7
25. Thanks Seth
No snifter valve. The new agent is determined I put one in but since it’s been working in the same place without one for over 60 years I know that is not the answer! I’ll recheck all the pipes
and connections this weekend and let you know how I go.
Thanks again
1. Do let me know what you find. To work That long and just now have an issue does it have a valve that has warn out?
26. Hi Seth
Yes, I think you are probably right – I have not been able to find leaks in or around the drive or delivery pipes. The ram (Billabong/ Danks hydraulic ram) is old-style with a cast iron dome (air
chamber) within which is a spring-loaded delivery valve with a valve disc and rubber. I think the rubber may finally have gone. Another possibility is a worn gasket where the dome is bolted onto
the “body” of the pump. There may be a tiny leak here but somehow I think it’s more than that. The waste valve is new. I’ve been reluctant to undo these 60 to 80 year old bolts for fear of things
falling apart but the time has come. Air is both leaking from the air chamber (pump works for a short while after I pump air back into the chamber), air is coming out of the waste valve stopping
the pump, and I can no longer get the pressure required to pump it up to the tank. I need to open the dome.
1. It sounds like you are on the right track. When you get those bolts of you might have to replace them sadly. If they have rusted out over that long a time. I have made gaskets from rubber
roof lining. I was able to get it from my local hardware store but you might be able to get some better gasket materiel. I have also seen a delivery valve like you say and I know they can
wear out too. BUT to be honest you have gotten a long long use out of that pump so dont be upset about replacing some parts. (except having to use less quality modern parts)
27. Seth, really helpful info on your site. I’ve built a ram pump and had it working well for a year or more but starting to get problems with it stopping after a while so am considering a few mods.
I have mine connected to a tank with float valve and was concerned that the float valve may have damaged the pump by shutting off the delivery while the pump was still running, but your
experience suggests not. I don’t think the non return is the issue as I’ve installed a commercial grade Crane duo check non-slam wafer valve which is supposed to be bullet proof, so my only other
potential issue is the tractor bladder in the pressure tank which may have lost pressure. I have a gauge on the tank and can get it up to 20psi but it doesn’t want to go any higher even if i keep
the delivery shut whereas before I could get it up to 40-50 before opening the delivery valve.
I have variable flow in the creek throughout the year but very little head of 1.5m. Being right on the lower limit with head I’ve compensated by making a small weir in the creek to get another
30cm head and use a 4″ steel drive pipe 36m long with commercial couplings (same stuff used for fire systems in multi storey buildings). I have a 4″ brass ‘flapper’ valve like the one you use but
mine is weighted down with steel washers and a nut to slow down stroke so its builds max speed before closing. I push through the crane 4″ non return into a 2″ T with the back end connected to a
60 litre gas cylinder with tractor bladder and the other side of the T connected to a reducing nipple connected to a 1″ delivery pipe. My setup is buried in a small creek bed and we get floods
each wet season. My design is a little different to yours and I’ve got the pressure tank at the back laying flat – this allows rocks and debris coming down the river to pass over the pump without
damage. I note your design has the pressure tank upright in the middle with the delivery pipe at the rear. I have two questions
1. Have you tried or considered laying the pressure tank horizontal with the tank, or even installing an elbow so it is parallel to the pump? This would minimise potential damage compared to the
“T” shape of your pump – especially if this is submerged
2. Have you tested the pump with the pressure tank at the rear (ie swap the position of the delivery pipe and pressure tank – but still have both behind a non return valve. I would think that
your system may be more efficient due to the linear flow, whereas the water flowing in my pump has to flow into the pressure vessel at the end, stop, then reverse direction until it hits the non
return and finally turn 90 degrees to push down the delivery pipe.
I’d be really interested to hear if you’ve experimented with swapping the position of the delivery and pressure tank, or played with orientation of the pressure tank, and if this has any impact
on flow rates or head. I’d try to swap my pump around but only the flapper is visible and I’ll need an excavator to uncover the rest of it!
1. A four inch pump is a big one!
Stopping the pump by the delivery pipe is actually of no consequence. At times when the pump comes back on it can surge and lose pressure but not if you have enough back pressure. So you
should be just fine there.
When I have tested the pressure tank on its side I have noticed that it fills with water and only the parts with the bladder are working as a pressure tank. But when the tank is upright it
has way more air in it and that allows for much better psi. It sounds to me that you have a bladder that has gotten old and sprung a leak. Since your tank is horizontal it has reduced the
available air space in the tank and thus the psi has gone way down.
I have seen those designs that use the delivery pipe out the back side of the pressure tank but I have not used them before. My thought process was the same as you mentioned. I figured the
the water would flow best in a direct shot approach. When I was first doing my research I looked at several designs and I like the one I build the best because it does allow the water to pass
right through the pump but under pressure.
Your tank is under a lot of rocks and stuff it sounds like but that is the first thing that I would try. Because every other component in the pump seems to be of good quality.
28. Hi Seth
Some follow-up – took the dome off to inspect the delivery valve – yep, the rubber’s badly worn. (The nuts though are brass – in perfect condition.) Also, much of the cast iron inside was layered
with clumps of rust; the delivery outlet rusted down to the size of a pin-hole. So right now it’s at the sandblasters (after I chipped out as much rust as I could) while I await some new parts
(rubber, gasket, spring).
Can now confirm that the pump is at least 70 years old – and I doubt if it was ever opened before.
Will let you know when it’s all back together.
(PS Do you know what pressure ranges are generated in the drive pipe?)
1. Hello
Brass nuts and bolts! thats awesome!
Considering that thing is 70 years old I am still impressed with the operation.
All of those things would effect the operation of the pump. I expect you to get better results than ever before once its clean and in good repair. Luckily there are not many things to change
on these pumps so you should be able to get it done rather quickly.
In my tests I have found a few values of psi. But I am not sure what the value gets up to when the pressure wave flows. There are a couple numbers you can work with to get the values.
For every 1ft of head you have .433psi. So if you have 7ft of head you have a little over 3psi at the pump in the drive pipe.
Now the pump works on a ratio of close to 1:7. I have found this to be true with my tests. For example if I have a pressure in the tank at 20psi and I have a hill that is 10ft tall. At the
top of the hill I place a psi gauge on the delivery and get 15psi.
20psi/.433= 46ft potential lift. If I subtract the 10 ft from the 46ft for the hill I have 36ft potential. So 36*.433=~15psi.
I know that was some cray jumbled math but basically you can expect to get psi in the drive pipe of greater than .433*ft. because of the pressure wave.
29. Hi Seth, At the driest time of year, my spring only delivers 1/2 gal / min. Is that too low a rate to power a ram pump? Would it help if I built one out of 1/2″ pvc rather than 3/4″? Also, am I
understanding correctly that my drive pipe should be the same inside diameter as the components of the ram? As you have said, the delivery pipe should be approx half of the diam of the drive
pipe. If I built a 1/2″ ram, I don’t believe there is a commonly available pipe smaller than 1/2″ to make the delivery pipe out of. Would it be ok to make both drive & delivery pipes 1/2″?
Thanks, Pete
1. Hello. You can make a 1/2″ ram pump just fine! The delivery can be 1/2″ just as the drive pipe. You do need to keep the drive pipe the same as the two check valves this is rather important.
In my testing the 1/2″ pump will use .5 to .8 gpm. So you might need to build a catchment tube to pull water from. I suggest the spring catchment from carolina water tank. Or make your own
with a uniseal. I hope this helps. You can look at my 1/2″ pumps for sale to get an idea of what the pump looks like.
30. Hi seth..i hope u can answer me with my given calculations..i have a 20 feet of head..with a required vertical lift of 100 feet..the delivery distance is about 150 feet..the climb to the hill is
approximately 130 degrees..is there a possibility to bring the water up with the use of a ram pump?..thanks seth in advance..im planning to build one..
1. Hello
Yes that is a good setup. With your flow rate I would suggest building a 1″ or 3/4″ pump to make sure you don’t have low flow issues. You will only need around 16 to 17 feet of head. It might
be helpful to add an inline check valve to the delivery pipe.
31. and the flow rate from the drive pipe will be 1 liter per 3 seconds..
1. What is the purpose of the in line check valve on the delivery pipe? is it to prevent the heavy back flow of water to the pump? what could happen if there is none? thanks seth for replying..i
want to buy your pump but it would cost much..i am from the philippines..
32. Hi seth! Does the size of the pressure chamber affect the output amount? And is there any computations on how the pressure in the chamber pushes the water? Thanks!
1. The tank size does effect the output. If the tank is too small the full potential won’t be reached. If the tank is too large it will take a while for it to reach pressure. But it won’t gain a
higher pressure than the pump can supply. As far as the math behind the flow I don’t have those details. 🙂
1. Hi seth..I already built my ram pump..i followed your advice on building a 1inch ram..however my pressure tank is 4inch for 3 feet long..is that too big?what could could be the best size
and length of pressure tank for a 1inch ram pump?..
Next is,we have made it working and pumping..the water got higher and higher..and when it reached to a quite higher point..the waste valve gets locked..and I forgot to put an inner tube
inside the pressure tank..is it water logged? i suspect it is the back pressure of the water on the delivery pipe..
thank you so much seth in advance for entertaining our questions..
1. Hello
The 1″ pumps that I sell have a tank size of 3″ x approximately 22″ tall. This seems to be an ideal size for that ram pump. You can use the larger tank size it just takes a little
longer to prime the system.
As long as your system is upright (not leaning to the side) you should be able to get a few months of use without the tank getting waterlogged. Can you give me the numbers of your
system? How many feet of head and how high you are lifting water? It does sound like you have an issue with back pressure. If the water pressure out the delivery pipe gets low because
to much water has gone out then the check valve will close.
1. Thanks seth..i really appreciate your responses..
I have made some changes on the location of the pump.. guessing 20-22 feet of head..25 feet long drive pipe..not steel or pvc..just a pex pipe same as yours..vertical height lift
is about 100-120 feet..and delivery distance is more than 120 meters..
I did some back reading on some questions here..and i think what closes the waste valve could also be the too much water pressure from the drive pipe?because of the light swing
check valve?and when the waste valve starts to click its really fast..
2. Pex is fine as long as the inner diameter is the same size as the valves. I think the black pipe I use is called “flex pipe”.
You can get water up that high with that feet of head. Should not have an issue.
Yes if you have too much head pressure and too low pressure on the delivery end the valve will close and stop. You can slightly close the drive pipe ball valve or reduce the head
pressure. Or you can add weight to the waste valve.
33. Ok seth..I’ll explore more options with your suggestions this week end..Thanks much..:)
1. Hi seth..just an update..we tested our pump yesterday with a pressure gauge..we only had 22 feet of drive..feet of head is just 6.5 to 7 feet..and we were able to pump close to 50 feet
up..almost halfway to the top..and we only had 19 psi per cycle..our next plan is extend the drive pipe down to a total of about 60 feet to gain at least 15 feet of head and i hope the psi
will increase..that is actually our next plan..are we on the right track?..i am really much excited on our next plan..
34. Seth
We have a problem with our ram pump brass check valve closest to the drive line. After it initially closes it does not re-open and cycle like it should. The check valve is not sticking or
defective. Our pump is the same as your design except we use an in-line spring check valve instead of a brass swing check valve for the valve located closest to the pressure tank. If I manually
open the swing valve repeatedly, the pump builds pressure and pumps water out our delivery line at a 10′ height.
We think our problem is that the pressure wave dissipates in the current drive line arrangement. We are using an existing 4″ PVC pipe from the nearby stream with an 7′ head as our drive line. It
runs about 120 ft. underground from our stream and has no standpipe. We step this 4″ line down to a 2″ ball valve, then again through a reducer to the 1.25″ input line to the pump.
Our proposed solution is to place a 2″ standpipe of sufficient height after our 2″ ball valve, then a 2″ to 1.25″ reducer and then run a 20′ section of 1.25″ PVC to the pump. We’ll add a ball
valve at the pump to control flow from the drive line. We would like your view on whether our proposed solution will work before we buy more stuff.
1. Hello. Nice to hear you are working with ram pumps!
I have two thoughts for you. First: yes reduce that 4″ to 1.25″ after a stand pipe. What happens with such a large drive pipe is there is too much water pressure for the pressure wave to
counteract when the check valve closes. A 1.25″ wave gets lost in a 4″ pipe. Second thought that will cause the valve to close and not reopen : 7 feet of head gives you the potential for
around 50 feet of lift. With only 10 feet of lift in your system the pump will depressurize the tank and the pump will stop. I recommend reducing the head pressure or increasing delivery
hight or closing the delivery valve half way.
I hope this helps
1. Seth
Thanks for your response and the additional thoughts about managing the inflow and outflow head and flow rate. We’ll implement this in a few weeks and let you know how it turns out.
All the best
1. You are welcome. Let me know how things go and if you have any more issues feel free to ask.
35. can i use ram pump for my aquaponic system??? also can i submerge the pump in the pond, if so wil the water flow into the pump????(this is because i dont want to lose the water that is coming out
from the swing valve)
pls do reply
1. The water loss is an issue because you will lose anywhere from 50 to 90% of the water from the waste check valve. The pump will work underwater as long as the source is above the location of
the pump in other words you cannot use a ram pump in Stillwater. The ram pump is not effective in aquaponics because you lose the water quickly.
1. hi seth..can I ask some suggestions?..my ram pump works well producing 50 psi..but it has not yet reached my storage tank..it needs to climb up 30 feet up more..however if I add the
delivery hose up, the water still climbs but when reached to the added height, the check valve closes..can u give me an idea on how to add weight on the swing check valve?..bcause i think
the waste valve needs more weight to open and continue pumping..
1. I have not used weight on the valve yet in my tests. I was thinking you could use a drill to make a hole in the flap then add a couple washers on a nut and bolt to add weight. Seems
like it would work but I have not tried it.
36. Hello ,
I live in Palau and I have actually installed the first ever ram pump here. Watching your videos has been most helpful in getting started. We are currently in a severe drought and we are so lucky
to have installed this pump just in time to save our little farm. There is a steady flow of water in a waterfall near by no matter what the weather is like.
I have made this pump myself since the one I got on ebay broke in just a few days as it was pvc and apparently way too weak for the water flow we have.
Now I have one out of steel parts and a pvc air tank with an innertube inside it. Its 4 inches pvc and 4 feet tall. the drive pipe is 1 inch steel and the fall is about 20 feet ( the waterfall is
taller but I tried to keep the pipe short , about 80 feet long). the waterfall is a cascade of falls that stretches over 150 feet at least so there is potential to make the fall larger if needed.
The delivery line is 1/2 inch steel for 100 feet then the rest is pvc and a total length of 1100 feet up a hill to a height of 150 feet above the pump elevation.
The question I have is this. the flow is rather high, about 5 gallons every 10 to 15 seconds at the drive pipe, yet I barely get a trickle at the tank. I noticed recently that even small dips in
the delivery pipe near the tanks cause the water to stop flowing altogether, meanwhile the pump operates just fine and there is flow at a lower elevation where I installed a faucet to discharge
water to a pond in case we will not need the water at the top tanks. The psi gauge stays at about 100psi so I think there is enough pressure there at the pump…
I am wondering if the issue is sizes of check valves I used ( they are 1.5 inches as they didn’t have 1 inch ones in stores here…
Or could it be the placement of the delivery pipe which is laying down on the slope and in places it dips and goes up and down a few times before it goes up the way steeper slope near the farm.
Could trapped air in the delivery pipe be the problem more than anything? if so should I try to relocate the pipe so that it is more in an upward sloping position at all times or should the
pressure be enough to clear all air bubbles out of it? I also wanna install a check valve in the delivery line since reading a lot about it in this forum. maybe this will help?
I heard there are valves to bleed the air out of the high points but is that reliable? please advise as to the best solution and if there is any way you can tell me the approximate flow I should
expect at the tanks with my setup.
Thanks so much!
1. Its awesome that you have a ram pump! They can be so very helpful especially in a drought.
If you were working with a lower head pressure i would say the first thing to consider is getting check valves that are the same size as the drive pipe or getting a drive pipe that is the
same size as the check valves. This difference in size makes the pump much less efficient. Basically the smaller amount of water in the drive pipe has to close that large check valve and
efficiency is lost. Now because you have so much head pressure you are still getting results but if your pump used the same size throughout you would see an increase in pressure. Because you
have the extra hight on the waterfall you might can just increase the head by another 1 foot. If the head pressure is to high it will hold the check valves closed.
As for the drive pipe I would likely start there. The way in witch it is laying is no issue. I would install the check valve about half way up to allow the water weight to be reduced on the
pump itself.
You might also have an issue with the tank being to small for the amount of pressure you are working with. Might try another 1 foot hight and see if it makes a difference. It is also worth
looking to see if the pressure tank needs more than one innertube to fill the void.
It has been my experience that 150 feet from a ram pump is about all you are going to get with the swing check valves. You might be able to change the waste valve to a modified spring valve
and it will have more weight.
38. Seth,
Thank you so much for all your helpful info and taking time to answer everyone’s questions. If you have the time, I’d much appreciate some help too. Here’s my situation.
I have 3 feet 7 inches of fall over 250 feet of stream running through a 2 inch pvc pipe. (shot with professional surveying equipment) Due to the shallow angle we found we had to have most of the
delivery pipe at 2 inches to get a solid 1 inch supply at the bottom. That 250 feet of pipe feeds into a 2 inch stand pipe that is more than 3feet 7 inches high and then a 1 inch poly pipe line.
three feet later, this comes into a 1 inch water ram I built. The delivery line out of the pump is 1/2 inch poly pipe and I need to lift water to 20 feet over about 100 feet of travel to get to
my garden.
My flow to the stand pipe is sufficient to maintain a standing head at 3 feet 7 inches even with the 1 inch pipe wide open. The water runs through the pump when connected with no problems. The
swinging valves clack consistently, but faster than most other ram pumps I’ve seen videos of. Water will flow into the delivery line to a head of about 7-8 feet and then stop. When I fill the
full 100 feet of delivery line with water and then lay it up the hill, it will stay full if the pump is off. As soon as you start opening the valve to the delivery line (very very slowly, and
after letting the pump clack for a while to pressurize) all the water that is in the line back feeds down to the same 7-8 feet of head level. I can not get water to pump higher than that. I think
I should be able to get a lift of at least 25 or so feet from a little over 3.5 feet of fall coming into the pump. Why is the water able to backfeed though the pump as is clacks?
Things we’ve already tried. There are no air bubbles in the drive line. The volume is sufficient to maintain the head in the stand pipe. There are no air leaks in the pressure chamber. (3 feet by
3 inch pressure chamber) There is no dirt in the line, we have a very good screen over the intake. The pump is sitting level and solidly secured to a very heavy chunk of cedar.
Do you have any suggestions on what is wrong? I and my friend who’s helping me have been working on this for a week now and are loosing serious sleep over it. Any ideas would be appreciated!
Thanks for your time,
1. Hello. Thank you for the kind words. It does sound like you have done your research on the ram and you have given this a good run. You should have the head pressure to get the water to a
height of at least 20 feet.
Here is my first thought. You are doing the right thing to have the supply line feed the stand pipe to a height above the 3 foot 7 inch mark. But the way I understand it you have a drive pipe
that is only 3 feet long? Meaning the pipe that comes out of the stand pipe and feeds the pump is only 3 feet long? If this is the case then I would say make that drive pipe at least 15 feet
long. What happens is the valve closes and sends the pressure wave back up the drive pipe but it returns so fast that the system does not have time to recover. So you are getting secondary
waves in the pressure tank and that is keeping the second check valve open to let water out.
So make that drive pipe around 15 to 20 feet long and see if that does not slow the pump down and let the pressure wave have its full effect. If you have extra pipe around then I would even
make the drive pipe 40 or 50 foot. I have not tested the values but it seems as though 50 to 75 foot long is ideal for pressure wave.
Let me know if this helps and we will go from there.
1. Thank you so much for replying! Yes, I believe we had the drive pipe way to short. I didn’t really understand that adding the stand pipe made everything uphill of that not a part of the
drive pipe anymore. We were just trying to get enough volume close to the pump and had been having a problem with that due to our long shallow angle on the creek, so we took the 2 inch
pipe right up to the pump.
We have now replaced the last piece of 2 inch pvc with a 21 foot galvanized 1 inch pipe. And moved the stand pipe to being behind that. And it’s working!
We also pulled the whole pump apart and re put it together. The one check valve may have been sticking as well. The inline one that we could not see working seemed to be sticky when we
had all the pieces apart. There was nothing in it that we could see, but after playing with it for a while it started to move more freely. So when we reassembled the pump, we placed that
valve as the waste valve so we could see it’s operation.
So wether it was a drive pipe issue, or a sticky valve, or a combination of both, I now have a nice little trickle of water flowing to my garden. Thank you again!
1. Any time.
There are a number of things that can cause the pump to work oddly but once you figure them out the ram pump is an awesome tool!
I am happy you have figured out what the problem was and you are now pumping water! If you have other questions just let me know.
Nice Tiny House you have.
1. Aww thanks! I do love my tiny house. And I’m really glad to not need to carry water up the hill to my garden now. Thanks again for your help!
39. hey seth
pleae help i build the ramp 1” inch sorcse and 1/2 ” inch delivery pipe but the presure is not so much to lfow the water up to hight my pressure chamber is of 2” inch is it ideal for that please
help me out for ideal pressure chamber build for this ram pump
1. Hello. In my 1″ pumps I use a 3″ pressure tank ~20″ tall.
What is the head pressure of your source? And what height are you pumping to?
40. hey seth
thanks for the valuable relay my head pressure is 20 feet and have to pump the water up 20 feet please suggest and also for all assembley of 1 inch ram pump
1. 20 feet of head pressure gives you the potential to pump to 140 feet so you should have more than enough to get water to your destination. I would recommend that you only use 4 feet of head
to get the water to 20 feet. There is a free ebook here on the site that you can use to build your 1″ pump. Just reduce the components by 1/4″ The pressure tank I would use 3″ pipe.
41. Hey Seth,
Great website. I really appreciate the effort you have put in here to create a comprehensive guide to ram pumps. By far the best site on the net for hydraulic ram pumps.
I built one this weekend that did not work. The pump cycled, but nothing ever came out the delivery pipe. I’m only dealing with 2 or 3 feet of head and a 100 foot drive pipe with 1 1/4 flex hose.
When I bought the check valves they sold me Y pattern brass swing valves instead of the standard ones (T-shape) you use in your pumps. Looking at them in cross-section I suspect that that the Y
pattern check valve might make the whole system significantly less efficient, especially on the waste valve. What are your thoughts?
1. Thank you for checking out my site!
I have actually never tried the y-shaped valves. It does seem like your assumption would be correct that efficiency would be reduced because of that shape. But first I have to ask what is the
height of your delivery pipe? With 3 feet of head pressure you should be able to get approximately 20 feet of lift. If you are pumping higher than that you won’t see any water come out the
42. Seth,
Kudos to you for allowing this info & free to use!
I have always wanted to build a a Ram Pump , ever since the 80’s.
But, never could get concise info on it.
can NOT wait to assemble the parts!
Steve Mcc
1. I am happy to share the knowledge! If you run into issues with your build send me a message and I will try my best to help.
43. Seth thanks for the great information. I have a unique situation I have a stream approximately 200 feet long that is S shaped And limited incline upstream. I need to push water uphill a disgrace
of 250 ft and approx. vertical 100 ft. Can you suggest a system?
1. I enjoy these pumps a lot. So to get water up to 100 feet you will need approximately 15 feet of head pressure. The horizontal distance is of little concern is that vertical that gets you. To
find out how much head pressure you have connect a few garden hoses together and place it in the creek. Bring the low end up until the water stops flowing and measure from the ground to the
hose. Obviously it will be hard to measure 15′ like this. … unless you are really tall. You can split the hose up into sections and add the previous to the new value.
44. Hi Seth!
Thanks for all the info on your site! It helped me a lot in building my pump, but still it is not yet how it should be. Could you please help me out? Here’s the data:
Drivepipe in PVC 1 inch, 1 bend of 60 degrees, length 19 feet
Head 5 feet, flow about 15 gallons per minute
I used brass flapper valves 1 inch all valves and piping for the pump.
Pressure chamber 2 feet long, 4 inch wide PVC tube
Delivery line 3/4 inch (should finally deliver over about 600 feet long, 35 feet elevation)
I get the pump to run easily, the waste valve running at about once a second. But I only get the water to go up to 10 feet above the pump. .. My guess is that I don’t have enough pressure. Should
I put more length on the drive pipe? Or could I fix it with some sort of weight on the waste valve? Or… do I seem to have an other problem here?
Hope you have the answer! Regards,
Wilco (Dutch living in France)
1. Hello. That drive by it is a bit short but if you were cycling at about 1 time per second that is not bad. If you have 5 feet of head pressure you should be able to pump to almost 50 feet. So
it sounds like you have a leak in a pressure tank or a leak in the delivery line. It may still be something else but that’s the first thing that comes to mind. Are you using threaded
connectors for your pressure tank? That is oftentimes an issue
45. Thanks for your quick reply! Would the drop in pressure be that severe that you go from a lift of 50 feet to only 10 from even a small leak? The drive pipe and the pressure chamber are glued, so
I don’t expect it leaking there… But the bras fittings and gavanised pipes are threaded. I did use teflon tape in screwing everything together, but it might be there. I’ll check again tomorrow
(now dark here in Europe).
So you don’t think the problem is in the pressure build from the waste valve?
Thanks again and I’ll report back!
1. My apologies I mean you should be able to pump to 35 feet with 5 feet of head pressure. As the volume of air inthetank decreases the ability to pump higher also decreases. Can you send a pump
picture to my email? That will help me see how things are built. Landtohouse at gmail.
1. Sure! Anything to get it right! You’re the best. I’ll send everything round noon your time.
46. Great information, thank you!
I need help working out if my source flow is 1,4gpm and the drop to the pump is 12m, with the 7:1 ratio it’s no problem to get the water up to 80m high, but I’m wondering what water flow should I
roughly expect to get at 80m height?
1. In most of my Ram pump setups I have not heard of going above 60 meters. There is one fellow who is going up to 85 meters I’m not sure entirely what he has done. With that much pressure the
check valves do not last long. Because your flow rate is so small you would need to use a 1/2 inch pump. I am uncertain of the pumps production at this height.
48. Hi Seth. I have a waterfall garden feature with a shallow flow about 15 mtrs in length ending with a pond. The water has to rise 4ft to the waterfall and the pond is 2ft deep. Can I use a ram
pump to cycle the water to the waterfall and let it come back down into the pond.
1. Hello.
The ram pump can work within these parameters quite well the only issue is you have to have a constant supply of water and your closed system will not be able to supply that. Because the ram
pump only has an efficiency of 40, roughly, you will soon find that all of the water is in the lower Pond and the pump has stopped. I would say you are much better off trying a solar pump or
maybe even 2 to keep your cycle going.
49. My first flapper value won’t shut when water is applied to the system? Water just pours out the open valve and no pressure can be built? Any ideas?
1. That almost always means that there is air in the drive pipe. Getting this air out can be a real pain. I often lift the lower end of the pipe and then set it down quickly to flush the air
out. Other things could be the pipe is not the same size as the check valves or the pipe is compressed at some point or there is a critter that has gotten into the drive pipe.
50. Thanks for all of the instruction. I have constructed my first ram pump. It’s made with 1 1/4 pipe and then goes down to 3/4 on the output to a garden hose. The head is probably only 18 to 24
inches above the pump. The input is coming through corrugated pipe then into 4″ PVC for a total of 40′ and a lot of volume of water. The stand pipe is a 24″ long, 4″ PVC pipe and a bicycle tube
inside. I am trying to pump the water through a 150′ garden hose up about 15′ to 20′. I am getting water coming out of the pump.
My question is…what’s the optimal size and contents inside the stand pipe? Tube or noodle? 2″, 3″ or 4″ diameter? Height?
1. Thank you for building a pump! IF you would like to see in improvement in your system I would recommend removing the corrugated pipe. A smooth pipe will allow a better pressure wave. I prefer
the bike tube inside the pressure tank because it has better flexibility. As the water is surged into the tank the tube will have better compression than the noodle. For a 1-1/4″ pump I would
go with 4″ pipe at around 19 to 20″ long.
1. Thanks for the reply. I have my bicycle tube pretty inflated. Any recommendations on just how firm of an inflation on that tube?
To clarify, the water source runs through 40 ft of 4″ corrugated and then into a 4″ 10 ft PVC pipe before being condensed down to 1 1/4″ piping for the pump.
Due to the lack of drop from the head I’m considering a stand pipe. I’m going to insert it between the 4″ PVC and the 1 1/4″ pipe for the pump. I’m hopeful that will create more pressure
to push the water coming out of the pump. It won’t pump all the way up the hill and it shuts the ram pump down under the pressure. The check valve just won’t open up.
51. Hi! Are you still answering questions on this page?
1. Yes I sure am. Ask away.
1. Ok, my dad built a RAM to irrigate our farm with. We put in 220 feet of drive pipe. The first 200 feet is 4 inch, then we dropped down to 20 feet of 2 inch. However, the pipe is in a
fairly straight line- there is only about 3 feet of drop even with that much pipe. So we got everything laid out and hooked up the pipe and the pump won’t work. There are air bubbles
coming out the entrance to the drive pipe. So my question is this- do we just not have enough pressure because we don’t have enough drop for the pump to work? I told him we need to dig a
hole at least 8 feet deep and put the pump in the bottom and run the pipe to it. Please help! Lol!
1. There might be a few things to consider here. First what size is your ram pump? The two check valves are what determine the size. The drive pipe needs to be the same size as the pump.
So if your pump is 1″ then the drive pipe also needs to be 1″. Is your 4″ pipe smooth walled or corrugated? You can still use that 4″ pipe but you will need to install a stand pipe
between 75 and 100 feet away from the pump and use a drive pipe that matches the pump size and make sure the drive pipe is not corrugated. 3 feet of head pressure (drop) will allow
you to pump to a height of 21 feet above the pump. haha no 8 foot holes. They will only be filled with water. Let me know some of these things listed and I will see about getting your
pump going.
1. Ok let’s see. His pump is 1″. The drive pipe is smooth walled. And we need to pump the water up a hill with at least a 50 foot lift….
2. With a 1″ ram pump you will need to have a 1″ drive pipe. you will need at least 8 feet of head pressure to get water to that height.
3. Ok so how long should the drive pipe be?
4. The drive pipe needs to be 100 feet or less but if you cannot get the 8 feet of head pressure in that distance you will need to install a stand pipe. Check out this video of mine
showing the stand pipe. https://www.youtube.com/watch?v=NPovo-QUUaM
5. Ok so no minimum distance as long as it has 8 feet of drop then?
6. I highly recommend that the drive pipe be at least 20 feet long. This will allow the pressure wave time to get to the end of the pipe and back to the pump without interfering with
the cycle of the check valve.
52. Seth,
Have you ever experimented with the efficiency of a “Spring check valve” vs. a “Swing check valve” as the in-line check valve? I just added a spring check valve and it seems like I am getting
twice the water out of the Ram.
Maybee you can do an Adventure?
1. That is a fun idea. I used the double swing valves in the early pumps I made. The version that I sell now has a swing valve for the waste valve and a spring valve as the inline. I do like the
inline valve a lot better.
53. Hi, I noticed there are 2 way to put the swing check valve. Which one better? after or before the air chamber? My drive pipe and swing check valve are 1″
1. The waste valve (swing valve) must be on the drive pipe side of the pump. Now there is a design where the waste valve goes past the pressure tank with a T and the pressure wave goes back to
the pressure tank with each cycle. It seems like this second design is less efficient but I have no proof of that. I prefer the design that I make because I prefer the inline design.
54. Hi!
Just built a PVC ram pump of your design and need some help trouble shooting.
The pump has 100′ of 1 1/4″ intake line at approximately 8′ of head. The outline is 3/4″ at approximately 410′ and maybe 75′ of head.
At first, the pump ran with both ball valves open, but nothing was coming out of the out pipe.
So I shut it off and tried again, opened the in valve to 90% open, purge the in of air, open the out about 50%, and it started to do something different. It would open and close quickly and then
start to slow, and get stuck with the brass valve open, still no water coming out the end. I spent alot of time manually moving the brass check valve, still not much changed.
What gives?
Any suggestions?
1. Hello. To get water to a height of 75 feet lift you will need a minimum head pressure of 11 feet. With only 8 feet of head pressure you can expect to get water to 56 feet.
Often when the pump opens and closes quickly that means there is air in the drive pipe.
55. Hello. I have a river beside my property that fluctuates drastically. I need to move water approx. 40 feet in height, and a distance of close to 400 feet. My question is can I draw water from the
river while using a water storage tank to power the ram? The ram pump would need to be located near the water storage tank above the rivers flood stage.
1. The ram pump needs head pressure to operate. Is your river flat or does it have sizable rapids? To get water up to 40 feet you will need a head pressure of 6 feet but 7 or 8 would be best.
The 400 feet is not problem. You can install the ram pump out of the water as long as you have sufficient head pressure still going to the pump.
1. Yes, the river is flat. With no rapids. A 2″ downpour can bring the river up 10′ or more.
1. The Ram pump is not going to be a practical pump for your river. Without head pressure the pump will not start. You might want to look into a river pump. I have started testing a DIY
version but it has not worked yet. Here is a link to that pump if you would like to check it out:
http://www.riferam.com/pumps.html Sadly they are rather expensive.
