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What is a single measurement of spirit?
Spirits used to be commonly served in 25ml measures, which are one unit of alcohol, many pubs and bars now serve 35ml or 50ml measures. Large wine glasses hold 250ml, which is one third of a bottle.
It means there can be nearly three units or more in just one glass.
What is a measure of gin in a pub?
By the glass
Port, sherry or other fortified wine 50ml, 70ml, multiples of 50ml or 70ml
Gin, rum, vodka and whisky Either 25ml and multiples of 25ml, or 35ml and multiples of 35ml (not both on the same premises)
Draught beer and cider Third, half, two-thirds of a pint and multiples of half a pint
How many ml is a pub measure of spirits in Ireland?
Some examples of a standard drink in Ireland are: a pub measure of spirits (35.5ml)
What is the standard measure for spirits in the UK?
25 ml
Spirits. Most spirits sold in the United Kingdom have 40% ABV or slightly less. In England, a single pub measure (25 ml) of a spirit contains one unit. However, a larger 35 ml measure is increasingly
used (and in particular is standard in Northern Ireland), which contains 1.4 units of alcohol at 40% ABV.
What is a single measure of gin?
A gin and tonic made with a single 25ml measure of 37.5% Alcohol by Volume (ABV) gin contains 0.9 units. So drinking 16 gin and tonics made with this same amount of alcohol means you will exceed the
guidelines. And remember if you drink doubles you’ll be over the guidelines with half the number of drinks.
What is a UK pub measure of gin?
What Is A Single Measure In The Uk? Pubs and bars have switched from serving 25ml measures (one unit of alcohol) of spirits to 35ml or 50ml measures of the spirit.
How many ml is a pub measure of spirits in England?
A pub measure of spirits (35.5ml)
What is a pub measure of whiskey?
Before the country went metric, most measures of whisky were one-fifth of a gill or one fluid ounce, although some pubs took great pride in serving the larger and older quarter-gill. Now the normal
measure is 25ml, which is smaller than the 28.4ml that was the fifth of a gill.
What is a measure of gin in Ireland?
How much is in a measure?
A “measure” is simply the proportion of alcohol to other ingredients in a drink. Using a conventional shot glass as the measure, 4 shots of brandy would be 6 oz, or 180 ml.
What is a measure of spirits in Ireland?
A pub measure of spirits is served as a single (35.5ml) or as a double (71.0ml). In the off-licence sector, there is a much greater variability in alcohol container sizes for sale. Beers and ciders
are generally for sale in the small can/bottle (330ml) and large can/bottle (500ml) sizes.
How many ml is a UK pub measure?
How much is single measure of gin?
What is an Irish measure of spirits?
What is a pub measure in Scotland?
Pubs in Scotland are failing to ensure spirit are served in correct measures of 25ml and 35ml. Between July 31 to August 18 trading standards… Pubs in Scotland are failing to ensure spirit are served
in correct measures of 25ml and 35ml.
How much is a single measure of gin?
How much is a single gin measure?
What is a single measure of whiskey?
How many ml is a gin shot?
43 ml (1.5 oz) shot of 40% hard liquor (vodka, rum, whisky, gin etc.)
What is a single measure of spirits in CL UK?
A 37 gin and tonic made with just a single 25ml measure. 5% alcohol by volume (ABV) gin contains 0 alcohol by volume (ABV) gin contains 0 alcohol by volume (ABV) 9 units….How Many Ml Is A Measure Of
The Drink The Strength The Amount
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In mathematics, especially measure theory, a set function is a function whose domain is a family of subsets of some given set and that (usually) takes its values in the extended real number line ${\
displaystyle \mathbb {R} \cup \{\pm \infty \},}$ which consists of the real numbers ${\displaystyle \mathbb {R} }$ and ${\displaystyle \pm \infty .}$
A set function generally aims to measure subsets in some way. Measures are typical examples of "measuring" set functions. Therefore, the term "set function" is often used for avoiding confusion
between the mathematical meaning of "measure" and its common language meaning.
If ${\displaystyle {\mathcal {F}}}$ is a family of sets over ${\displaystyle \Omega }$ (meaning that ${\displaystyle {\mathcal {F}}\subseteq \wp (\Omega )}$ where ${\displaystyle \wp (\Omega )}$
denotes the powerset) then a set function on ${\displaystyle {\mathcal {F}}}$ is a function ${\displaystyle \mu }$ with domain ${\displaystyle {\mathcal {F}}}$ and codomain ${\displaystyle [-\infty ,
\infty ]}$ or, sometimes, the codomain is instead some vector space, as with vector measures, complex measures, and projection-valued measures. The domain of a set function may have any number
properties; the commonly encountered properties and categories of families are listed in the table below.
Families ${\displaystyle {\mathcal {F}}}$ of sets over ${\displaystyle \Omega }$
Is necessarily true of ${\
displaystyle {\mathcal {F}} Directed ${\ ${\ ${\ ${\displaystyle ${\displaystyle A_ ${\displaystyle A_ ${\displaystyle \ ${\displaystyle \
\colon }$ by ${\ displaystyle displaystyle displaystyle \Omega \setminus {1}\cap A_{2}\cap \ {1}\cup A_{2}\cup \ Omega \in {\mathcal varnothing \in {\ F.I.P.
or, is ${\displaystyle {\ displaystyle A\cap B}$ A\cup B}$ B\setminus A} A}$ cdots }$ cdots }$ {F}}}$ mathcal {F}}}$
mathcal {F}}}$ closed \,\supseteq }$ $
Semiring Never
Semialgebra (Semifield) Never
only if ${\ only if ${\
Monotone class displaystyle A_{i}\ displaystyle A_{i}
searrow }$ earrow }$
only if only if ${\
${\ displaystyle A_{i}
𝜆-system (Dynkin System) displaystyle earrow }$ or Never
A\subseteq B} they are disjoint
Ring (Order theory)
Ring (Measure theory) Never
δ-Ring Never
𝜎-Ring Never
Algebra (Field) Never
𝜎-Algebra (𝜎-Field) Never
Dual ideal
${\displaystyle \
Filter Never Never varnothing ot \in {\
mathcal {F}}}$
${\displaystyle \
Prefilter (Filter base) Never Never varnothing ot \in {\
mathcal {F}}}$
${\displaystyle \
Filter subbase Never Never varnothing ot \in {\
mathcal {F}}}$
Open Topology (even arbitrary ${\ Never
displaystyle \cup }$
Closed Topology (even arbitrary ${\ Never
displaystyle \cap }$
Is necessarily true of ${\
displaystyle {\mathcal {F}} complements contains ${\ contains ${\ Finite
\colon }$ directed finite finite relative in ${\ countable countable displaystyle \Omega displaystyle \ Intersection
or, is ${\displaystyle {\ downward intersections unions complements displaystyle \ intersections unions }$ varnothing }$ Property
mathcal {F}}}$ closed Omega }$
Additionally, a semiring is a π-system where every complement ${\displaystyle B\setminus A}$ is equal to a finite disjoint union of sets in ${\displaystyle {\mathcal {F}}.}$
A semialgebra is a semiring where every complement ${\displaystyle \Omega \setminus A}$ is equal to a finite disjoint union of sets in ${\displaystyle {\mathcal {F}}.}$
${\displaystyle A,B,A_{1},A_{2},\ldots }$ are arbitrary elements of ${\displaystyle {\mathcal {F}}}$ and it is assumed that ${\displaystyle {\mathcal {F}}eq \varnothing .}$
In general, it is typically assumed that ${\displaystyle \mu (E)+\mu (F)}$ is always well-defined for all ${\displaystyle E,F\in {\mathcal {F}},}$ or equivalently, that ${\displaystyle \mu }$ does
not take on both ${\displaystyle -\infty }$ and ${\displaystyle +\infty }$ as values. This article will henceforth assume this; although alternatively, all definitions below could instead be
qualified by statements such as "whenever the sum/series is defined". This is sometimes done with subtraction, such as with the following result, which holds whenever ${\displaystyle \mu }$ is
finitely additive:
Set difference formula: ${\displaystyle \mu (F)-\mu (E)=\mu (F\setminus E){\text{ whenever }}\mu (F)-\mu (E)}$ is defined with ${\displaystyle E,F\in {\mathcal {F}}}$ satisfying ${\displaystyle E
\subseteq F}$ and ${\displaystyle F\setminus E\in {\mathcal {F}}.}$
Null sets
A set ${\displaystyle F\in {\mathcal {F}}}$ is called a null set (with respect to ${\displaystyle \mu }$ ) or simply null if ${\displaystyle \mu (F)=0.}$ Whenever ${\displaystyle \mu }$ is not
identically equal to either ${\displaystyle -\infty }$ or ${\displaystyle +\infty }$ then it is typically also assumed that:
• null empty set: ${\displaystyle \mu (\varnothing )=0}$ if ${\displaystyle \varnothing \in {\mathcal {F}}.}$
Variation and mass
The total variation of a set ${\displaystyle S}$ is ${\displaystyle |\mu |(S)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\sup\{|\mu (F)|:F\in {\mathcal {F}}{\text{ and }}F\subseteq S\}}$ where
${\displaystyle |\,\cdot \,|}$ denotes the absolute value (or more generally, it denotes the norm or seminorm if ${\displaystyle \mu }$ is vector-valued in a (semi)normed space). Assuming that ${\
displaystyle \cup {\mathcal {F}}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F\in {\mathcal {F}},}$ then ${\displaystyle |\mu |\left(\cup {\
mathcal {F}}\right)}$ is called the total variation of ${\displaystyle \mu }$ and ${\displaystyle \mu \left(\cup {\mathcal {F}}\right)}$ is called the mass of ${\displaystyle \mu .}$
A set function is called finite if for every ${\displaystyle F\in {\mathcal {F}},}$ the value ${\displaystyle \mu (F)}$ is finite (which by definition means that ${\displaystyle \mu (F)eq \infty }$
and ${\displaystyle \mu (F)eq -\infty }$ ; an infinite value is one that is equal to ${\displaystyle \infty }$ or ${\displaystyle -\infty }$ ). Every finite set function must have a finite mass.
Common properties of set functions
A set function ${\displaystyle \mu }$ on ${\displaystyle {\mathcal {F}}}$ is said to be
• non-negative if it is valued in ${\displaystyle [0,\infty ].}$
• finitely additive if ${\displaystyle \textstyle \sum \limits _{i=1}^{n}\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{n}F_{i}\right)}$ for all pairwise disjoint finite
sequences ${\displaystyle F_{1},\ldots ,F_{n}\in {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{n}F_{i}\in {\mathcal {F}}.}$
□ If ${\displaystyle {\mathcal {F}}}$ is closed under binary unions then ${\displaystyle \mu }$ is finitely additive if and only if ${\displaystyle \mu (E\cup F)=\mu (E)+\mu (F)}$ for all
disjoint pairs ${\displaystyle E,F\in {\mathcal {F}}.}$
□ If ${\displaystyle \mu }$ is finitely additive and if ${\displaystyle \varnothing \in {\mathcal {F}}}$ then taking ${\displaystyle E:=F:=\varnothing }$ shows that ${\displaystyle \mu (\
varnothing )=\mu (\varnothing )+\mu (\varnothing )}$ which is only possible if ${\displaystyle \mu (\varnothing )=0}$ or ${\displaystyle \mu (\varnothing )=\pm \infty ,}$ where in the latter
case, ${\displaystyle \mu (E)=\mu (E\cup \varnothing )=\mu (E)+\mu (\varnothing )=\mu (E)+(\pm \infty )=\pm \infty }$ for every ${\displaystyle E\in {\mathcal {F}}}$ (so only the case ${\
displaystyle \mu (\varnothing )=0}$ is useful).