56. Hello Seth,
First of all, thank you very much for sharing your knowledge about Ram pumps. It is very informative. I am involve in outreach work with indigenous Aeta’s here in the Philippines. One of their
problems is hygeine and we have decided to build a community toilet for them. For their water source, we initially thought of going solar until I realized that there is a stream with about 8′ of
head just about 100′ from their community toilets. I have read, check, reread and reviewed your build guide and I found it quite informative. I have decided to purchase the components mentioned
in your build guide. Unfortunately in our place, it is very hard to purchase 1 1/4 spring check valve but the 1 1/4 swing check valve is readily available. Can I use 2 swing valves instead of a
swing valve and a spring valve? If yes, what are the dis-advantages of using 2 swing valves. Thank you very much for your reply. As of now the community is very excited to finally have water in
their place without having to walk a far distance. 🙂
1. My apologies for the reply delay. It is just fine to use two swing valves. In my early tests I used two of them per pump. There are two main reasons that I use the spring valves. 1 they are a
little less expensive. 2. When tightening the pump together it is easier to align the parts with the spring valve. When you use the two swing valves you will want the second one to have the
flap swinging from the top. You can see in my early test that I have two swing valves: https://youtu.be/JvqA6zb1k_U?t=3m14s
57. Hi Seth,
It’s very kind of you to share this information.
I’m in New Zealand and unfortunately we have a lot of whale strandings. Incredibly people still resort to carrying buckets of water from the sea to keep the whales wet and cool, but it seems much
more logical to have some type of pump to get the water from the sea to the stranded whales. I understand the water for these pumps has to be falling, so I’m wondering if wave action would be
enough? If not, do you have any thoughts about what other type of pump might do the trick for whale strandings? I’d like to build or purchase some pumps to be located at our worst areas for whale
strandings to give the whales a better chance at survival.
1. Hello. That is a fun idea but sadly the ram pump has to have a constant flow of falling water. The wave idea would not allow the pump to access the head pressure with every cycle of the pump.
Sometime the pump would try to pull water and there would be no wave. I wonder if a solar pump would be ideal for this situation? Years ago I watched a video of a bicycle powered pump that
was used to bring water up for building sandcastles. I cant find the video though.
58. Is it possible to pump a water on ground level up to 20 m high? for a mid-high storey building?
Please reply! I gotta know! – thankyou
1. The ram pump needs head pressure to operate. To get water to 20m you would need a head pressure of 3m. This means that you would need the water on the ground to fall from the source to the
pump by 3 m. This would get water to 20m hight. I hope this helps.
59. Hello, i have a question. I am about to set up a ram pump, now i do not need to send water up basically my system is always going down hill. I was wondering if by using a ram pump i will get a
more steady water flow and a better water pressure as the tank starts running out of water. I will be using a 1500 gallon tank. Thank you for your help in advance.
1. The ram pump is actually only good for pumping uphill. In a situation where you are taking water down hill only you are better off just siphoning and letting gravity take care of the rest.
When you place a ram pump in a situation where the water only goes downhill the pump will stop. The water will bypass the pump and go right out the other end.
60. Hi Seth,
Been viewing your videos and i made a Ram pump that has 1 inch drive pipe, 2 inch tank and 1/2inch delivery pipe. From a height of 2 meters of water with 8 meters of drive pipe, i was able to
reach a pressure of 30 PSI. My question is if I make my drive pipe bigger than the check valve say a 1 1/4 inch drive pipe for a 1 inch check valve would it do any improvement in the pressure
build up? With my current set up, I was able to lift water up to around 40 meters with water flow rate at the delivery pipe of 150 liters/24 hrs @ 29-30 PSI.
1. Thank you for watching the videos! It is actually best to keep the drive pipe the same size as the waste valves. This is because the pressure wave needs to be the same size going and coming
from the pump to source. To increase the pressure you can use a ridged drive pipe such as PVC or you can increase the head pressure.
61. Can you send me or attach a supply just for new pump version please.
1. Are you looking for a list of components?
62. Hi Seth;
We have a small river at the bottom of our property, but it is flat (there is no fall to it). I want to pump the water 150′ uphill, and the approximate run is 700′. Because there is no head to
speak of, I am curious if I were to use a gas or solar powered pump to lift the water out of the river into a cistern placed higher up on the bank, then ran the water down from an outlet in the
cistern to a ram pump placed approximately 21′ below at the edge of the river would that be sufficient to push the water up 150′? How ample must the flow from the cistern be to allow this to work
(I am assuming that a solar powered pump would have minimal flow). Or is there some other solution to my problem?
1. Hello. To get water up the hill at 150 feet of lift you would need to have 21 feet of head pressure. BUT this is just to get the water to that point. You will want a little more head pressure
to get usable flow at the top. The flow needed to keep a 1-1/4″ ram pump going is 8gpm. The small 1/2″ pump only needs 3gpm to operate. If your solar pump can handle this flow rate you can
use the ram pump like this. One thing to consider is that 21 feet of head pressure will cause the ram pump waste valve to wear quickly.
63. Hello Seth ! What happens if delivery pipe outlet is lower than the pump? does it still add pressure on the output?
1. Hello. Sadly if the delivery pipe is lower than the pump the system will stop and the water will flow out of the pump by siphon. There is no extra pressure increase.
64. Hi, I’ve noticed that you sometimes close the valve at the output of the pump to get it started, and the pump functions fine, just pumping out the waste line, then you gradually open it. My
question is this: Rather than using a storage container, would it be possible to have a check valve in line near the elevated usage area, and have a shut off valve at the end of the delivery
line, and use this shut off valve to start and stop the flow each day. Would the pump continue to run while the valve was closed, and have the output flow return after the valve was opened back
up? Or would the pump need to be re-primed each day?
1. Hello. It is very possible to have a shutoff valve at the delivery end of the pump. You can turn off the water at the final location with valve and the pump will continue to work without
outputting water.
There is an issue that can come about though. If the pressure builds too high in the delivery pipe there is a chance that opening the valve will cause the pump to stop because too much
pressure is lost at the pump. If you hill is high enough this is not an issue though.
65. Seth, I need this very badly for several different reasons. 1. To water livestock 2. To help the water evaporation in a small pond and 3. To help water plants. I have a running water source, the
embankment is about 6 feet high depending on the water level. Which one of these do I need to buy?
1. The ram pump can lift water out of a 6 foot ditch with ease. To do this you will need around 2 feet of head pressure (drop in water)
Each ram pump size can lift water to this height. The difference is the amount of water that is required to run the pump as well as the amount out at the top. Do you know how much water is
flowing if your water source? The small pump needs 2GPM while the large pump needs 8GPM to operate.
66. Im contemplating my options for pumping water from a good flowing stream (some small rapids) up to my homesite. The water will need to gain 300 ft elevation and a distance of 450 feet.
Unfortunately, building intermittent holding ponds is not an option (as you can see it is very steep) and it would need to make it in one go. I would also prefer no to use gasoline/diesel or and
fossil fuel.
I’ve been try to understand these terms of standing pipe, head pressures and so forth but not sure I grasp it all totally…
The amount of water that comes out at the top isn´t very important as I get about 5 gallons per hour we should be good. I will also have a catchment system (we get consistent rains) and I plan to
have it into a gravity fed holding tank with any runoff ultimately going into irrigation and back down to the stream.
Is a hydraulic ram a realistic option with wear and tear on moving the parts?
Solar pump?
1. The ram pump is great for pumping out of a creek but there are a few limitations that you need to consider. The max lift height that I have every pumped with the ram pump is 185 feet. This is
due to the weight of the waste valve. To get water to this height you would need around 25 feet of head pressure (drop in water at the source to the pump)
In this case I do not think that a ram pump is going to get water to the top. Now you might be able to get water up a good distance and then use a solar pump to get the water the rest of the
way up. I have very little experience with solar pumps.
67. Seth,
I will be 65 in March 2021, and have watch your videos. They are really well done. I built my first pump with my Dad in 1975 in central TN. Fantastic results ! But the real reason I am writing to
you is that I wanted to tell you about seeing my first pump in 1966. It was made from a log. I was only 10, but this was fascinating ! But, I was never able to find the original pump when I was
old enough to want to replicate it. Several people remembered seeing it, but didn’t know the details. I thought it was incredible, and wanted to ask if you have seen one in your studies.
1. Oh wow a ram pump built out of a log! That is very interesting. I have seen some very creative ideas and designs but never one made from a log!
68. I have a spring approximately 200feet below my homesite. About 5gpm output. Will a water ram pump the 200feet vertical elevation? Any suggestions on size?
1. Hello. The ram pump that I have here on Land To house has a max lift of around 180 feet. This requires 25 feet of input head pressure. Any more than this and the waste valve will not close. | {"url":"https://www.landtohouse.com/rampump/","timestamp":"2024-11-10T03:11:35Z","content_type":"text/html","content_length":"436440","record_id":"<urn:uuid:620e5f14-30c7-4ec5-8442-12003dcdcefc>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00274.warc.gz"} |
Free Multiplication Chart Printable Paper Trail Design | Multiplication Chart Printable
Free Multiplication Chart Printable Paper Trail Design
Free Multiplication Chart Printable Paper Trail Design
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What is Multiplication Chart Printable?
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While specific tables work for presenting portions of information, a complete page chart makes it much easier to evaluate realities that have already been understood.
The multiplication chart will typically feature a leading row and also a left column. When you want to discover the product of 2 numbers, select the very first number from the left column and the 2nd
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Multiplication charts are practical discovering tools for both kids and grownups. Children can utilize them at home or in college. Times Table Chart Printable are available on the Internet and also
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Why Do We Use a Multiplication Chart?
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Add Multiple Trend Lines Google Charts 2024 - Multiplication Chart Printable
Add Multiple Trend Lines Google Charts
Add Multiple Trend Lines Google Charts – The Multiplication Graph Series might help your college students creatively signify numerous very early math principles. However, it must be used as a
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Multiples of 4 are 4 clear of each other
The style for determining multiples of 4 is usually to put each quantity to on its own and discover its other numerous. For instance, the 1st five multiples of 4 are: 4, 16, 12 and 8 and 20. And they
are four away from each other on the multiplication chart line, this trick works because all multiples of a number are even. Moreover, multiples of several are even numbers naturally.
Multiples of 5 are even
If they end in or 5, You’ll find multiples of 5 on the multiplication chart line only. Put simply, you can’t grow a variety by 2 or 3 to get a level quantity. If the number ends in five or , you can
only find a multiple of five! Thankfully, you can find tricks that will make getting multiples of five even much easier, like utilizing the multiplication chart collection to get the several of 5.
Multiples of 8 are 8 clear of the other
The design is apparent: all multiples of 8 are two-digit phone numbers and multiples of several-digit numbers are two-digit numbers. Each selection of 10 posesses a several of 8-10. 8 is even, so
that all its multiples are two-digit numbers. Its style persists approximately 119. Next time the thing is a variety, be sure to search for a several of seven in the first place.
Multiples of 12 are 12 from the other person
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about the multiplication graph or chart collection.
Multiples of 20 are 20 clear of the other person
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Multiples of 30 are 30 away from the other
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Multiples of 40 are 40 far from the other person
You may know that there are multiples of 40 on a multiplication chart line, but do you know how to find them? To achieve this, you could add from the outside-in. For instance, 10 12 14 = 40, and the
like. Similarly, 15 eight = 20. In this instance, the telephone number about the remaining of 10 is definitely an even quantity, even though the one in the proper is undoubtedly an unusual amount.
Multiples of 50 are 50 clear of the other
Utilizing the multiplication graph line to ascertain the sum of two phone numbers, multiples of 50 are identical length away from each other around the multiplication chart. They have got two perfect
factors, 80 and 50. For the most part, every phrase is different by 50. Another element is 50 on its own. Listed below are the most popular multiples of 50. A typical a number of will be the several
of a presented amount by 50.
Multiples of 100 are 100 away from each other
Allow me to share the numerous phone numbers that are multiples of 100. An optimistic pair is really a several of just one 100, although a negative set is actually a a number of of twenty. These 2
types of figures are not the same in many ways. The 1st way is to separate the number by successive integers. In such a case, the amount of multiples is a, thirty, twenty and ten and forty.
Gallery of Add Multiple Trend Lines Google Charts
How To Create A Line Chart In Google Sheets Step By Step 2020
Create Scatter Chart In Google Sheets Example Charts
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Neural Networks Breakdown Part II
Neural Networks
Neural Networks Breakdown Part II
Neural Networks
You may want to check <a href="/post/33">this</a> first, if you haven't. <h2><b>Learning Step</b></h2> This is where the "tricky" math will come in, which is actually not that tough. The main concept
that you must be aware of is calculating the derivative. The derivative basically means, calculating the change in some value due to a change in some other value. More specifically, it is the amount
of change which occurs in a variable, when another variable changes by the slightest amount possible, also called infinitesimally small. Just to get a grip on it, you can check out this great Khan
Academy video explaining derivatives. <iframe style="width:100%; height: 400px;" src="https://www.youtube.com/embed/rAof9Ld5sOg" frameborder="0" allowfullscreen></iframe> Calculating the derivatives
actually helps us know whether the neural network is learning or not. But derivatives of what? How does it help in learning? How does it know if it is learning? These are the probable questions which
can arise. This is what we will get into. The neural network traverses the neural network two times. Once from the fornt and the second time from the back, forward propagation and backward
propagation. The forward propagation helps in calculating the output of the entire network, whilst, the backward propagation helps in telling us how much the weights (the connections between each
layer), should be adjusted so as to get an output which is closer to the expected output. The backpropagation step is where the real learning happens. When we get an output, we can tell whether, it
is right or wrong, as we just need to compare it with the expected output or label which was assigned with the data (0 or 1). The artificial neural network also needs to know how incorrect the output
is, with respect to the labelled data. For this purpose, we use a loss function to calculate the amount of loss that will take place when we use the weights which have been currently assigned to the
neural network. There are many types of loss functions, namely, cross-entropy error function, mean squared error function, Gaussian log likelihood function, etc. For starters, we can use the mean
squared error function, for calculating the loss incurred by a neural network. The mean squared error looks like: <center><font size="+3"><b>loss = (1/n) x Σ(d<sub>i</sub>-y<sub>i</sub>)<sup>2</
sup></b></font></center> Here, n = number of examples used for training, d<sub>i</sub> = the label (0 or 1 for the dog example) for the i<sup>th</sup> example, y<sub>i</sub> = output of the i<sup>th
</sup> example from the neural network. The objective of the neural network during the back-propagation is to reduce the loss function value after each iteration of the back-propagation. How will it
do it? Adjust the weights. The neural networks adjust the weight, so that the loss function is reduced to the point that it can't be reduced anymore. The adjustment is the actual <b>"learning"</b>
the neural network does. How does the neural network adjust the weights so that the error is reduced? Recap the derivatives part and click <a href="/post/36">here</a>.
- Shubham Anuraj, 11:56PM, 16 July, 2018 | {"url":"https://www.s-tronomic.in/post/35","timestamp":"2024-11-11T14:27:43Z","content_type":"text/html","content_length":"10500","record_id":"<urn:uuid:06b026b0-79d7-4e7b-9e49-ec1f5f71c9e9>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00787.warc.gz"} |
CERTIFICATE OF SECONDARY EDUCATION EXAMINATION
041 BASIC MATHEMATICS
(For Both School and Private Candidates)
Time: 3 Hours Year: 2020
1.This paper consists of sections A and B with a total of fourteen (14) questions.
2.Answer all questions in sections A and B.
3.Each question in section A carries six (06) marks while each question in section B carries ten (10) marks.
4.All necessary working and answers for each question must be shown clearly.
5.NECTA mathematical tables and non-programmable calculator may be used.
6.All communication devices and any unauthorized materials are not allowed in the examination room.
7.Write your Examination Number on every page of your answer booklet(s).
SECTION A (60 Marks)
Answer all questions in this section.
1. (a) Simplify the expression
View Ans
(b) (i) Mr. Magani set an examination weighing a total of 96 marks with the following distribution: 20% of the marks were awarded for reading, 40% for writing, 15% for practical and the remaining
percentage for spelling. Find the marks that were awarded for spelling.
View Ans
(ii) Three airplanes arrived at Kilimanjaro International Airport (KIA) at the intervals of 30 minutes, 40 minutes and 55 minutes. If all three airplanes arrived at KIA at 2:00 p.m. on Saturday, when
and at what time would they arrive together again?
View Ans
2. (a) If
View Ans
(b) (i) Solve equation
View Ans
(ii) Given log2=0.3010 and log3 =0.4771, find value of
View Ans
3. In a certain school, 40 students were asked about whether they like tennis or football or both. It was found that the number of students who like both tennis and football was three times the
number of students who like tennis only. Furthermore, the number of students who like football only was 6 more than twice the number of students who like tennis only. However, 4 students like neither
tennis nor football.
(a) Represent this information in a Venn diagram, letting x be the number of students who like tennis only. (b) Use the results obtained in part (a) to determine the probability that a student
selected at random likes;
(i)Football only.
(ii)Both football and tennis.
View Ans
4. (a) (i) A line whose gradient is 3/ 2 has the x-intercept of - 3. Find the equation of the line in the form y=mx + c , where m and c are constants.
View Ans
(ii) Find the length of a line segment joining the points (3,-2) and (15,3) .
View Ans
(b) A boat sails due to North at speed of 120km/h and wind blows at speed of 40km/h due to east. Find actual speed of the boat. Use
View Ans
5.(a) In the following triangle ABC, AB=8cm, BC=11.3cm and angle ABC=30^0. Find area of triangle ABC.
View Ans
(b)(i) Find he perimeter of a regular hexagon inscribed in a circle whose radius is 100 m.
View Ans
(ii) Given that
where AB,BT and TA are sides of triangle ABT and KL, LC and CK are sides of triangle KLC. What does this information imply?
View Ans
6(a) The variables t and z in the following table are related by the formula z = at^n where a is a constant and n is a positive integer.
(i)Use the data from the table to determine the values of a and n.
(ii)Use the values of a and n obtained in part (a) (i) to complete the following table
View Ans
(b) If v varies directly as the square of x and inversely as
View Ans
7.(a)(i) A school has 2,000 students, of whom 1,500 are boys. What is the ratio of boys to girls in the school?
View Ans
(ii) Matiku bought a book for Tshs. 120,000. A year later, he sold the book at a profit of 20%. What was the selling price of the book?
View Ans
(b) Halima started a business o 1^st September, 2018 with a capital of Tshs. 25,0001= in cash.
September 2, bought goods for cash 15,0001=
3, sold goods for cash 3,0001=
5, sold goods for cash 5,0001=
6, paid carriage on goods 5001=
9,sold goods for cash 14,000/=
15, bought goods for cash I ,000/=
19 , paid rent 2,0001=
20, purchased goods 6,0001=
27, paid wages 5,000/=
28, sold goods on credit 1,000/=
By using these transactions, prepare the cash account
View Ans
8. (a) Find the first term and the common difference of an arithmetic progression whose 5^th term is 21 and 8^th term is 30.
View Ans
(b) Find the 10^th term of a sequence whose first three consecutive terms are 5, 15 and 45. (Leave the answer in exponent form).
View Ans
9. (a) In the following figure, AP is perpendicular to BC , AB= 13cm, BP = 5 cm and AC =.15 cm.
Calculate the lengths of AP and CP .
View Ans
(b) From the top of a building 75 m high, John sees a lorry and a minibus along the road, both being on one side of the building at the angles of depression of 30^0 and 60^0 respectively.
(i)Sketch a diagram representing this information.
(ii)Determine the distance between the cars, leaving the answer in surd form.
View Ans
10. (a) Rachel is three years older than her brother John. Three years to come, the product of their ages will be 130 years. Formulate a quadratic equation representing this information. Hence, by
using the quadratic formula, find their present ages.
View Ans
(b) The sum of square of two consecutive positive number is 61. Find the numbers.
View Ans
SECTION B(40 Marks)
Answer All Questions In This Section
11.The following data represent the marks scored by 36 students of a certain school in geography examination.
(a)Prepare a frequency distribution table representing the given data by using the class intervals: 50 - 54, 55 -59, 60 - 64, and so on.
(b)Use the frequency distribution table obtained in part (a) to:
(i)Draw a histogram.
(ii)Calculate the median. Write the answer correct to 2 decimal places.
View Ans
12. (a) Two towns, A and B, are located at (10^0S, 38^0E) and (10^0S, 43^0 E) respectively.
(i)Find the distance between the two towns in kilometers. (Use radius of the Earth, R = 6400 km and
(ii)Suppose a ship is sailing at 50 km/h from town A to town B. Using the answer obtained in part (a) (i), find how long will the ship take to reach town B.
View Ans
(b) The following figure represents a rectangular prism in which PQ =12 cm, QR=8cm and RY=4cm.
(i)The total surface area.
(ii)The angle between the planes PTZW and QRZW .
View Ans
(c) Calculate the volume of a cone whose base radius is 12 cm and slant height is 20 cm. (Use
View Ans
13. The inverse of matrix A is
Find matrix A.
View Ans
(b) Amani and Asha bought Coca-cola and Pepsi drinks for a farewell party. Amani spent Tshs. 9950 to buy 12 bottles of Coca-cola and 5 bottles of Pepsi drinks. Asha spent Tshs. 8150 to buy 9 bottles
of Coca-cola and 5 bottles of Pepsi drinks. Formulate a system of linear equations and hence apply the matrix method to find the price of one bottle of each type of the drinks.
View Ans
(c) Point A(4, 2) is reflected in the line y + x = 0 followed by an anticlockwise rotation through 90^° about the origin. Find the final image of point A.
View Ans
14.(a) Suppose a function f is defined by f (x) = (x + 2)^2 , find the domain and range of the inverse of the function f .
View Ans
(b) A businessman plans to buy at most 210 sacks of Irish and sweet potatoes. Irishpotatoes cost shs. 30,000 per sack and sweet potatoes cost shs. 5,000 per sack. He can spend up to shs. 2,500,000
for his business. The profit on a single sack of Irish potatoes is shs. 12,000 and for sweet potatoes is shs. 10,000. How many sacks of each type of potatoes the businessman will buy in order to
realize the maximum profit?
View Ans | {"url":"http://learninghubtz.co.tz/form4-necta-qns-ans.php?sub=bWF0aGVtYXRpY3M%3D&yr=MjAyMA%3D%3D","timestamp":"2024-11-02T15:24:55Z","content_type":"text/html","content_length":"106947","record_id":"<urn:uuid:afb90591-5121-4279-bef8-110dab2c867d>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00555.warc.gz"} |
Boundedness theorem | JustToThePointBoundedness theorem
Calculus is the most powerful weapon of thought yet devised by the wit of man, Wallace B. Smith
The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change, the ratio of
the instantaneous change in the dependent variable to that of the independent variable.
Definition. A function f(x) is differentiable at a point “a” of its domain, if its domain contains an open interval containing “a”, and the limit $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$ exists, f’
(a) = L = $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then |L-$\
frac {f(a+h)-f(a)}{h}$|< ε.
The Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝ^n has a convergent subsequence.
Theorem. If $\{x_n\}_{n=1}^\infty$ is a convergent sequence and a ≤ x[n] ≤ b ∀n ∈ ℕ, then a ≤ $\lim_{n \to ∞} x_n = c$ ≤ b.. In other words, the limit of $\{x_n\}_{n=1}^\infty$ must lie between a and
For the sake of contradiction, assume that c ∉ [a, b]. Without loss of generality, suppose that c < a.
Select ε = $\frac{a-c}{2}, \lim_{n \to ∞} x_n = c$ ⇒ ∃N: ∀n ≥ N, |x[n] - c| < ε ↭ -ε < x[n] - c < ε. We only need x[n] -c < ε = $\frac{a-c}{2}$.
However x[n] ≥ a, x[n] -c ≥ a - c = 2ε >[ε> 0] ε ⊥
Boundedness theorem
Definition. A function is bounded if for all x in its domain there is a real number M such that |f(x)| ≤ M.
Boundedness theorem. Let I = [a, b] be a closed bounded interval, let f: I → R be continuous on I. Then, f is bounded on I.
Proof. Suppose for the sake of contradiction that the function is not bounded on [a,b] ⇒ for any n ∈ ℝ, ∃x[n] ∈ [a,b] (we are abusing notation for the sake of brevity) such that |f(x[n])| > n.
Because [a,b] is obviously bounded, $\{x_n\}_{n=1}^\infty$ is bounded ⇒
[By the Bolzano–Weierstrass theorem] it implies there exists a convergent subsequence {x[nk]} of $\{x_n\}_{n=1}^\infty$ with {x[nk]} → x[0].
Theorem. If $\{x_n\}_{n=1}^\infty$ is a convergent sequence and if a ≤ x[n] ≤ b ∀n ∈ ℕ, then a ≤ $\lim_{n \to ∞} x_n$ ≤ b. ⇒ As [a,b] is closed, it contains x[0].
Since f is continuous, f is continuos at x[0], and {x[nk]} → x[0] ⇒[Theorem. If f: ℝ → ℝ is continuos at a point x ∈ ℝ and {x[n]} is a real sequence converging to x, the sequence {f(x[n])} converges
to f(x)] {f(x[nk])} → f(x[0]) and every convergent sequence is bounded.
However, |f(x[n])| > n ⇒ [and this obviously applies to every subsequence] |f(x[nk])| > nk, which implies {f(x[nk])} diverges ⊥
• Counterexample. f needs to be continuous on [a, b], $f(x) = \begin{cases} \frac{1}{x}, &0 < x < 1 \\ 0, &x = 0 \end{cases}$
f(x) is not bounded on (0, 1) -Figure 1.g.- because f is not continuos at x = 0, hence f manages to escape to infinity as 0 is not there to keep f in check.
Therefore, we can deduce that the conclusion of the Boundedness Theorem does not necessarily follow.
• f: [0, 4] → ℝ, defined by f(x) = x^2. f is continuous on all real numbers ⇒ f is also continuous on the closed bounded interval [0, 4] ⇒ f is is bounded. Furthermore, f’(x) = 2x > 0 on [0, 4] ⇒ f
is an increasing function ⇒ |f(x)| ≤ f(4) = 4^2 = 16.
• f: [$\frac{-π}{2}, \frac{π}{2}$] → ℝ, defined by f(x) = arctan(x). f is continuous on all real numbers ⇒ f is also continuous on the closed bounded interval [$\frac{-π}{2}, \frac{π}{2}$] ⇒ f is
is bounded. Furthermore, f’(x) = $\frac{1}{x^2+1}$ > 0 ⇒ f is an increasing function ⇒ |f(x)| ≤ $arctan(\frac{π}{2})$ ≈ 1.003. Besides, arctan(x) is bounded with $\frac{-π}{2} < y < \frac{π}{2}$,
$\lim_{x \to -∞} arctan(x) = \frac{-π}{2}, \lim_{x \to ∞} arctan(x) = \frac{π}{2}.$
• f: [-2π, 2π] → ℝ, defined by f(x) = sin(x). f is continuous on all real numbers ⇒ f is also continuous on the closed interval [0, 4] ⇒ f is bounded. Furthermore, f is bounded since |sin(x)| ≤ 1
∀x ∈ ℝ.
• f: [0, ∞) → ℝ defined by f(x) = 7x + 4 does not satisfy the conditions of the boundedness theorem because f is continuous, but the interval [0, ∞) is not bounded. $\lim_{x \to ∞} 7x + 4 = +∞$ ⇒ f
is not bounded.
• f: (-1, 1) → ℝ defined by $\frac{2x}{1-x^2}$, (-1, 1) is not a closed interval ⇒ the conclusion of the Boundedness Theorem does not necessarily follow. Futhermore, $\lim_{x \to 1⁻} \frac{2x}{1-x^
2} = \frac{2}{0⁺} = ∞, \lim_{x \to -1⁺} \frac{2x}{1-x^2} = \frac{-2}{0⁺} = -∞$ ⇒ f is not bounded, it has two vertical asymptotes at x = -1 and x = 1.
This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007. | {"url":"https://justtothepoint.com/calculus/boundednesstheorem/","timestamp":"2024-11-14T05:44:45Z","content_type":"text/html","content_length":"20108","record_id":"<urn:uuid:f117e6bb-7cef-4bc8-aef4-fbf5ee1d5743>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00152.warc.gz"} |
This looks much better! To see if the peak location makes sense, remember that the mean of the binomial distribution is $\langle n_H \rangle = Np = 1.8$ for our parameter choices ($N=9$ and $p=0.2$).
Indeed, the peak is around 2 which passes our sanity check.
As a last comment, notice that the $y$-axis of the histogram we made represents the number of trials for the given outcome, whereas the binomial distribution is a probability distribution that gives
the probability of each outcome. We can therefore divide the values on the $y$-axis by the number of trials to yield an actual probability distribution. | {"url":"http://www.rpgroup.caltech.edu/aph161/assets/tut/t2/binomial_partitioning.html","timestamp":"2024-11-06T14:04:50Z","content_type":"text/html","content_length":"317133","record_id":"<urn:uuid:611558a2-5846-4ecf-bb09-ac586ce20718>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00590.warc.gz"} |
Exploring Mathematical Optimization: A New Tool for Data Scientists
Written on
Chapter 1: Introduction to Optimization
In the current landscape of Data Science, there seems to be an overemphasis on machine learning techniques. It's akin to the old adage: to someone wielding a hammer, everything looks like a nail.
Consequently, modern Data Scientists often view every challenge through the lens of machine learning. This singular focus is unfortunate because it overlooks a plethora of valuable approaches within
the data science domain.
This article aims to shed light on an essential yet often neglected aspect of Data Science: Mathematical Optimization, particularly Constraint Programming. By incorporating these techniques into your
skill set, you can significantly enhance your career prospects, even if your mathematical background isn't particularly strong. I, for instance, studied Geography but found it surprisingly accessible
to dive into Mathematical Optimization using Google's open-source library, OR-Tools, which will be introduced in this beginner-friendly guide.
If you're eager to broaden your Data Science toolkit and acquire this in-demand skill, let's get started!
Section 1.1: What is Optimization?
Optimization encompasses a range of techniques designed to determine the best possible solution from a vast array of alternatives. This might involve identifying the optimal solution to a problem or
simply listing all feasible solutions. Consider a scenario where you're part of the Data Science team at an Amazon distribution center. You have 100 packages to deliver and three drivers, all needing
to complete their deliveries within a two-hour timeframe. This presents an optimization problem, necessitating the creation of the most efficient delivery schedule for each driver.
Alternatively, think about a teacher planning a group activity for ten students. The teacher needs to divide the students into three groups but faces specific constraints: (1) Timmy cannot be grouped
with Jimmy, (2) Billy must be in the same group as Willy, and (3) each group must include at least one of Mickey, Dickie, Ricky, or Vicky. This represents a constraint programming problem, where the
objective is to find a viable grouping that adheres to all established constraints.
These scenarios highlight the nature of optimization problems, where the goal is to sift through an immense number of potential solutions to find the most viable or optimal one.
Section 1.2: Understanding Constraint Programming
To illustrate the concept, let’s delve into the student grouping example discussed earlier. We’ll work with the following conditions:
• Students 1 and 2 cannot be in the same group.
• Student 3 must be grouped with Student 8.
• Each group must consist of at least one of the following students: 7, 8, 9, or 10.
Additionally, we have standard constraints:
• Each group must contain a minimum of three students.
• Each student must be assigned to exactly one group.
These constraints create a complex situation that is challenging to resolve through mental calculations alone. Fortunately, Constraint Programming can help us derive a possible solution using
OR-Tools, an open-source library developed by Google.
Chapter 2: Implementing OR-Tools for Optimization
To begin, install the OR-Tools Python library with the following command:
pip install ortools
Next, import the constraint programming module and initialize a CP Model:
from ortools.sat.python import cp_model
model = cp_model.CpModel()
Now, let's add decision variables and constraints to our model. Each student must be assigned to one of the three groups. We can express this as a series of linear equations.
The first constraint can be represented as follows:
student1_group1 = model.NewBoolVar("student1_group1")
student1_group2 = model.NewBoolVar("student1_group2")
student1_group3 = model.NewBoolVar("student1_group3")
model.Add(student1_group1 + student1_group2 + student1_group3 == 1)
This implies that if Student 1 is assigned to Group 1 (i.e., student1_group1 = 1), then the other two variables must be 0.
The implementation continues with additional constraints for all students, ensuring that each constraint is articulated as a linear equation to maintain clarity.
The video, "Predict The Stock Market With Machine Learning And Python," serves as an illustrative example of applying machine learning techniques, while this article highlights the complementary role
of optimization.
Section 2.1: Adding More Constraints
Once the decision variables are established, we can integrate additional constraints, such as ensuring each group has a minimum of three members. This can be expressed through the following
model.Add(student1_group1 + student2_group1 + student3_group1 + student4_group1 + student5_group1 + student6_group1 + student7_group1 + student8_group1 + student9_group1 + student10_group1 >= 3)
We will replicate this for each group to guarantee compliance with the minimum requirement.
Section 2.2: Solving the Model
With all constraints and decision variables in place, we can instruct OR-Tools to solve the problem:
solver = cp_model.CpSolver()
status = solver.Solve(model)
If a solution exists, we can print the group assignments for each student.
The output might look something like this:
Student 1: Group 1
Student 2: Group 2
Conclusion: The Value of Optimization
Optimization techniques are not new, yet they often remain underutilized in Data Science compared to analytics and machine learning. I believe it’s crucial for Data Scientists to embrace these
methodologies. I hope you found this overview useful, and I'm eager to hear about any innovative use cases you might have!
For further insights, I also manage the SQLgym and publish a free newsletter titled "AI in Five," where I share weekly updates on AI developments, coding strategies, and career advice for Data
Scientists and Analysts. Subscribe if you're interested!
Thank you for reading, and I welcome you to connect with me on Twitter or LinkedIn! 😊 | {"url":"https://ronwdavis.com/exploring-mathematical-optimization-tool.html","timestamp":"2024-11-02T21:58:02Z","content_type":"text/html","content_length":"13861","record_id":"<urn:uuid:832f921f-7150-44fb-b498-9a8e3f25eff0>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00690.warc.gz"} |
Complete convergence theorem for a competition model
Publication , Journal Article
Durrett, R; Møller, AM
Published in: Probability Theory and Related Fields
In this paper we consider a hierarchical competition model. Durrett and Swindle have given sufficient conditions for the existence of a nontrivial stationary distribution. Here we show that under a
slightly stronger condition, the complete convergence theorem holds and hence there is a unique nontrivial stationary distribution. © 1991 Springer-Verlag.
Duke Scholars
Published In
Probability Theory and Related Fields
Publication Date
Start / End Page
Related Subject Headings
• Statistics & Probability
• 4905 Statistics
• 4904 Pure mathematics
• 0104 Statistics
• 0102 Applied Mathematics
• 0101 Pure Mathematics
Published In
Probability Theory and Related Fields
Publication Date
Start / End Page
Related Subject Headings
• Statistics & Probability
• 4905 Statistics
• 4904 Pure mathematics
• 0104 Statistics
• 0102 Applied Mathematics
• 0101 Pure Mathematics | {"url":"https://scholars.duke.edu/publication/767567","timestamp":"2024-11-13T22:36:59Z","content_type":"text/html","content_length":"61413","record_id":"<urn:uuid:3061acce-8a2a-4320-ad1a-4c3b0c8d2472>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00340.warc.gz"} |
Online LCM calculation of two integers - Calculating the Least Common Multiple - Solumaths
A binomial coefficient calculator that allows you to calculate a binomial coefficient from two integers.
Definition of the binomial coefficient
In mathematics, the binomial coefficient of two integers n and k is the number `(n!)/(k!(n-k)!`, with `k<=n`. This number can be noted `((n),(k))` or `C_n^k`.
Binomial Coefficient Calculator
The binomial coefficient calculator allows you to calculate a binomial coefficient from two integers. To calculate the binomial coefficient of two numbers n and k, the calculator uses the following
formula: `(n!)/(k!(n-k)!`. The steps of the calculation are specified.
For example, to calculate the binomial coefficient of the following two integers 5 and 3, simply enter binomial_coefficient(`5;3`), and the calculator returns the result, which is 10.
The binomial coefficients intervene in particular in the expansion of algebraic expression with Newton's binomial formula or in probability with combinatorics or combinations. One of the specific
features of the binomial coefficients calculator is to indicate the different calculation steps that allow to find the result.
Syntax :
binomial_coefficient(n;k), n and k are integers.
Examples :
binomial_coefficient(5;3), returns 10 | {"url":"https://www.solumaths.com/en/calculator/calculate/binomial_coefficient","timestamp":"2024-11-02T08:02:11Z","content_type":"text/html","content_length":"60989","record_id":"<urn:uuid:2de54ab7-ff5c-4baf-930e-cdd35781da59>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00174.warc.gz"} |
Asymptotic behaviour of a nonlinear parabolic equation with gradient absorption and critical exponent | EMS Press
Asymptotic behaviour of a nonlinear parabolic equation with gradient absorption and critical exponent
• Razvan Gabriel Iagar
Universidad Autónoma de Madrid, Spain
• Philippe Laurençot
Université de Toulouse, Toulouse, France
• Juan Luis Vázquez
Universidad Autónoma de Madrid, Spain
We study the large-time behaviour of solutions of the evolution equation involving nonlinear diffusion and gradient absorption,
We consider the problem posed for and with nonnegative and compactly supported initial data. We take the exponent which corresponds to slow -Laplacian diffusion. The main feature of the paper is that
the exponent takes the critical value which leads to interesting asymptotics. This is due to the fact that in this case both the Hamilton-Jacobi term and the diffusive term have a similar size for
large times. The study performed in this paper shows that a delicate asymptotic equilibrium happens, so that the large-time behaviour of the solutions is described by a rescaled version of a suitable
self-similar solution of the Hamilton-Jacobi equation , with logarithmic time corrections. The asymptotic rescaled profile is a kind of sandpile with a cusp on top, and it is independent of the space
Cite this article
Razvan Gabriel Iagar, Philippe Laurençot, Juan Luis Vázquez, Asymptotic behaviour of a nonlinear parabolic equation with gradient absorption and critical exponent. Interfaces Free Bound. 13 (2011),
no. 2, pp. 271–295
DOI 10.4171/IFB/258 | {"url":"https://ems.press/journals/ifb/articles/4574","timestamp":"2024-11-07T10:23:26Z","content_type":"text/html","content_length":"92961","record_id":"<urn:uuid:ffc406be-4563-41c5-be2f-c7208ccc7b42>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00250.warc.gz"} |
StableSwap-NG Pools: Overview
A Curve pool is essentially a smart contract that implements the StableSwap invariant, housing the logic for exchanging stable tokens. While all Curve pools share this core implementation, they come
in various pool flavors.
In its simplest form, a Curve pool is an implementation of the StableSwap invariant involving two or more tokens, often referred to as a 'plain pool.' Alternatively, Curve offers more complex pool
variants, including pools with rebasing tokens and metapools. Metapools facilitate the exchange of one or more tokens with those from one or more underlying tokens.
New features:
• price and D oracles
• dynamic fees
Supported Assets¶
Stableswap-NG pools supports the following asset types:
Asset Type Description
0 Standard ERC20 token with no additional features
1 Oracle - token with rate oracle (e.g. wstETH)
2 Rebasing - token with rebase (e.g. stETH)
3 ERC4626 - token with convertToAssets method (e.g. sDAI)
Consequently, supported tokens include:
• ERC20 support for return True/revert, return True/False, return None
• ERC20 tokens can have arbitrary decimals (<=18)
• ERC20 tokens that rebase (either positive or fee on transfer)
• ERC20 tokens that have a rate oracle (e.g. wstETH, cbETH, sDAI, etc.) Oracle precision must be 10^18
• ERC4626 tokens with arbitrary percision (<=18) of Vault token and underlying asset
Rebasing Tokens¶
Rebasing Tokens
Pools including rebasing tokens work a bit differently compared to others. The internal **_balance()** function - which is used to calculate the coin balances within the pool - makes sure that LP's
keep all rebases.
def _balances() -> DynArray[uint256, MAX_COINS]:
@notice Calculates the pool's balances _excluding_ the admin's balances.
@dev If the pool contains rebasing tokens, this method ensures LPs keep all
rebases and admin only claims swap fees. This also means that, since
admin's balances are stored in an array and not inferred from read balances,
the fees in the rebasing token that the admin collects is immune to
slashing events.
result: DynArray[uint256, MAX_COINS] = empty(DynArray[uint256, MAX_COINS])
balances_i: uint256 = 0
for i in range(MAX_COINS_128):
if i == N_COINS_128:
if 2 in asset_types:
balances_i = ERC20(coins[i]).balanceOf(self) - self.admin_balances[i]
balances_i = self.stored_balances[i] - self.admin_balances[i]
return result
Dynamic Fees¶
Stableswap-NG introduces dynamic fees. The use of the offpeg_fee_multiplier allows the system to dynamically adjust fees based on the pool's state.
The internal _dynamic_fee() function calculates the fee based on the balances and rates of the tokens being exchanged. If the balances of the tokens being exchanged are highly imbalanced or
significantly differ from its peg, the fee is adjusted using the offpeg_fee_multiplier.
Dynamic Fee Formula¶
If the formulas below do not render, please make sure to refresh the site. A solution is being worked on.
Let's define some terms and variables for clarity:
• Let \(fee\) represent the fee, as retrieved by the method StableSwap.fee()
• Let \(fee_m\) denote the off-peg fee multiplier, sourced from StableSwap.offpeg_fee_multiplier()
• FEE_DENOMINATOR is a constant with a value of \(10^{10}\), representing the precision of the fee
• The terms \(rate_{i}\) and \(balance{i}\) refer to the specific rate and balance for coin \(i\), respectively, and similarly, \(rate_j\) and \(balance_j\) for coin \(j\)
• \(PRECISION_{i}\) and \(PRECISION_{j}\) are the precision constants for the respective coins
Given these, we define:
\(xp_{i} = \frac{{rate_{i} \times balance_{i}}}{{PRECISION_{i}}}\)
\(xp_{j} = \frac{{rate_{j} \times balance_{j}}}{{PRECISION_{j}}}\)
\(xp_{i}\) and \(xp_{j}\) are the token balances of the pool adjusted for decimals and the pool's internal rates (stored in stored_rates).
And we also have:
\(xps2 = (xp_{i} + xp_{j})^2\)
The dynamic fee is calculated by the following formula:
\[\text{dynamic fee} = \frac{{fee_{m} \times fee}}{\frac{(fee_{m} - 10^{10}) \times 4 \times xp_{i} \times xp_{j}}{xps2}+ 10^{10}}\]
dynamic_fee method
A_PRECISION: constant(uint256) = 100
MAX_COINS: constant(uint256) = 8
PRECISION: constant(uint256) = 10 ** 18
FEE_DENOMINATOR: constant(uint256) = 10 ** 10
def dynamic_fee(i: int128, j: int128, pool:address) -> uint256:
@notice Return the fee for swapping between `i` and `j`
@param i Index value for the coin to send
@param j Index value of the coin to recieve
@return Swap fee expressed as an integer with 1e10 precision
N_COINS: uint256 = StableSwapNG(pool).N_COINS()
fee: uint256 = StableSwapNG(pool).fee()
fee_multiplier: uint256 = StableSwapNG(pool).offpeg_fee_multiplier()
rates: DynArray[uint256, MAX_COINS] = empty(DynArray[uint256, MAX_COINS])
balances: DynArray[uint256, MAX_COINS] = empty(DynArray[uint256, MAX_COINS])
xp: DynArray[uint256, MAX_COINS] = empty(DynArray[uint256, MAX_COINS])
rates, balances, xp = self._get_rates_balances_xp(pool, N_COINS)
return self._dynamic_fee(xp[i], xp[j], fee, fee_multiplier)
def _dynamic_fee(xpi: uint256, xpj: uint256, _fee: uint256) -> uint256:
_offpeg_fee_multiplier: uint256 = self.offpeg_fee_multiplier
if _offpeg_fee_multiplier <= FEE_DENOMINATOR:
return _fee
xps2: uint256 = (xpi + xpj) ** 2
return (
(_offpeg_fee_multiplier * _fee) /
((_offpeg_fee_multiplier - FEE_DENOMINATOR) * 4 * xpi * xpj / xps2 + FEE_DENOMINATOR)
def _get_rates_balances_xp(pool: address, N_COINS: uint256) -> (
DynArray[uint256, MAX_COINS],
DynArray[uint256, MAX_COINS],
DynArray[uint256, MAX_COINS],
rates: DynArray[uint256, MAX_COINS] = StableSwapNG(pool).stored_rates()
balances: DynArray[uint256, MAX_COINS] = StableSwapNG(pool).get_balances()
xp: DynArray[uint256, MAX_COINS] = empty(DynArray[uint256, MAX_COINS])
for idx in range(MAX_COINS):
if idx == N_COINS:
xp.append(rates[idx] * balances[idx] / PRECISION)
return rates, balances, xp
Interactive Graph¶
The embedded graph has limited features, such as the inability to modify the axis. However, by clicking the "edit graph on desmos" button at the bottom right, one is redirected to the main Desmos
site. There, a wider range of functionalities is available, allowing for further adjustments and detailed exploration of the graph.
The new generation (NG) of stableswap introduces two new pool-built-in oracles:
• price oracle (spot and moving-average price)
• moving average D oracle
More on oracles here.
This new function allows the exchange of tokens without actually transfering tokens in, as the exchange is based on the change of the coins balances within the pool.
Users of this method are dex aggregators, arbitrageurs, or other users who do not wish to grant approvals to the contract. They can instead send tokens directly to the contract and call
Explore the exchange_received function's role in streamlining swaps without approvals, its efficiency benefits, and security considerations in a succinct article. Learn more about this innovative
feature for cost-effective, secure trading through Curve pools: How to Do Cheaper, Approval-Free Swaps. | {"url":"https://curve-docs-git-lending-curvedocs.vercel.app/stableswap-exchange/stableswap-ng/pools/overview/","timestamp":"2024-11-09T10:20:51Z","content_type":"text/html","content_length":"107252","record_id":"<urn:uuid:09045034-a806-4355-984c-c0ac32395d48>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00162.warc.gz"} |
Two Tank System: C MEX-File Modeling of Time-Continuous SISO System
This example shows how to perform IDNLGREY modeling based on C MEX model files. It uses a simple system where nonlinear state space modeling really pays off.
A Two Tank System
The objective is to model the liquid level of the lower tank of a laboratory scale two tank system, schematically shown in Figure 1.
Figure 1: Schematic view of a two tank system.
Input-Output Data
We start the modeling job by loading the available input-output data, which was simulated using the below IDNLGREY model structure, with noise added to the output. The twotankdata.mat file contains
one data set with 3000 input-output samples, generated using a sampling rate (Ts) of 0.2 seconds. The input u(t) is the voltage [V] applied to a pump, which generates an inflow to the upper tank. A
rather small hole at the bottom of this upper tank yields an outflow that goes into the lower tank, and the output y(t) of the two tank system is then the liquid level [m] of the lower tank. We
create an IDDATA object z to hold the tank data. For bookkeeping and documentation purposes we also specify channel names and units. This step is optional.
z = iddata(y, u, 0.2, 'Name', 'Two tanks');
set(z, 'InputName', 'Pump voltage', 'InputUnit', 'V', ...
'OutputName', 'Lower tank water level', 'OutputUnit', 'm', ...
'Tstart', 0, 'TimeUnit', 's');
The input-output data that will be used for estimation are shown in a plot window.
figure('Name', [z.Name ': input-output data']);
Figure 2: Input-output data from a two tank system.
Modeling the Two Tank System
The next step is to specify a model structure describing the two tank system. To do this, let x1(t) and x2(t) denote the water level in the upper and the lower tank, respectively. For each tank,
fundamental physics (mass balance) states that the change of water volume depends on the difference between in- and outflow as (i = 1, 2):
d/dt (Ai*xi(t)) = Qini(t) - Qouti(t)
where Ai [m^2] is the cross-sectional area of tank i and Qini(t) and Qouti(t) [m^3/s] are the inflow to and the outflow from tank i at time t.
For the upper tank, the inflow is assumed to be proportional to the voltage applied to the pump, i.e., Qin1(t) = k*u(t). Since the outlet hole of the upper tank is small, Bernoulli's law can be
applied, stating that the outflow is proportional to the square root of the water level, or more precisely that:
Qout1(t) = a1*sqrt(2*g*x1(t))
where a1 is the cross-sectional area of the outlet hole and g is the gravity constant. For the lower tank, the inflow equals the outflow from the upper tank, i.e., Qin2(t) = Qout1(t), and the outflow
is given by Bernoulli's law:
Qout2(t) = a2*sqrt(2*g*x2(t))
where a2 is the cross-sectional area of the outlet hole.
Put altogether these facts lead to the following state-space structure:
d/dt x1(t) = 1/A1*(k*u(t) - a1*sqrt(2*g*x1(t)))
d/dt x2(t) = 1/A2*(a1*sqrt(2*g*x1(t)) - a2*sqrt(2*g*x2(t)))
y(t) = x2(t)
Two Tank C MEX Model File
These equations are next put into a C MEX-file with 6 parameters (or constants), A1, k, a1, g, A2 and a2. The C MEX-file is normally a bit more involved than the corresponding file written using
MATLAB® language, but C MEX modeling generally gives a distinct advantage in terms of execution speed, especially for more complex models. A template C MEX-file is provided (see below) to help the
user to structure the code. For most applications, it suffices to define the number of outputs and to enter the code lines that describe dx and y into this template. An IDNLGREY C MEX-file should
always be structured to return two outputs:
dx: the right-hand side(s) of the state-space equation(s)
y: the right-hand side(s) of the output equation(s)
and it should take 3+Npo(+1) input arguments specified as follows:
t: the current time
x: the state vector at time t ([] for static models)
u: the input vector at time t ([] for time-series models)
p1, p2, ..., pNpo: the individual parameters (which can be real
scalars, column vectors or 2-dimensional matrices); Npo is here
the number of parameter objects, which for models with scalar
parameters coincide with the number of parameters Np
FileArgument: optional inputs to the model file
In our two tank system there are 6 scalar parameters and hence the number of input arguments to the C MEX modeling file should be 3+Npo = 3+6 = 9. The trailing 10:th argument can here be omitted as
no optional FileArgument is employed in this application.
Writing a C MEX modeling file is normally done in four steps:
1. Inclusion of C-libraries and definitions of the number of outputs.
2. Writing the function computing the right-hand side(s) of the state
equation(s), compute_dx.
3. Writing the function computing the right-hand side(s) of the output
equation(s), compute_y.
4. Writing the main interface function, which includes basic error
checking functionality, code for creating and handling input and
output arguments, and calls to compute_dx and compute_y.
Let us view the C MEX source file (except for some comments) for the two
tank system and based on this discuss these four items in some more
Figure 3: C MEX source code for the two tank system.
1. Two C-libraries mex.h and math.h are normally included to provide access to a number of MEX-related as well as mathematical functions. The number of outputs is also declared per modeling file
using a standard C-define:
/* Include libraries. */
#include "mex.h"
#include "math.h"
/* Specify the number of outputs here. */
#define NY 1
2-3. Next in the file we find the functions for updating the states, compute_dx, and the output, compute_y. Both these functions hold argument lists, with the output to be computed (dx or y) at
position 1, after which follows all variables and parameters required to compute the right-hand side(s) of the state and the output equations, respectively.
The first step in these functions is to unpack the model parameters that will be used in the subsequent equations. Any valid variable name (except for those used in the input argument list) can be
used to provide physically meaningful names of the individual parameters.
As is the case in C, the first element of an array is stored at position 0. Hence, dx[0] in C corresponds to dx(1) in MATLAB (or just dx in case it is a scalar), the input u[0] corresponds to u (or u
(1)), the parameter A1[0] corresponds to A1, and so on.
The two tank model file involves square root computations. This is enabled through the inclusion of the mathematical C library math.h. The math library realizes the most common trigonometric
functions (sin, cos, tan, asin, acos, atan, etc.), exponential (exp) and logarithms (log, log10), square root (sqrt) and power of functions (pow), and absolute value computations (fabs). The math.h
library must be included whenever any math.h function is used; otherwise it can be omitted. See "Tutorials on Nonlinear Grey Box Model Identification: Creating IDNLGREY Model Files" for further
details about the C math library.
4. The main interface function should almost always have the same content and for most applications no modification whatsoever is needed. In principle, the only part that might be considered for
changes is where the calls to compute_dx and compute_y are made. For static systems, one can leave out the call to compute_dx. In other situations, it might be desired to only pass the variables and
parameters referred in the state and output equations. For example, in the output equation of the two tank system, where only one state is used, one could very well shorten the input argument list
void compute_y(double *y, double *x)
and call compute_y as:
The input argument lists of compute_dx and compute_y might also be extended to include further variables inferred in the interface function, like the number of states and the number of parameters.
Once the model source file has been completed it must be compiled, which can be done from the MATLAB command prompt using the mex command; see "help mex". (This step is omitted here.)
When developing model specific C MEX-files it is often useful to start the work by copying the IDNLGREY C MEX template file. This template contains skeleton source code as well as detailed
instructions on how to customize the code for a particular application.
type IDNLGREY_MODEL_TEMPLATE.c
/* Copyright 2005-2006 The MathWorks, Inc. */
/* Template file for IDNLGREY model specification.
Use this file to create a MEX function that specifies the model
structure and equations. The MEX file syntax is
[dx, y] = mymodel(t, x, u, p1, p2, ..., pn, auxvar)
* t is the time (scalar).
* x is the state vector at time t (column vector).
* u is the vector of inputs at time t (column vector).
* p1, p2,... pn: values of the estimated parameters specified
in the IDNLGREY model.
* auxvar: a cell array containing auxiliary data in any format
* dx is the vector of state derivatives at time t (column vector).
* y is the vector of outputs at time t.
To create the MEX file "mymodel", do the following:
1) Save this template as "mymodel.c" (replace "mymodel" by the
name of your choice).
2) Define the number NY of outputs below.
3) Specify the state derivative equations in COMPUTE_DX below.
4) Specify the output equations in COMPUTE_Y below.
5) Build the MEX file using
>> mex mymodel.c
/* Include libraries. */
#include "mex.h"
#include "math.h"
/* Specify the number of outputs here. */
#define NY 2
/* State equations. */
void compute_dx(
double *dx, /* Vector of state derivatives (length nx). */
double t, /* Time t (scalar). */
double *x, /* State vector (length nx). */
double *u, /* Input vector (length nu). */
double **p, /* p[j] points to the j-th estimated model parameters (a double array). */
const mxArray *auxvar /* Cell array of additional data. */
Define the state equation dx = f(t, x, u, p[0],..., p[np-1], auvar)
in the body of this function.
Accessing the contents of auxvar:
Use mxGetCell to fetch pointers to individual cell elements, e.g.:
mxArray* auxvar1 = mxGetCell(auxvar, 0);
extracts the first cell element. If this element contains double
data, you may obtain a pointer to the double array using mxGetPr:
double *auxData = mxGetPr(auxvar1);
See MATLAB documentation on External Interfaces for more information
about functions that manipulate mxArrays.
/* Example code from ODE function for DCMOTOR example
used in idnlgreydemo1 (dcmotor_c.c) follows.
double *tau, *k; /* Estimated model parameters. */
tau = p[0];
k = p[1];
dx[0] = x[1]; /* x[0]: Angular position. */
dx[1] = -(1/tau[0])*x[1]+(k[0]/tau[0])*u[0]; /* x[1]: Angular velocity. */
/* Output equations. */
void compute_y(
double *y, /* Vector of outputs (length NY). */
double t, /* Time t (scalar). */
double *x, /* State vector (length nx). */
double *u, /* Input vector (length nu). */
double **p, /* p[j] points to the j-th estimated model parameters (a double array). */
const mxArray *auxvar /* Cell array of additional data. */
Define the output equation y = h(t, x, u, p[0],..., p[np-1], auvar)
in the body of this function.
Accessing the contents of auxvar: see the discussion in compute_dx.
/* Example code from ODE function for DCMOTOR example
used in idnlgreydemo1 (dcmotor_c.c) follows.
y[0] = x[0]; /* y[0]: Angular position. */
y[1] = x[1]; /* y[1]: Angular velocity. */
/*----------------------------------------------------------------------- *
DO NOT MODIFY THE CODE BELOW UNLESS YOU NEED TO PASS ADDITIONAL
INFORMATION TO COMPUTE_DX AND COMPUTE_Y
To add extra arguments to compute_dx and compute_y (e.g., size
information), modify the definitions above and calls below.
void mexFunction(int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[])
/* Declaration of input and output arguments. */
double *x, *u, **p, *dx, *y, *t;
int i, np, nu, nx;
const mxArray *auxvar = NULL; /* Cell array of additional data. */
if (nrhs < 3) {
"At least 3 inputs expected (t, u, x).");
/* Determine if auxiliary variables were passed as last input. */
if ((nrhs > 3) && (mxIsCell(prhs[nrhs-1]))) {
/* Auxiliary variables were passed as input. */
auxvar = prhs[nrhs-1];
np = nrhs - 4; /* Number of parameters (could be 0). */
} else {
/* Auxiliary variables were not passed. */
np = nrhs - 3; /* Number of parameters. */
/* Determine number of inputs and states. */
nx = mxGetNumberOfElements(prhs[1]); /* Number of states. */
nu = mxGetNumberOfElements(prhs[2]); /* Number of inputs. */
/* Obtain double data pointers from mxArrays. */
t = mxGetPr(prhs[0]); /* Current time value (scalar). */
x = mxGetPr(prhs[1]); /* States at time t. */
u = mxGetPr(prhs[2]); /* Inputs at time t. */
p = mxCalloc(np, sizeof(double*));
for (i = 0; i < np; i++) {
p[i] = mxGetPr(prhs[3+i]); /* Parameter arrays. */
/* Create matrix for the return arguments. */
plhs[0] = mxCreateDoubleMatrix(nx, 1, mxREAL);
plhs[1] = mxCreateDoubleMatrix(NY, 1, mxREAL);
dx = mxGetPr(plhs[0]); /* State derivative values. */
y = mxGetPr(plhs[1]); /* Output values. */
Call the state and output update functions.
Note: You may also pass other inputs that you might need,
such as number of states (nx) and number of parameters (np).
You may also omit unused inputs (such as auxvar).
For example, you may want to use orders nx and nu, but not time (t)
or auxiliary data (auxvar). You may write these functions as:
compute_dx(dx, nx, nu, x, u, p);
compute_y(y, nx, nu, x, u, p);
/* Call function for state derivative update. */
compute_dx(dx, t[0], x, u, p, auxvar);
/* Call function for output update. */
compute_y(y, t[0], x, u, p, auxvar);
/* Clean up. */
Also see "Creating IDNLGREY Model Files" example for more details on IDNLGREY C MEX model files.
Creating a Two Tank IDNLGREY Model Object
The next step is to create an IDNLGREY object describing the two tank system. For convenience we also set some bookkeeping information about the inputs and outputs (name and units).
FileName = 'twotanks_c'; % File describing the model structure.
Order = [1 1 2]; % Model orders [ny nu nx].
Parameters = {0.5; 0.0035; 0.019; ...
9.81; 0.25; 0.016}; % Initial parameters.
InitialStates = [0; 0.1]; % Initial value of initial states.
Ts = 0; % Time-continuous system.
nlgr = idnlgrey(FileName, Order, Parameters, InitialStates, Ts, ...
'Name', 'Two tanks');
set(nlgr, 'InputName', 'Pump voltage', 'InputUnit', 'V', ...
'OutputName', 'Lower tank water level', 'OutputUnit', 'm', ... ...
'TimeUnit', 's');
We continue to add information about the names and the units of the states and the model parameters via the commands SETINIT and SETPAR. Furthermore, both states x1(t) and x2(t) are tank levels that
cannot be negative, and thus we also specify that x1(0) and x2(0) >= 0 via the 'Minimum' property. In fact, we also know that all model parameters ought to be strictly positive. We therefore set the
'Minimum' property of all parameters to some small positive value (eps(0)). These settings imply that constraint estimation will be carried out in the upcoming estimation step (i.e., the estimated
model will be a model such that all entered constraints are honored).
nlgr = setinit(nlgr, 'Name', {'Upper tank water level' 'Lower tank water level'});
nlgr = setinit(nlgr, 'Unit', {'m' 'm'});
nlgr = setinit(nlgr, 'Minimum', {0 0}); % Positive levels!
nlgr = setpar(nlgr, 'Name', {'Upper tank area' ...
'Pump constant' ...
'Upper tank outlet area' ...
'Gravity constant' ...
'Lower tank area' ...
'Lower tank outlet area'});
nlgr = setpar(nlgr, 'Unit', {'m^2' 'm^3/(s*V)' 'm^2' 'm/(s^2)' 'm^2' 'm^2'});
nlgr = setpar(nlgr, 'Minimum', num2cell(eps(0)*ones(6,1))); % All parameters > 0!
The cross-sectional areas (A1 and A2) of the two tanks can rather accurately be determined. We therefore treat these and g as constants and verify that the 'Fixed' field is properly set for all 6
parameters through the command GETPAR. All in all, this means that 3 of the model parameters will be estimated.
nlgr.Parameters(1).Fixed = true;
nlgr.Parameters(4).Fixed = true;
nlgr.Parameters(5).Fixed = true;
getpar(nlgr, 'Fixed')
ans=6×1 cell array
Performance of the Initial Two Tank Model
Before estimating the free parameters k, a1 and a2 we simulate the system using the initial parameter values. We use the default differential equation solver (a Runge-Kutta 45 solver with adaptive
step length adjustment) and set the absolute and relative error tolerances to rather small values (1e-6 and 1e-5, respectively). Notice that the COMPARE command, when called with two input arguments,
as default will estimate all initial state(s) regardless of whether any initial state has been defined to be 'Fixed'. In order to only estimate the free initial state(s), call COMPARE with a third
and a fourth input argument as follows: compare(z, nlgr, 'init', 'm'); as both initial states of the tank model by default are 'Fixed', no initial state estimation will be performed by this command.
nlgr.SimulationOptions.AbsTol = 1e-6;
nlgr.SimulationOptions.RelTol = 1e-5;
compare(z, nlgr);
Figure 4: Comparison between true output and the simulated output of the initial two tank model.
The simulated and true outputs are shown in a plot window, and as can be seen the fit is not so impressive.
Parameter Estimation
In order to improve the fit, the 3 free parameters are next estimated using NLGREYEST. (Since, by default, the 'Fixed' fields of all initial states are false, no estimation of the initial states will
be done in this call to the estimator.)
nlgr = nlgreyest(z, nlgr, nlgreyestOptions('Display', 'on'));
Performance of the Estimated Two Tank Model
To investigate the performance of the estimated model, a simulation of it is performed (the initial states are here reestimated).
Figure 5: Comparison between true output and the simulated output of the estimated two tank model.
The agreement between the true and the simulated outputs is quite good. A remaining question is, however, if the two tank system can be accurately described using a simpler and linear model
structure. To answer this, let us try to fit the data to some standard linear model structures, and then use COMPARE to see how well these models capture the dynamics of the tanks.
nk = delayest(z);
arx22 = arx(z, [2 2 nk]); % Second order linear ARX model.
arx33 = arx(z, [3 3 nk]); % Third order linear ARX model.
arx44 = arx(z, [4 4 nk]); % Fourth order linear ARX model.
oe22 = oe(z, [2 2 nk]); % Second order linear OE model.
oe33 = oe(z, [3 3 nk]); % Third order linear OE model.
oe44 = oe(z, [4 4 nk]); % Fourth order linear OE model.
sslin = ssest(z); % State-space model (order determined automatically)
compare(z, nlgr, 'b', arx22, 'm-', arx33, 'm:', arx44, 'm--', ...
oe22, 'g-', oe33, 'g:', oe44, 'g--', sslin, 'r-');
Figure 6: Comparison between true output and the simulated outputs of a number of estimated two tank models.
The comparison plot clearly reveals that the linear models cannot pick up all dynamics of the two tank system. The estimated nonlinear IDNLGREY model on the other hand shows an excellent fit to the
true output. In addition, the IDNLGREY model parameters are also well in line with those used to generate the true output. In the following display computations, we are using the command GETPVEC,
which returns a parameter vector created from the structure array holding the model parameters of an IDNLGREY object.
disp(' True Estimated parameter vector');
True Estimated parameter vector
ptrue = [0.5; 0.005; 0.02; 9.81; 0.25; 0.015];
fprintf(' %1.4f %1.4f\n', [ptrue'; getpvec(nlgr)']);
0.5000 0.5000
0.0050 0.0049
0.0200 0.0200
9.8100 9.8100
0.2500 0.2500
0.0150 0.0147
The prediction errors obtained using PE are small and look very much like random noise.
Figure 7: Prediction errors obtained for the estimated IDNLGREY two tank model.
Let us also investigate what happens if the input voltage is increased from 5 to 6, 7, 8, 9, and 10 V in a step-wise manner. We do this by calling STEP with different specified step amplitudes
starting from a fixed offset of 5 Volts. The step response configuration is facilitated by a dedicated option-set created by stepDataOptions:
figure('Name', [nlgr.Name ': step responses']);
t = (-20:0.1:80)';
Opt = stepDataOptions('InputOffset',5,'StepAmplitude',6);
step(nlgr, t, 'b', Opt);
hold on
Opt.StepAmplitude = 7; step(nlgr, t, 'g', Opt);
Opt.StepAmplitude = 8; step(nlgr, t, 'r', Opt);
Opt.StepAmplitude = 9; step(nlgr, t, 'm', Opt);
Opt.StepAmplitude = 10; step(nlgr, t, 'k', Opt);
grid on;
legend('5 -> 6 V', '5 -> 7 V', '5 -> 8 V', '5 -> 9 V', '5 -> 10 V', ...
'Location', 'Best');
Figure 8: Step responses obtained for the estimated IDNLGREY two tank model.
By finally using the PRESENT command, we get summary information about the estimated model:
nlgr =
Continuous-time nonlinear grey-box model defined by 'twotanks_c' (MEX-file):
dx/dt = F(t, x(t), u(t), p1, ..., p6)
y(t) = H(t, x(t), u(t), p1, ..., p6) + e(t)
with 1 input(s), 2 state(s), 1 output(s), and 3 free parameter(s) (out of 6).
u(1) Pump voltage(t) [V]
States: Initial value
x(1) Upper tank water level(t) [m] xinit@exp1 0 (fixed) in [0, Inf]
x(2) Lower tank water level(t) [m] xinit@exp1 0.1 (fixed) in [0, Inf]
y(1) Lower tank water level(t) [m]
Parameters: Value Standard Deviation
p1 Upper tank area [m^2] 0.5 0 (fixed) in ]0, Inf]
p2 Pump constant [m^3/(s*V)] 0.00488584 0.0259032 (estimated) in ]0, Inf]
p3 Upper tank outlet area [m^2] 0.0199719 0.0064682 (estimated) in ]0, Inf]
p4 Gravity constant [m/(s^2)] 9.81 0 (fixed) in ]0, Inf]
p5 Lower tank area [m^2] 0.25 0 (fixed) in ]0, Inf]
p6 Lower tank outlet area [m^2] 0.0146546 0.0776058 (estimated) in ]0, Inf]
Name: Two tanks
Termination condition: Change in cost was less than the specified tolerance..