• countably additive or σ-additive if in addition to being finitely additive, for all pairwise disjoint sequences ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ such
that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}},}$ all of the following hold:
a. ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$
☆ The series on the left hand side is defined in the usual way as the limit ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)~{\stackrel {\scriptscriptstyle {\
text{def}}}{=}}~{\displaystyle \lim _{n\to \infty }}\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right).}$
☆ As a consequence, if ${\displaystyle \rho :\mathbb {N} \to \mathbb {N} }$ is any permutation/bijection then ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=
\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right);}$ this is because ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}=\textstyle \bigcup \limits _{i=1}
^{\infty }F_{\rho (i)}}$ and applying this condition (a) twice guarantees that both ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \
bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ and ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \
left(F_{\rho (i)}\right)}$ hold. By definition, a convergent series with this property is said to be unconditionally convergent. Stated in plain English, this means that rearranging/
relabeling the sets ${\displaystyle F_{1},F_{2},\ldots }$ to the new order ${\displaystyle F_{\rho (1)},F_{\rho (2)},\ldots }$ does not affect the sum of their measures. This is desirable
since just as the union ${\displaystyle F~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{i\in \mathbb {N} }F_{i}}$ does not depend on the order of these
sets, the same should be true of the sums ${\displaystyle \mu (F)=\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots }$ and ${\displaystyle \mu (F)=\mu \left(F_{\rho (1)}\right)+\mu \
left(F_{\rho (2)}\right)+\cdots \,.}$
b. if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ is not infinite then this series ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}
\right)}$ must also converge absolutely, which by definition means that ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|}$ must be finite. This is
automatically true if ${\displaystyle \mu }$ is non-negative (or even just valued in the extended real numbers).
☆ As with any convergent series of real numbers, by the Riemann series theorem, the series ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)={\displaystyle \lim
_{N\to \infty }}\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots +\mu \left(F_{N}\right)}$ converges absolutely if and only if its sum does not depend on the order of its terms (a
property known as unconditional convergence). Since unconditional convergence is guaranteed by (a) above, this condition is automatically true if ${\displaystyle \mu }$ is valued in ${\
displaystyle [-\infty ,\infty ].}$
c. if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$ is infinite then it is also required that
the value of at least one of the series ${\displaystyle \textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)>0}}\mu \left(F_{i}\right)\;{\text{ and }}\;\textstyle \
sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)<0}}\mu \left(F_{i}\right)\;}$ be finite (so that the sum of their values is well-defined). This is automatically true if ${\
displaystyle \mu }$ is non-negative.
• a pre-measure if it is non-negative, countably additive (including finitely additive), and has a null empty set.
• a measure if it is a pre-measure whose domain is a σ-algebra. That is to say, a measure is a non-negative countably additive set function on a σ-algebra that has a null empty set.
• a probability measure if it is a measure that has a mass of ${\displaystyle 1.}$
• an outer measure if it is non-negative, countably subadditive, has a null empty set, and has the power set ${\displaystyle \wp (\Omega )}$ as its domain.
• a signed measure if it is countably additive, has a null empty set, and ${\displaystyle \mu }$ does not take on both ${\displaystyle -\infty }$ and ${\displaystyle +\infty }$ as values.
• complete if every subset of every null set is null; explicitly, this means: whenever ${\displaystyle F\in {\mathcal {F}}{\text{ satisfies }}\mu (F)=0}$ and ${\displaystyle N\subseteq F}$ is any
subset of ${\displaystyle F}$ then ${\displaystyle N\in {\mathcal {F}}}$ and ${\displaystyle \mu (N)=0.}$
□ Unlike many other properties, completeness places requirements on the set ${\displaystyle \operatorname {domain} \mu ={\mathcal {F}}}$ (and not just on ${\displaystyle \mu }$ 's values).
• 𝜎-finite if there exists a sequence ${\displaystyle F_{1},F_{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \mu \left(F_{i}\right)}$ is finite for every index
${\displaystyle i,}$ and also ${\displaystyle \textstyle \bigcup \limits _{n=1}^{\infty }F_{n}=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F.}$
• decomposable if there exists a subfamily ${\displaystyle {\mathcal {P}}\subseteq {\mathcal {F}}}$ of pairwise disjoint sets such that ${\displaystyle \mu (P)}$ is finite for every ${\displaystyle
P\in {\mathcal {P}}}$ and also ${\displaystyle \textstyle \bigcup \limits _{P\in {\mathcal {P}}}\,P=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F}$ (where ${\displaystyle {\mathcal {F}}=\
operatorname {domain} \mu }$ ).
□ Every 𝜎-finite set function is decomposable although not conversely. For example, the counting measure on ${\displaystyle \mathbb {R} }$ (whose domain is ${\displaystyle \wp (\mathbb {R} )}$
) is decomposable but not 𝜎-finite.
• a vector measure if it is a countably additive set function ${\displaystyle \mu :{\mathcal {F}}\to X}$ valued in a topological vector space ${\displaystyle X}$ (such as a normed space) whose
domain is a σ-algebra.
□ If ${\displaystyle \mu }$ is valued in a normed space ${\displaystyle (X,\|\cdot \|)}$ then it is countably additive if and only if for any pairwise disjoint sequence ${\displaystyle F_{1},F_
{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}},}$ ${\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right)-\mu \left(\textstyle \bigcup \limits _{i=
1}^{\infty }F_{i}\right)\right\|=0.}$ If ${\displaystyle \mu }$ is finitely additive and valued in a Banach space then it is countably additive if and only if for any pairwise disjoint
sequence ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}},}$ ${\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{n}\cup F_{n+1}\cup F_{n+2}\cup \cdots \right)\
• a complex measure if it is a countably additive complex-valued set function ${\displaystyle \mu :{\mathcal {F}}\to \mathbb {C} }$ whose domain is a σ-algebra.
□ By definition, a complex measure never takes ${\displaystyle \pm \infty }$ as a value and so has a null empty set.
• a random measure if it is a measure-valued random element.
Arbitrary sums
As described in this article's section on generalized series, for any family ${\displaystyle \left(r_{i}\right)_{i\in I}}$ of real numbers indexed by an arbitrary indexing set ${\displaystyle I,}$ it
is possible to define their sum ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ as the limit of the net of finite partial sums ${\displaystyle F\in \operatorname {FiniteSubsets} (I)\mapsto \
textstyle \sum \limits _{i\in F}r_{i}}$ where the domain ${\displaystyle \operatorname {FiniteSubsets} (I)}$ is directed by ${\displaystyle \,\subseteq .\,}$ Whenever this net converges then its
limit is denoted by the symbols ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ while if this net instead diverges to ${\displaystyle \pm \infty }$ then this may be indicated by writing ${\
displaystyle \textstyle \sum \limits _{i\in I}r_{i}=\pm \infty .}$ Any sum over the empty set is defined to be zero; that is, if ${\displaystyle I=\varnothing }$ then ${\displaystyle \textstyle \sum
\limits _{i\in \varnothing }r_{i}=0}$ by definition.
For example, if ${\displaystyle z_{i}=0}$ for every ${\displaystyle i\in I}$ then ${\displaystyle \textstyle \sum \limits _{i\in I}z_{i}=0.}$ And it can be shown that ${\displaystyle \textstyle \sum
\limits _{i\in I}r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}=0}}r_{i}+\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}eq 0}}r_{i}=0+\textstyle \sum \limits _{\stackrel {i\in I,}{r_
{i}eq 0}}r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}eq 0}}r_{i}.}$ If ${\displaystyle I=\mathbb {N} }$ then the generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_
{i}}$ converges in ${\displaystyle \mathbb {R} }$ if and only if ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }r_{i}}$ converges unconditionally (or equivalently, converges absolutely) in
the usual sense. If a generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ converges in ${\displaystyle \mathbb {R} }$ then both ${\displaystyle \textstyle \sum \limits _{\
stackrel {i\in I}{r_{i}>0}}r_{i}}$ and ${\displaystyle \textstyle \sum \limits _{\stackrel {i\in I}{r_{i}<0}}r_{i}}$ also converge to elements of ${\displaystyle \mathbb {R} }$ and the set ${\
displaystyle \left\{i\in I:r_{i}eq 0\right\}}$ is necessarily countable (that is, either finite or countably infinite); this remains true if ${\displaystyle \mathbb {R} }$ is replaced with any normed
space.^[proof 1] It follows that in order for a generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ to converge in ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb
{C} ,}$ it is necessary that all but at most countably many ${\displaystyle r_{i}}$ will be equal to ${\displaystyle 0,}$ which means that ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}~=~\
textstyle \sum \limits _{\stackrel {i\in I}{r_{i}eq 0}}r_{i}}$ is a sum of at most countably many non-zero terms. Said differently, if ${\displaystyle \left\{i\in I:r_{i}eq 0\right\}}$ is uncountable
then the generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ does not converge.
In summary, due to the nature of the real numbers and its topology, every generalized series of real numbers (indexed by an arbitrary set) that converges can be reduced to an ordinary absolutely
convergent series of countably many real numbers. So in the context of measure theory, there is little benefit gained by considering uncountably many sets and generalized series. In particular, this
is why the definition of "countably additive" is rarely extended from countably many sets ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ (and the usual countable series $
{\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$ ) to arbitrarily many sets ${\displaystyle \left(F_{i}\right)_{i\in I}}$ (and the generalized series ${\displaystyle \
textstyle \sum \limits _{i\in I}\mu \left(F_{i}\right)}$ ).
Inner measures, outer measures, and other properties
A set function ${\displaystyle \mu }$ is said to be/satisfies
• monotone if ${\displaystyle \mu (E)\leq \mu (F)}$ whenever ${\displaystyle E,F\in {\mathcal {F}}}$ satisfy ${\displaystyle E\subseteq F.}$
• modular if it satisfies the following condition, known as modularity: ${\displaystyle \mu (E\cup F)+\mu (E\cap F)=\mu (E)+\mu (F)}$ for all ${\displaystyle E,F\in {\mathcal {F}}}$ such that ${\
displaystyle E\cup F,E\cap F\in {\mathcal {F}}.}$
• submodular if ${\displaystyle \mu (E\cup F)+\mu (E\cap F)\leq \mu (E)+\mu (F)}$ for all ${\displaystyle E,F\in {\mathcal {F}}}$ such that ${\displaystyle E\cup F,E\cap F\in {\mathcal {F}}.}$
• finitely subadditive if ${\displaystyle |\mu (F)|\leq \textstyle \sum \limits _{i=1}^{n}\left|\mu \left(F_{i}\right)\right|}$ for all finite sequences ${\displaystyle F,F_{1},\ldots ,F_{n}\in {\
mathcal {F}}}$ that satisfy ${\displaystyle F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{n}F_{i}.}$
• countably subadditive or σ-subadditive if ${\displaystyle |\mu (F)|\leq \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|}$ for all sequences ${\displaystyle F,F_{1},F_
{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ that satisfy ${\displaystyle F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}.}$
□ If ${\displaystyle {\mathcal {F}}}$ is closed under finite unions then this condition holds if and only if ${\displaystyle |\mu (F\cup G)|\leq |\mu (F)|+|\mu (G)|}$ for all ${\displaystyle
F,G\in {\mathcal {F}}.}$ If ${\displaystyle \mu }$ is non-negative then the absolute values may be removed.