Number of iterations: 8, Number of function evaluations: 9
Estimated using Solver: ode45; Search: lsqnonlin on time domain data "Two tanks".
Fit to estimation data: 97.35%
FPE: 2.419e-05, MSE: 2.414e-05
More information in model's "Report" property.
In this example we have shown:
1. how to use C MEX-files for IDNLGREY modeling, and
2. provided a rather simple example where nonlinear state-space
modeling shows good potential | {"url":"https://uk.mathworks.com/help/ident/ug/two-tank-system-c-mex-file-modeling-of-time-continuous-siso-system.html","timestamp":"2024-11-12T10:17:11Z","content_type":"text/html","content_length":"104115","record_id":"<urn:uuid:0b066758-26f4-4169-820d-fa6a8cc86260>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00107.warc.gz"} |
Shape Flashcards
16 Shape Flashcards
Download Free Shapes Flashcards here >>>
Learn geometric shapes with your children & toddlers.
You’ll find the following shape cards in this printable PDF file:
Circle, Triangle, Rectangle, Square,
Oval / Ellipse, Right triangle, Heart, Diamond,
Star, Parallelogram, Trapezoid, Crescent,
Pentagon, Arrow, Semicircle, Octagon.
All the cards have a colored picture of a shape and its name beneath. Some of the cards are basic, you can teach them to any small students. Others are the advanced shapes (e.g., Parallelogram,
Trapezoid, Crescent). Tell about them only to those who already know all simple shapes.
Save the document, print the colored flashcards, and use them in the classroom or at home to memorize shapes with kids.
3 thoughts on “16 Shape Flashcards”
1. uhh … that’s a hexagon (6 sides) not a pentagon (5 sides)
1. fixed that, thanks!
2. The names for the square and rectangle are on the wrong shapes. The square is labelled ‘rectangle’ and vice versa. | {"url":"https://flashcard.online/shapes-flashcards/","timestamp":"2024-11-12T17:04:56Z","content_type":"text/html","content_length":"48476","record_id":"<urn:uuid:9f7b54fd-d407-4a42-9a2d-a151356eccb8>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00871.warc.gz"} |
Exercise-10: Three methods of allocating joint costs - Accounting For Management
Exercise-10: Three methods of allocating joint costs
Sun Inc. produces four joint products – product A, product B, product C and product D. The joint production cost at split-off point is $70,000. The data for the month of April is given below:
Required: Allocate joint production cost using following three methods:
1. The market value method
2. The average unit cost method
3. The weighted average method
1. The market value method:
*The joint cost is 70% of the hypothetical market value. Hypothetical market value at split-off point is equal to ultimate market value less processing cost incurred after split-off point. For
product A, this value is $26,500 (= $27,500 – $1,500).
2. Average unit cost method:
*Joint cost/Total number of units produced
= $70,000/50,000
= $1.40 per unit
3. Weighted average method:
Under weighted average method, we need to use the weighted number of units to allocate joint cost. For this purpose, we would simply multiply the number of units produced by the weighted factors
given in the problem. An appropriate format for the solution is given below:
*Joint cost/Total number of weighted units
= $70,000/140,000
= $.50 per weighted unit
Help us grow by sharing our content ♡ | {"url":"https://www.accountingformanagement.org/problem-4-jpabp/","timestamp":"2024-11-05T14:01:09Z","content_type":"text/html","content_length":"44780","record_id":"<urn:uuid:aa4e23ff-c0b9-4a0f-baf1-475b194a7bf3>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00710.warc.gz"} |
Why Einstein Could Not Solve The EPR Paradox Though He Could Have
Usually, the Einstein-Podolsky-Rosen paradox is presented as if it potentially conflicts with the theory of relativity. This is because the correlation between Alice’s and Bob’s measurements seems to
travel with superluminal speeds in one real world. The solution [1] of the EPR paradox shows that this view is upside-down, which is the reason it took so long to solve it satisfactorily.
Scientists and philosophers where trying to understand the quantum mechanics involved without bothering with relativity, because the problematic seems to conflict with relativity. However, in that
way, the actual solution looks suspect:
A non-relativistic universe would need to quantum split everywhere into many different ones all the time (every moment, infinite universes infinitely often). Why the hell should it do so? What makes
it split? How could this ever be later turned into a relativistic, let alone general relativistic model where the splits would have to have a certain shape through curved space-time? For all these
and more reasons, the many worlds concept is often ridiculed.
What has been missed consistently by the community interested in the philosophy of physics, partially due to its obsession with "hyperspace foliations" (~ slices of the real world out there), is that
special relativity already deconstructs the world into a collection of different observers’ past light cones (~histories, memory contents, minds) in a sort of ‘temporal modal realism’.
Everett relativity in the EPR setup does not conflict with special relativity; on the contrary, it is suspect without special relativity and natural with it. Branching only needs to occur at the
observation events. The funny thing is: This really is already obvious from special relativity basically since 1915. If you understand special relativity, there are only two options:
1)You either believe in one determined “block universe” where everything is predetermined, which is ridiculous lest you think that the very fundament of nature and the whole universe is there to let
you on some Tuesday afternoon throw a coin and get heads instead of tails,
2) or you know that totality must in some way ultimately be describable by all the possible past light cones, i.e. once you are on a truly fundamental level, the you that throws tails is exactly as
real as the one that gets heads, and the difference between those two propagates at most with the speed of light through the model that describes spatial relations (the universe).
If Einstein would have been less of a direct realist, he could have been twice the genius he was already and come up with Everett-relativity long before Everett. I know, this is misleading;
Einstein’s, Feynman’s, and many other physicist's successes are due to them being originally direct realists*, because that makes you focus on “real stuff” like elevators falling in gravity fields.
Nevertheless, my point is that taking special relativity seriously renders the many world concept natural. For example the question about where the splits of the world branchings are supposed to be
located becomes trivial: Along the light cones of course. If you know special relativity, you then also know that the “split” is thereby of zero eigen-length and happens in a sense (I said: in a
sense!) instantaneously.
Quantum decoherence “dislocates” [2] via interactions and therefore at the speed of light. The model that resolves the EPR paradox works therefore without superluminal velocities. The last step that
turns the model quantum physical is a local branching that destroys the very grounds on which absolute actualization makes sense. Einstein locality stays; realism is modified.
Similar conclusions have been drawn before. The Heisenberg representation of the many world interpretation is local [3], but there are no models. The simplicity of the new model and the fact that a
single, local modification turns it into quantum physics while destroying its realism shows that not every many-worlds model is a quantum world and quantum physics is not synonymous with multiverses
or modal realism.
This further corroborates that the many-world aspect of the universe, which is philosophically of course self-evident,
should be understood as a relativistic rather than quantum physical phenomenon
The gist is that relativity indicates modal realism and that quantum = necessarily modal! The main point is not that the world is local. The model is local! I do not care about locality much, because
only if space is viewed as being ‘out there’ does it even make sense to defend its locality. Once the world is not out there anyway (more in our heads, in a sense, with much poetic license), it does
not matter whether its consistent description involves locality or non-locality, or whether the world is best consistently described as made out of green cheese!
* This is an important aspect that naturally also hinders physicists from progressing. Without realism, you are a lousy physicist right from the start, but if you cannot shake realism along the way,
you are going to be a lousy physicist in the end.
[1] S. Vongehr: "Many Worlds Model resolving the Einstein Podolsky Rosen paradox via a Direct Realism to Modal Realism Transition that preserves Einstein Locality." arXiv:1108.1674v1 [quant-ph]
(2011). UPDATE: This reference is the first paper on the possibility of such models, but the models have now actually been constructed and are much better explained in S. Vongehr: “Against Absolute
Actualization: Three "Non-Localities" and Failure of Model-External Randomness made easy with Many-Worlds Models including Stronger Bell-Violation and Correct QM Probability” http://arxiv.org/abs/
1311.5419 (2013)
[2] H. Dieter Zeh: “Quantum discreteness is an illusion.” arxiv:0809.2904
[3] David Deutsch, Patrick Hayden: “Information Flow in Entangled Quantum Systems.” Proc. R. Soc. London A456, 1759-1774 (1999) | {"url":"https://www.science20.com/alpha_meme/why_einstein_could_not_solve_epr_paradox_though_he_could_have-81602","timestamp":"2024-11-09T11:15:25Z","content_type":"text/html","content_length":"39848","record_id":"<urn:uuid:ff61e724-6bf1-4db1-a022-5c847979ff20>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00170.warc.gz"} |
1. Exponential Smoothing
20 Exponential Smoothing
Learn about the Exponential Smoothing algorithm.
• About Exponential Smoothing
Exponential smoothing is a forecasting method for time series data. It is a moving average method where exponentially decreasing weights are assigned to past observations.
• Data Preparation for Exponential Smoothing Models
Prepare your data for exponential smoothing by providing input data, aggregation methods, and model build parameters.
20.1 About Exponential Smoothing
Exponential smoothing is a forecasting method for time series data. It is a moving average method where exponentially decreasing weights are assigned to past observations.
Exponential smoothing methods have been widely used in forecasting for over half a century. A forecast is a prediction based on historical data and patterns. preIt has applications at the strategic,
tactical, and operation level. For example, at a strategic level, forecasting is used for projecting return on investment, growth and the effect of innovations. At a tactical level, forecasting is
used for projecting costs, inventory requirements, and customer satisfaction. At an operational level, forecasting is used for setting targets and predicting quality and conformance with standards.
In its simplest form, exponential smoothing is a moving average method with a single parameter which models an exponentially decreasing effect of past levels on future values. With a variety of
extensions, exponential smoothing covers a broader class of models than other well-known approaches, such as the Box-Jenkins auto-regressive integrated moving average (ARIMA) approach. Oracle Machine
Learning for SQL implements exponential smoothing using a state of the art state space method that incorporates a single source of error (SSOE) assumption which provides theoretical and performance
Exponential smoothing is extended to the following:
• A matrix of models that mix and match error type (additive or multiplicative), trend (additive, multiplicative, or none), and seasonality (additive, multiplicative, or none)
• Models with damped trends.
• Models that directly handle irregular time series and time series with missing values.
• Multiple time series models
See Also:
Ord, J.K., et al, Time Series Forecasting: The Case for the Single Source of Error State Space Approach, Working Paper, Department of Econometrics and Business Statistics, Monash University, VIC
3800, Australia, April 2, 2005.
20.1.1 Exponential Smoothing Models
Exponential Smoothing models are a broad class of forecasting models that are intuitive, flexible, and extensible.
Members of this class include simple, single parameter models that predict the future as a linear combination of a previous level and a current shock. Extensions can include parameters for linear or
non-linear trend, trend damping, simple or complex seasonality, related series, various forms of non-linearity in the forecasting equations, and handling of irregular time series.
Exponential smoothing assumes that a series extends infinitely into the past, but that influence of past on future, decays smoothly and exponentially fast. The smooth rate of decay is expressed by
one or more smoothing constants. The smoothing constants are parameters that the model estimates. The assumption is made practical for modeling real world data by using an equivalent recursive
formulation that is only expressed in terms of an estimate of the current level based on prior history and a shock to that estimate dependent on current conditions only.The procedure requires an
estimate for the time period just prior to the first observation, that encapsulates all prior history. This initial observation is an additional model parameter whose value is estimated by the
modeling procedure.
Components of ESM such as trend and seasonality extensions, can have an additive or multiplicative form. The simpler additive models assume that shock, trend, and seasonality are linear effects
within the recursive formulation.
20.1.2 Simple Exponential Smoothing
Simple exponential smoothing assumes the data fluctuates around a stationary mean, with no trend or seasonal pattern.
In a simple Exponential Smoothing model, each forecast (smoothed value) is computed as the weighted average of the previous observations, where the weights decrease exponentially depending on the
value of smoothing constant α. Values of the smoothing constant, α, near one, put almost all weight on the most recent observations. Values of α near zero allows the distant past observations to have
a large influence.
20.1.3 Models with Trend but No Seasonality
The preferred form of additive (linear) trend is sometimes called Holt’s method or double exponential smoothing.
Models with trend add a smoothing parameter γ and optionally a damping parameter φ. The damping parameter smoothly dampens the influence of past linear trend on future estimates of level, often
improving accuracy.
20.1.4 Models with Seasonality but No Trend
When the time series average does not change over time (stationary), but is subject to seasonal fluctuations, the appropriate model has seasonal parameters but no trend.
Seasonal fluctuations are assumed to balance out over periods of length m, where m is the number of seasons, For example, m=4 might be used when the input data are aggregated quarterly. For models
with additive errors, the seasonal parameters must sum to zero. For models with multiplicative errors, the product of seasonal parameters must be one.
20.1.5 Models with Trend and Seasonality
Holt and Winters introduced both trend and seasonality in an Exponential Smoothing model.
The original model, also known as Holt-Winters or triple exponential smoothing, considered an additive trend and multiplicative seasonality. Extensions include models with various combinations of
additive and multiplicative trend, seasonality and error, with and without trend damping.
20.1.6 Prediction Intervals
To compute prediction intervals, an Exponential Smoothing (ESM) model is divided into three classes.
The simplest class is the class of linear models, which include, among others, simple ESM, Holt’s method, and additive Holt-Winters. Class 2 models (multiplicative error, additive components) make an
approximate correction for violations of the Normality assumption. Class 3 modes use a simple simulation approach to calculate prediction intervals.
20.2 Data Preparation for Exponential Smoothing Models
Prepare your data for exponential smoothing by providing input data, aggregation methods, and model build parameters.
To build an ESM model, you must supply the following :
• Input data
• An aggregation level and method, if the case id is a date type
• Partitioning column, if the data are partitioned
In addition, for a greater control over the build process, the user may optionally specify model build parameters, all of which have defaults:
• Model
• Error type
• Optimization criterion
• Forecast Window
• Confidence level for forecast bounds
• Missing value handling
• Whether the input series is evenly spaced
20.2.1 Input Data
Time series analysis requires ordered input data. Hence, each data row must consist of an [index, value] pair, where the index specifies the ordering.
When you create an Exponential Smoothing (ESM) model using the CREATE_MODEL or the CREATE_MODEL2 procedure, the CASE_ID_COLUMN_NAME and the TARGET_COLUMN_NAME parameters are used to specify the
columns used to compute the input indices and the observed time series values, respectively. The time column bears Oracle number, or Oracle date, timestamp, timestamp with time zone, or timestamp
with local time zone. When the case id column is of type Oracle NUMBER, the model considers the input time series to be equally spaced. Only the ordinal position matters, with a lower number
indicating a later time. In particular, the input time series is sorted based on the value of case_id (time label). The case_id column cannot contain missing values. To indicate a gap, the value
column can contain missing values as NULL. The magnitude of the difference between adjacent time labels is irrelevant and is not used to calculate the spacing or gap size. Integer numbers passed as
CASE_ID are assumed to be non-negative.
ESM also supports partitioned models and in such cases, the input table contains an extra column specifying the partition. All [index, value] pairs with the same partition ID form one complete time
series. The Exponential Smoothing algorithm constructs models for each partition independently, although all models use the same model settings.
Data properties may result in a warning notice, or settings may be disregarded. If the user sets a model with a multiplicative trend, multiplicative seasonality, or both, and the data contains values
Y[t]<= 0, the model type is set to default. If the series contains fewer values than the number of seasons given by the user, then the seasonality specifications are ignored and a warning is issued.
If the user has selected a list of predictor series using the parameter EXSM_SERIES_LIST, the input data can also include up to twenty additional time series columns.
20.2.2 Accumulation
Use accumulation procedures for date-type columns to generate equally spaced time series data.
For the Exponential Smoothing algorithm, the accumulation procedure is applied when the column is a date type (date, datetime, timestamp, timestamp with timezone, or timestamp with local timezone).
The case id can be a NUMBER column whose sort index represents the position of the value in the time series sequence of values. The case id column can also be a date type. A date type is accumulated
in accordance with a user specified accumulation window. Regardless of type, the case id is used to transform the column into an equally spaced time series. No accumulation is applied for a case id
of type NUMBER. As an example, consider a time series about promotion events. The time column contains the date of each event, and the dates can be unequally spaced. The user must specify the spacing
interval, which is the spacing of the accumulated or transformed equally spaced time series. In the example, if the user specifies the interval to be month, then an equally spaced time series with
profit for each calendar month is generated from the original time series. Setting EXSM_INTERVAL is used to specify the spacing interval. The user must also specify a value for EXSM_ACCUMULATE, for
example, EXSM_ACCU_MAX, in which case the equally spaced monthly series would contain the maximum profit over all events that month as the observed time series value.
20.2.3 Missing Value
Handle missing values effectively in your time series data for reliable exponential smoothing models.
Input time series can contain missing values. A
entry in the target column indicates a missing value. When the time column is of the type datetime, the accumulation procedure can also introduce missing values. The setting
can be used to specify how to handle missing values. The special value
indicates that, if the series contains missing values it is to be treated as an irregular time series.
Missing value handling setting must be compatible with model setting, otherwise an error is thrown.
20.2.4 Prediction
Specify the prediction window for your exponential smoothing model to generate accurate forecasts.
Setting EXSM_PREDICTION_STEP can be used to specify the prediction window. The prediction window is expressed in terms of number of intervals (setting EXSM_INTERVAL), when the time column is of the
type datetime. If the time column is a number then the prediction window is the number of steps to forecast. Regardless of whether the time series is regular or irregular, EXSM_PREDICTION_STEP
specifies the prediction window.
The term hyperparameter is also interchangeably used for model setting.
20.2.5 Parallellism by Partition
Enhance performance by processing time series data in parallel, using partitioning for efficient model building.
For example, a user can choose PRODUCT_ID as one partition column and can generate forecasts for different products in a model build. Although a distinct smoothing model is built for each partition,
all partitions share the same model settings. For example, if setting EXSM_MODEL is set to EXSM_SIMPLE, all partition models will be simple Exponential Smoothing models. Time series from different
partitions can be distributed to different processes and processed in parallel. The model for each time series is built serially.
20.2.6 Initial Value Optimization
Optimize initial values for long seasonal cycles for improved performance.
This is in contrast to standard ESM optimization, in which the initial values are adjusted during the optimization process to minimize error. Optimizing only the level, trend, and seasonality
parameters rather than the initial values can result in significant performance improvements and faster optimization convergence. When domain knowledge indicates that long seasonal variation is a
significant contributor to an accurate forecast, this approach is appropriate. Despite the performance benefits, Oracle does not recommend disabling the optimization of the initial values for typical
short seasonal cycles because it may result in model overfitting and less reliable confidence bounds. | {"url":"https://docs.oracle.com/en/database/oracle/machine-learning/oml4sql/23/dmcon/exponential-smoothing.html#DMCON-GUID-04F53FA3-467F-4A87-B59E-D8BE034A0D4C","timestamp":"2024-11-09T18:03:25Z","content_type":"text/html","content_length":"51065","record_id":"<urn:uuid:ea4d7616-c5d2-45da-9fc7-6da7ac198e78>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00622.warc.gz"} |
For all values of A,B,C and P,Q,R, show that∣∣cos(A−P)cos(B−P)... | Filo
For all values of and , show that
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Question Text For all values of and , show that
Topic Determinants
Subject Mathematics
Class Class 12
Answer Type Text solution:1
Upvotes 123 | {"url":"https://askfilo.com/math-question-answers/for-all-values-of-a-b-c-and-p-q-r-show-thatleftbeginarrayccccos-a-p-cos-a-q-cos","timestamp":"2024-11-12T06:57:17Z","content_type":"text/html","content_length":"781504","record_id":"<urn:uuid:09a2f208-664a-49da-8eb6-f1f65d0bfb72>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00546.warc.gz"} |
to Arpent
Rod to Arpent Converter
β Switch toArpent to Rod Converter
How to use this Rod to Arpent Converter π €
Follow these steps to convert given length from the units of Rod to the units of Arpent.
1. Enter the input Rod value in the text field.
2. The calculator converts the given Rod into Arpent in realtime β using the conversion formula, and displays under the Arpent label. You do not need to click any button. If the input changes,
Arpent value is re-calculated, just like that.
3. You may copy the resulting Arpent 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 Rod to Arpent?
The formula to convert given length from Rod to Arpent is:
Length[(Arpent)] = Length[(Rod)] / 11.63636362625536
Substitute the given value of length in rod, i.e., Length[(Rod)] in the above formula and simplify the right-hand side value. The resulting value is the length in arpent, i.e., Length[(Arpent)].
Calculation will be done after you enter a valid input.
Consider that a boundary fence is 40 rods long.
Convert this length from rods to Arpent.
The length in rod is:
Length[(Rod)] = 40
The formula to convert length from rod to arpent is:
Length[(Arpent)] = Length[(Rod)] / 11.63636362625536
Substitute given weight Length[(Rod)] = 40 in the above formula.
Length[(Arpent)] = 40 / 11.63636362625536
Length[(Arpent)] = 3.4375
Final Answer:
Therefore, 40 rd is equal to 3.4375 arpent.
The length is 3.4375 arpent, in arpent.
Consider that a farmer marks a field boundary using 25 rods.
Convert this distance from rods to Arpent.
The length in rod is:
Length[(Rod)] = 25
The formula to convert length from rod to arpent is:
Length[(Arpent)] = Length[(Rod)] / 11.63636362625536
Substitute given weight Length[(Rod)] = 25 in the above formula.
Length[(Arpent)] = 25 / 11.63636362625536
Length[(Arpent)] = 2.1484
Final Answer:
Therefore, 25 rd is equal to 2.1484 arpent.
The length is 2.1484 arpent, in arpent.
Rod to Arpent Conversion Table
The following table gives some of the most used conversions from Rod to Arpent.
Rod (rd) Arpent (arpent)
0 rd 0 arpent
1 rd 0.08593750007 arpent
2 rd 0.1719 arpent
3 rd 0.2578 arpent
4 rd 0.3438 arpent
5 rd 0.4297 arpent
6 rd 0.5156 arpent
7 rd 0.6016 arpent
8 rd 0.6875 arpent
9 rd 0.7734 arpent
10 rd 0.8594 arpent
20 rd 1.7188 arpent
50 rd 4.2969 arpent
100 rd 8.5938 arpent
1000 rd 85.9375 arpent
10000 rd 859.375 arpent
100000 rd 8593.75 arpent
A rod is a unit of length used in land measurement and surveying. One rod is equivalent to 16.5 feet or approximately 5.0292 meters.
The rod is defined as 16.5 feet, providing a measurement that is useful for various applications in land surveying, agriculture, and construction.
Rods are commonly used in tasks such as property measurement, plotting land, and agricultural practices. The unit provides a practical measurement for shorter distances and has historical
significance in land surveying.
An arpent is a historical unit of length used primarily in French-speaking regions and in land measurement. One arpent is approximately equivalent to 192.75 feet or 58.66 meters.
The arpent was used in various regions, including France and the former French colonies, to measure land and property. Its length could vary slightly depending on the specific region and historical
Arpents were used in land surveying and agriculture, particularly in historical and regional contexts. Although less common today, the unit provides historical insight into land measurement practices
and regional variations in measurement standards.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Rod to Arpent in Length?
The formula to convert Rod to Arpent in Length is:
Rod / 11.63636362625536
2. Is this tool free or paid?
This Length conversion tool, which converts Rod to Arpent, is completely free to use.
3. How do I convert Length from Rod to Arpent?
To convert Length from Rod to Arpent, you can use the following formula:
Rod / 11.63636362625536
For example, if you have a value in Rod, you substitute that value in place of Rod in the above formula, and solve the mathematical expression to get the equivalent value in Arpent. | {"url":"https://convertonline.org/unit/?convert=rods-arpents","timestamp":"2024-11-09T14:19:54Z","content_type":"text/html","content_length":"90438","record_id":"<urn:uuid:70b54daf-c57c-4c05-95fd-0f5e63da2259>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00739.warc.gz"} |
Impact of different estimations of the background-error covariance matrix on climate reconstructions based on data assimilation
Articles | Volume 15, issue 4
© Author(s) 2019. This work is distributed under the Creative Commons Attribution 4.0 License.
Impact of different estimations of the background-error covariance matrix on climate reconstructions based on data assimilation
Data assimilation has been adapted in paleoclimatology to reconstruct past climate states. A key component of some assimilation systems is the background-error covariance matrix, which controls how
the information from observations spreads into the model space. In ensemble-based approaches, the background-error covariance matrix can be estimated from the ensemble. Due to the usually limited
ensemble size, the background-error covariance matrix is subject to the so-called sampling error. We test different methods to reduce the effect of sampling error in a published paleoclimate data
assimilation setup. For this purpose, we conduct a set of experiments, where we assimilate early instrumental data and proxy records stored in trees, to investigate the effect of (1) the applied
localization function and localization length scale; (2) multiplicative and additive inflation techniques; (3) temporal localization of monthly data, which applies if several time steps are estimated
together in the same assimilation window. We find that the estimation of the background-error covariance matrix can be improved by additive inflation where the background-error covariance matrix is
not only calculated from the sample covariance but blended with a climatological covariance matrix. Implementing a temporal localization for monthly resolved data also led to a better reconstruction.
Received: 04 Dec 2018 – Discussion started: 20 Dec 2018 – Revised: 05 Jun 2019 – Accepted: 26 Jun 2019 – Published: 02 Aug 2019
Estimating the state of the atmosphere in the past is important to enhance our understanding of the natural climate variability, the underlying mechanisms of past climate changes and their impacts.
To infer past climate states, two basic sources of information are available: observations and numerical models. Climate models constrained with realistic, time-dependent external forcings provide
fields that are consistent with these forcings and the model physics. Observations can be instrumental meteorological measurements, which are mainly available from the mid-19th century. Prior to this
time, information from proxies stored in natural archives (like trees, speleothems, marine sediments, ice cores) or documentary data can be exploited. Observations provide important local
information; however, their spatial and temporal coverage is sparse.
In recent years, a novel technique, the data assimilation (DA) approach, has been adapted for paleoclimatological research. DA creates a framework to combine information from different sources. If
information from observations is optimally blended with climate model simulations, the result is the best estimate of the climatic state, given the observations, given the external forcings and given
the known climate physics. The field of paleoclimate data assimilation (PDA) has undergone profound developments, and many DA techniques have been implemented to reconstruct past climate states, such
as forcing singular vectors and pattern nudging (Widmann et al., 2010), selection of ensemble members (Goosse et al., 2006; Matsikaris et al., 2015), particle filters (e.g., Goosse et al., 2010), the
variation approach (Gebhardt et al., 2008), the Kalman filter and its modifications (e.g., Bhend et al., 2012; Hakim et al., 2016; Franke et al., 2017a; Steiger et al., 2018). However, there are
still unresolved problems and thus a need for improvements of how to best combine observations with climate model simulations.
One popular DA method is the Kalman filter (KF; Kalman, 1960). In standard applications, the processes of the KF can be summarized in two main steps (Ide et al., 1997). In the update step, the
background state and the uncertainty of the background state provided by the model simulation are adjusted by assimilating new observations. In the forecast step, the updated state, called the
analysis, and the uncertainty of the analysis are propagated forward in time. These processes are repeated when new observations become available. However, in PDA, the forecast step is usually
neglected; that is, the filter is used offline (e.g., Franke et al., 2017a). Because the process is not cycled, the background state is obtained from a precomputed model simulation. In some previous
PDA studies, the background state is constructed once from the model simulation, and later, the same state is used in every assimilation window (Steiger et al., 2018, and references therein); we
refer to them as stationary (forcing-independent) offline DA methods. In other PDA studies, the background state is specific for the current assimilation window; that is, the state changes in each
assimilation window according to the forcings (Bhend et al., 2012; Franke et al., 2017a); we call them transient (forcing-dependent) offline DA methods.
An essential component of the KF is the uncertainty of the background state. In ensemble-based approaches, an ensemble of the background state provides estimation of the truth, represented by the
ensemble mean, and the perturbations from the mean are used to estimate the uncertainty, represented by the background-error covariance matrix. Ensemble-based KFs are approximations of the KF,
because the true state is usually sampled with a few tens to a few hundreds of ensemble members. The limited ensemble size leads to errors in the estimation of the background-error covariance matrix.
This effect is known as the sampling error.
Two methods are commonly used in online ensemble-based KF approaches to reduce the negative effect of sampling error: inflation (e.g., Anderson and Anderson, 1999) and localization (e.g., Hamill
et al., 2001) of the background-error covariance matrix. A simple inflation technique is the multiplicative inflation (Anderson and Anderson, 1999), which compensates for potential underestimation of
the analysis error. Multiplicative inflation helps to maintain a more realistic distribution of the ensemble members by increasing the deviation of the members from the ensemble mean at each DA cycle
(Anderson and Anderson, 1999). However, the underestimation of the analysis error is of minor importance in offline approaches, because the ensemble members are not propagated forward in time.
Covariance inflation, besides reducing the sampling error, can also account for underestimated model error. In the additive inflation technique, the covariances are inflated by, e.g., adding an
additional error term to the background-error covariances (Houtekamer et al., 2005). Covariance localization removes long-range spurious covariances in the background-error covariance matrix that
occur by chance due to a limited sample size. Several localization techniques have been proposed, from a simple cut-off radius approach (Houtekamer and Mitchell, 1998) to more sophisticated ones (
Houtekamer and Mitchell, 2001; Hamill et al., 2001). By applying covariance localization methods, the elements of the background-error covariance matrix are modified, and in the standard approach the
covariances are forced to approach zero at a certain separation length from the location of the observation. This is achieved by multiplying the background-error covariance matrix element-wise with a
distance-dependent function. In practice, this function is often estimated by a Gaussian localization function, recommended by Gaspari and Cohn (1999).
In stationary offline PDA studies, the time-dependent background-error covariance matrix is replaced by a constant covariance matrix (e.g., Steiger et al., 2014). By using a constant
background-error covariance matrix in the update step, the dependence on the climate state is lost. However, it is possible to estimate the covariance matrix from a much larger ensemble size, which
reduces the sampling error. If the constant covariance matrix is built from a large enough sample size, representing different climate states, it can be successfully used in the assimilation process
(Steiger et al., 2014).
Covariance inflation and localization techniques are used and under improvement in weather forecasting (e.g., Bowler et al., 2017), but have not yet been sufficiently explored for PDA. In this paper,
we discuss three possibilities to improve the estimates of background error, relevant to our PDA method:
• The first possibility involves using a two-dimensional multivariate Gaussian function as a horizontal localization function to test the hypothesis of longer correlation length scales in zonal
than meridional direction.
• The second method is by applying covariance inflation techniques. In the multiplicative inflation technique, a constant factor is used to inflate the deviations from the ensemble mean. In the
additive method, the background-error covariance matrix is calculated as the sum of the sample covariance matrix plus a climatological background matrix, where the climatological background is
based on all ensemble members of multiple years. This larger sample size decreases the chances of spurious correlations.
• The third possibility is adding temporal localization to the background-error covariance matrix. Multiple time steps are combined in one assimilation window to efficiently assimilate seasonal
paleoclimate data. In the case of monthly observations, covariances between the months have been used to update all 6 months (Franke et al., 2017a).
This paper is structured as follows: an overview of our PDA approach, introducing the model, the observational network and the offline DA technique is given in Sect. 2. Section 3 describes the
experimental framework. In Sect. 4, the results are presented and each experiment followed directly by a discussion. We summarize our experiments in Sect. 5.
2Ensemble Kalman fitting framework
2.1Model simulation: CCC400
We start from an existing DA system, which is described in Bhend et al. (2012) and Franke et al. (2017a). We use the same atmospheric model simulation as in the previous studies. The model
simulation, termed as Chemical Climate Change over the Past 400 years (CCC400), has 30 ensemble members that are used as background to reconstruct monthly climate states between 1600 and 2005.
Simulations were performed with the ECHAM5.4 climate model (Roeckner et al., 2003) at a resolution of T63 with 31 levels in the vertical. The 30 ensemble members were forced and driven with the same
external forcings and with the same boundary conditions. For sea-surface temperatures (SSTs), which have a particularly large effect on the simulations, the reconstruction by Mann et al. (2009) was
used. At the time when the model simulation was run, this was the only available global gridded SST reconstruction that dated back until 1600. The surface temperature reconstruction by Mann et al. (
2009) is based on a multiproxy network and was produced by a climate field reconstruction method. The SST reconstruction by design captures interdecadal variations (Mann et al., 2009); hence,
intra-annual variability dependent on a El Niño–Southern Oscillation reconstructions (Cook et al., 2008) was added to the SST fields. Further forcings include solar irradiance, volcanic activity and
greenhouse gas concentrations (for more details, see Bhend et al., 2012; Franke et al., 2017a). The land-use reconstruction by Pongratz et al. (2008) was used to derive the land-surface parameters.
The 6-hourly output fields provided by the model were transformed to monthly means. To reduce the computational burden, only every second grid point in latitude and longitude was selected. We limit
the analysis in this study to 2m temperature, precipitation and sea-level pressure.
2.2Observational network
In this study, we use the same observational network of tree-ring proxies, documentary data and early instrumental measurements as described in Franke et al. (2017a) (Fig. 1), but we only assimilate
tree-ring proxies and instrumental data. The temporal resolution of the instrumental air temperature and sea-level pressure measurements is monthly. The tree-ring proxy records have annual
resolution. Trees respond to a locally varying growing seasons. We consider temperature from May to August and precipitation from April to June to possibly affect tree-ring width data. The maximum
latewood density proxies were considered to be affected by temperature over May until August. The observations were quality checked before the assimilation, and outliers which were more than 5
standard deviations away from the calculated 71-year running mean were discarded, both for instrumental and proxy data.