□ If ${\displaystyle \mu }$ is a measure then this condition holds if and only if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\leq \textstyle \sum \limits _
{i=1}^{\infty }\mu \left(F_{i}\right)}$ for all ${\displaystyle F_{1},F_{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}.}$ If ${\displaystyle \mu }$ is a probability measure then this
inequality is Boole's inequality.
□ If ${\displaystyle \mu }$ is countably subadditive and ${\displaystyle \varnothing \in {\mathcal {F}}}$ with ${\displaystyle \mu (\varnothing )=0}$ then ${\displaystyle \mu }$ is finitely
• superadditive if ${\displaystyle \mu (E)+\mu (F)\leq \mu (E\cup F)}$ whenever ${\displaystyle E,F\in {\mathcal {F}}}$ are disjoint with ${\displaystyle E\cup F\in {\mathcal {F}}.}$
• continuous from above if ${\displaystyle \lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ for all non-increasing sequences of sets $
{\displaystyle F_{1}\supseteq F_{2}\supseteq F_{3}\cdots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}}}$
with ${\displaystyle \mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ and all ${\displaystyle \mu \left(F_{i}\right)}$ finite.
□ Lebesgue measure ${\displaystyle \lambda }$ is continuous from above but it would not be if the assumption that all ${\displaystyle \mu \left(F_{i}\right)}$ are eventually finite was omitted
from the definition, as this example shows: For every integer ${\displaystyle i,}$ let ${\displaystyle F_{i}}$ be the open interval ${\displaystyle (i,\infty )}$ so that ${\displaystyle \lim
_{n\to \infty }\lambda \left(F_{i}\right)=\lim _{n\to \infty }\infty =\infty eq 0=\lambda (\varnothing )=\lambda \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ where ${\
displaystyle \textstyle \bigcap \limits _{i=1}^{\infty }F_{i}=\varnothing .}$
• continuous from below if ${\displaystyle \lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ for all non-decreasing sequences of sets $
{\displaystyle F_{1}\subseteq F_{2}\subseteq F_{3}\cdots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}}.}$
• infinity is approached from below if whenever ${\displaystyle F\in {\mathcal {F}}}$ satisfies ${\displaystyle \mu (F)=\infty }$ then for every real ${\displaystyle r>0,}$ there exists some ${\
displaystyle F_{r}\in {\mathcal {F}}}$ such that ${\displaystyle F_{r}\subseteq F}$ and ${\displaystyle r\leq \mu \left(F_{r}\right)<\infty .}$
• an outer measure if ${\displaystyle \mu }$ is non-negative, countably subadditive, has a null empty set, and has the power set ${\displaystyle \wp (\Omega )}$ as its domain.
• an inner measure if ${\displaystyle \mu }$ is non-negative, superadditive, continuous from above, has a null empty set, has the power set ${\displaystyle \wp (\Omega )}$ as its domain, and ${\
displaystyle +\infty }$ is approached from below.
• atomic if every measurable set of positive measure contains an atom.
If a binary operation ${\displaystyle \,+\,}$ is defined, then a set function ${\displaystyle \mu }$ is said to be
• translation invariant if ${\displaystyle \mu (\omega +F)=\mu (F)}$ for all ${\displaystyle \omega \in \Omega }$ and ${\displaystyle F\in {\mathcal {F}}}$ such that ${\displaystyle \omega +F\in {\
mathcal {F}}.}$
If ${\displaystyle \tau }$ is a topology on ${\displaystyle \Omega }$ then a set function ${\displaystyle \mu }$ is said to be:
• a Borel measure if it is a measure defined on the σ-algebra of all Borel sets, which is the smallest σ-algebra containing all open subsets (that is, containing ${\displaystyle \tau }$ ).
• a Baire measure if it is a measure defined on the σ-algebra of all Baire sets.
• locally finite if for every point ${\displaystyle \omega \in \Omega }$ there exists some neighborhood ${\displaystyle U\in {\mathcal {F}}\cap \tau }$ of this point such that ${\displaystyle \mu
(U)}$ is finite.
□ If ${\displaystyle \mu }$ is a finitely additive, monotone, and locally finite then ${\displaystyle \mu (K)}$ is necessarily finite for every compact measurable subset ${\displaystyle K.}$
• ${\displaystyle \tau }$ -additive if ${\displaystyle \mu \left({\textstyle \bigcup }\,{\mathcal {D}}\right)=\sup _{D\in {\mathcal {D}}}\mu (D)}$ whenever ${\displaystyle {\mathcal {D}}\subseteq \
tau \cap {\mathcal {F}}}$ is directed with respect to ${\displaystyle \,\subseteq \,}$ and satisfies ${\displaystyle {\textstyle \bigcup }\,{\mathcal {D}}~{\stackrel {\scriptscriptstyle {\text
{def}}}{=}}~\textstyle \bigcup \limits _{D\in {\mathcal {D}}}D\in {\mathcal {F}}.}$
□ ${\displaystyle {\mathcal {D}}}$ is directed with respect to ${\displaystyle \,\subseteq \,}$ if and only if it is not empty and for all ${\displaystyle A,B\in {\mathcal {D}}}$ there exists
some ${\displaystyle C\in {\mathcal {D}}}$ such that ${\displaystyle A\subseteq C}$ and ${\displaystyle B\subseteq C.}$
• inner regular or tight if for every ${\displaystyle F\in {\mathcal {F}},}$ ${\displaystyle \mu (F)=\sup\{\mu (K):F\supseteq K{\text{ with }}K\in {\mathcal {F}}{\text{ a compact subset of }}(\
Omega ,\tau )\}.}$
• outer regular if for every ${\displaystyle F\in {\mathcal {F}},}$ ${\displaystyle \mu (F)=\inf\{\mu (U):F\subseteq U{\text{ and }}U\in {\mathcal {F}}\cap \tau \}.}$
• regular if it is both inner regular and outer regular.
• a Borel regular measure if it is a Borel measure that is also regular.
• a Radon measure if it is a regular and locally finite measure.
• strictly positive if every non-empty open subset has (strictly) positive measure.
• a valuation if it is non-negative, monotone, modular, has a null empty set, and has domain ${\displaystyle \tau .}$
Relationships between set functions
If ${\displaystyle \mu }$ and ${\displaystyle u }$ are two set functions over ${\displaystyle \Omega ,}$ then:
• ${\displaystyle \mu }$ is said to be absolutely continuous with respect to ${\displaystyle u }$ or dominated by ${\displaystyle u }$ , written ${\displaystyle \mu \ll u ,}$ if for every set ${\
displaystyle F}$ that belongs to the domain of both ${\displaystyle \mu }$ and ${\displaystyle u ,}$ if ${\displaystyle u (F)=0}$ then ${\displaystyle \mu (F)=0.}$
□ If ${\displaystyle \mu }$ and ${\displaystyle u }$ are ${\displaystyle \sigma }$ -finite measures on the same measurable space and if ${\displaystyle \mu \ll u ,}$ then the Radon–Nikodym
derivative ${\displaystyle {\frac {d\mu }{du }}}$ exists and for every measurable ${\displaystyle F,}$ ${\displaystyle \mu (F)=\int _{F}{\frac {d\mu }{du }}du .}$
□ ${\displaystyle \mu }$ and ${\displaystyle u }$ are called equivalent if each one is absolutely continuous with respect to the other. ${\displaystyle \mu }$ is called a supporting measure of
a measure ${\displaystyle u }$ if ${\displaystyle \mu }$ is ${\displaystyle \sigma }$ -finite and they are equivalent.^[4]
• ${\displaystyle \mu }$ and ${\displaystyle u }$ are singular, written ${\displaystyle \mu \perp u ,}$ if there exist disjoint sets ${\displaystyle M}$ and ${\displaystyle N}$ in the domains of $
{\displaystyle \mu }$ and ${\displaystyle u }$ such that ${\displaystyle M\cup N=\Omega ,}$ ${\displaystyle \mu (F)=0}$ for all ${\displaystyle F\subseteq M}$ in the domain of ${\displaystyle \mu
,}$ and ${\displaystyle u (F)=0}$ for all ${\displaystyle F\subseteq N}$ in the domain of ${\displaystyle u .}$
Examples of set functions include:
The Jordan measure on ${\displaystyle \mathbb {R} ^{n}}$ is a set function defined on the set of all Jordan measurable subsets of ${\displaystyle \mathbb {R} ^{n};}$ it sends a Jordan measurable set
to its Jordan measure.
Lebesgue measure
The Lebesgue measure on ${\displaystyle \mathbb {R} }$ is a set function that assigns a non-negative real number to every set of real numbers that belongs to the Lebesgue ${\displaystyle \sigma }$
Its definition begins with the set ${\displaystyle \operatorname {Intervals} (\mathbb {R} )}$ of all intervals of real numbers, which is a semialgebra on ${\displaystyle \mathbb {R} .}$ The function
that assigns to every interval ${\displaystyle I}$ its ${\displaystyle \operatorname {length} (I)}$ is a finitely additive set function (explicitly, if ${\displaystyle I}$ has endpoints ${\
displaystyle a\leq b}$ then ${\displaystyle \operatorname {length} (I)=b-a}$ ). This set function can be extended to the Lebesgue outer measure on ${\displaystyle \mathbb {R} ,}$ which is the
translation-invariant set function ${\displaystyle \lambda ^{\!*\!}:\wp (\mathbb {R} )\to [0,\infty ]}$ that sends a subset ${\displaystyle E\subseteq \mathbb {R} }$ to the infimum ${\displaystyle \
lambda ^{\!*\!}(E)=\inf \left\{\sum _{k=1}^{\infty }\operatorname {length} (I_{k}):{(I_{k})_{k\in \mathbb {N} }}{\text{ is a sequence of open intervals with }}E\subseteq \bigcup _{k=1}^{\infty }I_{k}
\right\}.}$ Lebesgue outer measure is not countably additive (and so is not a measure) although its restriction to the 𝜎-algebra of all subsets ${\displaystyle M\subseteq \mathbb {R} }$ that satisfy
the Carathéodory criterion: ${\displaystyle \lambda ^{\!*\!}(M)=\lambda ^{\!*\!}(M\cap E)+\lambda ^{\!*\!}(M\cap E^{c})\quad {\text{ for every }}S\subseteq \mathbb {R} }$ is a measure that called
Lebesgue measure. Vitali sets are examples of non-measurable sets of real numbers.
Infinite-dimensional space
As detailed in the article on infinite-dimensional Lebesgue measure, the only locally finite and translation-invariant Borel measure on an infinite-dimensional separable normed space is the trivial
measure. However, it is possible to define Gaussian measures on infinite-dimensional topological vector spaces. The structure theorem for Gaussian measures shows that the abstract Wiener space
construction is essentially the only way to obtain a strictly positive Gaussian measure on a separable Banach space.
Finitely additive translation-invariant set functions
The only translation-invariant measure on ${\displaystyle \Omega =\mathbb {R} }$ with domain ${\displaystyle \wp (\mathbb {R} )}$ that is finite on every compact subset of ${\displaystyle \mathbb {R}
}$ is the trivial set function ${\displaystyle \wp (\mathbb {R} )\to [0,\infty ]}$ that is identically equal to ${\displaystyle 0}$ (that is, it sends every ${\displaystyle S\subseteq \mathbb {R} }$
to ${\displaystyle 0}$ ) However, if countable additivity is weakened to finite additivity then a non-trivial set function with these properties does exist and moreover, some are even valued in ${\
displaystyle [0,1].}$ In fact, such non-trivial set functions will exist even if ${\displaystyle \mathbb {R} }$ is replaced by any other abelian group ${\displaystyle G.}$
Theorem — If ${\displaystyle (G,+)}$ is any abelian group then there exists a finitely additive and translation-invariant^[note 1] set function ${\displaystyle \mu :\wp (G)\to [0,1]}$ of mass ${\
displaystyle \mu (G)=1.}$
Extending set functions
Extending from semialgebras to algebras
Suppose that ${\displaystyle \mu }$ is a set function on a semialgebra ${\displaystyle {\mathcal {F}}}$ over ${\displaystyle \Omega }$ and let ${\displaystyle \operatorname {algebra} ({\mathcal
{F}}):=\left\{F_{1}\sqcup \cdots \sqcup F_{n}:n\in \mathbb {N} {\text{ and }}F_{1},\ldots ,F_{n}\in {\mathcal {F}}{\text{ are pairwise disjoint }}\right\},}$ which is the algebra on ${\displaystyle \
Omega }$ generated by ${\displaystyle {\mathcal {F}}.}$ The archetypal example of a semialgebra that is not also an algebra is the family ${\displaystyle {\mathcal {S}}_{d}:=\{\varnothing \}\cup \
left\{\left(a_{1},b_{1}\right]\times \cdots \times \left(a_{1},b_{1}\right]~:~-\infty \leq a_{i}<b_{i}\leq \infty {\text{ for all }}i=1,\ldots ,d\right\}}$ on ${\displaystyle \Omega :=\mathbb {R} ^
{d}}$ where ${\displaystyle (a,b]:=\{x\in \mathbb {R} :a<x\leq b\}}$ for all ${\displaystyle -\infty \leq a<b\leq \infty .}$ Importantly, the two non-strict inequalities ${\displaystyle \,\leq \,}$
in ${\displaystyle -\infty \leq a_{i}<b_{i}\leq \infty }$ cannot be replaced with strict inequalities ${\displaystyle \,<\,}$ since semialgebras must contain the whole underlying set ${\displaystyle
\mathbb {R} ^{d};}$ that is, ${\displaystyle \mathbb {R} ^{d}\in {\mathcal {S}}_{d}}$ is a requirement of semialgebras (as is ${\displaystyle \varnothing \in {\mathcal {S}}_{d}}$ ).
If ${\displaystyle \mu }$ is finitely additive then it has a unique extension to a set function ${\displaystyle {\overline {\mu }}}$ on ${\displaystyle \operatorname {algebra} ({\mathcal {F}})}$
defined by sending ${\displaystyle F_{1}\sqcup \cdots \sqcup F_{n}\in \operatorname {algebra} ({\mathcal {F}})}$ (where ${\displaystyle \,\sqcup \,}$ indicates that these ${\displaystyle F_{i}\in {\
mathcal {F}}}$ are pairwise disjoint) to: ${\displaystyle {\overline {\mu }}\left(F_{1}\sqcup \cdots \sqcup F_{n}\right):=\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right).}$ This extension ${\
displaystyle {\overline {\mu }}}$ will also be finitely additive: for any pairwise disjoint ${\displaystyle A_{1},\ldots ,A_{n}\in \operatorname {algebra} ({\mathcal {F}}),}$ ${\displaystyle {\
overline {\mu }}\left(A_{1}\cup \cdots \cup A_{n}\right)={\overline {\mu }}\left(A_{1}\right)+\cdots +{\overline {\mu }}\left(A_{n}\right).}$
If in addition ${\displaystyle \mu }$ is extended real-valued and monotone (which, in particular, will be the case if ${\displaystyle \mu }$ is non-negative) then ${\displaystyle {\overline {\mu }}}$
will be monotone and finitely subadditive: for any ${\displaystyle A,A_{1},\ldots ,A_{n}\in \operatorname {algebra} ({\mathcal {F}})}$ such that ${\displaystyle A\subseteq A_{1}\cup \cdots \cup A_
{n},}$ ${\displaystyle {\overline {\mu }}\left(A\right)\leq {\overline {\mu }}\left(A_{1}\right)+\cdots +{\overline {\mu }}\left(A_{n}\right).}$
Extending from rings to σ-algebras
If ${\displaystyle \mu :{\mathcal {F}}\to [0,\infty ]}$ is a pre-measure on a ring of sets (such as an algebra of sets) ${\displaystyle {\mathcal {F}}}$ over ${\displaystyle \Omega }$ then ${\
displaystyle \mu }$ has an extension to a measure ${\displaystyle {\overline {\mu }}:\sigma ({\mathcal {F}})\to [0,\infty ]}$ on the σ-algebra ${\displaystyle \sigma ({\mathcal {F}})}$ generated by $
{\displaystyle {\mathcal {F}}.}$ If ${\displaystyle \mu }$ is σ-finite then this extension is unique.
To define this extension, first extend ${\displaystyle \mu }$ to an outer measure ${\displaystyle \mu ^{*}}$ on ${\displaystyle 2^{\Omega }=\wp (\Omega )}$ by ${\displaystyle \mu ^{*}(T)=\inf \left\
{\sum _{n}\mu \left(S_{n}\right):T\subseteq \cup _{n}S_{n}{\text{ with }}S_{1},S_{2},\ldots \in {\mathcal {F}}\right\}}$ and then restrict it to the set ${\displaystyle {\mathcal {F}}_{M}}$ of ${\
displaystyle \mu ^{*}}$ -measurable sets (that is, Carathéodory-measurable sets), which is the set of all ${\displaystyle M\subseteq \Omega }$ such that ${\displaystyle \mu ^{*}(S)=\mu ^{*}(S\cap M)+
\mu ^{*}(S\cap M^{\mathrm {c} })\quad {\text{ for every subset }}S\subseteq \Omega .}$ It is a ${\displaystyle \sigma }$ -algebra and ${\displaystyle \mu ^{*}}$ is sigma-additive on it, by
Caratheodory lemma.
Restricting outer measures
If ${\displaystyle \mu ^{*}:\wp (\Omega )\to [0,\infty ]}$ is an outer measure on a set ${\displaystyle \Omega ,}$ where (by definition) the domain is necessarily the power set ${\displaystyle \wp (\
Omega )}$ of ${\displaystyle \Omega ,}$ then a subset ${\displaystyle M\subseteq \Omega }$ is called ${\displaystyle \mu ^{*}}$ –measurable or Carathéodory-measurable if it satisfies the following
Carathéodory's criterion: ${\displaystyle \mu ^{*}(S)=\mu ^{*}(S\cap M)+\mu ^{*}(S\cap M^{\mathrm {c} })\quad {\text{ for every subset }}S\subseteq \Omega ,}$ where ${\displaystyle M^{\mathrm {c} }:=
\Omega \setminus M}$ is the complement of ${\displaystyle M.}$
The family of all ${\displaystyle \mu ^{*}}$ –measurable subsets is a σ-algebra and the restriction of the outer measure ${\displaystyle \mu ^{*}}$ to this family is a measure.
See also
1. ^ Kallenberg, Olav (2017). Random Measures, Theory and Applications. Probability Theory and Stochastic Modelling. Vol. 77. Switzerland: Springer. p. 21. doi:10.1007/978-3-319-41598-7. ISBN
2. ^ Kolmogorov and Fomin 1975
1. ^ The function ${\displaystyle \mu }$ being translation-invariant means that ${\displaystyle \mu (S)=\mu (g+S)}$ for every ${\displaystyle g\in G}$ and every subset ${\displaystyle S\subseteq G.}
1. ^ Suppose the net ${\textstyle \textstyle \sum \limits _{i\in I}r_{i}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\textstyle \lim \limits _{A\in \operatorname {Finite} (I)}}\ \textstyle \
sum \limits _{i\in A}r_{i}=\lim \left\{\textstyle \sum \limits _{i\in A}r_{i}\,:A\subseteq I,A{\text{ finite }}\right\}}$ converges to some point in a metrizable topological vector space ${\
displaystyle X}$ (such as ${\displaystyle \mathbb {R} ,}$ ${\displaystyle \mathbb {C} ,}$ or a normed space), where recall that this net's domain is the directed set ${\displaystyle (\
operatorname {Finite} (I),\subseteq ).}$ Like every convergent net, this convergent net of partial sums ${\displaystyle A\mapsto \textstyle \sum \limits _{i\in A}r_{i}}$ is a Cauchy net, which
for this particular net means (by definition) that for every neighborhood ${\displaystyle W}$ of the origin in ${\displaystyle X,}$ there exists a finite subset ${\displaystyle A_{0}}$ of ${\
displaystyle I}$ such that ${\textstyle \textstyle \sum \limits _{i\in B}r_{i}-\textstyle \sum \limits _{i\in C}r_{i}\in W}$ for all finite supersets ${\displaystyle B,C\supseteq A_{0};}$ this
implies that ${\displaystyle r_{i}\in W}$ for every ${\displaystyle i\in I\setminus A_{0}}$ (by taking ${\displaystyle B:=A_{0}\cup \{i\}}$ and ${\displaystyle C:=A_{0}}$ ). Since ${\displaystyle
X}$ is metrizable, it has a countable neighborhood basis ${\displaystyle U_{1},U_{2},\ldots }$ at the origin, whose intersection is necessarily ${\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\}}$
(since ${\displaystyle X}$ is a Hausdorff TVS). For every positive integer ${\displaystyle n\in \mathbb {N} ,}$ pick a finite subset ${\displaystyle A_{n}\subseteq I}$ such that ${\displaystyle
r_{i}\in U_{n}}$ for every ${\displaystyle i\in I\setminus A_{n}.}$ If ${\displaystyle i}$ belongs to ${\displaystyle (I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\
cup A_{2}\cup \cdots \right)}$ then ${\displaystyle r_{i}}$ belongs to ${\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\}.}$ Thus ${\displaystyle r_{i}=0}$ for every index ${\displaystyle i\in I}$
that does not belong to the countable set ${\displaystyle A_{1}\cup A_{2}\cup \cdots .}$ ${\displaystyle \blacksquare }$
Further reading | {"url":"https://www.knowpia.com/knowpedia/Set_function","timestamp":"2024-11-02T05:13:29Z","content_type":"text/html","content_length":"968192","record_id":"<urn:uuid:24db71ca-4f24-4a2e-bf80-3fa9b4aa7dba>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00186.warc.gz"} |
Basic Circuit Laws
Understanding the fundamental laws that govern electric circuits is crucial for anyone delving into the field of electronics or electrical engineering. Just like the laws of physics govern the
universe, the basic circuit laws form the backbone of circuit analysis and design. This article aims to demystify these laws, provide practical insights, and equip you with a solid understanding of
circuit behavior.