2.3Assimilation method
In our paleoclimate reconstruction, we combine the CCC400 model simulation with the observations as described above by implementing a modified version of the ensemble square root filter (EnSRF;
Whitaker and Hamill, 2002). This ensemble-based DA method is called ensemble Kalman fitting (EKF; Franke et al., 2017a). In fact, the EKF is an offline version of the EnSRF, and the update step of
the EKF remains the same as of the EnSRF. EKF is described in more detail in Bhend et al. (2012) and Franke et al. (2017a). Here, we shortly highlight the most important aspects of the EKF. The
update step in the EnSRF scheme has two parts: updating the mean ($\stackrel{\mathrm{‾}}{\mathbit{x}}$), and for each member, the deviation from the mean (x^′). They are calculated as
$\begin{array}{}\text{(1)}& & {\stackrel{\mathrm{‾}}{\mathbit{x}}}^{\mathrm{a}}={\stackrel{\mathrm{‾}}{\mathbit{x}}}^{\mathrm{b}}+\mathbf{K}\left(\mathbit{y}-\mathbf{H}{\stackrel{\mathrm{‾}}{\mathbit
{x}}}^{\mathrm{b}}\right)\text{(2)}& & {\mathbit{x}}^{\prime \mathrm{a}}={\mathbit{x}}^{\prime \mathrm{b}}+\stackrel{\mathrm{̃}}{\mathbf{K}}\left(\mathbf{H}{\mathbit{x}}^{\prime \mathrm{b}}\right),\
where K and $\stackrel{\mathrm{̃}}{\mathbf{K}}$ are
$\begin{array}{}\text{(3)}& \mathbf{K}={\mathbf{P}}^{\mathrm{b}}{\mathbf{H}}^{T}{\left({\mathbf{HP}}^{\mathrm{b}}{\mathbf{H}}^{T}+\mathbf{R}\right)}^{-\mathrm{1}}\text{(4)}& \begin{array}{rl}\
stackrel{\mathrm{̃}}{\mathbf{K}}& ={\mathbf{P}}^{\mathrm{b}}{\mathbf{H}}^{T}{\left({\left(\sqrt{{\mathbf{HP}}^{\mathrm{b}}{\mathbf{H}}^{T}+\mathbf{R}}\right)}^{-\mathrm{1}}\right)}^{T}\\ & ×{\left(\
The background state vector (x^b) contains the variables of interest from CCC400 (Table 1). In the EKF, the length of the assimilation window is 6 months (October–March and April–September), which
was adapted to the Southern and Northern Hemisphere growing seasons to effectively incorporate the proxy records stored in trees. Due to the 6-monthly assimilation window, x^b contains the variables
of 6 months. x^a stands for the analysis state vector. H is the forward operator that maps the model state to the observation space (here, it is linear). H differs depending on the type of
observation being assimilated. In the case of tree-ring width data, H extracts temperature between May and August and precipitation between April and June from the model; these fields are transformed
to observational space by using a multiple regression approach (for more details, see Franke et al., 2017a). y represents the observations. K is the Kalman gain matrix, and $\stackrel{\mathrm{̃}}{\
mathbf{K}}$ is the reduced Kalman gain matrix. P^b is the background-error covariance matrix, estimated from the 30 ensemble members. A common assumption is to treat the observation-error covariance
matrix (R) as a diagonal matrix: it is presumed that the elements of R are uncorrelated. Therefore, the observations can be processed serially. We set the error variances of instrumental temperature
observations to 0.9K^2 and of instrumental pressure data to 10hPa^2. The error variances are rough estimates that include, for instance, measurement uncertainties, temporal inhomogeneities and the
fact that a station is not representative for a grid cell (see Frei, 2014; Franke et al., 2017a). The errors of tree-ring proxy data are calculated as the variance of the multiple regression
residuals of the H operator. The assimilation is conducted on the anomaly level: we subtract both from model and from observational data their 71-year running mean in order to deal with the biases
related to systematic model errors and inconsistent low-frequency variability in the paleoclimate data.
The use of DA in an offline manner is typical in paleoclimate reconstructions (e.g., Dee et al., 2016). Bhend et al. (2012) argue that the assimilation step is too long for initial conditions to
matter, whereas there is some predictability from the boundary conditions. In addition, Matsikaris et al. (2015) found that both online and offline DA methods perform similarly in their paleoclimate
reconstruction setup. Furthermore, the offline DA is advantageous as it allows using the precomputed simulations. In our case, we can use CCC400 (Bhend et al., 2012) and test the method without
having to repeat the simulations.
2.4Spatial localization
As R is a diagonal matrix, the EKF can be used to assimilate the observations one by one; that is, after the first observation is assimilated and the analysis is obtained, this analysis field becomes
the background state for the next observation (see the arrow pointing from x^a to x^b on Fig. 2). This serial implementation makes the calculation of P^b simpler. H becomes a vector (not a matrix) of
the same length as x^b. It is zero everywhere except for a few elements (those required to model the observation). This translates to only a few columns of P^b that are actually required. HP^bH^T and
R are then scalars (Whitaker and Hamill, 2002). This procedure also makes the localization simpler, as it needs to be applied only to those columns. In the original setup, the elements of P^b were
multiplied (Schur product) with a distance-dependent function (see Eq. 7 in Franke et al., 2017a). For all the variables in the state vector, the same Gaussian function was used but with different
localization length scale parameters (Table 1). The localization length scale parameters are defined based on the spatial correlation of the variables in the monthly CCC400 model simulation fields.
For the cross-covariances between two variables, the smaller localization length scale of the two variables is applied. With the serial implementation, the calculation and localization of P^b is
significantly simplified.
Franke et al. (2017a) produced a monthly global paleoclimatological data set by using the EKF method. We leave most of the original setup unchanged and mainly focus on the estimation of P^b. To
investigate the performance of the EKF, some aspects involving localization and estimation of the P^b were tested. An overview of all experiments conducted in this study is given in Table 2. The
results of the various experiments are evaluated in terms of performance measures, which then compared to those obtained with the original setup.
3.1Spatial localization
In most of the studies, the localization function is implemented in an isotropic manner. In the original setup, the same horizontally isotropic localization function was used with different
localization parameters. However, such spatial symmetries may not be realistic. In the real atmosphere, correlation lengths might be longer in the zonal than in the meridional direction, due to the
prevailing winds and the weaker large-scale temperature gradients in this direction. On multi-annual to multi-decadal timescales, multiple processes act in the meridional direction, e.g., a widening/
shrinking of the Hadley cell, shifts of the Intertropical Convergence Zone or changes in atmospheric modes like the Atlantic Multi-Decadal Oscillation or the North Atlantic Oscillation. These can
shift the zonal circulation northward or southward, but the zonal coherence will be less effected. Hence, instead of using a circular Gaussian function, we conducted an experiment with a spatially
anisotropic localization function
$\begin{array}{}\text{(5)}& C=\mathrm{exp}\left(-\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{d}_{z}^{\mathrm{2}}}{{L}_{\mathrm{z}}^{\mathrm{2}}}+\frac{{d}_{m}^{\mathrm{2}}}{{L}_{\mathrm{m}}^{\mathrm
where d[z] and d[m] are the distances from the selected grid box in the zonal and meridional directions, respectively. L[z] and L[m] are the length scale parameters used for localization in the zonal
and meridional directions, respectively. As a first experiment, we tested a 2:1 ratio for L[z]:L[m]. We used the values from Table 1 in the meridional direction and doubled them in the zonal
direction. Thus, the resulting localization function has an elliptical shape.
3.2Inflation techniques
Covariance inflation techniques are another possible method to compensate for errors in the DA system (Whitaker et al., 2008). The multiplicative inflation technique uses a small factor γ(γ>1) with
which the x^′b is multiplied (Anderson and Anderson, 1999). This type of covariance inflation accounts for filter divergence due to sampling error (Whitaker and Hamill, 2002) but can be also applied
to take into account system errors (Whitaker et al., 2008). We conducted some experiments using multiplicative inflation, although in our offline approach, filter divergence is not the main concern
as ensemble members are not propagated in time.
The other methodology that we adapt shows similarities with the additive inflation technique (e.g., Houtekamer and Mitchell, 2005) and with the hybrid DA scheme (e.g., Clayton et al., 2013). In both
methods, P^b is modified by either adding model error (Houtekamer and Mitchell, 2005) or a so-called climatological covariance matrix (Clayton et al., 2013) to P^b. This has given rise to the idea of
generating a climatological ensemble in order to alleviate the effect of the small ensemble size. In the original setup, P^b is approximated from only 30 members. Here, we additionally build a
climatological state vector (x^clim) from randomly selected ensemble members from our 400-year long model simulation. The number of ensemble members should be higher than the original ensemble size,
but still computationally affordable. The climatological state vector is created as follows: (1) define the ensemble size (n) of x^clim; (2) select n random years between 1601 and 2005; (3) every
year has 30 members from which one member is randomly selected and kept; (4) the chosen members are combined in x^clim. x^clim is randomly resampled after every second assimilation cycle. Using x^
clim in the assimilation leads to increased computational cost, which partly comes from the creation of x^clim. The other time-consuming part comes from the updating of the climatological part after
each observation is assimilated (the standard way when observations are assimilated serially). We tested numbers between 100 and 500. From x^clim, a climatological background-error covariance matrix
(P^clim) can be obtained by using the ensemble perturbations. The background-error covariance matrix used in the blending experiments (P^blend) is built as a linear combination of the sample
covariance matrix (P^b) and the climatological covariance matrix (P^clim):
$\begin{array}{}\text{(6)}& {\mathbf{P}}^{\mathrm{blend}}={\mathit{\beta }}_{\mathrm{1}}{\mathbf{P}}^{\mathrm{b}}+{\mathit{\beta }}_{\mathrm{2}}{\mathbf{P}}^{\mathrm{clim}},\end{array}$
where β[1], β[2] are the weights given to the covariance matrices. The sum of the weights is unity.
Figure 2 shows the main steps of the blending assimilation process. First, the covariance matrices were localized separately; then, we blended them according to the given weights. We conducted
several experiments to tune the ratio between the two covariance matrices while using different localization length scale parameters (L) (Table 2). We used the same L values for localizing P^b in
most of our experiments to evaluate improvements in comparison with the original setup. For this study, we calculated the latitudinal dependency of correlation of the state variables from a bigger
ensemble of the model than in Franke et al. (2017a). The result suggested that longer L values can be applied in the tropics and the L of precipitation is probably too strict. Based on the rather
strict L values in the previous study and the assumption that the covariances can be better estimated from a bigger ensemble, we used doubled length scale parameters (2L) in some of the experiments
for localizing the climatological covariances. In this case, the L for temperature is 3000km, which means that the correlation is decreased close to zero approximately 6000km away from the
Since observations are assimilated serially, we also update x^clim after an observation is assimilated with the same Kalman gain matrices as x^b. Thus, in the assimilation process, we update 30+n
ensemble members.
3.3Temporal localization
Localizing observations in time is a special feature of the EKF due to its 6-month assimilation window. Having the state vector in half-year format, every month within the October–March or
April–September time window is updated by each single observation. To test whether the covariances between a single observation and the multivariate climate fields are correctly captured, we ran an
instrumental-only experiment with temporal localization. We set covariances between different months to zero.
3.4Skill scores
The EKF method is tested with different localization functions and with a set of mixed background-error covariance matrices as described above. We have performed the experiments by assimilating
either only proxy records (proxy-only experiment) or only instrumental data (instrumental-only experiment). The proxy-only experiments were carried out between 1902 and 1959, because many proxy
records already end in the 1960s, while the instrumental-only experiments were tested over the 1902–2002 period. We separated the different observation types to see whether different settings perform
better depending on the type of data being assimilated. We do not compare proxy-only results with instrumental-only results; hence, the difference in time periods used does not matter; we simply use
the longest possible time period. To evaluate the reconstructions, we examined two verification measures: correlation coefficient and reduction of error (RE) skill score (Cook et al., 1994). We use
the Climate Research Unit (CRU) TS 3.10 dataset (Harris et al., 2014) for reference in the validation process. The presented verification measures are functions of time. Correlation is calculated
between the absolute values of the ensemble mean of the analysis and the reference series at each grid point. The RE compares, in our case, the reconstruction with the model simulation, both
expressed as deviations from a reference.
$\begin{array}{}\text{(7)}& \mathrm{RE}=\mathrm{1}-\frac{\sum \left({\mathbit{x}}_{i}^{\mathrm{u}}-{\mathbit{x}}_{i}^{\mathrm{ref}}{\right)}^{\mathrm{2}}}{\sum \left({\mathbit{x}}_{i}^{\mathrm{f}}-{\
where x^u is the ensemble mean of the analysis, x^f is the ensemble mean of the model background state, x^ref is the reference dataset, and i refers to the time step. The RE skill scores are computed
based on anomalies with respect to the 71-year running climatologies. Note that x^f comes from a forced model simulation; therefore, it already has skill compared with a climatological state vector.
The RE is 1 if the x^u is equal to x^ref. Negative RE values indicate that the background state is closer to the reference series than the analysis.
To test which experiments have significantly different skill compared with the original skill, we carried out a permutation test following the method described in DelSole and Tippett (2014).
Permutation was performed 10000 times. If the difference between the median of the experiment and the median of the original data falls outside of the 95% confidence level of the interval
calculated from permuted data, then the experiment is significantly different from the original data.
In the next section, we will focus on analyzing the result of the experiments mainly over the extratropical Northern Hemisphere (ENH), because most of the data are located in this region. The skill
scores refer to seasonal averages of the ensemble mean.
4Results and discussions
4.1Localization function
We compared the original setup applying an isotropic localization function and the experiment in which an anisotropic localization function was used, to test whether we can obtain a more skillful
reconstruction by implementing anisotropic localization method. As an example of the spatial reconstruction skill, we show the RE values of temperature (Fig. 3). The figures reveal that the type of
localization function only resulted in small differences in both experiments. Nonetheless, there are larger areas of negative RE values (Greenland, Siberia) with the anisotropic localization function
in the proxy-only experiment (Fig. 3). In the instrumental-only experiment, the decrease of RE values occur in the northern high latitudes and in the Tibetan Plateau in both seasons (Fig. 3). To have
a better overview of how the skill scores changed, we summarize their distributions with the help of box plots. Figure 4 shows the differences of skill scores between the aniso experiment and the
original skill for the three variables (temperature, precipitation and sea-level pressure) in the ENH region. In the instrumental-only experiment, correlation values of temperature and sea-level
pressure decreased in both seasons, while for precipitation they remained mostly unchanged. The RE values show that the experiments with anisotropic localization function reduced the skill of the
reconstructions, but the extent of the reduction varies with the variables and with the seasons (Fig. 4). In general, the same holds for the proxy-only experiment (Fig. 4).
In a previous ozone reconstruction study, a seasonally and latitudinally varying localization method was tested which mostly positively affected the analysis (Brönnimann et al., 2013). Here, we
increased the zonal distances to see if we can use the information of the observations for a larger region. However, the verification measures are shifted more to the negative direction. We assume
that the degraded skill of the reconstruction is due to the choice of too-long L[z]; hence, spurious correlations were not removed. Using anisotropic localization (doubling the L values only in the
zonal direction) consistently makes the reconstruction worse.
4.2Inflation experiments
The main problem of ensemble-based DA techniques is the computationally affordable limited ensemble size. Due to the finite ensemble size, the estimation of P^b suffers from sampling error. Applying
inflation techniques is one method to mitigate its effect (see Sect. 3.2).
Using the multiplicative inflation method, the deviations from the ensemble mean are multiplied with a small factor (γ). To find the optimal γ, a set of experiment runs is required. We used γ=1.02
and γ=1.12 in our experiments, where only instrumental data were assimilated. We chose γ from a range that was previously tested by Whitaker and Hamill (2002). Multiplying the deviations from the
ensemble mean with γ=1.02 in the assimilation process hardly affected the skill of the reconstruction over the ENH region (not shown). When we increased the value of γ to 1.12, the RE values slightly
decreased (not shown). We did not carry out further experiments since, based on the results, randomly increasing the error in background field did not lead to improvement.
In the other set of experiments, we used P^blend in the update equation (Eq. 6). The experiments were run with using β[2] equal to 0.25, 0.50, 0.75 and 1 to estimate the P^blend (denoted 25c, 50c,
etc.). Besides the varying weight given to P^clim, the applied L values on P^b and P^clim differed as well. Three L values were used: no localization (termed “no”), applying L values as in Table 1 (L
) and doubling these numbers (2L). Different combinations of the fraction of P^clim and L values were termed accordingly (e.g., 50c_PbL_Pc2L).
We expect that estimating the covariances from a bigger ensemble size (n=100–500) instead of 30 members leads to a more accurate background matrix. In most of our experiment, n is 250. Hence, P^
clim is likely less affected by the sampling error implying that long-range spurious correlations are less prominent, which makes localization less needed. We presume that using P^blend helps to
better reconstruct areas which were characterized with lower skill score values in the original setup and to improve the estimation of unobserved climate variables. The reconstruction skill of the
blending experiments is always calculated from x^a (Fig. 2).
For the ENH region, we present how the verification measures changed by replacing P^b with P^blend in the assimilation process. We conducted an experiment without localizing P^clim and using L values
from Table 1 on P^b in the construction of P^blend. However, the skill of the reconstruction was largely reduced, implying that 250 members are not enough to avoid localization altogether (not
Figures 5 and 6 show the distribution of the differences of the skill scores between the experiments and the original analysis for correlation coefficients and RE values, respectively. Depending on
the variables and the data type being assimilated, different setups perform better. In the case of assimilating only instrumental data, one of the largest increases of median for temperature
reconstruction was obtained from the 100c_PcL experiment in both seasons (Figs. 5, 6). Precipitation records were not assimilated; thus, a reasonable estimation of the cross-variable covariances is
essential. The skill of the precipitation reconstruction with the original setup, in terms of correlation, is better than the forced simulation (not shown); however, the RE values are negative over
large regions in the ENH (Fig. 7). Using P^blend in the assimilation, with, e.g., the settings of 50c_PbL_Pc2L experiment, led to more positive RE skill (Fig. 7). The biggest improvement, in terms of
RE skill score, was found in Europe (Fig. 7). The 50c_PbL_Pc2L analysis also has higher skill in North-America, especially in the summer season (Fig. 7). The skill of the sea-level pressure
reconstruction also improved in the 50c_PbL_Pc2L experiment (Figs. 5, 6). In the proxy-only experiments, 75c_PbL_Pc2L is among the best performing experiments for all the variables (Figs. 5, 6).
We also investigated the effect of the ensemble size in the estimation of P^clim. To test whether further improvements can be achieved by doubling the ensemble size of x^clim, we ran an experiment
with the following setup: β[1] and β[2] are equally weighted, and L and 2L are applied on P^b and P^clim, respectively (Table 2). In the experiment, we assimilated only instrumental data. The skill
scores of x^a (corr, RE) from the 500-ensemble-member experiment showed no marked improvement compared with the same experiment with 250 ensemble members. An additional experiment was carried out
with the same setup but using only 100 ensemble members in the construction of x^clim. The verification measures of the 50c_PbL_Pc2L_100m experiment are higher than the original one, and the
distribution of the skill scores over the ENH region is very similar to what we obtain by using 250 members in P^clim for temperature and precipitation. However, the sea-level pressure fields from
the 50c_PbL_Pc2L experiment have higher skill than those in the 50c_PbL_Pc2L_100m experiment (not shown).
Furthermore, we conducted two experiments in which only x^b was updated after an observation was assimilated, and x^clim was kept constant in the assimilation window. However, the ensemble members of
x^clim were randomly reselected for each year (October–September). The advantage of this setup compared to the setup described in Sect. 3.2 is that it is computationally less demanding since only the
original 30 members keep being updated with the observations. In the first test, we give β[2]=0.75 weight to P^clim with 2L values. In the second test, β[2]=1; that is, only P^clim used for updating
x^b and the L values in Table 1 were applied for the localization. By comparing the skill of the reconstructions without and with updating the climatological part, we see that the skill scores are
higher when the climatological part is also updated with the information from the observations (Fig. 8). The only exception is the correlation values of sea-level pressure: when keeping the
climatological part constant, they are slightly higher in both seasons (Fig. 8). Nonetheless, by keeping the climatological part static in one assimilation window, the experiments still outperform
the original reconstruction (Fig. 8).
We have tested a number of configurations of the mixed covariance matrix P^blend to evaluate the effect of the sampling error. In numerical weather predication (NWP) applications, various methods
have been designed to better estimate the errors of the background state. In hybrid DA systems, the advantages of variational and ensemble Kalman filter techniques are combined (Hamill and Snyder,
2000; Lorenc, 2003). In another method, the background-error covariances are obtained from an ensemble of assimilation experiments performed by a variational assimilation system (Pereira and Berre,
2006). In an additive inflation experiment, a term is added to the x^a to account for the errors of the DA system (Whitaker et al., 2008).
In our implementation, P^blend is calculated from x^b and x^clim. Using P^blend in the assimilation process improved on the reconstruction performed with the original setup. The skill scores show the
largest improvement in the sea-level pressure reconstruction. Moreover, the skill of the precipitation reconstruction also improved, indicating that P^clim helps to better estimate the
cross-covariances of the background errors between the variables. In general, increasing the weight of P^clim in forming P^blend positively affected the skill of the analysis. The 100c_PcL
experiment, in which P^blend is equal to P^clim, is similar to the DA technique used in the last millennium climate reanalysis (LMR) project (Hakim et al., 2016). In the LMR, 100 randomly chosen
ensemble members form a climatological state vector, which is used in each assimilation window and is updated with the observations. In this study, x^clim is randomly resampled every year and
primarily used in the estimation of P^blend. The settings used in the 100c_PcL experiment led to one of the largest increases in the median for temperature reconstruction when only instrumental
measurements are assimilated. However, other settings resulted in larger increase of median for different variables and observation types. By applying no localization on P^clim in the 50c_PbL_PcnoL
experiment, we obtained a less skillful reconstruction than by using the other two localization schemes. The skills reduced especially over the areas where no local observations were assimilated.
Using 2L values for localizing the covariances of P^clim in the instrumental-only experiments resulted in higher correlation values of sea-level pressure (50c_PbL_Pc2L) and helped to obtain higher
correlation scores of precipitation in summer. Among the proxy-only experiments, 75c_PbL_Pc2L shows the largest increase of median for pressure reconstruction. Here, pressure data are not
assimilated, and the result suggests that by applying longer L values, the cross-variable covariances are treated better. We tested whether the skill of the experiments performed with various
settings is significantly different from the skill of the original analysis. We compared the median value of the skill scores from the experiments and the original data, and with most of the settings
a significant difference was obtained for all the variables. The results of the experiments show that with a mixed covariance matrix implementation, a major drawback of the ensemble-based DA system,
due to the limited ensemble size, can be improved.
4.3Localization in time
Since 6-monthly time steps were combined in one state vector (one assimilation window), covariances between different months also need to be considered. An additional experiment was conducted in
which the (localized) P^b was multiplied with a temporal localization function when instrumental data were assimilated. This is a specific experiment due to the structure of EKF. The assimilation
window in the EKF is 6 months; hence, a single observation is enabled to adjust all the meteorological variables in x^b in a half-year time window. In the temporal localization experiment, the
information from a given observation can only modify the different climate fields in its current month, while leaving all other fields of the 5 months unchanged (Table 2). In general, the skill
scores indicate an improvement. The differences of RE values between the temp_loc and original experiments are mostly positive over the northern high-latitude areas (Fig. 9).
The higher skill scores with temporal localization (Fig. 9) indicate that the cross-covariances in time were not correctly represented by P^b. Hence, it is likely that in the original setup some
non-physical covariances were taken into account. Applying the same assimilation scheme to another problem (estimating the two-dimensional ozone distribution from an ensemble of chemistry–climate
models and historical observations), Brönnimann et al. (2013) used a localization timescale of 3 months based on empirical studies. It may be worth considering or allowing for temporal covariance in
specific cases (e.g., in the case of ozone concentrations) which vary on longer timescales.
In this study, a transient offline data assimilation approach was used to test the effect of the estimation of the background-error covariance matrix in a climate reconstruction. Several experiments
were evaluated with different validation measures to see which background-error covariance matrix estimation techniques improve the skill of the reconstruction. The evaluation of the presented
techniques suggests the following: (1) applying an anisotropic localization function on the sample covariance matrix did not improve the reconstruction; (2) most of the settings, which make use of
covariance estimates from a larger climatological sample, result in significantly improved skills compared to an estimation from the 30-member ensemble; (3) assimilating early instrumental data with
temporal localization leads to a better analysis. To which extent the different techniques helped in the estimation of the background-error covariance matrix varies geographically and also depends on
the climate variable being reconstructed. The cross-variable covariances of the background-error covariance matrix can provide information from unobserved climate variables. Including climatological
information in the estimation of precipitation has led to a better reconstruction, especially in Europe. Estimating sea-level pressure with the blended P^blend matrix also improved the skill of the
reconstruction. For instance, the 50c_PbL_Pc2L experiment performs constantly better than the original setup. This study shows that results can be improved by better specifying the background-error
covariance matrix. In the future, we will combine all the techniques that lead to more skillful analyses to produce a climate reconstruction over the last 400 years.
The EKF400 reanalysis is available at the World Data Center for Climate at Deutsches Klimarechenzentrum (DKRZ) in Hamburg, Germany (https://cera-www.dkrz.de/WDCC/ui/cerasearch/entry?acronym=
EKF400_v1.1; Franke et al., 2017b). The sensitivity experiments analyzed in this study are available upon request: veronika.valler@giub.unibe.ch, joerg.franke@giub.unibe.ch.
All authors were involved designing the study and contributed to writing the paper. VV conducted the experiments and performed most of the analyses. JF developed the original code and helped with the
The authors declare that they have no conflict of interest.
The CCC400 simulation was performed at the Swiss National Supercomputing Centre CSCS. The comments of the two anonymous reviewers are gratefully acknowledged.
This research has been supported by the Swiss National Science Foundation (grant no. 162668) and the European Commission – Horizon 2020 (grant no. 787574).
This paper was edited by Bjørg Risebrobakken and reviewed by two anonymous referees.
Anderson, J. L. and Anderson, S. L.: A Monte Carlo implementation of the nonlinear filtering problem to produce ensemble assimilations and forecasts, Mon. Weather Rev., 127, 2741–2758, https://
doi.org/10.1175/1520-0493(1999)127<2741:AMCIOT>2.0.CO;2, 1999.a, b, c, d
Bhend, J., Franke, J., Folini, D., Wild, M., and Brönnimann, S.: An ensemble-based approach to climate reconstructions, Clim. Past, 8, 963-976, https://doi.org/10.5194/cp-8-963-2012, 2012.a, b, c, d
, e, f, g
Bowler, N., Clayton, A., Jardak, M., Jermey, P., Lorenc, A., Wlasak, M., Barker, D., Inverarity, G., and Swinbank, R.: The effect of improved ensemble covariances on hybrid variational data
assimilation, Q. J. Roy. Meteor. Soc., 143, 785–797, https://doi.org/10.1002/qj.2964, 2017.a
Brönnimann, S., Bhend, J., Franke, J., Flückiger, S., Fischer, A. M., Bleisch, R., Bodeker, G., Hassler, B., Rozanov, E., and Schraner, M.: A global historical ozone data set and prominent features
of stratospheric variability prior to 1979, Atmos. Chem. Phys., 13, 9623–9639, https://doi.org/10.5194/acp-13-9623-2013, 2013.a, b
Clayton, A. M., Lorenc, A. C., and Barker, D. M.: Operational implementation of a hybrid ensemble/4D-Var global data assimilation system at the Met Office, Q. J. Roy. Meteor. Soc., 139, 1445–1461,
https://doi.org/10.1002/qj.2054, 2013.a, b
Cook, E. R., D'Arrigo, R., and Anchukaitis, K.: ENSO reconstructions from long tree-ring chronologies: Unifying the differences?, talk presented at a special workshop on “Reconciling ENSO
Chronologies for the Past 500 Years”, 2–3 April 2008, Moorea, French Polynesia, 2008.a
Cook, E. R., Briffa, K. R., and Jones, P. D.: Spatial regression methods in dendroclimatology: a review and comparison of two techniques, Int. J. Climatol., 14, 379–402, https://doi.org/10.1002/
joc.3370140404, 1994.a
Dee, S. G., Steiger, N. J., Emile-Geay, J., and Hakim, G. J.: On the utility of proxy system models for estimating climate states over the common era, J. Adv. Model. Earth Sy., 8, 1164–1179, https://
doi.org/10.1002/2016MS000677, 2016.a
DelSole, T. and Tippett, M. K.: Comparing forecast skill, Mon. Weather Rev., 142, 4658–4678, https://doi.org/10.1175/MWR-D-14-00045.1, 2014.a
Franke, J., Brönnimann, S., Bhend, J., and Brugnara, Y.: A monthly global paleo-reanalysis of the atmosphere from 1600 to 2005 for studying past climatic variations, Scientific data, 4, 170076,
https://doi.org/10.1038/sdata.2017.76, 2017a.a, b, c, d, e, f, g, h, i, j, k, l, m, n
Franke, J., Brönnimann, S., Bhend, J., and Brugnara, Y.: Ensemble Kalman Fitting Paleo-Reanalysis Version 1.1, World Data Center for Climate (WDCC) at DKRZ, available at: http://cera-www.dkrz.de/WDCC
/ui/Compact.jsp?acronym=EKF400_v1.1 (last access: 19 July 2019), 2017b.a
Frei, C.: Interpolation of temperature in a mountainous region using nonlinear profiles and non-Euclidean distances, Int. J. Climatol., 34, 1585–1605, 2014.a
Gaspari, G. and Cohn, S. E.: Construction of correlation functions in two and three dimensions, Q. J. Roy. Meteor. Soc., 125, 723–757, https://doi.org/10.1002/qj.49712555417, 1999.a
Gebhardt, C., Kühl, N., Hense, A., and Litt, T.: Reconstruction of Quaternary temperature fields by dynamically consistent smoothing, Clim. Dynam., 30, 421–437, https://doi.org/10.1007/
s00382-007-0299-9, 2008.a
Goosse, H., Renssen, H., Timmermann, A., Bradley, R. S., and Mann, M. E.: Using paleoclimate proxy-data to select optimal realisations in an ensemble of simulations of the climate of the past
millennium, Clim. Dynam., 27, 165–184, https://doi.org/10.1007/s00382-006-0128-6, 2006.a
Goosse, H., Crespin, E., de Montety, A., Mann, M., Renssen, H., and Timmermann, A.: Reconstructing surface temperature changes over the past 600 years using climate model simulations with data
assimilation, J. Geophys. Res.-Atmos., 115, D09108, https://doi.org/10.1029/2009JD012737, 2010.a
Hakim, G. J., Emile-Geay, J., Steig, E. J., Noone, D., Anderson, D. M., Tardif, R., Steiger, N., and Perkins, W. A.: The last millennium climate reanalysis project: Framework and first results, J.