1. Introduction to Circuit Laws
When we talk about circuit laws, we're primarily referring to two foundational principles: Ohm's Law and Kirchhoff's Laws. These principles help us analyze and predict how electric circuits behave
under various conditions.
Ohm's Law, introduced by Georg Simon Ohm in 1827, establishes a direct relationship between voltage, current, and resistance in a circuit. Kirchhoff's Laws, formulated by Gustav Kirchhoff in the 19th
century, delve deeper into how currents and voltages behave in complex circuits.
Whether you're a novice looking to build your first circuit or a seasoned engineer refining a design, understanding these laws is essential.
Why Should You Care?
You might be wondering: why is it important to learn these laws? Here’s why:
• Foundation for Circuit Analysis: These laws provide the basic framework for analyzing electrical circuits.
• Real-World Applications: From designing household electronics to understanding automotive electrical systems, circuit laws are everywhere.
• Problem-Solving Skills: They enable you to troubleshoot issues, whether you're facing resistance, voltage drops, or current overloads.
2. Ohm's Law Explained
Ohm's Law is arguably the most fundamental equation in electronics. It can be stated simply as:
V = I × R
• V is the voltage (in volts, V)
• I is the current (in amperes, A)
• R is the resistance (in ohms, Ω)
Understanding Each Component
• Voltage (V): Think of voltage as the "pressure" that pushes electric charges through a circuit. It's similar to water pressure in a pipe—higher pressure results in more flow.
• Current (I): This is the flow of electric charge through a circuit. Imagine it as the amount of water flowing through that pipe at any moment. It’s measured in amperes (A).
• Resistance (R): Resistance hinders the flow of current, much like a narrowing in a pipe would slow down water flow. It's determined by the material and dimensions of the conductor.
Practical Applications of Ohm's Law
Let's take a moment to explore some practical applications of Ohm’s Law:
• Calculating Current: If you have a circuit with a 12V battery and a resistor of 4Ω, you can find the current flowing through the circuit. Using Ohm’s Law:
I = V/R = 12V/4Ω = 3A.
• Voltage Drops: In a series circuit, if you know the current and resistance of each component, you can determine the voltage drop across each component, allowing for precise design and
Limitations of Ohm's Law
While Ohm's Law is a powerful tool, it has its limitations:
• It applies only to ohmic materials, where the relationship between voltage and current is linear.
• In non-linear components like diodes or transistors, the relationship can vary based on the operating conditions.
3. Kirchhoff's Laws: Current and Voltage
To deepen our understanding of circuits, we introduce Kirchhoff's Laws: the Current Law (KCL) and the Voltage Law (KVL).
Kirchhoff's Current Law (KCL)
KCL states that the total current entering a junction must equal the total current leaving that junction. In other words, charge is conserved at any node in an electrical circuit.
Mathematically, this is expressed as:
ΣI_in = ΣI_out
Practical Example of KCL
Consider a junction with three wires. If 3A and 2A are flowing into the junction and 1A is flowing out, then according to KCL:
• Total current entering = 3A + 2A = 5A
• Total current leaving = 1A
This tells us something isn't right, as it implies there's a 4A accumulation in the junction—an impossibility under normal conditions. Such discrepancies often signal a need for troubleshooting.
Kirchhoff's Voltage Law (KVL)
KVL, on the other hand, states that the sum of the voltages around any closed loop in a circuit must equal zero. This includes the voltage drops and rises across circuit components.
In formula form, KVL can be written as:
ΣV = 0
Practical Example of KVL
Consider a simple circuit with a battery and two resistors in series:
• Battery voltage = 12V
• Resistor 1 (R1) = 4Ω
• Resistor 2 (R2) = 8Ω
According to KVL, if we travel around the loop, we would see:
• Rise in voltage from the battery (12V)
• Voltage drop across R1: V1 = I × R1 (with I determined from Ohm's Law)
• Voltage drop across R2: V2 = I × R2
The sum of these values will equal zero when we complete the loop:
12V - V1 - V2 = 0.
The Importance of Kirchhoff's Laws
These laws are vital for analyzing complex circuits, particularly those involving multiple components:
• Circuit Analysis: They allow engineers to systematically analyze how current and voltage behave in interconnected circuits.
• Design Efficiency: Understanding KCL and KVL helps in optimizing circuit designs to ensure reliability and performance.
4. Practical Applications of Circuit Laws
Now that we have a solid grasp of the basic circuit laws, let’s look at how they are applied in real-world scenarios.
4.1. Electrical Engineering
In electrical engineering, these laws are pivotal for designing and analyzing circuits in devices like smartphones, computers, and home appliances. For instance, when designing a power supply
circuit, engineers use Ohm's Law to ensure that the voltage and current ratings match the load requirements.
4.2. Troubleshooting Circuits
When a device fails to operate as expected, technicians often apply KCL and KVL to diagnose issues. By measuring currents and voltages at various points in the circuit, they can pinpoint where a
failure may have occurred—much like a detective solving a mystery.
4.3. Educational Tools
In educational settings, these laws serve as foundational concepts in physics and engineering courses. Laboratories often utilize circuit simulations, allowing students to visualize how altering
resistance, voltage, or current affects the overall circuit. This hands-on experience solidifies understanding and application of the concepts.
5. Circuit Analysis Techniques
Beyond the fundamental laws, several techniques for analyzing circuits can enhance our understanding and application of circuit laws.
5.1. Series and Parallel Circuits
Understanding the differences between series and parallel circuits is crucial:
• Series Circuits: In a series circuit, components are connected end-to-end. The total resistance increases as more components are added, and the same current flows through all components.
Example: If we have two resistors in series, R1 = 3Ω and R2 = 2Ω, the total resistance (R_total) is:
R_total = R1 + R2 = 3Ω + 2Ω = 5Ω.
• Parallel Circuits: In a parallel configuration, components are connected across the same voltage source, resulting in the same voltage across each component, but the current can vary. The total
resistance decreases as more branches are added.
Example: For two resistors in parallel, R1 = 6Ω and R2 = 3Ω, the total resistance can be calculated using:
1/R_total = 1/R1 + 1/R2
1/R_total = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 ⇒ R_total = 2Ω.
5.2. Nodal Analysis
Nodal analysis is a systematic method to determine the voltage at each node in a circuit using KCL. By defining a reference node (ground) and applying KCL to other nodes, we can create a set of
equations that can be solved simultaneously.
5.3. Mesh Analysis
Mesh analysis focuses on loops within a circuit, applying KVL to determine unknown currents. By defining mesh currents and writing KVL equations for each loop, we can derive values for all currents
in the circuit.
6. Conclusion
In conclusion, mastering the basic circuit laws—Ohm's Law and Kirchhoff's Laws—equips you with the tools necessary to analyze and design electrical circuits. These laws not only form the foundation
for circuit analysis but also offer insight into the behavior of electrical components in various configurations. Whether you're an aspiring engineer, a DIY enthusiast, or simply someone curious
about how electronic devices function, understanding these principles will significantly enhance your knowledge and capabilities.
1. What is Ohm's Law in simple terms?
Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.
2. How does Kirchhoff's Current Law work?
Kirchhoff's Current Law asserts that the total current entering a junction in an electrical circuit must equal the total current leaving that junction, ensuring charge conservation.
3. What is the difference between series and parallel circuits?
In series circuits, components are connected one after another, resulting in the same current through each component but a higher total resistance. In parallel circuits, components are connected
across the same voltage source, allowing different currents through each branch and resulting in lower total resistance.
4. Can Ohm's Law be applied to all materials?
No, Ohm's Law is only applicable to ohmic materials where the relationship between voltage and current is linear. In non-linear devices, such as diodes or transistors, the relationship can change.
5. Why are circuit laws important in real-world applications?
Circuit laws help engineers design and troubleshoot electronic devices, ensuring they function correctly and efficiently. Understanding these principles is essential for anyone working in electronics
or electrical engineering. | {"url":"https://theglobalpresence.com/post/basic-circuit-laws","timestamp":"2024-11-03T21:29:58Z","content_type":"text/html","content_length":"95932","record_id":"<urn:uuid:a4816e72-c8e4-4d65-b001-2c88b1aea507>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00142.warc.gz"} |
Bonnie Gold of Math Topics | Question AI
<div id="mw-content-wrapper"><div id="mw-content"><div id="content" class="mw-body" role="main"><div id="bodyContentOuter"><div class="mw-body-content" id="bodyContent"><div id="mw-content-text"
class="mw-body-content mw-content-ltr" lang="en" dir="ltr"><div class="mw-parser-output"><p><b>Bonnie Gold</b> (born 1948)<sup id="cite_ref-born_1-0" class="reference"><span></span></sup> is an
American mathematician, <span title="Mathematical logic">mathematical logician</span>, <span title="Philosophy:Philosophy of mathematics">philosopher of mathematics</span>, and <span title=
"Mathematics education">mathematics educator</span>. She is a professor emerita of mathematics at <span title="Organization:Monmouth University">Monmouth University</span>.<sup id=
"cite_ref-monmouth_2-0" class="reference"><span></span></sup> </p><h2 id="qai_title_1"><span class="mw-headline" id="Education_and_career">Education and career</span></h2><p>Gold completed her Ph.D.
in 1976 at <span title="Organization:Cornell University">Cornell University</span>, under the supervision of Michael D. Morley.<sup id="cite_ref-mgp_3-0" class="reference"><span></span></sup> </p><p>
She was the chair of the mathematics department at <span title="Organization:Wabash College">Wabash College</span> before moving to Monmouth, where she also became department chair.<sup id=
"cite_ref-prestigious_4-0" class="reference"><span></span></sup> </p><h2 id="qai_title_2"><span class="mw-headline" id="Contributions">Contributions</span></h2><p>The research from Gold's
dissertation, <i>Compact and <span style="opacity:.5">$\displaystyle{ \omega }$</span>-compact formulas in <span style="opacity:.5">$\displaystyle{ L_{\omega_{1},\omega} }$</span></i>,<sup id=
"cite_ref-mgp_3-1" class="reference"><span></span></sup> was later published in the journal <i>Archiv für Mathematische Logik und Grundlagenforschung</i>, and concerned <span title=
"Philosophy:Infinitary logic">infinitary logic</span>.<sup id="cite_ref-nebres_5-0" class="reference"><span></span></sup> </p><p>With Sandra Z. Keith and William A. Marion she co-edited <i>Assessment
Practices in Undergraduate Mathematics</i>, published by the Mathematical Association of America (MAA) in 1999.<sup id="cite_ref-lesser_6-0" class="reference"><span></span></sup> With Roger A.
Simons, Gold is also the editor of another book, <i>Proof and Other Dilemmas: Mathematics and Philosophy</i> (MAA, 2008).<sup id="cite_ref-pod_7-0" class="reference"><span></span></sup> </p><p>Her
essay "How your philosophy of mathematics impacts your teaching" was selected for inclusion in <i>The Best Writing on Mathematics 2012</i>. In it, she argues that the <span title=
"Philosophy:Philosophy of mathematics">philosophy of mathematics</span> affects the teaching of mathematics even when the teacher's philosophical principles are implicit and unexamined.<sup id=
"cite_ref-best_8-0" class="reference"><span></span></sup> </p><h2 id="qai_title_3"><span class="mw-headline" id="Recognition">Recognition</span></h2><p>In 2012, Gold became the winner of the 22nd
Louise Hay Award of the Association for Women in Mathematics for her contributions to <span title="Mathematics education">mathematics education</span>. The award citation noted her work in <span
title="Educational assessment">educational assessment</span> for undergraduate study in mathematics.<sup id="cite_ref-hay_9-0" class="reference"><span></span></sup> </p></div></div></div></div></div> | {"url":"https://www.questionai.com/knowledge/kl6pZLkxKx-bonnie-gold","timestamp":"2024-11-04T11:58:41Z","content_type":"text/html","content_length":"60459","record_id":"<urn:uuid:b1c471be-58f3-429b-bc1e-f1658fa5ba9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00862.warc.gz"} |
Adjectives for theorems | Adjective1.com
Adjectives for theorems
Theorems adjectives are listed in this post. Each word below can often be found in front of the noun theorems in the same sentence. This reference page can help answer the question what are some
adjectives commonly used for describing THEOREMS.
above, basic, certain, corresponding
elementary, following, fundamental, general
geometric, geometrical, important, main
many, mathematical, new, other
preceding, several, such, various
Hope this word list had the adjective used with theorems you were looking for. Additional describing words / adjectives that describe / adjectives of various nouns can be found in the other pages on
this website. | {"url":"http://adjective1.com/for-theorems/","timestamp":"2024-11-05T15:59:50Z","content_type":"application/xhtml+xml","content_length":"65547","record_id":"<urn:uuid:cbeea4b1-f910-40f7-b7be-326a6d9f70f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00373.warc.gz"} |
Heron's Formula to find Area of Triangle
Before you understand Heron's Formula, you are advised to read:
How to find Area of Triangle ?
What is Perimeter ?
What is Square-Root ?
As you know that:
Area of Triangle = 1/2 X Base X Altitude
But you may encounter some situation where you do not have the value of altitude. So in such situation, where altitude is unknown, Heron's formula is used to calculate Area of Triangle.
Heron's Formula was given by a famous Egyptian Mathematician Heron in about 10AD and therefore this formula was also named after him.
Heron's Formula = √ s (s-a) (s-b) (s-c)
In the above formula:
a, b and c are the three sides of a triangle
s is semi-perimeter i.e. half of the perimeter
or we can write it as:
Example: Find the area of following triangle:
Solution: As per the given question:
Three sides of given Triangle ABC are AB, BC and CA
AB = 28 cm = a
BC = 15 cm = b
CA = 41 cm = c
Semi-perimeter of Triangle ABC =
Apply the values of AB, BC & CA and we get:
Solve addition in the numerator and we get:
Solve division expression and we get:
= 42
So, s = 42
Apply Heron's Formula = √ s (s-a) (s-b) (s-c)
Apply the values of s, a, b & c and we get:
= √ 42 (42-28) (42-15) (42-41)
Solve brackets and we get:
= √ 42 (14) (27) (1)
Solve square-root and we get:
= 126
Hence, Area of Given Triangle ABC = 12cm^2 | {"url":"https://www.algebraden.com/herons-forumla.htm","timestamp":"2024-11-13T01:58:30Z","content_type":"text/html","content_length":"14604","record_id":"<urn:uuid:9b0c5863-ab0d-4149-98d8-0d1d6c190291>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00075.warc.gz"} |
Logging and Analysis of Lift Journeys Using an Accelerometer - Peters Research
Anna Peters, Richard Peters
Peters Research Ltd
This paper was presented at The 10th Symposium on Lift & Escalator Technology (CIBSE Lifts Group, The University of Northampton and LEIA) (2019). This web version © Peters Research Ltd 2019
Keywords: Kinematics, accelerometer, performance, logging, passenger demand
Abstract. Data measured with an accelerometer in or on a lift car can be very useful. Using an accelerometer to measure individual trips allows engineers to confirm that a lift is working as
specified. Further analysis of extended measurements can also provide an understanding of lift passenger demand, useful in planning new buildings and addressing traffic problems in existing
buildings. Accelerometers can also be used as part of lift monitoring systems, collecting data about the lifts without the need for interfacing with lift controllers, which can be expensive due to
the use of proprietary protocols. In this paper the authors address the analysis of accelerometer data for a multi trip scenario. With real as opposed to ideal data, the analysis procedure must
account for accelerometer drift, noise and other data anomalies. The final analysis software provides an idealised version of the measured data including the distance, velocity, acceleration and jerk
for each trip. The distance measurements combined provide a spatial plot of lift position.
1 Introduction
The logging of lift motion is valuable when measuring lift performance, analysing lift traffic and lift monitoring. Accelerometers can be used as part of lift monitoring systems, collecting data
without the need for interfacing with lift controllers, which can be expensive due to the use of proprietary protocols [1].
Software has been developed to process data collected by a low-cost computer and accelerometer placed on top of a lift car. By analysing a sequence of individual lift journeys, the software is able
to provide a summary of lift stops analysed by floor and time of day.
1.1 Motivation
Estimates of lift passenger demand are required in the planning of new lift installations and when addressing lift traffic problems in existing buildings. By recording and processing the output of an
accelerometer, a spatial plot of lift motion can be produced. An indication of lift passenger demand can then be determined without the need for human observers.
This is possible as the spatial plot data recorded by the accelerometer software can be applied in the development of mathematical models to extrapolate passenger demand from stops. This work is
outside the scope of this paper, but a range of methods have been proposed by several researchers over many years [2], [3] with recent work showing how good passenger demand predictions can be
estimated with limited data sets [4].
The software developed can also support the monitoring of lift installations by providing a connection-free solution. Because the application of interfacing standards is rare, third party monitoring
of lift installations can be expensive.
1.2 Ideal Lift Kinematics
It is possible to derive equations to represent the ideal motion of a lift, which can be plotted as continuous functions that represent the optimum displacement (D), velocity (V), acceleration (A)
and jerk (J) profiles, see Figure 1.
Modern variable speed drives can be programmed to match these ideal lift kinematics curves closely. Since it is necessary to model each lift trip as accurately as possible, a good approach is to fit
the measured accelerometer data to the idealised kinematics plots.
Figure 1 [5]: Ideal lift kinematics for: (A) lift reaches full speed; (B) lift reaches full acceleration, but not full speed; (C) lift does not reach full speed or acceleration
Ideal lift kinematics represents a lift acceleration profile by a series of straight lines. The software therefore applies a linear regression method to fit raw measured data to an ideal plot.
2 Coding methodology
2.1 Raw Accelerometer Data
The software is required to process a full set of raw accelerometer data as shown in Figure 2.
Figure 2: Multiple Lift Trip Raw Accelerometer Plot
To simplify the problem, the data set is isolated into a series of single up and down lift journeys (trips) that can each be analysed individually. An example of an isolated single trip is
illustrated in Figure 3.
Figure 3: Single Lift Trip Raw Accelerometer Plot
2.2 Language Selection
The software was written in the C++ programming language and is object orientated. Object orientated programming combines groups of related variables and functions into a class. Properties and
methods can be hidden inside the class making the software easier to use, understand and maintain [6].
2.3 Processing Ideal Data
The basic methodology of the software was created, and initially tested on self-generated ideal lift journeys. This ensured that the code could correctly follow an expected journey before tackling
real world data. These ideal journeys followed a profile identical to ideal lift kinematics curves, therefore the software outputted an identical profile.
Each isolated single trip can be separated into two phases, one of acceleration and the other deceleration. Since it is assumed that the accelerometer data will start and finish with a stationary
lift, there will be an even number of acceleration and deceleration phases. Therefore, for each single trip, the following analysis is carried out twice, once for each phase.
It must be decided whether another phase exists. A new phase occurs when the modulus of the acceleration reaches a threshold value, 50% of the maximum acceleration identified. The next phase is the
first case in the remaining data set that this threshold is reached, as shown in Figure 4.
If a phase is identified, it must be determined whether this is an acceleration or deceleration phase. A positive acceleration determines an acceleration phase and a negative value determines a
deceleration phase.
For each phase, linear regression analysis is carried out on the two sections where acceleration is non-constant:
(a) Modulus of the acceleration rising
(b) Modulus of the acceleration falling
In anticipation of noise that will be present in real data, linear regression analysis is not carried out over the full length of the phase sections. A reasonable assumption is to carry out analysis
on the segments of the sections that fall within 20% and 60% of the maximum acceleration.
Figure 4: Identification of an Acceleration Phase Within an Up Trip
Linear regression analysis is carried out on all the data that falls between the four calculated limits. Two linear regression lines are identified that minimise the sum of the squares of the errors
between the lines and the raw data, with results shown in Figure 5.
Figure 5: Linear Regression Fit of Acceleration Profile Between Calculated Limits
The software identifies the length of the regression lines (that are to be joined with a horizontal line in specific cases) that minimise the sum of the squares of the errors between the approximated
and the raw accelerometer data. Figure 6 demonstrates the process of extending the regression line to the length of minimum error.
Figure 6: Identification of the Correct Regression Line Length
The series of time and acceleration coordinates to represent each phase are stored in a vector. Zero acceleration coordinates are added from the start of the test data to the beginning of the
acceleration/deceleration phase. The regression analysis is then repeated for the second phase of the trip. Figure 7 plots the approximated single up trip profile against the ideal up trip,
confirming that the software could accurately represent the data prior to testing on real data.
Figure 7: Ideal vs. Approximated Acceleration Profiles of a Single Up Trip
An approximation of the ideal jerk profile is generated via a central difference differentiation of the approximated acceleration profile.
A trapezium rule integration is carried out on the approximated acceleration profile, generating an approximated velocity profile. From ideal lift kinematics, it is known that the integral of a single
trip should equal zero, since the lift starts and ends stationary. A scaling factor is determined from the difference between the integrals of the acceleration and deceleration phases. This scaling
factor is applied to the acceleration profile such that the integral will equal zero.
To determine the approximated displacement profile, a trapezium rule integration is carried out on the approximated velocity profile as shown in Figure 8. The end value of the displacement represents
the total vertical distance moved between the floor on the single trip and is stored separately for later use in multiple trip analysis.
Figure 8: Ideal vs. Approximated Displacement Profiles of a Single Up Trip
A data set containing multiple trips is simply a series of linked single trips. To carry out multiple trip analysis, single trip analysis is looped until the end of the data set is reached. Since the
software is capable of outputting the final displacement of each single trip, these can be stored allowing a spatial plot of the lift’s motion to be plotted over time. At this point, it is possible to
begin to see the effects of accelerometer drift as the approximated positions can be compared to building data.
2.4 Processing Real Data
The introduction of real data introduced a series of issues to be tackled. The solution to each identified issue was integrated into the existing software such that it was capable of correctly
representing the raw data.
2.5 Time Section Representation
Existing ideal lift kinematics curves are defined by a set of equations which are divided into time sections, identified by a change in jerk. To tackle an issue introduced with real data, a similar
approach was taken. Rather than plotting continuous lines, five significant points are identified for each phase with respect to changes in acceleration, shown in Figure 9. A benefit of recording
significant points rather than continuous data is the reduction in file size.