Geophys. Res.-Atmos., 121, 6745–6764, https://doi.org/10.1002/2016JD024751, 2016.a, b
Hamill, T. M. and Snyder, C.: A hybrid ensemble Kalman filter – 3D variational analysis scheme, Mon. Weather Rev., 128, 2905–2919, https://doi.org/10.1175/1520-0493(2000)128<2905:AHEKFV>2.0.CO;2,
Hamill, T. M., Whitaker, J. S., and Snyder, C.: Distance-dependent filtering of background error covariance estimates in an ensemble Kalman filter, Mon. Weather Rev., 129, 2776–2790, https://doi.org/
10.1175/1520-0493(2001)129<2776:DDFOBE>2.0.CO;2, 2001.a, b
Harris, I., Jones, P. D., Osborn, T. J., and Lister, D. H.: Updated high-resolution grids of monthly climatic observations–the CRU TS3. 10 Dataset, Int. J. Climatol., 34, 623–642, https://doi.org/
10.1002/joc.3711, 2014.a
Houtekamer, P. L. and Mitchell, H. L.: Data assimilation using an ensemble Kalman filter technique, Mon. Weather Rev., 126, 796–811, https://doi.org/10.1175/1520-0493(1998)126<0796:DAUAEK>2.0.CO;2,
Houtekamer, P. L. and Mitchell, H. L.: A sequential ensemble Kalman filter for atmospheric data assimilation, Mon. Weather Rev., 129, 123–137, https://doi.org/10.1175/1520-0493(2001)129<0123:ASEKFF>
2.0.CO;2, 2001.a
Houtekamer, P. L. and Mitchell, H. L.: Ensemble Kalman filtering, Q. J. Roy. Meteor. Soc., 131, 3269–3289, https://doi.org/10.1256/qj.05.135, 2005.a, b
Houtekamer, P. L., Mitchell, H. L., Pellerin, G., Buehner, M., Charron, M., Spacek, L., and Hansen, B.: Atmospheric data assimilation with an ensemble Kalman filter: Results with real observations,
Mon. Weather Rev., 133, 604–620, https://doi.org/10.1175/MWR-2864.1, 2005. a
Ide, K., Courtier, P., Ghil, M., and Lorenc, A. C.: Unified Notation for Data Assimilation: Operational, Sequential and Variational, J. Meteorol. Soc. Jpn. Ser. II, 75, 181–189, https://doi.org/
10.2151/jmsj1965.75.1B_181, 1997.a
Kalman, R. E.: A new approach to linear filtering and prediction problems, J. Basic Eng.-T. ASME, 82, 35–45, 1960.a
Lorenc, A. C.: The potential of the ensemble Kalman filter for NWP—a comparison with 4D-Var, Q. J. Roy. Meteor. Soc., 129, 3183–3203, https://doi.org/10.1256/qj.02.132, 2003.a
Mann, M. E., Zhang, Z., Rutherford, S., Bradley, R. S., Hughes, M. K., Shindell, D., Ammann, C., Faluvegi, G., and Ni, F.: Global signatures and dynamical origins of the Little Ice Age and Medieval
Climate Anomaly, Science, 326, 1256–1260, https://doi.org/10.1126/science.1177303, 2009.a, b, c
Matsikaris, A., Widmann, M., and Jungclaus, J.: On-line and off-line data assimilation in palaeoclimatology: a case study, Clim. Past, 11, 81–93, https://doi.org/10.5194/cp-11-81-2015, 2015.a, b
Pereira, M. B. and Berre, L.: The use of an ensemble approach to study the background error covariances in a global NWP model, Mon. Weather Rev., 134, 2466–2489, https://doi.org/10.1175/MWR3189.1,
Pongratz, J., Reick, C., Raddatz, T., and Claussen, M.: A reconstruction of global agricultural areas and land cover for the last millennium, Global Biogeochem. Cy., 22, GB3018, https://doi.org/
10.1029/2007GB003153, 2008.a
Roeckner, E., Bäuml, G., Bonaventura, L., Brokopf, R., Esch, M., Giorgetta, M., Hagemann, S., Kirchner, I., Kornblueh, L., Manzini, E., Rhodin, A., Schlese, U., Schulzweida, U., and Tompkins, A.: The
atmospheric general circulation model ECHAM5 Part I: model description, Tech. rep. 349, Max Planck Institute for Meteorology, 2003.a
Steiger, N. J., Hakim, G. J., Steig, E. J., Battisti, D. S., and Roe, G. H.: Assimilation of time-averaged pseudoproxies for climate reconstruction, J. Climate, 27, 426–441, https://doi.org/10.1175/
JCLI-D-12-00693.1, 2014.a, b
Steiger, N. J., Smerdon, J. E., Cook, E. R., and Cook, B. I.: A reconstruction of global hydroclimate and dynamical variables over the Common Era, Scientific Data, 5, 180086, https://doi.org/10.1038/
sdata.2018.86, 2018.a, b
Whitaker, J. S. and Hamill, T. M.: Ensemble data assimilation without perturbed observations, Mon. Weather Rev., 130, 1913–1924, https://doi.org/10.1175/1520-0493(2002)130<1913:EDAWPO>2.0.CO;2,
2002.a, b, c, d
Whitaker, J. S., Hamill, T. M., Wei, X., Song, Y., and Toth, Z.: Ensemble data assimilation with the NCEP global forecast system, Mon. Weather Rev., 136, 463–482, https://doi.org/10.1175/
2007MWR2018.1, 2008.a, b, c
Widmann, M., Goosse, H., van der Schrier, G., Schnur, R., and Barkmeijer, J.: Using data assimilation to study extratropical Northern Hemisphere climate over the last millennium, Clim. Past, 6,
627–644, https://doi.org/10.5194/cp-6-627-2010, 2010.a | {"url":"https://cp.copernicus.org/articles/15/1427/2019/cp-15-1427-2019.html","timestamp":"2024-11-02T15:51:46Z","content_type":"text/html","content_length":"294122","record_id":"<urn:uuid:555fff87-2a50-4207-94e0-ecc0b32495bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00489.warc.gz"} |
Tau Day
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Make it happy by adding an image to it… You know you want to… You see the Upload Button…
On 28 June it is Tau (τ) Day, where we celebrate the new mathematical constant that has began a movement in modern mathematics. The Tau Movement suggests that actually, π can be a bit of a handful
when doing Trigonometry, and that instead, a constant value of 2π, also known as Tau (τ).
The reason this weird value was chosen actually makes more sense when you look at a circle in Radians. A Radian is a measurement of an angle based on its Arc Length size compared to its radius. At
Heart, the circumference of a circle is 2πr, which means that in radians, a circle has an angle of 2π.
Therefore τ, roughly 6.28318, is equal to 2π and defines the full cycle of a circle in radians. Tau is equal to a circle’s Circumference (C) divided by its radius (r), instead of Pi’s Circumference /
One of the challenges Tau faces is that π is more useful for day-to-day calculations using circles, while those who use radians on a regular basis would find Tau far more useful.
Tau day is held on the 28th of June, which appears to look like 6/28 in the Month-First date system. This is similar to Tau’s numerical value of 6.28318. | {"url":"https://math-soc.com/calendar/tau-day/","timestamp":"2024-11-14T17:07:45Z","content_type":"text/html","content_length":"42066","record_id":"<urn:uuid:1b3361e1-9d14-445a-a7a0-d0c6b21f88e6>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00111.warc.gz"} |
Theoretical and Experimental Investigation of N-Bit Reconfigurable Retrodirective Metasurface
Article information
J. Electromagn. Eng. Sci. 2024;24(1):51-56
Received 2022 December 28; Revised 2023 April 14; Accepted 2023 August 1.
The PIN diode-based N-bit reconfigurable retrodirective metasurface (N-bit RRDM) is a next-generation retro-reflector that offers the advantages of effective electronical control of the
retro-reflection angle, low loss, and thin planar structure. However, since the unit cell of an N-bit RRDM is controlled by a quantized N-bit phase (360°/2^N), it encounters operational errors, such
as beam gain reduction and spurious beams. This can be a fatal disadvantage in military radar or satellite communication, which requires accurate beam tracking. This paper theoretically analyzes the
operation of the N-bit RRDM by utilizing generalized Snell’s law and array factor theory. The analysis results present the design criteria for an N-bit RRDM that eliminates issues related to beam
gain reduction and spurious beam errors. Furthermore, to verify the theoretical analysis results, High-Frequency Structure Simulator (HFSS) full-wave simulation and experimentation are conducted
using the 1-bit RRDM.
Retro-reflectors are devices or surfaces with attractive characteristics that reflect incident waves in an incoming direction as opposed to a specular direction. Since this characteristic serves to
increase the target’s monostatic radar cross section (RCS), retro-reflectors are extensively used in fields dealing with millimeter waves and microwaves, such as the military and civilian industries
[1–4]. The types of retro-reflectors include 3D structure types, such as the corner cube reflector and cat’s eye retroreflector, and 2D structure types, such as the Van Atta array and metasurfaces [5
–7]. Among these, the 2D structure retrodirective metasurface (RDM) has been gaining attention as a next-generation retro-reflector due to its advantages, such as its thin planar structure, light
weight, and easy fabrication [5–7]. However, the RDM has a limited retro-reflection angle since it is a passive retro-reflector. Therefore, it is difficult to use it in radar in real time or in a
wireless power system with multiple targets [1–4].
Notably, some research on reconfigurable metasurface (RMS) using PIN diodes has recently been conducted [8–10]. PIN diodes exhibit several distinct advantages, including fast switching speed, low
loss, and a simple structure. In this context, it is anticipated that the PIN diode-based N-bit reconfigurable retrodirective metasurface (N-bit RRDM), as an application of RMS, can overcome the
above-mentioned challenges. The PIN diode-based N-bit RRDM is a retro-reflector that freely determines the angle of retro-reflection by electronically controlling the reflection phase of the unit
cell and the phase gradient of the surface through the PIN diode, which is mounted on the unit cell. In such a case, however, the reflection phase of the unit cell is controlled by a quantized N-bit
phase (360°/2^N), which is based on the number (N) of mounted PIN diodes [8, 9].
Therefore, the N-bit RRDM does not fully implement the phase gradient required for retro-reflection. These characteristics may sometimes cause problems such as beam gain reduction and spurious beam
errors. In particular, spurious beams occurring in 1-bit RRDM relate to a completely different concept from that of the grating lobe.
This paper proposes a PIN diode-based N-bit RRDM whose characteristics are theoretically analyzed to propose the criteria for an N-bit RRDM design that eliminates the problems of beam gain reduction
and spurious beam errors. The proposed metasurface is analyzed using generalized Snell’s law [11] and array factor theory [12]. Furthermore, the theoretical analysis is validated through full-wave
simulation and experimentation using the 1-bit metasurface.
II. THEORY AND ANALYSIS OF N-BIT RRDM
The PIN diode-based N-bit RRDM is a metasurface composed of N PIN diodes mounted on each of its M unit cells, as shown in Fig. 1. The N-bit RRDM controls the phase of the unit cell related to the
N-bit by switching the PIN diodes on/off. As a result, the quantized phase is considered 360°/2^N. If a plane incident wave with an incident angle θ[i] retro-reflects at this N-bit RRDM, the
reflective angle of θ[r] will be equal to − θ[i]. Effectively, the ideal reflection phase gradient ( dΦIdealdx) required for this operation can be calculated as Eq. (1) using generalized Snell’s law.
Additionally, Φ[Ideal](m), which is the ideal reflection phase value of the m^th unit cell, can be calculated using Eq. (2), as noted below:
(1) λ02πn0dΦIdealdx=sinθi-sinθr=2sinθi,
(2) ΦIdeal(m)=(m-1)×dΦIdeal.
Here, n[0] indicates the refractive index of free space, λ[0] refers to the wavelength in free space, and d[x] represents the distance between adjacent cells. Following this, the N-bit RRDM
implements the reflection phase acquired from Eq. (2) using the quantized N-bit phase. Therefore, the reflection phase (Φ[N–bit](m)) of the m^th unit cell in the N-bit RRDM can be expressed as Eq.
(3) ΦN-bit(m)=360°N×R(ΦIdeal(m)360°/2N).
Here, R(X) refers to an integer value obtained by rounding off X. The phase of the unit cell is quantized to only 0° and 180° when N is 1. Subsequently, as N increases, the quantized phase difference
becomes smaller. At the same time, a continuous retrodirective metasurface (continuous RRDM), which controls the phase continuously, can implement a perfect reflection phase gradient, with the
controlled reflection phase (Φ[Continuos](m)) of the m^th unit cell being equal to Φ[Ideal](m). Due to this difference between the N-bit and continuous RRDMs, they operate differently. Furthermore,
the array factor (AF(θ)) of the N-bit and continuous RRDMs can be calculated by changing N from 1 to 4. Fig. 2 shows the normalized value of AF^2 on using 1, 2, 3, and 4-bit RRDMs, with the
retroreflection angles being 5° and 25°. The cell distance is 0.35λ[0] and the number of cells is 20.
Moreover, since the magnitude of AF(θ)^2 and the gain of the beam are proportional to each other, it is possible to assess the reduction in gain caused by the phase quantization by comparing the AF(θ
)^2 of the N-bit and continuous RRDMs. The ratio of the maximum peak value (Peak) of the N-bit and continuous RRDMs can be considered the quantization efficiency (η[Quantization]), as shown in Eq.
(4) [10]:
(4) ηQuantization=PeakN-bitPeakContinuous.
In accordance with Fig. 2, the value of η[Quantization] converges to 1 as N increases, indicating an improvement in the gain reduction.
Meanwhile, in the case of 1-bit (N = 1) RRDM, Fig. 2(a) shows a spurious beam error at the same level as the main lobe occurring in an undesired direction (θ=−15.5°). This spurious beam was analyzed
using array factor theory, according to which the AF(θ) of the N-bit RRDM (AF[N–bit](θ)) can be denoted as Eqs. (5) and (6), noted below:
(5) ΨN-bit(m,θ)=Σk=1mψN-bit(k,θ),(ψN-bit(m,θ)=k0dxsinθ-k0dxsinθi+ΔΦN-bit(m),ΔΦN-bit(1)=0),
(6) AFN-bit(θ)=Σk=1MejΨN-bit(k,θ).
Here, k[0] (=2π/λ[0]) indicates a propagation constant in free space and ΔΦ[N–bit](m) denotes a difference in the reflection phase between the m^th and m–1^th cells when retro-reflection occurs in
the N-bit RRDM. In addition, Ψ[N–bit](m,θ) refers to a progressive phase—the relative phase between the m^th and m–1^th cells in the direction θ°.
With regard to the N-bit RRDM, let us consider defining Ψ[N–bit](m,θ[re]) and AF[N–bit](θ[re]) as the progressive phase and the array factor in the θ[re](=−θ[i]) direction, respectively. Furthermore,
let us consider defining Ψ[N–bit](m,θ[s]) and AF[N–bit] (m,θ[s]) as the progressive phase and the array factor in the θ[s] direction—in which the spurious beam error occurs— respectively. Considering
these assumptions, according to Fig. 2, since the spurious beam error is of the same level as the main lobe (retro-reflection), it can be accomplished that AF[N–bit](θ[re]) =AF[N–bit] (m,θ[s]). This
can be expressed as Eqs. (7) and (8), since AF[N–bit](θ) represents the even function of Ψ[N–bit](m,θ).
(7) ψN-bit(m,θre)=-ψN-bit(m,θs)+2aπ,
(8) ψN-bit(m,θre)=ψN-bit(m,θs)+2aπ.
Furthermore, since θ[re] is equal to − θ[i], Eqs. (7) and (8) can be formulated as Eqs. (9) and (10), respectively, when retroreflection occurs.
(9) k0dxsinθs=-3k0dxsinθre-2ΔΦN-bit(m)+2aπ,
(10) k0dxsinθs=k0dxsinθre+2aπ.
Here, a is an integer. Notably, Eq. (9) has a quantization term (ΔΦ[N–bit](m)), which is satisfied for all m only when N = 1. Therefore, Eq. (9) represents the spurious beam error information in the
1-bit RRDM. Meanwhile, since Eq. (10) does not contain the quantization term (ΔΦ[N–bit](m)), it can be satisfied, regardless of N. Effectively, this is similar to the conditions under which the
grating lobe occurs.
The average value of η[Quantization], calculated by changing the dx and θ[re], were found to be 43.6% for 1-bit, 82.2% for 2-bit, 95.3% for 3-bit, and 98.8% for 4-bit. The number of spurious beam
errors is presented in Fig. 3 in terms of the dx and θ[re] by considering the results obtained from Eqs. (9) and (10). For instance, in the 1-bit RRDM, no spurious beam error occurs if the cell
distance is 0.35λ[0] and the retro-reflection angle is 25°, as shown in Fig. 3(a). Meanwhile, when the retroreflection angle is 30° and the cell distance is 0.75λ[0] in the 2- bit RRDM, one spurious
beam error occurs, as shown in Fig. 3(b), with the angle derived using Eq. (10) being −56.4°.
Fig. 4 presents schematic diagrams of the simulations and the measurement method employed to verify the analyzed results. Notably, this study adopted the ring patch structural unit cell used in [10]
as the 1-bit RRDM’s unit cell. The measurement and simulations were conducted at 10.1 GHz for the 1-bit RRDM, arranged as 12 × 12 (126 mm × 126 mm). Each unit cell (10.5 mm × 10.5 mm) was mounted
with one PIN diode [13], whose phase was controlled by a 1-bit by switching the PIN diode on/off. In Simulation 1, as depicted in Fig. 4(a), the bistatic RCS result is obtained by simulating a case
in which the incident wave is a plane incident wave moving in the direction (25°, 0°). In Simulation 2, the incident wave is radiated from the Tx horn antenna fixed at (60 cm, 25°, 0°), resulting in
scattered power patterns that indicate the relative power value of the wave scattered in the 1-bit RRDM. This phenomenon bears the same meaning as the RCS, which expresses scattered power in an area.
In both Simulation 3 and the measurement, the Tx horn antenna is fixed at (60 cm, 25°, 0°), while the Rx horn antenna is moved to the arc of (60 cm, θ°, 0°), ultimately resulting in S[21].
Subsequently, the bistatic RCS was calculated using S[21] and the radar range equation. It is noted that the retro-reflected wave can be measured from S[1] using only one Tx horn, while the other
waves can be measured using the Tx and Rx horn antennas.
The bistatic RCS and the scattered power pattern obtained from the three simulations and the measurement were normalized and plotted, as depicted in Fig. 5. For normalization, each simulation and
measurement was performed again using the conducting plate (126 mm × 126 mm). Subsequently, the resulting RCS values in the specular direction were used as the normalization factor. As observed in
Fig. 5, the results of the simulations and the measurement conducted for the different methods exhibit similar shapes. The main lobe is formed at around 25°, indicating a retro-reflection angle, with
no spurious beams. Meanwhile, the beam around −25° is in the specular direction, exhibiting a lower level than the main lobe. Overall, the measured results indicate relatively good agreement with the
simulated results, thus verifying the analysis and theory of RRDM presented in this study.
This paper proposed a PIN diode-based N-bit RRDM that can electronically control the retro-reflection angle. The array factor of the N-bit RRDM achieved a lower value than that the continuous RRDM
due to phase quantization, resulting in gain reduction. However, undesired spurious beam errors still occurred. This phenomenon was theoretically analyzed and arranged using generalized Snell’s law
and the array factor theory. The obtained results show that the gain reduction can be controlled by the selection of N. In particular, it was found that when N increased from 1 to 2, the quantization
efficiency increased dramatically from 43.6% to 82.2%. Furthermore, it was established that the occurrence of a spurious beam can be controlled by the selection of d[x] and θ[re]. Moreover, the
theoretical results were successfully verified through simulation and measurement.
This work was supported by the National Research Foundation of Korea, funded by the Ministry of Education through the Basic Science Research Program (Grant No. 2015R1A6A1A03031833).
2. Li Y., Jandhyala V.. Design of retrodirective antenna arrays for short-range wireless power transmission. IEEE Transactions on Antennas and Propagation 60(1):206–211. 2012;
4. Fairouz M., Saed M. A.. A complete system of wireless power transfer using a circularly polarized retrodirective array. Journal of Electromagnetic Engineering and Science 20(2):139–144. 2020;
5. Wong A. M., Christian P., Eleftheriades G. V.. Binary Huygens’ metasurfaces: experimental demonstration of simple and efficient near-grazing retroreflectors for TE and TM polarizations. IEEE
Transactions on Antennas and Propagation 66(6):2892–2903. 2018;
6. Hoang T. V., Lee C. H., Lee J. H.. Two-dimensional efficient broadband retrodirective metasurface. IEEE Transactions on Antennas and Propagation 68(3):2451–2456. 2020;
8. Huang C., Sun B., Pan W., Cui J., Wu X., Luo X.. Dynamical beam manipulation based on 2-bit digitally-controlled coding metasurface. Scientific Reports 7article no. 42302. 2017;
9. Yang H., Yang F., Cao X., Xu S., Gao J., Chen X., Li M., Li T.. A 1600-element dual-frequency electronically reconfigurable reflectarray at X/Ku-band. IEEE Transactions on Antennas and Propagation
65(6):3024–3032. 2017;
10. Lee S. G., Nam Y. H., Kim Y., Kim J., Lee J. H.. A wide-angle and high-efficiency reconfigurable reflectarray antenna based on a miniaturized radiating element. IEEE Access 10:103223–103229.
11. Yu N., Genevet P., Kats M. A., Aieta F., Tetienne J. P., Capasso F., Gaburro Z.. Light propagation with phase discontinuities: generalized laws of reflection and refraction. Science 334
(6054):333–337. 2011;
12. Balanis C. A.. Antenna Theory: Analysis and Design Hoboken, NJ: John Wiley & Sons; 2016.
13. Jung H. B., Lee S. G., Lee J. H.. Extraction method for X-band PIN diode equivalent circuit parameters based on waveguide measurement. The Journal of Korean Institute of Electromagnetic
Engineering and Science 33(8):585–590. 2022;
Hae-Bin Jung, https://orcid.org/0000-0002-3254-231X received his B.S. and M.S. degrees in electronics and electrical engineering from Hongik University, South Korea, in 2021 and 2023, respectively.
He is currently a research engineer at LIG Nex1, Yongin, South Korea. His research interests include array antennas and metasurfaces.
Jeong-Hae Lee, https://orcid.org/0000-0002-5135-6360 received his B.S. and M.S. degrees in electrical engineering from Seoul National University, South Korea, in 1985 and 1988, respectively, and his
Ph.D. in electrical engineering from the University of California at Los Angeles, Los Angeles, USA, in 1996. From 1993 to 1996, he was a visiting scientist at General Atomics, San Diego, CA, USA,
where his major research initiatives were the development of a millimeter-wave diagnostic system and studying plasma wave propagation. Since 1996, he has been working at Hongik University, Seoul,
South Korea, as a professor in the Department of Electronic and Electrical Engineering. He was the president of the Korea Institute of Electromagnetic Engineering and Science in 2019. He is currently
the director of the Metamaterial Electronic Device Center. He has more than 120 articles published in journals and 70 patents. His current research interests include metamaterial radio frequency
devices and wireless power transfer.
Article information Continued
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Platelet-rich plasma (PRP) has been applied in a number of clinical Platelet-rich plasma (PRP) has been applied in a number of clinical
Cone beam calculated tomography (CBCT) imaging is a key step in image guided radiation therapy (IGRT) to improve tumor targeting. to reduce the imaging dose. To be compared with other
state-of-the-art spatial interpolation (called inpainting) methods in terms of signal-to-noise ratio (SNR) on a Catphan and head phantoms IPI increases SNR from 15. 3dB and 12. 7dB to 29. 0dB and 28.
1dB respectively. The SNR of IPI on sparse-view CBCT reconstruction can achieve from 28dB to 17dB for undersample projection sets with gantry angle interval varying from 1 to 3 degrees for both
phantoms. between and is calculated as: is the local U coordinate. The sign of the U coordinate is the same as the Z coordinate of equidistantly and find the ‘abrupt’ point defined as the depth of
blocked pixel is the width of the detector in pixels) can not be guaranteed. Thus we built up an energy function: is the sum of the data cost from the whole scanline and is the sum of the difference
of depths between two adjacent pixels in the scanline while adjusts the weight between the data and smooth terms. This energy can be minimized by rewriting the energy in recursive format and applying
dynamic programming [20]:
(7) In this way the sum of energy along the scanline is split into three parts: the first part is the cost of the first pixel defined in formula 4; the second part is the difference of depths between
the second and first pixel; the third part is the sum of energy along the scanline expected for the first pixel. Iteratively the sum of the energy could be split into a group of K-252a the first part
and a group of the second part. Starting from the last pixel we could buy Alogliptin Benzoate trace back the depths of pixels over the scanline with minimized strength. After picking out the
corresponding couple based on the optimized absolute depths the lacking pixel level is believed as the mean of this intensity of this corresponding combined projection px. 2 . 4 SMOG Simulation and
Evaluation of the IPI Technique A Catphan phantom and a head phantom were scanned using the onboard CBCT system on a Varian? Trilogy? machine using a half fan mode with the X-ray tube voltage of 120
kVp. In each scan approximately 650 projections were acquired and each projection’s buy Alogliptin Benzoate dimensions were 1024 × 768 with resolutions of 0. 388mm × 0. 388mm. The reconstruction
images contain 384 × 384 × 64 voxels with resolutions of 0. 651mm × 0. 651mm ×. | {"url":"http://www.brain-tumor-cancer-information.com/2016/02/23/platelet-rich-plasma-prp-has-been-applied-in-a-number-of-clinical-platelet-rich-plasma-prp-has-been-applied-in-a-number-of-clinical/","timestamp":"2024-11-14T23:58:17Z","content_type":"text/html","content_length":"38908","record_id":"<urn:uuid:befce719-b47d-4e7a-b6bb-249f6c007edd>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00077.warc.gz"} |
Consider two vectors A=3i - 1j and B = - i - 5j, how do you calculate A + B? | Socratic
Consider two vectors #A=3i - 1j# and #B = - i - 5j#, how do you calculate #A + B#?
1 Answer
You can simply add the components in each direction:
$A + B = \left(3 i - 1 j\right) + \left(- i - 5 j\right) = 2 i - 6 j$
I'm not sure this question needs much more explanation, but there is value for you in drawing a diagram showing the two vectors, then sliding $B$ from the origin until its 'tail' is at the 'nose' of
$A$. The resultant vector is then from the origin to the 'nose' of $B$. That will give you a visual reference for why it is possible to simply add the components.
Impact of this question
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Why do Coverage Gradient Strategies work so effectively in Cooperative MARL? Proof from Coverage Illustration - Techmaggie
In cooperative multi-agent reinforcement studying (MARL), resulting from its on-policy nature, coverage gradient (PG) strategies are usually believed to be much less pattern environment friendly than
worth decomposition (VD) strategies, that are off-policy. Nonetheless, some latest empirical research show that with correct enter illustration and hyper-parameter tuning, multi-agent PG can obtain
surprisingly robust efficiency in comparison with off-policy VD strategies.
Why may PG strategies work so effectively? On this publish, we are going to current concrete evaluation to indicate that in sure eventualities, e.g., environments with a extremely multi-modal reward
panorama, VD could be problematic and result in undesired outcomes. In contrast, PG strategies with particular person insurance policies can converge to an optimum coverage in these instances. As
well as, PG strategies with auto-regressive (AR) insurance policies can study multi-modal insurance policies.
Determine 1: completely different coverage illustration for the 4-player permutation sport.
CTDE in Cooperative MARL: VD and PG strategies
Centralized coaching and decentralized execution (CTDE) is a well-liked framework in cooperative MARL. It leverages international info for simpler coaching whereas protecting the illustration of
particular person insurance policies for testing. CTDE could be applied through worth decomposition (VD) or coverage gradient (PG), main to 2 various kinds of algorithms.
VD strategies study native Q networks and a mixing perform that mixes the native Q networks to a world Q perform. The blending perform is often enforced to fulfill the Particular person-World-Max (
IGM) precept, which ensures the optimum joint motion could be computed by greedily selecting the optimum motion domestically for every agent.
In contrast, PG strategies immediately apply coverage gradient to study a person coverage and a centralized worth perform for every agent. The worth perform takes as its enter the worldwide state
(e.g., MAPPO) or the concatenation of all of the native observations (e.g., MADDPG), for an correct international worth estimate.
The permutation sport: a easy counterexample the place VD fails
We begin our evaluation by contemplating a stateless cooperative sport, particularly the permutation sport. In an $N$-player permutation sport, every agent can output $N$ actions ${ 1,ldots, N }$.
Brokers obtain $+1$ reward if their actions are mutually completely different, i.e., the joint motion is a permutation over $1, ldots, N$; in any other case, they obtain $0$ reward. Observe that
there are $N!$ symmetric optimum methods on this sport.
Determine 2: the 4-player permutation sport.
Determine 3: high-level instinct on why VD fails within the 2-player permutation sport.
Allow us to deal with the 2-player permutation sport now and apply VD to the sport. On this stateless setting, we use $Q_1$ and $Q_2$ to indicate the native Q-functions, and use $Q_textrm{tot}$ to
indicate the worldwide Q-function. The IGM precept requires that
We show that VD can’t signify the payoff of the 2-player permutation sport by contradiction. If VD strategies have been capable of signify the payoff, we’d have
[Q_textrm{tot}(1, 2)=Q_textrm{tot}(2,1)=1quad text{and}quad Q_textrm{tot}(1, 1)=Q_textrm{tot}(2,2)=0.]
If both of those two brokers has completely different native Q values (e.g. $Q_1(1)> Q_1(2)$), we now have $argmax_{a^1}Q_1(a^1)=1$. Then in accordance with the IGM precept, any optimum joint motion
satisfies $a^{1star}=1$ and $a^{1star}neq 2$, so the joint motion $(a^1,a^2)=(2,1)$ is sub-optimal, i.e., $Q_textrm{tot}(2,1)<1$.
In any other case, if $Q_1(1)=Q_1(2)$ and $Q_2(1)=Q_2(2)$, then
[Q_textrm{tot}(1, 1)=Q_textrm{tot}(2,2)=Q_textrm{tot}(1, 2)=Q_textrm{tot}(2,1).]
In consequence, worth decomposition can’t signify the payoff matrix of the 2-player permutation sport.
What about PG strategies? Particular person insurance policies can certainly signify an optimum coverage for the permutation sport. Furthermore, stochastic gradient descent can assure PG to converge
to one among these optima underneath gentle assumptions. This means that, regardless that PG strategies are much less widespread in MARL in contrast with VD strategies, they are often preferable in
sure instances which might be frequent in real-world functions, e.g., video games with a number of technique modalities.
We additionally comment that within the permutation sport, as a way to signify an optimum joint coverage, every agent should select distinct actions. Consequently, a profitable implementation of PG
should be certain that the insurance policies are agent-specific. This may be achieved by utilizing both particular person insurance policies with unshared parameters (known as PG-Ind in our paper),
or an agent-ID conditioned coverage (PG-ID).
PG outperforms current VD strategies on widespread MARL testbeds
Going past the easy illustrative instance of the permutation sport, we prolong our research to widespread and extra lifelike MARL benchmarks. Along with StarCraft Multi-Agent Problem (SMAC), the
place the effectiveness of PG and agent-conditioned coverage enter has been verified, we present new leads to Google Analysis Soccer (GRF) and multi-player Hanabi Problem.
Determine 4: (left) successful charges of PG strategies on GRF; (proper) greatest and common analysis scores on Hanabi-Full.
In GRF, PG strategies outperform the state-of-the-art VD baseline (CDS) in 5 eventualities. Curiously, we additionally discover that particular person insurance policies (PG-Ind) with out parameter
sharing obtain comparable, typically even larger successful charges, in comparison with agent-specific insurance policies (PG-ID) in all 5 eventualities. We consider PG-ID within the full-scale
Hanabi sport with various numbers of gamers (2-5 gamers) and examine them to SAD, a robust off-policy Q-learning variant in Hanabi, and Worth Decomposition Networks (VDN). As demonstrated within the
above desk, PG-ID is ready to produce outcomes corresponding to or higher than the perfect and common rewards achieved by SAD and VDN with various numbers of gamers utilizing the identical variety of
surroundings steps.
Past larger rewards: studying multi-modal conduct through auto-regressive coverage modeling
In addition to studying larger rewards, we additionally research study multi-modal insurance policies in cooperative MARL. Let’s return to the permutation sport. Though we now have proved that PG
can successfully study an optimum coverage, the technique mode that it lastly reaches can extremely rely on the coverage initialization. Thus, a pure query will probably be:
Can we study a single coverage that may cowl all of the optimum modes?
Within the decentralized PG formulation, the factorized illustration of a joint coverage can solely signify one explicit mode. Subsequently, we suggest an enhanced technique to parameterize the
insurance policies for stronger expressiveness — the auto-regressive (AR) insurance policies.
Determine 5: comparability between particular person insurance policies (PG) and auto-regressive insurance policies (AR) within the 4-player permutation sport.
Formally, we factorize the joint coverage of $n$ brokers into the type of
[pi(mathbf{a} mid mathbf{o}) approx prod_{i=1}^n pi_{theta^{i}} left( a^{i}mid o^{i},a^{1},ldots,a^{i-1} right),]
the place the motion produced by agent $i$ relies upon by itself statement $o_i$ and all of the actions from earlier brokers $1,dots,i-1$. The auto-regressive factorization can signify any joint
coverage in a centralized MDP. The solely modification to every agent’s coverage is the enter dimension, which is barely enlarged by together with earlier actions; and the output dimension of every
agent’s coverage stays unchanged.
With such a minimal parameterization overhead, AR coverage considerably improves the illustration energy of PG strategies. We comment that PG with AR coverage (PG-AR) can concurrently signify all
optimum coverage modes within the permutation sport.
Determine: the heatmaps of actions for insurance policies discovered by PG-Ind (left) and PG-AR (center), and the heatmap for rewards (proper); whereas PG-Ind solely converge to a selected mode
within the 4-player permutation sport, PG-AR efficiently discovers all of the optimum modes.
In additional complicated environments, together with SMAC and GRF, PG-AR can study fascinating emergent behaviors that require robust intra-agent coordination that will by no means be discovered by
Determine 6: (left) emergent conduct induced by PG-AR in SMAC and GRF. On the 2m_vs_1z map of SMAC, the marines hold standing and assault alternately whereas guaranteeing there is just one attacking
marine at every timestep; (proper) within the academy_3_vs_1_with_keeper situation of GRF, brokers study a “Tiki-Taka” model conduct: every participant retains passing the ball to their teammates.
Discussions and Takeaways
On this publish, we offer a concrete evaluation of VD and PG strategies in cooperative MARL. First, we reveal the limitation on the expressiveness of widespread VD strategies, exhibiting that they
might not signify optimum insurance policies even in a easy permutation sport. In contrast, we present that PG strategies are provably extra expressive. We empirically confirm the expressiveness
benefit of PG on widespread MARL testbeds, together with SMAC, GRF, and Hanabi Problem. We hope the insights from this work may gain advantage the group in the direction of extra normal and extra
highly effective cooperative MARL algorithms sooner or later.
This publish relies on our paper: Revisiting Some Frequent Practices in Cooperative Multi-Agent Reinforcement Studying (paper, web site). | {"url":"https://techmaggie.com/why-do-coverage-gradient-strategies-work-so-effectively-in-cooperative-marl-proof-from-coverage-illustration/","timestamp":"2024-11-14T11:57:42Z","content_type":"text/html","content_length":"399891","record_id":"<urn:uuid:9a79daf1-c583-4d6c-9a77-7be6a91ed87b>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00036.warc.gz"} |
Math Remediation for the College Bound: How Teachers Can Close the Gap, from the Basics through Algebra
Math Remediation for the College Bound
How Teachers Can Close the Gap, from the Basics through Algebra
Daryao Khatri
Algebra is the language that must be mastered for any course that uses math because it is the gateway for entry into any science, technology, engineering, and mathematics (STEM) discipline. This book
fosters mastery of critical math and algebraic concepts and skills essential to all of the STEM disciplines and some of the social sciences.