Figure 9: Representation of Time Sections For an Acceleration Phase
2.6 Spatial Plot Generation
The software creates two spatial plots; approximated and calibrated. The approximated spatial plot combines the displacement values stored for each single trip and adjusts the values at points where
it is assumed that identical floors have been met. Access to real floor positions from building data allows the calibrated spatial plot to be generated. The approximated spatial plot is adjusted to
the correct floor positions such that it can correctly plot the lift motion by floor and time of day.
3 Results
The first set of results processed were from data collected in a high-rise building in Central London, with an express zone using an accelerometer included in a low-cost consumer tablet. The
existence of an express zone in the high-rise building resulted in the inability to generate a calibrated spatial plot due to the lack of accelerometer precision. The analysis was repeated using data
collected in a low-rise office building. This three-story building does not have an express zone and therefore it is possible to calibrate and test the results to determine the accuracy of the
software approximation.
3.1 High Rise Building
Figure 10 plots the raw displacement data against the approximated spatial plot generated by the software for the 39 floor building. The effects of drift are clear, and it is visible how the software
has managed to tackle this problem.
Figure 10: Calibrated vs. Approximated Displacement Profiles (High Rise Building)
3.2 Low Rise Building
Figure 11 plots the raw displacement profile without correction against the approximated spatial plot created by the software. It is clear from the scale of the raw data the significance of drift
when modelling continuous lift motion.
Figure 11: Calibrated vs. Approximated Displacement Profiles (Low Rise Building)
The lack of express zone in the low rise building allowed calibration to be carried out against real building data. Figure 12 plots the approximated spatial plot generated by the software against the
adjusted spatial plot once calibration has been carried out.
Figure 12: Approximated vs. Calibrated Displacement Profiles (Low Rise Building)
Table 1 shows a comparison between the approximated floor positions found by the software and the real floor positions provided from building data.
Level 1 Level 2 Level 3
Approximated Floor Position 0.00 2.83 6.08
Real Floor Position 0.00 3.00 6.29
% Error 0.00% 5.61% 3.38%
Table 1: Comparison of Approximated and Real Floor Positions
4 Discussion and conclusions
Sensors are not ideal and solving a task with idealised data does not necessarily provide a real-world solution.
The high-rise data was collected by an accelerometer integrated in a budget tablet. The sampling frequency was inconsistent over the data set and the data contained significant noise. This made it
particularly challenging to process, but ultimately led to a more robust software processing technique.
Using the accelerometer provided with an existing lift performance measurement tool [7] on a low-rise building, the floor positions were identified reliably with a floor position error of up to 5.6%.
Given that in a commercial building, the floor to floor height is at least three meters, these errors do not inhibit the floor positions being identified. However, in the instance of a 100-meter
express zone, a 5.6% error corresponds to 5.6 meters which is greater than a typical floor height. A more accurate sensor would be required to address this issue.
The software can be applied in lift monitoring applications, and the development of mathematical models to extrapolate passenger demand from stops [3]. It could be extended to work in three
dimensions to be applied to other positioning monitoring applications.
There is a relationship between noise and sampling frequency, i.e. it is possible to reduce noise level by lowering the sampling frequency [8]. An investigation into the optimal sampling frequency
and minimum resolution in relation to this application would be worthwhile. That is to minimise noise without compromising the identification of the key phases of the acceleration profile.
1. CIBSE, CIBSE Guide D: Transportation systems in buildings. Chartered Institution of Building Services Engineers London, 2015.
2. L. Al-Sharif, ‘New Concepts in Lift Traffic Analysis: The Inverse S-P (I-S-P) Method’, in Proceedings of the International Conference of Elevator Technology, Amsterdam, 1992.
3. R. D. Peters, ‘Lift Passenger Traffic Patterns: Applications, Current Knowledge and Measurement’. [Online]. Available: https://www.peters-research.com/index.php/support/articles-and-papers/
50-lift-passenger-traffic-patterns-applications-current-knowledge-and-measurement. [Accessed: 22-Mar-2019].
4. R. Basagoiti, M. Beamurgia, R. D. Peters, and S. Kaczmarczyk, ‘Origin Destination Matrix Estimation and Prediction in Vertical Transportation’, in Proceedings of The 2nd Symposium on Lift &
Escalator Technology (CIBSE Lifts Group and The University of Northampton), 2012.
5. R. D. Peters, ‘Ideal Lift Kinematics’. [Online]. Available: https://www.peters-research.com/index.php/support/articles-and-papers/53-ideal-lift-kinematics. [Accessed: 22-Mar-2019].
6. R. M. Asha, M. D. Kavana, S. J. Parvathy, and C. M. Shreelakshmi, ‘Object-Oriented Programming and its Concepts’, IJSRD – Int. J. Sci. Res. Dev., vol. 5, no. 09, 2017.
7. Peters Research, ‘Elevate Perform’, 2018. [Online]. Available: https://www.peters-research.com/. [Accessed: 26-Mar-2019].
8. ‘RMS noise of accelerometers and gyroscopes’, Knowledge Base. [Online]. Available: http://base.xsens.com/hc/en-us/articles/115000224125-RMS-noise-of-accelerometers-and-gyroscopes. [Accessed:
Anna Peters is a Research Assistant at Peters Research Ltd and a fourth-year engineering student studying for a MEng in Aeronautics and Astronautics at the University of Southampton. This paper is a
shortened form of her third-year dissertation, supported by Professor Alexander I J Forrester (academic supervisor).
Richard Peters has a degree in Electrical Engineering and a Doctorate for research in Vertical Transportation. He is a director of Peters Research Ltd and a Visiting Professor at the University of
Northampton. He has been awarded Fellowship of the Institution of Engineering and Technology, and of the Chartered Institution of Building Services Engineers. Dr Peters is the author of Elevate,
elevator traffic analysis and simulation software. | {"url":"https://peters-research.com/index.php/papers/logging-and-analysis-of-lift-journeys-using-an-accelerometer/","timestamp":"2024-11-14T11:57:28Z","content_type":"text/html","content_length":"92250","record_id":"<urn:uuid:8231f103-8cb4-46fe-ac05-ce6d1e054c70>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00600.warc.gz"} |
Extension Spring Calculator for Fatigue Loading
Extension spring calculator to calculate fatigue resistance of ordinary extension springs with full twisted end. The points at extension spring hook where maximum stresses occurred (tensile stress at
point A and shear stress at pont B) are shown in the following figure. In addition to these points, high shear stresses are occured at the body of the extension springs. This calculator can be used
to check these critical points fatigue resistance against dynamic loading. Gerber and Goodman failure criterias are used in this calculator for the fatigue evaluation of the spring.
For the fatigue resistance calculations of the points where shear stresses occur (body of the spring and point B at extension spring end), Zimmerli's data [Ref 2] are used to calculate shear
endurance limit value. Zimmerli's data is based on torsion in spring so it has not been used for the points where tensile stresses occur. See the "Definitions" section for more information about the
Zimmerli's Data.
Location of Maximum Bending and Torsion Stresses in Twisted Loops
For the point where maximum tensile stresses occur (at point-A) due to bending, fatigue calculations are done by using maximum allowable tensile stress value , which is entered as an input parameter
in the calculator. Values given in the "Supplement" section can be used as a reference for maximum allowable tensile stress.
The calculator is valid for dynamic loading case, un-peened spring steel material and ordinary extension spring with full twisted end as shown in the figure.
Extension Spring Design with Twisted End
For the extension spring design which works under dynamic loading, first define the design parameters with the "Dimensional Design of Extension Spring". Then use "Stress Analysis of Extension Spring
for Static Loading" calculator to check spring against yielding and use "Stress Analysis of Extension Spring for Fatigue Loading" calculator to check spring against fatigue.
Extension Spring Calculator for Fatigue Loading:
Parameter Value
Wire diameter [d]
Radius-1 [R[1]]
Radius-2 [R[2]]
Maximum working load [F[max]]
Minimum working load [F[min]]
Parameter Value
Material selection^x
Material tensile strength [S[ut]]
Allowable bending strength for the spring end for cycling loading (% of S[ut]) ^+ %
Design factor for dynamic loading [n[d]]^o ---
Note 1 : ^x Material properties are from Ref-2 except "User defined" selection.
Note 2 : ^+ See supplements for reference values.
Note 3 : ^o The design factor value that used for all of the points of interest ( Tensile stress at point-A, shear stress at point-B and shear stress at spring body).
Parameter Value
Factor of safety (According to Gerber) [fos[gerber]] ^+ --- ---
Factor of safety (Acc.to Goodman) [fos[goodman]]^+ ---
Shear stress amplitude [τ[a]] ---
Midrange shear stress [τ[m]] ---
Factor of safety @ B (According to Gerber) [fos[gerber]]^+ --- ---
Factor of safety @ B (According to Goodman) [fos[goodman]]^+ ---
Shear stress amplitude @ B [τ[a]] ---
Midrange shear stress @ B [τ[m]] ---
Factor of safety @ A (According to Gerber) [fos[gerber]] ^+ --- ---
Factor of safety @ A (According to Goodman) [fos[goodman]] ^+ ---
Tensile stress amplitude @ A [σ[a]] ---
Midrange tensile stress @ A [σ[m]] ---
Ultimate tensile strength of material [S[ut]] ---
Shearing ultimate strength [S[su]] ---
Material ASTM No. ---
Note 1 : ^+ Green color means, fos ≥ n[d], red color means fos ≤ n[d]
Design factor (nd): The ratio of failure stress to allowable stress. The design factor is what the item is required to withstand .The design factor is defined for an application (generally provided
in advance and often set by regulatory code or policy) and is not an actual calculation.
Dynamic Loading: A loading which varies with time with a number of load cycles over 10^4 and torsional stress range greater than 10 % of fatigue strength (or endurance strength) at:
• Constant torsional stress range
• Variable torsional stress range
Extension spring: Extension / tension springs are coil springs which work under tensile loading. In order to carry and transfer tensile loads, extension springs require special ends in the form of
hooks or loops. These special ends are generally produced by using the last coil of the spring or a separate component like screwed inserts. Generally, extension springs are connected to other
component via these ends. If there is a motion to extend extension spring, it exerts force to component to move it back.
Extension springs are usually coiled with an initial tension which keeps the extension spring coils closed. Due to initial tension incorporated into spring, spring can’t be extended theoretically
until a force that is greater than initial tension. In practice, extension springs extends slightly with smaller forces than initial tension due to deflection of end loops.
Tension springs are generally used to return back the component to its default position by providing return force.
Factor of Safety (Safety Factor): The ratio of failure stress to actual/expected stress. The difference between the factor of safety (safety factor) and design factor is: The factor of safety gives
the safety margin of designed part against failure. The design factor gives the requirement value for the design. Safety factor shall be greater than or equal to design factor.
Gerber fatigue criteria: A fatigue failure criteria with characteristics shown in the figure.
Goodman fatigue criteria: A fatigue failure criteria with characteristics shown in the figure.
Spring index: Ratio of spring mean diameter to coil diameter.
Spring rate: Parameter which shows relation between applied force and deflection. In other words, reaction force per unit deflection or spring resistance to length change.