This book is written by practitioners whose primary teaching subject is not math but who use math extensively in their courses in STEM disciplines, social science statistics, and their own research.
Moreover, in the writing of this book, the authors have used the teaching principles of anchoring, overlearning, pruning the course to its essentials, and using simple and familiar language in word
Math Remediation for the College Bound
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ipgmr.com Amortization Calculator
iPgmr.com Amortization Calculator AMORTZcgi
AMORTZ Command Overview
Buying a house, a car, a boat, an RV? Chances are that purchase will require a loan. How does the loan work? How much will the purchase actually cost? How do repayment options affect the cost of the
loan? That is what an amortization schedule will tell you. It provides a month by month accounting of the the loan repayment and the distribution of principal and interest for each payment.
To view an amortization schedule, enter enter the amount of the loan, the annual interest rate, the payoff period in years, and click Submit. Payment amount is optional and will be computed if not
An expanded version of this loan calculator is included with the iPgmr.com Integrated Accounting System (IAS) in the form of an iSeries command, AMORTZ. See the link above for an overview of the
command and its usage.
Every loan comes with a number of terms and conditions. The ammortization schedule can help you see how each of these affect the repayment of the loan.
SIMPLE INTEREST The amortization table is based on simple interest. That means that interest is computed as a fixed amount based on the number of days elapsed between payments. Enter the quoted
interest rate and submit the request.
COMPOUNDED INTEREST Compounded interest computes the interest each day and adds the interest to the balance before computing the next day's interest. APR (annual percentage rate) is higher than the
quoted rate where interest is compounded. Change the quoted interest rate to the computed APR and resubmit the request.
NOTE: APR, in addition to compounded interest, can include other costs such as points, processing fees, etc. The Truth in Lending Act requires the listing of the APR in addition to the nominal
interest rate on all consumer loans.
BALLOON PAYMENT A balloon payment is used to provide short term financing. The loan is structured to make minimum payments for a period of time, during which other financing can be arranged. The
balloon is the payment due at the end of the term and represents the outstanding balance of principal and interest at that time. Change the payback years to shorten the term, but leave payment amount
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NOTE: It is possible that the payments have been structured in such a way that the payments do not even cover the interest each month resulting in a balloon payment at the end of the term that is
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ACCELERATED PAYMENTS Accelerated payments are payments that are above and beyond the required monthly loan payment. Accelerated payment amounts increase the monthly principal payment and in doing so,
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Volume in high dimensions
Dimension 5 isn’t so special
Lately I’ve been reading The Best Writing on Mathematics 2012. I’d like to present a alternative perspective on one of the articles.
In his article “An Adventure in the Nth Dimension,” Brian Hayes explores how in high dimensions, balls have surprisingly little volume. As the dimension n increases, the volume of a ball of radius 1
increases until n = 5. Then for larger n the volume steadily decreases. Hayes asks
What is it about five-dimensional space that allows a unit 5-ball to spread out more expansively than any other n-ball?
He says that it all has to do with the value of π and that if π were different, the unit ball would have its maximum value for a different dimension n. While that is true, it seems odd to speculate
about changing the value of π. It seems much more natural to speculate about changing the radius of the balls.
The volume of a ball of radius r in dimension n is
If we fix r at 1 and let n vary, we get a curve like this:
But for different values of r, the plot will have its maximum at different values of n. For example, here is the curve for balls of radius 2:
Let’s think of n in our volume formula as a continuous variable so we can differentiate with respect to n. It turns out to be more convenient to work with the logarithm of the volume. This makes no
difference: the logarithm of a function takes on its maximum exactly where the original function does since log is an increasing function.
We can tell from this equation that volume (eventually) decreases as a function of n because ψ is an unbounded increasing function. The derivative has a unique zero, and we can move the location of
that zero out by increasing r. So for any dimension n, we can solve for a value of r such that a ball of radius r has its maximum volume in that dimension:
Related: High dimensional integration
14 thoughts on “Dimension 5 isn’t so special”
1. Does not compute. V(n,r) is [length^n]. Comparing length to area, area to volume, and so forth. Does it make sense?
What makes sense is how much smaller is the sphere w.r.t. the cube in the same dimension. And to tell that it’s better to deal with unit-cube, i.e. r=0.5 and n in Z, n > 0, then V(n,r=0.5) is
decreasing monotonically. That is the higher n, the lower fraction of volume we can fill with spheres.
Can you give an interpretation of V(n,r) for a real n?
2. @S. Just so, those plots are comparing meters with liters.
3. That last formula doesn’t seem right. It doesn’t give something close to 1 for n=5, or something close to 2 for n=23.
4. This is super interesting. Though something smells fishy. You are comparing the volumes across dimensions, yet this comparison depends on some arbitrary unit of measurement.
In other words, is a circle or a sphere (each with the same radius) “bigger”? If the answer depends on whether we use inches or centimeters, then that is not a well-posed question. Just because
the definition “volume” has been generalized across dimensions does not mean we can do such comparisons.
5. Yes, it is fishy. If you don’t look at the volume relative to some other volume, you can get different results for different r’s.
6. Along the same lines as the other comments: I’d say it’s NOT about the radius of the ball, but rather the units we use.
7. Several of these results can be found in Steven Krantz’s article at http://www.maa.org/joma/volume7/krantz/higher.pdf, along with the very cool result that, in the limit as N goes to infinity,
all of the volume of the N-dimensional unit ball lies arbitrarily close to the surface of the ball.
8. I think it’s intuitive:
A ball with radius 1 in n=2 IS a lot smaller than a ball with r=1 in n=100.
In order for the squared coordinates to add up to 1, a lot of them have to be small in the high-dimensional case.
One could imagine a process whereby we extrude a line 1 unit on each side, getting a square. They you have to cut a bit off to get a circle. Then you extrude it into a cylinder. To get a unit
sphere, you again need to cut a bit off. Etc.
9. It’s obvious that the ball is largest when the dimension is 1 and
decreases in volume with increasing dimension. Just evaluate the volume
of the ball with fits inside a unit n-dimensional cube (i.e., r = 1/2).
On the contrary, it’s obvious that the ball is smallest when the
dimension is 1 and increases in volume with increasing dimension. Just
evaluate the volume of the ball which just surrounds a unit
n-dimensional cube (i.e., r = sqrt(n)/2).
Bottom line is it all depends on what you are comparing. The only thing
you can be sure about is that the volume of a zero-dimensional ball is
always 1.
10. Your last formula has two typos. It should be
or simply
11. Veky: Thanks. I’ve updated the post.
12. This isn’t just about a choice of units. Imagine that you have a set of cubes in different dimensions, each containing the largest possible sphere. Now imagine choosing random points in the box
and compare the odds of those points also being within the sphere. Your odds are greatest in dimension 5. (Actually in the fractional dimension 5.2569464147072722). I still don’t really have a
sense of why there is a local maximum in this curve.
13. Melinda, the way you posed the problem, the odds are maximum when the dimension is 1 and fall off monotonically from there. (See my comment above.)
14. I see your point. I guess it really is about units. Still, this is a very unexpected curve even if we are comparing apples with oranges. Is there an intuitive way of seeing why this curve peaks
near dimension 5? | {"url":"https://www.johndcook.com/blog/2012/10/23/dimension-5-isnt-so-special/","timestamp":"2024-11-05T16:59:08Z","content_type":"text/html","content_length":"74142","record_id":"<urn:uuid:b8ce4763-d23a-46ca-8cc3-2213a90a0a07>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00784.warc.gz"} |
Digital Math Resources
Display Title
Math Example: Comparing Area: Example 2
Math Example: Comparing Area: Example 2
In this example, we compare the areas of two rectangles, A and B, drawn on a grid. Rectangle A has dimensions of 3 units by 5 units, while Rectangle B measures 2 units by 6 units. Calculating their
areas, we find that Rectangle A has an area of 3 * 5 = 15 square units, and Rectangle B has an area of 2 * 6 = 12 square units. Comparing these results, we can conclude that A > B, meaning Rectangle
A has a larger area than Rectangle B.
This example is part of a series designed to teach students about comparing areas of different geometric shapes. By presenting various scenarios with rectangles of different dimensions, students
learn to calculate areas and make comparisons. This approach helps reinforce the concept that area is not solely determined by a shape's perimeter or visual size, but by the product of its
Exposing students to multiple worked-out examples is essential for developing a deep understanding of area comparison. Each example provides a unique scenario, allowing students to apply the same
principles to different dimensions and shapes. This repetition helps solidify the concept and improves students' ability to tackle various area-related problems independently.
Teacher Script: "Now, let's look at these two new rectangles, A and B. They have different dimensions from our previous example. Can you calculate their areas for me? What do you notice about the
results? Even though Rectangle B is longer, Rectangle A is wider. This shows us that both length and width play important roles in determining area. What conclusion can we draw about comparing areas
of different rectangles?"
For a complete collection of math examples related to Geometry click on this link: Math Examples: Comparing Areas Collection. | {"url":"https://www.media4math.com/library/math-example-comparing-area-example-2","timestamp":"2024-11-15T01:31:14Z","content_type":"text/html","content_length":"51923","record_id":"<urn:uuid:1e597ed8-3f27-431e-bb40-8ecfb8b2985f>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00581.warc.gz"} |
Letters In Math
Math Letters Mathematics Themed PNG Letters Instant Etsy
Letters In Math. Web mathematical operators and supplemental mathematical operators. Web upper case letter lower case letter greek letter name english equivalent letter name.
Math Letters Mathematics Themed PNG Letters Instant Etsy
Web upper case letter lower case letter greek letter name english equivalent letter name. Web mathematical alphanumeric symbols is a unicode block comprising styled forms of latin and greek letters
and decimal digits that enable mathematicians to denote different notions with different letter. You can help by adding missing items. Web mathematical operators and supplemental mathematical
operators. Using letters allows us to represent quantities that could change or that we don't know the value of yet. Arrow (symbol) and miscellaneous symbols and arrows and. (january 2011) latin and
greek letters are used in mathematics, science, engineering, and. Web in algebra, we often want to work with general concepts instead of specific numbers. Web list of letters used in mathematics,
science, and engineering.
You can help by adding missing items. You can help by adding missing items. Web mathematical operators and supplemental mathematical operators. Using letters allows us to represent quantities that
could change or that we don't know the value of yet. Arrow (symbol) and miscellaneous symbols and arrows and. Web list of letters used in mathematics, science, and engineering. Web in algebra, we
often want to work with general concepts instead of specific numbers. Web mathematical alphanumeric symbols is a unicode block comprising styled forms of latin and greek letters and decimal digits
that enable mathematicians to denote different notions with different letter. (january 2011) latin and greek letters are used in mathematics, science, engineering, and. Web upper case letter lower
case letter greek letter name english equivalent letter name. | {"url":"https://math.nckl.gov.kh/letters-in-math.html","timestamp":"2024-11-04T05:29:08Z","content_type":"text/html","content_length":"20824","record_id":"<urn:uuid:63b07258-f9e3-4b5d-98ad-a02ecccba9c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00533.warc.gz"} |
How to Recover Data from an NTFS Partition on Windows [2022 Guide]
Reading Time: 5 minutesNew Technology File System (NTFS), is the file system that has been made standard for the Windows ecosystem. Since NTFS forms the basis for the way Windows’ file system works,
any issues with its functioning can lead to seemingly catastrophic data storage problems.
There are, however, dependable ways to recover NTFS files that have been deleted or corrupted due to the file system going bad. We’re going to take a look at how the underlying technology works and
the NTFS data recovery process.
What Is an NTFS File System
When your computer stores any new data that you provide it as input, the way that it decides how to process and store that data is guided by its file system technology. Everything from how you can
name files, where you can store them on the hard disk, and the ways in which you can retrieve data are all the domain of the file system architecture.
NTFS is a file system technology that Windows introduced in its computers in 1993. It came as an update to the FAT32 file system, which is what was standard for the Windows operating system before
The robust NTFS technology came with many advancements in how data is stored and retrieved in Windows computers. That includes:
• File-level encryption so that the data in individual folders and files can be protected easily.
• A b-tree data structure for directories, making it easier to store and sort through folders.
• A log system for changes made in the file system. This makes it possible to reverse changes that you make to the file system.
• A Volume Shadow Copy Service as an in-built means to backup data.
• A naming convention for files that allows a greater number of character and wider character options.
Many of the features that most users might not even notice on a Windows computer now were the result of the upgrade to the NTFS file system. It was a significant upgrade from Microsoft’s FAT32 file
system, which didn’t come with encryption options and had a maximum file size of 4GB.
Like with any other storage technology, there are instances where the data in your NTFS partitions can become corrupted or deleted without your permission. Let’s take a look at how you can go about
recovering this data.
How to Recover Deleted Files From an NTFS Partition
Time is of the essence if you’re trying to get back data from an NTFS partition. The files or folders that you can’t access in an NTFS drive are not fully gone until it has been overwritten with
other data. So as long as that hasn’t happened, the data recovery process is fairly straightforward.
Disk Drill is a reliable, feature-rich software that you can use for NTFS data recovery. You can follow this guide if you’re trying to recover data from an exFAT drive instead.
Here are the steps to follow to recover NTFS partitions using Disk Drill.
Step 1: Download Disk Drill at this link and install it on your computer. The software requires administrator-level permissions to recover data so make sure you’re signed into an admin account on
your computer.
Step 2: Launch Disk Drill. You will be met with the following screen when the tool opens.
Step 3: Go ahead and select the NTFS partition which contained the data that was lost. Make sure that you select Search for lost partitions in the right hand column as shown above. Then click on
Search for lost data to let Disk Drill know to kick off the process.
Step 4: Once the tool shows you the results of its search, you can click on Review found items to look at the data that has been recovered and select the files that you want to restore to your
Step 5: Choose the partition where you want to store the recovered data. It is recommended that you select a different partition from your NTFS drive so that no data is overwritten.
If you’ve deleted an entire partition on Windows 10 that you want to recover, then follow the instructions laid out here.
How to Recover Deleted NTFS Partition Using TestDisk
TestDisk is a popular open-source tool that can help with repairing an NTFS file system. Here’s how you can go about doing that.
Step 1: Download TestDisk at the CGSecurity website here.
Step 2: Extract the files in the zipped folder that you’ve downloaded. Click on the testdisk_win application file in that folder.
Step 3: Now a terminal window will open and you will see a Create option, which is for creating new log files. Click on that.
Step 4: You will now see a list of drives on your computer. Select the NTFS drive that you want to recover from the list.
Step 5: TestDisk will then show you a list of partition table types to choose from. It will also display a hint at the bottom suggesting the partition table type it has detected on your computer. So
select that one if you’re not sure. Hit Enter.
Step 6: You will now see an Analyse option, which you should click on so that the tool can begin to search for lost NTFS partition data.
Step 7: Once TestDisk surfaces a list of lost partitions, you can select on the ones that you want to recover and hit Enter.
Step 8: You can now navigate to the Write option at the bottom of the screen and hit Enter to complete NTFS partition recovery.
You’ve successfully recovered the lost NTFS partition at this point. You might have to restart your computer so that the changes can take effect.
Losing data from an NTFS partition can be a scary proposition, especially if you’ve lost sensitive or important files. But as we’ve seen, you can use tools like Disk Drill or TestDisk to reliably
recover that data.
The most important thing to remember when dealing with lost NTFS partition data is that the data on the disk must not be overwritten. So you must start working on recovering the data as soon as you
realize that it is not accessible anymore. | {"url":"https://data-recovery.wiki/ntfs-file-recovery/","timestamp":"2024-11-02T11:28:59Z","content_type":"text/html","content_length":"59197","record_id":"<urn:uuid:087b4ed2-026f-42a7-af11-fe31cf02f53f>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00882.warc.gz"} |
Exercise 4.2 class 8 solution
NCERT exercise 4.2 class 8 solution -students can click here to download the entire pdf chapterwise .here free pdf can be download for NCERT solution class 8. all Exercise are available here. These
solutions are available in downloadable PDF format as well. it will help students in getting rid of all the doubts about those particular topics that are covered in the exercise. The NCERT textbook
provides plenty of questions for the students to solve and practise. Solving and practising is more than enough to score high in the Class 8 examinations. Moreover, students should make sure that
they practise every problem given in the textbook . exercise 4.2 class 8 solution Pdf can be downloaded free here.
the pdf and exercise preview available at bottom of the page
Exercise 4.2 class 8 solution-graphical method of representing the data.
Graphical methods are a powerful way to represent data visually, making it easier to understand patterns, trends, and relationships within the data. Here are some common graphical methods for
representing data:
Bar Chart: Bar charts are used to display categorical data. The categories are shown on the horizontal axis (x-axis), and the frequency, count, or values are represented by bars on the vertical axis
(y-axis). Bar charts can be either horizontal or vertical.
Histogram: A histogram is used to represent the frequency distribution of continuous data. It divides the data into intervals (bins) on the x-axis and shows the frequency or count of data points
falling within each interval on the y-axis. Histograms are useful for visualizing the shape of the data distribution.^1
Line Chart: Line charts are used to show trends and changes in data over time. They connect data points with lines, making it easy to see how values evolve. They are commonly used in time series
Pie Chart: A pie chart is a circular chart that divides a whole into sectors or “slices” to represent the proportions of different categories within the whole. It is often used to show the
composition of a whole in terms of percentages.
The choice of graphical method depends on the nature of the data, the research questions, and the insights you want to gain from the data. Different types of data may be best suited to different
types of graphs or charts.
Exercise 4.2 class 8 solution – histogram
A histogram is a graphical representation of the distribution of a dataset, particularly used for displaying the frequency or count of data points within specific intervals or bins. It’s a commonly
used tool in statistics and data analysis to understand the shape, central tendency, and variability of a dataset. Here’s how to create and interpret a histogram:
Creating a Histogram:
1. Data Collection: Gather your dataset, which may consist of a series of measurements on a continuous scale.
2. Data Range Determination: Determine the range of values in your dataset. This defines the lower and upper bounds of the data.
3. Class Intervals (Bins): Divide the data range into non-overlapping intervals or bins. The width and number of intervals depend on your preferences and the characteristics of the data. Common
methods for determining class intervals include the square root method, Sturges’ rule, and Scott’s normal reference rule.
4. Frequency Count: Count how many data points fall within each interval. This count represents the frequency for that interval.
5. Graphical Representation: Create a bar chart where the x-axis represents the intervals (bins), and the y-axis represents the frequency of data points within each interval. Each interval is
represented as a bar, and the height of the bar corresponds to the frequency.
Exercise 4.2 class 8 solution – exercise preview
here the exercise preview is given-
Exercise 4.2 class 8 solution – solution pdf
students can view or download the pdf from here.click at the bottom to scroll the pdf pages.we provide Exercise 4.2 class 8 solution, just to help student to achieve their efficiency.
for more solution visit-
Table of Contents | {"url":"https://cmaindiagroup.in/exercise-4-2-class-8-solution/","timestamp":"2024-11-04T11:03:23Z","content_type":"text/html","content_length":"179199","record_id":"<urn:uuid:11e08337-e11f-4fc4-b8f3-42eaa60df45b>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00070.warc.gz"} |
The $\bf T=0$ $\bf 2k_F$ density wave phase transition in a two dimensional Fermi liquid is first order
We study $T=0$ spin density wave transitions in two dimensional Fermi liquids in which the ordering wavevector $\bf Q$ is such that the tangents to the Fermi line at the points connected by $\bf Q$
are parallel (e.g. $Q=2p_F$ in a system with a circular Fermi line) and the Fermi line is not flat. We show that the transition is first order if the ordering wave vector $\bf Q$ is not commensurate
with a reciprocal lattice vector, $\bf G$, i.e. ${\bf Q} \neq {\bf G}/2$. If $\bf Q$ is close to ${\bf G}/2$ the transition is weakly first order and an intermediate scaling regime exists; in this
regime the $2p_F$ susceptibility and observables such as the NMR rates $T_1$ and $T_2$ have scaling forms which we determine.
arXiv e-prints
Pub Date:
April 1995
11 pages, 8 Postscript figures in a separate file | {"url":"https://ui.adsabs.harvard.edu/abs/1995cond.mat..4024A","timestamp":"2024-11-03T20:52:01Z","content_type":"text/html","content_length":"36913","record_id":"<urn:uuid:ba37eaf3-029e-43b5-b6f5-f2549e81ea54>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00747.warc.gz"} |
Explore Hierarchical Structure of Topological Space Types
Explore Hierarchical Structure of Topological Space Types
Topological spaces are classified based on a hierarchy of mathematical properties they satisfy. As a result, some space types are more specific cases of more general ones. This information is encoded
for "TopologicalSpaceType" entities with the "MoreGeneralClassifications" property.
For example, a Banach space is also a topological space of the following types.
The network of hierarchical relation implications can be visualized directly using the "RelationshipGraph" property, which returns the full network with the space in question highlighted.
The relationship graph is a Wolfram Language graph expression, so built-in commands can be used to create a more streamlined version in which nodes are shown as points with tooltips and directed
edges are more easily visible.
As can be seen from the graph, some nodes lie at the bottom of the implication chain and hence have no more general classifications. Such leaf nodes correspond precisely to topological space types
for which the "MoreGeneralClassifications" property is missing.
The same list can be obtained by identifying the nodes in the graph that have vertex out-degree of 0.
Of more interest than space types corresponding to leaf nodes are those that correspond to "central" nodes according to some centrality measure.
Picking a few standard centrality types, you can easily find the topological space types that are most central by these measures.
These results can be summarized visually by highlighting the relevant nodes in the relationship graph. | {"url":"https://www.wolfram.com/language/12/math-entities/explore-hierarchical-structure-of-topological-space-types.html.en?footer=lang","timestamp":"2024-11-14T08:19:12Z","content_type":"text/html","content_length":"41624","record_id":"<urn:uuid:c8cc95fb-6e1f-4b34-a841-14fdd5e11cfd>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00250.warc.gz"} |
Introducing Bokeh in Python
We need to begin by setting up the environment, so start with making a venv
$ python -m venv .venv
$ source .venv/bin/activate
Make a requirements.txt and place it in the root. Add bokeh to that file, and then run pip install -r requirements.txt
Making a Simple Line Chart in Bokeh
Let’s start by making a super simple line chart. This will show up some of the basic concepts of Bokeh.
We can start by importing the fundamentals
from bokeh.plotting import figure, show
And then we create some data
x = [1, 2, 3, 4, 5]
y = [6, 7, 2, 4, 5]
It’s important that these lists are the same length. The next step is to create the figure.
p = figure(title="Simple line example", x_axis_label="x", y_axis_label="y")
p.line(x, y, legend_label="Values", line_width=2)
Notice how we first create a figure, and then add a line to it. We also get a lot of options to customise the figure and line - we can add titles, labels and legends. Finally, we want to see this
plot so we can use
Bokeh is intended for the web, so when we run this it will open the chart on a page in your default web browser. Here’s the final chart.
Let’s add some more data - this is really easy to do! All we have to do is create more lines. Let’s start with some more data
1x = [1, 2, 3, 4, 5]
2y = [1, 2, 3, 2, 1]
3y1 = [2, 3, 4, 5, 6]
4y2 = [5, 4, 3, 2, 1]
And then we make more calls to line
1p = figure(title="Multiple lines", x_axis_label="x", y_axis_label="y")
2p.line(x, y, legend_label="Values", line_width=2, color="blue")
3p.line(x, y1, legend_label="More Values", line_width=2, color="red")
4p.line(x, y2, legend_label="Even More Values", line_width=2, color="purple")
And this is the chart it produces
So now we can see the basic workflow of Bokeh
1. Prepare some data, usually into lists or maybe a numpy array
2. Create a figure
3. As as many series as you have data
4. Show the plot
We can customise each step of this flow quite extensively, as we will see in this article
Mixing Glyphs in Bokeh Plots
Bokeh supports many different kinds of “glyphs” - that’s basically what Bokeh calls different items that can be displayed on the figure. Let’s explore some of these options.
1p.vbar(x=x, top=y, legend_label="Bar", color="blue", width=0.5, bottom=0)
2p.scatter(x, y1, legend_label="Scatter Crosses", color="red", size=16, marker="x")
3p.scatter(x, y2, legend_label="Scatter Circles", color="purple", size=16)
Here we use vbar and scatter. scatter works a lot like lines , but we can customise the size of the marker and the marker type. Circles and crosses are the most common, but there are others too. vbar
is a little more involved - we need to set the x and top as named arguments. You can customise the width of the bars, as well as from where the bottom starts (usually, you just want this to be 0).
And then when we display it we get the following chart
Using Annotations to Mark Standard Deviations
A really useful feature in Bokeh is annotations - they let you mark certain areas of the plot. To give a real life example of this, we’re going to plot some random data, as well as display the
standard deviation of that data. Let’s start by generating those numbers and standard deviations
1import random
2import statistics
4N = 30 # Number of random numbers to generate
6x = [i for i in range(N)]
7random_numbers = [random.random() for _ in range(N)]
8mean = statistics.mean(random_numbers)
Now we’ll create the basic line graph - it’s the same as we did before so this should be familiar.
p = figure(title="Standard Deviation Example", x_axis_label="x", y_axis_label="y")
p.line(x, random_numbers, line_width=2, color="black")
In order to add annotations to this plot, we need to import the following
from bokeh.models import BoxAnnotation
Then we can create the annotations. Since we want to annotate the areas within and outside the standard deviation of the data, we want the “inside” region to be between the mean plus and minus the
standard deviation. The middle box has two bounds - the top and bottom. The high and low box have no bottom and top bound respectively, meaning they extend to the end of the plot
1low = mean - std_dev
2high = mean + std_dev
4low_box = BoxAnnotation(top=low, fill_alpha=0.2, fill_color="red")
5mid_box = BoxAnnotation(bottom=low, top=high, fill_alpha=0.2, fill_color="green")
6high_box = BoxAnnotation(bottom=high, fill_alpha=0.2, fill_color="red")
We then simply add the three boxes to the plot with add_layout - and we get the following plot
K-Means Plot
An interesting plotting project we can use to show off some of Bokeh’s potential is plotting K-means. For this we need a few more dependencies, so add the following to the requirements.txt (and make
sure you run pip install -r requirements.txt)
Since this isn’t a K-means tutorial, we’ll skip over the details - but if you don’t know what K-means does, the basic idea is to group data into K groups. Here’s the code we’ll use for this
1import numpy as np
2from sklearn.cluster import KMeans
4data = np.vstack(
5 [
6 np.random.normal(loc=(0, 0), scale=1.0, size=(100, 2)),
7 np.random.normal(loc=(5, 5), scale=1.0, size=(100, 2)),
8 np.random.normal(loc=(0, 5), scale=1.0, size=(100, 2)),
9 ]
12kmeans = KMeans(n_clusters=3)
13pred = kmeans.fit_predict(data)
Now we have the groups we need to group the data in a more convenient way for us to plot.
1plotting_data = {}
3for i in range(N):
4 plotting_data[i] = []
6for point, group in zip(data, pred):
7 plotting_data[group].append(point.tolist())
This will be nice and generic so if we increase N in future it still works - we are essentially making a dictionary of group to a list of the coordinates in that group.
We can make the basic plot again with
p = figure(title="K-means", x_axis_label="x", y_axis_label="y")
The next thing to do is sort out the colours. For this kind of plot the best colour scheme to use would be viridis. In order to create the viridis colours we can do the following
from bokeh.palettes import Viridis256
colors = Viridis256[::len(Viridis256) // N]
This gives us a list of colours, which we can access. Fortunately, scikitlearn numbers the groups from 0 to N-1, which is exactly the same format as the colours we just generated! Therefore, we can
plot with the following
1for k in plotting_data:
2 v = plotting_data[k]
4 x = [row[0] for row in v]
5 y = [row[1] for row in v]
7 p.scatter(x, y, legend_label="Group: {}".format(k), size=8, color=colors[k])
And this generated the following graph
Save Plots to PDF with Bokeh
Bokeh doesn’t have a built in way to save to PDF. However, we can export to an SVG and then convert that into a PDF plot. We need a few other dependencies to do this, so add the following to the
requirements.txt (and make sure you run pip install -r requirements.txt)
We also need to have a webbrowser installed. According to the docs, FireFox or Chrome will work, but I couldn’t make it work with FireFox on my ArchLinux system. I just had to install Chromium and it
worked fine (sudo pacman -S chromium on Arch).
First, we need to import a few things
1from bokeh.io import export_svgs
2import svglib.svglib as svglib
3from reportlab.graphics import renderPDF
And then I turned saving to PDF into a simple function
1def save_to_pdf(p, name):
2 # Step 1: Save to SVG
3 p.output_backend = "svg"
4 export_svgs(p, filename=name + ".svg")
6 # Step 2: Read in SVG
7 svglib.register_font("helvetica", "/home/fonts/Helvetica.ttf")
8 svg = svglib.svg2rlg(name + ".svg")
10 # Step 3: Save as PDF
11 renderPDF.drawToFile(svg, name + ".pdf")
All you have to is to provide the plot and the name of the PDF (without the .pdf extension). An example usage looks like this
1x = [1, 2, 3, 4, 5]
2y = [1, 2, 3, 2, 1]
4p = figure(title="Save in PDF", x_axis_label="x", y_axis_label="y")
5p.line(x, y, line_width=2, color="blue")
7save_to_pdf(p, "pdf_test")
Also, this will keep the SVG saved on your system, which is helpful as you can also use that in many places where you might want to use a PDF!
In conclusion, Bokeh is a very powerful library for creating beautiful interactive plots. When it comes to the workflow of using it just remember the four steps
1. Prepare some data, usually into lists or maybe a numpy array
2. Create a figure
3. As as many series as you have data
4. Show the plot
The examples in this guide should be enough to get you started in most applications. There’s a huge amount of customisation which Bokeh supports, but too much to cover everything in this article. You
can find the full reference here. | {"url":"https://www.naurt.com/blog-posts/naurt-introducing-bokeh-in-python","timestamp":"2024-11-13T22:51:57Z","content_type":"text/html","content_length":"54906","record_id":"<urn:uuid:7d472551-124f-469c-b134-c7e3b09ebcee>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00678.warc.gz"} |
The following table shows the cumulative frequency distribution of marks of 800 students in an examination. Construct a frequency distribution table for the data above.
Here, we observe that 10 students have scored marks below 10 i.e., it lies between class interval 0 - 10. Similarly, 50 students have scored marks below 20. So, 50 – 10 = 40 students lie in the
interval 10 – 20 and so on. The table of a frequency distribution for the given data is: | {"url":"https://philoid.com/question/29416-the-following-table-shows-the-cumulative-frequency-distribution-of-marks-of-800-students-in-an-examination-construct-a-frequency","timestamp":"2024-11-05T16:00:36Z","content_type":"text/html","content_length":"33497","record_id":"<urn:uuid:17b672f4-b37e-4d5e-857d-ed1ae73e03ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00734.warc.gz"} |
Narsampet to Mahabubabad distance
Distance in KM
The distance from Narsampet to Mahabubabad is 48.756 Km
Distance in Mile
The distance from Narsampet to Mahabubabad is 30.3 Mile
Distance in Straight KM
The Straight distance from Narsampet to Mahabubabad is 38.5 KM
Distance in Straight Mile
The Straight distance from Narsampet to Mahabubabad is 23.9 Mile
Travel Time
Travel Time 0 Hrs and 48 Mins
Narsampet Latitud and Longitude
Latitud 17.9280982 Longitude 79.8945536
Mahabubabad Latitud and Longitude
Latitud 17.5975349 Longitude 80.00156879999997 | {"url":"http://www.distancebetween.org/narsampet-to-mahabubabad","timestamp":"2024-11-04T04:59:15Z","content_type":"text/html","content_length":"1781","record_id":"<urn:uuid:9c9db2d7-93a6-48ed-98cf-1e8f828b4163>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00576.warc.gz"} |
Modern Math - Permutation & Combination
Modern Math - Permutation & Combination - Previous Year CAT/MBA Questions
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XAT 2019 QADI | Modern Math - Permutation & Combination XAT Question
A bag contains marbles of three colours-red, blue and green. There are 8 blue marbles in the bag. There are two additional statement of facts available:
1. If we pull out marbles from the bag at random, to guarantee that we have at least 3 green marbles, we need to extract 17 marbles.
2. If we pull out marbles from the bag at random, to guarantee that we have at least 2 red marbles, we need to extract 19 marbles.
Which of the two statements above, alone or in combination shall be sufficient to answer the question "how many green marbles are there in the bag"?
• (a)
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.
• (b)
Each statement alone is sufficient to answer the question.
• (c)
Statements 1 and 2 together are not sufficient, and additional data is needed to answer the question.
• (d)
Statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question.
• (e)
Statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question.
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© 2024 | All Rights Reserved | Apti4All | {"url":"https://www.apti4all.com/cat-mba/topic-wise-preparation/qa/XAT/modern-math-permutation-combination","timestamp":"2024-11-11T05:18:46Z","content_type":"text/html","content_length":"104866","record_id":"<urn:uuid:5baf9adf-5ba4-47d6-9d63-7b14ccf14c96>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00323.warc.gz"} |
A numerical spectroscopic technique for analyzing combustor flowfields
A computer program which calculates the ultraviolet emission and absorption spectra of OH is presented for use in conjunction with numerical programs which predict combustor flow field properties.
Spatial distributions of OH number density and temperature, resulting from analytical flow field calculations, are used as input data for calculating the absolute intensities of the spectra. Of
particular interest is the ability to calculate the shapes of the intensity envelopes associated with the low resolution slit settings of a given spectrometer. Comparisons are made with actual
spectral data obtained with various degrees of spectral resolution. The computer program is also used to generate graphical inversion techniques for analyzing experimental spectra. An example is
given in which one such graphical technique is used to obtain average temperatures and number densities along the axis of an axisymmetric duct containing a supersonic diffusion flame. Another example
is presented to demonstrate the manner in which a second inversion technique can be used to obtain radial profiles of temperature and OH number density from radial scanning of an axisymmetric
combustor flow field. Two cases involving thermodynamic nonequilibrium are also discussed, one of which involves a hot vibrational band and the other an electronic nonequilibrium.
In AGARD Anal. and Numerical Methods for Invest. of Flow Fields with Chem. Reactions
Pub Date:
May 1975
□ Combustible Flow;
□ Computer Programs;
□ Flame Spectroscopy;
□ Flow Distribution;
□ Numerical Analysis;
□ Diffusion Flames;
□ Flame Temperature;
□ Flow Characteristics;
□ Prediction Analysis Techniques;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1975anni.agar.....N/abstract","timestamp":"2024-11-07T16:43:08Z","content_type":"text/html","content_length":"37224","record_id":"<urn:uuid:74172293-792d-4b0a-a40c-aefc6e1e6f15>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00594.warc.gz"} |
Game Theory - Mini Lectures | Lindau Mediatheque
Game theory enables rational insight into the basic principles of social interaction and has therefore become indispensable for economic and social sciences. Whether in politics, sports or medicine,
modelling problems as a strategic game helps in decision-making in a variety of fields. With lecture snippets of Reinhard Selten, Robert Aumann and Alvin Roth, this Mini Lecture introduces to the
mathematical beginnings of game theory, its socioscientific development and entrepreneurial integration. | {"url":"https://mediatheque.lindau-nobel.org/recordings/34653/2015-mini-lecture-game-theory-en","timestamp":"2024-11-14T15:16:04Z","content_type":"text/html","content_length":"39678","record_id":"<urn:uuid:2f955e67-964e-4ae1-8fa8-b6240b6fe9d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00688.warc.gz"} |
Machine Theory
Discrete Mathematics
ISBN: 9780198507178 / Angielski / Miękka / 440 str.
ISBN: 9780198507178/Angielski/Miękka/440 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
The long-awaited second edition of Norman Bigg's best-selling Discrete Mathematics, includes new chapters on statements and proof, logical framework, natural numbers, and the integers, in addition
to updated chapters from the previous edition. Carefully structured, coherent and comprehensive, each chapter contains tailored exercises and solutions to selected questions, and miscellaneous
exercises are presented throughout. This is an invaluable text for students seeking a clear introduction to discrete mathematics, graph theory, combinatorics, number theory and abstract algebra.
The long-awaited second edition of Norman Bigg's best-selling Discrete Mathematics, includes new chapters on statements and proof, logical fr...
cena: 335,61 zł
Topology and Category Theory in Computer Science
ISBN: 9780198537601 / Angielski / Twarda / 408 str.
ISBN: 9780198537601/Angielski/Twarda/408 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
This volume reflects the growing use of techniques from topology and category theory in the field of theoretical computer science. In so doing it offers a source of new problems with a practical
flavor while stimulating original ideas and solutions. Reflecting the latest innovations at the interface between mathematics and computer science, the work will interest researchers and advanced
students in both fields.
This volume reflects the growing use of techniques from topology and category theory in the field of theoretical computer science. In so doing it offe...
cena: 480,31 zł
This volume reflects the growing use of techniques from topology and category theory in the field of theoretical computer science. In so doing it offe...
The New Hacker's Dictionary, third edition
ISBN: 9780262680929 / Angielski / Miękka / 568 str.
ISBN: 9780262680929/Angielski/Miękka/568 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
This new edition of the hacker's own phenomenally successful lexicon includes more than 100 new entries and updates or revises 200 more. Historically and etymologically richer than its predecessor,
it supplies additional background on existing entries and clarifies the murky origins of several important jargon terms (overturning a few long-standing folk etymologies) while still retaining its
high giggle value.
Sample definition
hacker n. originally, someone who makes furniture with an axe] 1. A person who enjoys exploring the details of programmable systems and...
This new edition of the hacker's own phenomenally successful lexicon includes more than 100 new entries and updates or revises 200 more. Historical...
cena: 428,19 zł
This new edition of the hacker's own phenomenally successful lexicon includes more than 100 new entries and updates or revises 200 more. Historically and etymologically richer than its predecessor,
it supplies additional background on existing entries and clarifies the murky origins of several important jargon terms (overturning a few long-standing folk etymologies) while still retaining its
high giggle value.
hacker n. originally, someone who makes furniture with an axe] 1. A person who enjoys exploring the details of programmable systems and...
This new edition of the hacker's own phenomenally successful lexicon includes more than 100 new entries and updates or revises 200 more. Historical...
Parallel-Vector Equation Solvers for Finite Element Engineering Applications
ISBN: 9780306466403 / Angielski / Twarda / 344 str.
ISBN: 9780306466403/Angielski/Twarda/344 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Despite the ample number of articles on parallel-vector computational algorithms published over the last 20 years, there is a lack of texts in the field customized for senior undergraduate and
graduate engineering research. Parallel-Vector Equation Solvers for Finite Element Engineering Applications aims to fill this gap, detailing both the theoretical development and important
implementations of equation-solution algorithms. The mathematical background necessary to understand their inception balances well with descriptions of their practical uses. Illustrated with a
Despite the ample number of articles on parallel-vector computational algorithms published over the last 20 years, there is a lack of texts in the fie...
cena: 589,01 zł
Despite the ample number of articles on parallel-vector computational algorithms published over the last 20 years, there is a lack of texts in the fie...
Recent Progress in Computational and Applied Pdes: Conference Proceedings for the International Conference Held in Zhangjiajie in July 2001
ISBN: 9780306474200 / Angielski / Twarda / 432 str.
ISBN: 9780306474200/Angielski/Twarda/432 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
The International Symposium on Computational & Applied PDEs was held at Zhangjiajie National Park of China from July 1-7, 2001. The main goal of this conference is to bring together computational,
applied and pure mathematicians on different aspects of partial differential equations to exchange ideas and to promote collaboration. Indeed, it attracted a number of leading scientists in
computational PDEs including Doug Arnold (Minnesota), Jim Bramble (Texas A & M), Achi Brandt (Weizmann), Franco Brezzi (Pavia), Tony Chan (UCLA), Shiyi Chen (John Hopkins), Qun Lin (Chinese Academy
of Sciences),...
The International Symposium on Computational & Applied PDEs was held at Zhangjiajie National Park of China from July 1-7, 2001. The main goal of this ...
cena: 392,66 zł
The International Symposium on Computational & Applied PDEs was held at Zhangjiajie National Park of China from July 1-7, 2001. The main goal of this conference is to bring together computational,
applied and pure mathematicians on different aspects of partial differential equations to exchange ideas and to promote collaboration. Indeed, it attracted a number of leading scientists in
computational PDEs including Doug Arnold (Minnesota), Jim Bramble (Texas A & M), Achi Brandt (Weizmann), Franco Brezzi (Pavia), Tony Chan (UCLA), Shiyi Chen (John Hopkins), Qun Lin (Chinese Academy
of Sciences),...
The International Symposium on Computational & Applied PDEs was held at Zhangjiajie National Park of China from July 1-7, 2001. The main goal of this ...
Hierarchical Scheduling in Parallel and Cluster Systems
ISBN: 9780306477614 / Angielski / Twarda / 251 str.
ISBN: 9780306477614/Angielski/Twarda/251 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Multiple processor systems are an important class of parallel systems. Over the years, several architectures have been proposed to build such systems to satisfy the requirements of high performance
computing. These architectures span a wide variety of system types. At the low end of the spectrum, we can build a small, shared-memory parallel system with tens of processors. These systems
typically use a bus to interconnect the processors and memory. Such systems, for example, are becoming commonplace in high-performance graph ics workstations. These systems are called uniform
memory access...
Multiple processor systems are an important class of parallel systems. Over the years, several architectures have been proposed to build such systems ...
cena: 589,01 zł
Multiple processor systems are an important class of parallel systems. Over the years, several architectures have been proposed to build such systems to satisfy the requirements of high performance
computing. These architectures span a wide variety of system types. At the low end of the spectrum, we can build a small, shared-memory parallel system with tens of processors. These systems
typically use a bus to interconnect the processors and memory. Such systems, for example, are becoming commonplace in high-performance graph ics workstations. These systems are called uniform memory
Multiple processor systems are an important class of parallel systems. Over the years, several architectures have been proposed to build such systems ...
Automated Theorem Proving: Theory and Practice
ISBN: 9780387950754 / Angielski / Twarda / 231 str.
ISBN: 9780387950754/Angielski/Twarda/231 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
As the 21st century begins, the power of our magical new tool and partner, the computer, is increasing at an astonishing rate. Computers that perform billions of operations per second are now
commonplace. Multiprocessors with thousands of little computers - relatively little -can now carry out parallel computations and solve problems in seconds that only a few years ago took days or
months. Chess-playing programs are on an even footing with the world's best players. IBM's Deep Blue defeated world champion Garry Kasparov in a match several years ago. Increasingly computers are
expected to be...
As the 21st century begins, the power of our magical new tool and partner, the computer, is increasing at an astonishing rate. Computers that perform ...
cena: 589,01 zł
As the 21st century begins, the power of our magical new tool and partner, the computer, is increasing at an astonishing rate. Computers that perform billions of operations per second are now
commonplace. Multiprocessors with thousands of little computers - relatively little -can now carry out parallel computations and solve problems in seconds that only a few years ago took days or
months. Chess-playing programs are on an even footing with the world's best players. IBM's Deep Blue defeated world champion Garry Kasparov in a match several years ago. Increasingly computers are
expected to be...
As the 21st century begins, the power of our magical new tool and partner, the computer, is increasing at an astonishing rate. Computers that perform ...
Machine Beauty
ISBN: 9780465043163 / Angielski / Miękka / 176 str.
ISBN: 9780465043163/Angielski/Miękka/176 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
When something works well, you can feel it; there is a sense of rightness to it. We call that rightness beauty, and it ought to be the single most important component of design.This recognition is
at the heart of David Gelernter's witty argued essay, Machine Beauty, which defines beauty as an inspired mating of simplicity and power. You can see it in a Bauhaus chair, the Hoover Dam, or an
Emerson radio circa 1930. In contrast, too many contemporary technologists run out of ideas and resort to gimmicks and features; they are rarely capable of real, structural ingenuity.Nowhere is
this more...
When something works well, you can feel it; there is a sense of rightness to it. We call that rightness beauty, and it ought to be the single most imp...
cena: 86,06 zł
When something works well, you can feel it; there is a sense of rightness to it. We call that rightness beauty, and it ought to be the single most important component of design.This recognition is at
the heart of David Gelernter's witty argued essay, Machine Beauty, which defines beauty as an inspired mating of simplicity and power. You can see it in a Bauhaus chair, the Hoover Dam, or an Emerson
radio circa 1930. In contrast, too many contemporary technologists run out of ideas and resort to gimmicks and features; they are rarely capable of real, structural ingenuity.Nowhere is this more...
When something works well, you can feel it; there is a sense of rightness to it. We call that rightness beauty, and it ought to be the single most imp...
Arithmetic and Logic in Computer Systems
ISBN: 9780471469452 / Angielski / Twarda / 246 str.
ISBN: 9780471469452/Angielski/Twarda/246 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Arithmetic and Logic in Computer Systems provides a useful guide to a fundamental subject of computer science and engineering. Algorithms for performing operations like addition, subtraction,
multiplication, and division in digital computer systems are presented, with the goal of explaining the concepts behind the algorithms, rather than addressing any direct applications. Alternative
methods are examined, and explanations are supplied of the fundamental materials and reasoning behind theories and examples.
No other current books deal with this subject, and the author is a leading...
Arithmetic and Logic in Computer Systems provides a useful guide to a fundamental subject of computer science and engineering. Algorithms for performi...
cena: 559,67 zł
Arithmetic and Logic in Computer Systems provides a useful guide to a fundamental subject of computer science and engineering. Algorithms for performi...
the definitive guide to how computers do math: featuring the virtual diy calculator
ISBN: 9780471732785 / Angielski / Miękka / 464 str.
ISBN: 9780471732785/Angielski/Miękka/464 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
The Basics of Computer Arithmetic Made Enjoyable and Accessible-with a Special Program Included for Hands-on Learning
"The combination of this book and its associated virtual computer is fantastic Experience over the last fifty years has shown me that there's only one way to truly understand how computers work;
and that is to learn one computer and its instruction set-no matter how simple or primitive-from the ground up. Once you fully comprehend how that simple computer functions, you can easily
extrapolate to more complex machines."
-Fred Hudson, retired...
The Basics of Computer Arithmetic Made Enjoyable and Accessible-with a Special Program Included for Hands-on Learning
"The combination of...
cena: 271,38 zł
Discrete Mathematics: An Introduction for Software Engineers
ISBN: 9780521386227 / Angielski / Miękka / 332 str.
ISBN: 9780521386227/Angielski/Miękka/332 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
This book is designed to form the basis of a one-year course in discrete mathematics for first-year computer scientists or software engineers. The materials presented cover much of undergraduate
algebra with a particular bias toward the computing applications. Topics covered include mathematical logic, set theory, finite and infinite relations and mapping, graphs, graphical algorithms and
axiom systems. It concludes with implementations of many of the algorithms in Modula-2 to illustrate how the mathematics may be turned into concrete calculations. Numerous examples and exercises
are included...
This book is designed to form the basis of a one-year course in discrete mathematics for first-year computer scientists or software engineers. The mat...
cena: 205,83 zł
This book is designed to form the basis of a one-year course in discrete mathematics for first-year computer scientists or software engineers. The materials presented cover much of undergraduate
algebra with a particular bias toward the computing applications. Topics covered include mathematical logic, set theory, finite and infinite relations and mapping, graphs, graphical algorithms and
axiom systems. It concludes with implementations of many of the algorithms in Modula-2 to illustrate how the mathematics may be turned into concrete calculations. Numerous examples and exercises are
This book is designed to form the basis of a one-year course in discrete mathematics for first-year computer scientists or software engineers. The mat...
Categories and Computer Science
ISBN: 9780521419970 / Angielski / Twarda / 180 str.
ISBN: 9780521419970/Angielski/Twarda/180 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Category Theory has, in recent years, become increasingly important and popular in computer science, and many universities now introduce Category Theory as part of the curriculum for undergraduate
computer science students. Here, the theory is developed in a straightforward way, and is enriched with many examples from computer science.
Category Theory has, in recent years, become increasingly important and popular in computer science, and many universities now introduce Category Theo...
cena: 410,46 zł
Category Theory has, in recent years, become increasingly important and popular in computer science, and many universities now introduce Category Theo...
Randomized Algorithms
ISBN: 9780521474658 / Angielski / Twarda / 496 str.
ISBN: 9780521474658/Angielski/Twarda/496 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
For many applications, a randomized algorithm is either the simplest or the fastest algorithm available, and sometimes both. This book introduces the basic concepts in the design and analysis of
randomized algorithms. The first part of the text presents basic tools such as probability theory and probabilistic analysis that are frequently used in algorithmic applications. Algorithmic
examples are also given to illustrate the use of each tool in a concrete setting. In the second part of the book, each chapter focuses on an important area to which randomized algorithms can be
applied, providing...
For many applications, a randomized algorithm is either the simplest or the fastest algorithm available, and sometimes both. This book introduces the ...
cena: 360,51 zł
For many applications, a randomized algorithm is either the simplest or the fastest algorithm available, and sometimes both. This book introduces the basic concepts in the design and analysis of
randomized algorithms. The first part of the text presents basic tools such as probability theory and probabilistic analysis that are frequently used in algorithmic applications. Algorithmic examples
are also given to illustrate the use of each tool in a concrete setting. In the second part of the book, each chapter focuses on an important area to which randomized algorithms can be applied,
For many applications, a randomized algorithm is either the simplest or the fastest algorithm available, and sometimes both. This book introduces the ...
Algorithmic Information Theory
ISBN: 9780521616041 / Angielski / Miękka / 192 str.
ISBN: 9780521616041/Angielski/Miękka/192 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Chaitin, the inventor of algorithmic information theory, presents in this book the strongest possible version of Godel's incompleteness theorem, using an information theoretic approach based on the
size of computer programs. One half of the book is concerned with studying the halting probability of a universal computer if its program is chosen by tossing a coin. The other half is concerned
with encoding the halting probability as an algebraic equation in integers, a so-called exponential diophantine equation."
Chaitin, the inventor of algorithmic information theory, presents in this book the strongest possible version of Godel's incompleteness theorem, using...
cena: 322,70 zł
Chaitin, the inventor of algorithmic information theory, presents in this book the strongest possible version of Godel's incompleteness theorem, using...
Interior Point Approach to Linear, Quadratic and Convex Programming: Algorithms and Complexity
ISBN: 9780792327349 / Angielski / Twarda / 210 str.
ISBN: 9780792327349/Angielski/Twarda/210 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
This book describes the rapidly developing field of interior point methods (IPMs). An extensive analysis is given of path-following methods for linear programming, quadratic programming and convex
programming. These methods, which form a subclass of interior point methods, follow the central path, which is an analytic curve defined by the problem. Relatively simple and elegant proofs for
polynomiality are given. The theory is illustrated using several explicit examples. Moreover, an overview of other classes of IPMs is given. It is shown that all these methods rely on the same
notion as the...
This book describes the rapidly developing field of interior point methods (IPMs). An extensive analysis is given of path-following methods for linear...
cena: 196,31 zł
This book describes the rapidly developing field of interior point methods (IPMs). An extensive analysis is given of path-following methods for linear programming, quadratic programming and convex
programming. These methods, which form a subclass of interior point methods, follow the central path, which is an analytic curve defined by the problem. Relatively simple and elegant proofs for
polynomiality are given. The theory is illustrated using several explicit examples. Moreover, an overview of other classes of IPMs is given. It is shown that all these methods rely on the same notion
as the...
This book describes the rapidly developing field of interior point methods (IPMs). An extensive analysis is given of path-following methods for linear...
Applications of Continuous Mathematics to Computer Science
ISBN: 9780792347224 / Angielski / Twarda / 419 str.
ISBN: 9780792347224/Angielski/Twarda/419 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
This volume is intended to be used as a textbook for a special topic course in computer science. It addresses contemporary research topics of interest such as intelligent control, genetic
algorithms, neural networks, optimization techniques, expert systems, fractals, and computer vision. The work incorporates many new research ideas, and focuses on the role of continuous
Audience: This book will be valuable to graduate students interested in theoretical computer topics, algorithms, expert systems, neural networks, and software engineering.
This volume is intended to be used as a textbook for a special topic course in computer science. It addresses contemporary research topics of interest...
cena: 785,36 zł
This volume is intended to be used as a textbook for a special topic course in computer science. It addresses contemporary research topics of interest...
Random Generation of Trees: Random Generators in Computer Science
ISBN: 9780792395287 / Angielski / Twarda / 208 str.
ISBN: 9780792395287/Angielski/Twarda/208 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Random Generation of Trees is about a field on the crossroads between computer science, combinatorics and probability theory. Computer scientists need random generators for performance analysis,
simulation, image synthesis, etc. In this context random generation of trees is of particular interest. The algorithms presented here are efficient and easy to code. Some aspects of
Horton--Strahler numbers, programs written in C and pictures are presented in the appendices. The complexity analysis is done rigorously both in the worst and average cases.
Random Generation of...
Random Generation of Trees is about a field on the crossroads between computer science, combinatorics and probability theory. Computer scient...
cena: 785,36 zł
Error Detecting Codes: General Theory and Their Application in Feedback Communication Systems
ISBN: 9780792396291 / Angielski / Twarda / 249 str.
ISBN: 9780792396291/Angielski/Twarda/249 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Error detecting codes are very popular for error control in practical systems for two reasons. First, such codes can be used to provide any desired reliability of communication over any noisy
channel. Second, implementation is usually much simpler than for a system using error correcting codes. To consider a particular code for use in such a system, it is very important to be able to
calculate or estimate the probability of undetected error. For the binary symmetric channel, the probability of undetected error can be expressed in terms of the weight distribution of the code.
The first part of...
Error detecting codes are very popular for error control in practical systems for two reasons. First, such codes can be used to provide any desired re...
cena: 589,01 zł
Error detecting codes are very popular for error control in practical systems for two reasons. First, such codes can be used to provide any desired reliability of communication over any noisy
channel. Second, implementation is usually much simpler than for a system using error correcting codes. To consider a particular code for use in such a system, it is very important to be able to
calculate or estimate the probability of undetected error. For the binary symmetric channel, the probability of undetected error can be expressed in terms of the weight distribution of the code. The
first part of...
Error detecting codes are very popular for error control in practical systems for two reasons. First, such codes can be used to provide any desired re...
Neural Networks and Analog Computation: Beyond the Turing Limit
ISBN: 9780817639495 / Angielski / Twarda / 181 str.
ISBN: 9780817639495/Angielski/Twarda/181 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
The theoretical foundations of Neural Networks and Analog Computation conceptualize neural networks as a particular type of computer consisting of multiple assemblies of basic processors
interconnected in an intricate structure. Examining these networks under various resource constraints reveals a continuum of computational devices, several of which coincide with well-known
classical models. On a mathematical level, the treatment of neural computations enriches the theory of computation but also explicated the computational complexity associated with biological
networks, adaptive...
The theoretical foundations of Neural Networks and Analog Computation conceptualize neural networks as a particular type of computer consisting of ...
cena: 589,01 zł
The theoretical foundations of Neural Networks and Analog Computation conceptualize neural networks as a particular type of computer consisting of multiple assemblies of basic processors
interconnected in an intricate structure. Examining these networks under various resource constraints reveals a continuum of computational devices, several of which coincide with well-known classical
models. On a mathematical level, the treatment of neural computations enriches the theory of computation but also explicated the computational complexity associated with biological networks,
The theoretical foundations of Neural Networks and Analog Computation conceptualize neural networks as a particular type of computer consisting of ...
Acronyms and Abbreviations of Computer Technology and
ISBN: 9780824787479 / Angielski / Twarda / 304 str.
ISBN: 9780824787479/Angielski/Twarda/304 str.
Termin realizacji zamówienia: ok. 5-8 dni roboczych.
Catalogues approximately 7000 acronyms and abbreviations used in computer technology, telecommunications and related fields. The entries are organized in tabular form to enable readers to locate
any specific acronym easily.
Catalogues approximately 7000 acronyms and abbreviations used in computer technology, telecommunications and related fields. The entries are organized...
cena: 684,68 zł
Catalogues approximately 7000 acronyms and abbreviations used in computer technology, telecommunications and related fields. The entries are organized in tabular form to enable readers to locate any
specific acronym easily.
Catalogues approximately 7000 acronyms and abbreviations used in computer technology, telecommunications and related fields. The entries are organized... | {"url":"https://krainaksiazek.pl/ksiegarnia,m_products,bi_COM037000,Machine-Theory.html","timestamp":"2024-11-08T09:14:11Z","content_type":"text/html","content_length":"101572","record_id":"<urn:uuid:fea90705-bd73-44d8-ae4f-ad2d44e63b74>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00122.warc.gz"} |
Combined interest rate calculator
Interest rates may change as often as daily without prior notice. Fees may reduce earnings. 4. This is a tiered, interest earning variable rate account. All daily collected balances up to and
including $150,000 will earn interest based on the combined rate rewards. All daily collected balances greater than $150,000 will not earn interest. This ROI calculator (return-on-investment)
calculates an annualized rate-of-return using exact dates. Also known as ROR (rate-of-return), these financial calculators allow you to compare the results of different investments. A blended
interest rate is a combination of interest rates for different loans that gives the total amount of interest on the loans collected into one. This can give you a sense of what a person or company is
paying on its total debt and is used in certain formulas, such as calculating the interest rate on consolidated student loans.
If you can't secure a better interest rate, then a consolidation loan may not make and interest rate, into a single combined debt with one monthly payment. Interest rates are subject to change. Tools
to get you there. Repayment calculator . Calculate That is why rates go up and down when the fed changes rates. 1 comment In order to calculate simple interest use the formula: A=P.R.T/100. Where: A
= the So there's two ways folks will calculate the real interest rate, given the nominal interest rate and the inflation rate. The first way is an approximation, but it's very
Credit Card Interest Rate Calculator. Calculate weighted average interest rate on all credit card balances combined. IMPORTANT! This redesigned calculator
This blended-rate mortgage calculator helps determine the effective, or blended, interest rate if you use a first and a second mortgage to finance the purchase of a home. This calculator will help
you compute the average combined interest rate you are paying on up to fifteen of your outstanding debts. This can be very helpful when deciding whether or not to move the balances of several credit
cards to another card or to another form of debt (loans, etc.). Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined
figure for principal and compound interest. Subtract the principal if you want just the compound interest. Read more about the formula. The formula used in the compound interest calculator is A = P
(1+r/n) (nt) Compound Interest Formula. Compound interest - meaning that the interest you earn each year is added to your principal, so that the balance doesn't merely grow, it grows at an increasing
rate - is one of the most useful concepts in finance. It is the basis of everything from a personal savings plan to the long term growth of the stock market.
This loan calculator will help you determine the monthly payments on a loan. Simply enter the loan amount, term and interest rate in the fields below and click calculate to calculate your monthly
Free interest calculator to find the interest, final balance, and accumulation schedule using either a fixed starting principal and/or periodic contributions. Included are options for tax,
compounding period, and inflation. Also explore hundreds of other calculators addressing investment, finance math, fitness, health, and many more. This simple interest calculator calculates interest
between any two dates. Per Dictionary.com simple interest is "interest payable only on the principal." Interest is never earned or collected on previous interest. Because this calculator is date
sensitive, it is a suitable tool for calculating simple interest owed on any debt.. You can calculate the accrued interest from any point in time Simple Interest Calculator. Simple interest is money
you can earn by initially investing some money (the principal). A percentage (the interest) of the principal is added to the principal, making your initial investment grow! What amount of money is
loaned or borrowed?(this is the principal amount) $ What is the interest rate (in percent
This simple interest calculator calculates interest between any two dates. Per Dictionary.com simple interest is "interest payable only on the principal." Interest is never earned or collected on
previous interest. Because this calculator is date sensitive, it is a suitable tool for calculating simple interest owed on any debt.. You can calculate the accrued interest from any point in time
EMI chart. ReCalculate. Calculate EMI For. Principal. | 1L| 25L | 50L | 75L| 1Cr | 1.25Cr | 1.50Cr| 1.75Cr| 2Cr| 2.25Cr| 2.50Cr. Annual Rate of Interest. | 4%| 6% | Combining your student loans would
use your Weighted Average Interest Rate Calculator for your new loan for an instantaneous Consolidation. Credit Card Interest Rate Calculator. Calculate weighted average interest rate on all credit
card balances combined. IMPORTANT! This redesigned calculator Monthly Payments Per $1000 & Total Cost [Principal and Interest Combined] You can't reliably use the chart to calculate the monthly
payment for an Combine and pay off all your outstanding debt. One monthly payment. One interest rate. A debt-freedom date.
Calculate Home Loan EMI. Home Loan EMI Calculator With lower EMIs, ICICI Bank Home Loans are light on your wallet. Lower interest rate and repayment tenure
This calculator will help you compute the average combined interest rate you are paying on up to fifteen of your outstanding debts. This can be very helpful when deciding whether or not to move the
balances of several credit cards to another card or to another form of debt (loans, etc.). Multiply the principal amount by one plus the annual interest rate to the power of the number of compound
periods to get a combined figure for principal and compound interest. Subtract the principal if you want just the compound interest. Read more about the formula. The formula used in the compound
interest calculator is A = P(1+r/n) (nt) Compound Interest Formula. Compound interest - meaning that the interest you earn each year is added to your principal, so that the balance doesn't merely
grow, it grows at an increasing rate - is one of the most useful concepts in finance. It is the basis of everything from a personal savings plan to the long term growth of the stock market.
Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. Subtract the principal
if you want just the compound interest. Read more about the formula. The formula used in the compound interest calculator is A = P(1+r/n) (nt) Compound Interest Formula. Compound interest - meaning
that the interest you earn each year is added to your principal, so that the balance doesn't merely grow, it grows at an increasing rate - is one of the most useful concepts in finance. It is the
basis of everything from a personal savings plan to the long term growth of the stock market. Total Balance: $ Blended Rate : % Effective rate only correct if all loans paid off over same time | {"url":"https://optionebpuy.netlify.app/giannetto66142wid/combined-interest-rate-calculator-hyte.html","timestamp":"2024-11-11T10:19:53Z","content_type":"text/html","content_length":"33014","record_id":"<urn:uuid:bbe4744d-a38b-4ec6-8c27-a43e214d3633>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00332.warc.gz"} |
Testing the Hypothesis of no Fixed Main-Effects in Scheffe's Mixed Model
Ann. Math. Statist. 33(3): 1085-1095 (September, 1962). DOI: 10.1214/aoms/1177704474
Various approaches are available for the formulation of linear models for the analysis of variance. The mixed model in which one factor is a fixed-effects factor, and one factor is a random-effects
factor can be obtained for instance as a limiting case of the general models of Cornfield and Tukey [3], and of Wilk and Kempthorne [8]. An entirely different approach is that given by Scheffe [6],
and is the one considered in the present paper. Whatever the approach used, a hypothesis of interest will usually be the hypothesis of no fixed-effects. Consider a two-way layout in which $A$ denotes
a fixed-effects factor and $B$ a random-effects factor. Let $I$ and $J$ be the numbers of levels of factors $A$ and $B$ respectively at which measurements are taken $(I > 1, J > 1)$. Let $K$ be the
number of replications performed in each cell $(K > 1)$. In the light of $K$ be the number of replications performed in each cell $(K > 1)$. In the light of Table 1 of [3], Table 3 of [8], and
formulas (46) and (54) of [6], an adequate $F$-type statistic to use for testing the hypothesis $H_A$ that all main-effects corresponding to the levels of factor $A$ are zero appears to be \begin
{equation*}\tag{1.1}\mathscr{F} = (\mathrm{MS})_A/(\mathrm{MS})_{AB}.\end{equation*} The usual mean squares $(\mathrm{MS})_A$ and $(\mathrm{MS})_{AB}$ corresponding to factor $A$ and to $A \times B$
interactions are explicitly defined below. With the normal theory models which are commonly used in the case of two fixed-effects factors and in the case of two random-effects factors (e.g., in [7]),
the criterion (1.1) has under the hypothesis $H_A$ the $F$-distribution with $I - 1$ and $(I - 1)(J - 1)$ d.f. In Scheffe's mixed model [6], this is no longer the case. When $J \geqq I$, a Hotelling
$T^2$ statistic can then be constructed for the test of $H_A$. When multiplied by a constant factor, this statistic has the $F$-distribution with $I - 1$ and $J - I + 1$ d.f. While requiring a larger
amount of computational work, the $T^2$ test will have little power when $J - I + 1$ is small. It is therefore tempting to construct a test of $H_A$, based on the ratio (1.1), by assuming that the
law of $\mathscr{F}$ is not much different under $H_A$ from that of $F$ with $I - 1$ and $(I - 1)(J - 1)$ d.f. In Sub-section 4.1, we investigate the possible ill-effects of this assumption. They can
be considerable, and remedies are suggested in Subsections 4.2 and 4.3.
Download Citation
J. P. Imhof. "Testing the Hypothesis of no Fixed Main-Effects in Scheffe's Mixed Model." Ann. Math. Statist. 33 (3) 1085 - 1095, September, 1962. https://doi.org/10.1214/aoms/1177704474
Published: September, 1962
First available in Project Euclid: 27 April 2007
Digital Object Identifier: 10.1214/aoms/1177704474
Rights: Copyright © 1962 Institute of Mathematical Statistics
Vol.33 • No. 3 • September, 1962 | {"url":"https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-33/issue-3/Testing-the-Hypothesis-of-no-Fixed-Main-Effects-in-Scheffes/10.1214/aoms/1177704474.full","timestamp":"2024-11-14T18:48:44Z","content_type":"text/html","content_length":"143427","record_id":"<urn:uuid:c3289fe3-7ebd-4cde-8239-c5fb0b128c16>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00498.warc.gz"} |
Modelling, Dynamics and Control - Section one: Time series models
Chapter five
Discrete models and Z-transforms
Section one: time series models
This chapter is on the theme of discrete time linear models, for example:
y[k]+a[1]y[k-1] + ...+ a[n]y[k-n] = b[1]u[k-1] + ...+ b[n]u[k-n]
where y(t) is the output, u(t) the input and a[i ], b[i] are model parameters. The subscript 'k' denotes the sampling index.
This section focuses on an introduction to time series models, that is models which represent data which changes only at specific instants in time rather than continuously.
1. Introduction to time series models
Examples of simple first order time series models from economics and radioactive decay.
Introduction to time series (PDF, 593 KB)
Examples of how to estimate the parameters of a simple first order time series model from measured data.
Time series parameters from data (PDF, 608 KB)
3. Modelling from data with high order models
Examples of how to estimate the parameters of a high order time series model from measured data using least squares identification methods.
Time series parameters from data (PDF, 572 KB)
Tutorial sheets for chapter five
Online quizzes for chapter five | {"url":"https://controleducation.sites.sheffield.ac.uk/chapterdiscrete/sectiontimeseries","timestamp":"2024-11-13T15:54:26Z","content_type":"text/html","content_length":"110418","record_id":"<urn:uuid:10638736-a775-4b9c-9957-f70e495b72d8>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00200.warc.gz"} |