Static/Quasistatic Loading: Following loading cases are defined as Static/Quasistatic loading:
• A constant loading
• A cycling loading with torsional shear stress range up to 10 % of fatigue strength (or endurance strength)
• A cycling loading with torsional shear stress range more than 10 % of fatigue strength (or endurance strength) up to 10
Zimmerli's Data: Data reported in Ref-2 about the torsional endurance limits of spring steels. According to these data, spring steel material and its tensile strength has no effect on the torsional
endurance limit for the wire size under 3/8 in (10mm). The endurace strength components for infinite life are reported as follows:
Shot Peended S[sa] S[sm]
Unpeened 35 kpsi (241 MPa) 55 kpsi (379 MPa)
Peened 57.5 kpsi (398 MPa) 77.5 kpsi (534 MPa)
Link Usage
Spring Steels for Coil Springs List of spring steel materials given in the calculator.
Formulas For Extension Spring Fatigue Design List of formulas used in the calculator.
Allowable Stresses for Extension Springs Supplemantary tables about the material strength properties of helical extension springs.
• Budynas.R , Nisbett.K . (2014) . Shigley's Mechanical Engineering Design
• F. P. Zimmerli, “Human Failures in Spring Applications,” The Mainspring, no. 17, Associated Spring Corporation, Bristol, Conn., August–September 1957.
• EN 13906-1: 2002 - Cylindrical helical springs made from round wire and bar – Calculation and design – Part 1: Compression springs | {"url":"https://amesweb.info/MechanicalSprings/StressAnalysisOfExtensionSpringForFatigueLoading.aspx","timestamp":"2024-11-02T11:05:54Z","content_type":"application/xhtml+xml","content_length":"54307","record_id":"<urn:uuid:1a252743-c21c-4dde-8f68-873715947990>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00750.warc.gz"} |
How the competitive exclusion principle can be validated using optical density measurements collected on artificially reconstituted soil ecosystems
Microbial world is very complex by its taxonomic and functional diversity and by the ubiquity of the microorganisms spread in the different ecosystems at the surface of the earth. Such ubiquity
resulted from the various colonisation strategies of microbes based on their high ability to use a wide range of substrate in different physico-chemical niches. In environmental matrices, microbial
communities are considered as a complex assemblage of a huge taxonomic and functional diversity of populations and several studies describe population dynamics as well as their spatial distribution.
According to the occurrence of particular populations consecutive to environmental perturbations basic ecological attributes were used to explain the resulting population structure. Among them the
concepts of r - and K - strategists as well as oligotrophic and /or copiotrophic were classically used for explaining the dynamics of soil populations in response to modification of substrate
availability [1-3]. However, the fitness of each population, in terms of growth rate according to different levels and types of substrate, is generally invoked as determinant but again rather unknown
and modelled. Studies evaluating precisely the fitness of bacterial populations must now be conducted to better define the ecological attributes and to model the dynamics of populations in complex
and heterogeneous environments.
Only a few studies were dedicated in the modelling of bacterial growth in the soil. Because the soil is a very complex medium, to study and model bacterial growth process, it was decided to proceed
in several steps :
• In a first step, we studied the bacterial growth process of a limited number of pure strains in a liquid medium in batch cultures;
• In a second step, a number of assemblages of these pure strains were mixed together in order to artificially reconstitute complex ecosystems;
• Finally, the same assemblages were used in a complex medium (reactors fed with sand to create heterogeneity and operated in batch mode) to study the growth process in non homogeneous environments.
The aim of the present paper is to present the modelling results obtained within the first two steps : studying and modelling the growth kinetics for soil strains in liquid media realized in both
batch pure and mixed cultures. The considered strains are Peani baccilus, Pseudomonas syringae, Xanthomonas axonopodis, Rhodococcus and Bradyrhizobium japonicum. These cultures were isolated from
bacterial soil collections. Studied strains belong to different bacterial taxonomic groups and were chosen for the variability of their growth rate in synthetic culture media. In particular,
Cupriviadus or Pseudomonas are known for their rapid growth rate whereas Rhizobiacae exhibit slow growth. Hereafter, we propose a mathematical model as an extension of the Monod’s one, validated on
experimental data, and able to describe and predict the dynamics of pure as well as mixed cultures of bacteria growing on an essential limited substrate (glucose) in reactors operating in batch mode
(Bergersen medium). The general model takes into account the following processes : bacterial growth, microbial mortality, non-viable cell accumulation in the medium and partial dead cell recycling
into substrate over the time. The least square method is used to identify the model parameters. The model is evaluated on experimental data from pure cultures and is extended to mixed cultures. For
the particular mixed consortia considered, it is established that there is a simple competition for the limited substrate. In other terms, it seems that no complex interactions between bacteria
appear in the medium.
The paper is organized as follows. In section 2, the experiment process is described and the mathematical model is proposed with a detailed description, in particular the modelling of optical density
and parameters identification. In section 3, the model is validated in pure culture and then it is extended to mixed culture proving that there is only competition for the substrate.
Material and Method
2.1 Experimental process
2.1.1 Culture medium
The substrate solution was prepared by dissolving glucose (the only one carbon source) in the medium culture to the required concentration and the pH was adjusted to 6.8 due to its high
bio-degradation efficiency. The medium culture was sterilized in an autoclave at 110 C for 40 min. Steam sterilization procedures also were applied to all equipments. Biothine and thiamine are used
after autoclaving.
2.1.2 Soil strains:
The strains isolated from the soil, stored at -80 C in test tubes containing Luria Bertani modified (LB+) medium, used in a 50% glycerol solution. They were activated at 28 C in the nutrient medium,
into which 1 g/L of glucose was added.
2.1.3 Experimental process steps: The cells collected after centrifugation were re-suspended in the medium culture without glucose and re-centrifuged (washed twice). After cleaning, the activated
cells were inoculated into culture medium. The temperature was controlled at 28 C, and the optical density of the cell suspensions was measured at 600 nm, where the culture were automatically
continuously shaken at 150 rpm. The experiments were carried out in triplicate.
The biodegradation data of glucose, for an initial glucose concentration (So) of about 1 g/L, are measured with the correspondent optical density. Glucose concentrations were measured on-line using
an enzymatic proportioning by glucose oxidase. In this method, the D-glucose is oxidised by glucose oxidase (GOD) to give gluconic acid and hydrogen peroxide. Hydrogen peroxide reacts with
O-dianisidine in the presence of peroxidase to form a colored product. Oxidized o-dianisidine reacts with sulfuric acid to form a more stable colored product. The intensity of the pink color measured
at 450 nm is proportional to the original glucose concentration. The data were analyzed using Scilab and calculating the averages of the triplicates for all glucose concentration and for each
inocula. For different values of glucose concentration (0.1, 0.2, ..., 10 g/L), the averages were used to generate the growth curves, constructed as a function of the incubation time and the
absorbancy of the culture medium. For each strain, the duration of the lag phase is the same for all glucose concentration. The maximum optical density increases coupled with an increase of the
initial glucose concentration so.
The lag time variable depends on the condition of the innoculum [4] and it is easily obtained from the experimental data.
2.2 Mathematical model: an extension of the Monod’s one
In this section, we describe the proposed mathematical model which takes into account viable cell (χ) growth, substrate (s) consumption, non-viable cell (χd)accumulation in the medium and partial
dead cell recycling into substrate.
2.2.1 Specific growth rate: Under fixed environment conditions, the cell growth kinetics is given by
$x ˙ =μ(s)x−mx, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where μ is the specific growth rate of viable cell and m is the natural mortality rate.
The majority of kinetic models describing microbial growth are empirical and based on either Monod’s equation. Let
$μ(s)= μ max s k s +s , MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where µ[max] is the maximum growth rate and k[s] is the saturation constant.
2.2.2 Dead cell accumulation: In general, bacterial growth is monitored using optical density measurements which take into account viable and also non-viable cell that accumulated in the medium. The
non-viable cell accumulation, in the medium, has the following kinetics :
$x ˙ d=δmx, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where constant is the part of inactive cells that are not burst.
2.2.3 Substrate degradation: Substrate consumption depends on instantaneous viable cell growth rate µ and partial dead cell recycling into substrate with recycling conversion factor λ, then substrate
consumptions kinetics is given as follows :
$s ˙ =− μ(s) Y x+λ(1−δ)mx, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
such that $1 Y >λ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeWaaSaaa8aabaWdbiaaigdaa8aabaWdbiaadMfaaaGaeyOpa4Jaeq4UdWgaaa@40E6@$ which stones the point that substrate utilization yield coefficient $1 Y
OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeWaaSaaa8aabaWdbiaaigdaa8aabaWdbiaadMfaaaaaaa@3E2A@$ is greater than the substrate
recycling yield coefficient $λ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
3j0xXdbba91rFfpec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaeq4UdWgaaa@3DF7@$ .
2.2.4 Complete mathematical model: The proposed mathematical model consists of a set of ordinary differential equations taking into account bacterial growth, substrate utilization, bacterial
mortality, non-viable cell accumulation in the medium, and partial dead cell recycling into substrate with time.
${ s =− μ(s) Y x+λ(1−δ)mx, x ˙ =μ(s)x−mx, x ˙ d =δmx. (1) MathType@MTEF@5@5@+=
such that
$δ≤1 and λ< 1 Y . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
Assume that for t=0, there is no non-viable cell, then the following initial conditions :
$s(0)= s 0 >0, x d (0)=0 and x(0)= x 0 >0. MathType@MTEF@5@5@+=
2.3 Available data
2.3.1 Optical density and substrate measurments: The growth of innocula on different glucose concentrations was monitored by manual optical density measurements in a spectrometer at 600 nm, where the
Beer-Lambert law is applied by meaning that a linear relationship between optical density and concentration of species.
$O.D. =ε d C MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where ε is the wavelength-dependent molar absorptivity coefficient, d is the path length, and C is the species concentration. We verified, by dilution, the Beer-Lambert Law, showing that absorbance
is a direct function of concentration of viable and non-viable cells in the culture. Then optical density is a linear combination of viable and non-viable cell concentrations :
$z= O.D. = γ 1 x+ γ 2 x d optical density units , MathType@MTEF@5@5@+=
where $γ 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
and $γ 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
are respectively specific absorptivity coefficients for viable and non-viable cells. One can also measure on-line substrate concentration. Then available measurements, γ, are the substrate
concentration and the optical density, modelled as a linear combination of viable and non-viable cells concentrations
$y=( s z ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
2.3.2 Growth rate estimation: We suppose that, initially, there was no dead cell, and that at exponential phase, the natural mortality is negligible. We calculated, for each glucose concentration,
the regression coefficient at exponential phase, we obtained growth rate of all strains as functions of glucose concentrations. The best-fit results based on the Monod equation were obtained using
the “leastsq” routine available in Scilab software are presented hereafter (Figure 1).
The growth rate parameters are given hereafter (Table 1).
3.1 Pure cultures
3.1.1 Parameter identification: The least squares method is used to identify model parameters by minimizing the following criterion :
$J= σ 1 2 ∑ i=1 n s ( t i ) 2 − s e ( t i ) ) 2 + σ 2 2 ∑ i=1 n z e ( t i )− z e ( t i ) ) 2 MathType@MTEF@5@5@+=
where $t i MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
3j0xXdbba91rFfpec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamiDa8aadaWgaaWcbaWdbiaadMgaa8aabeaaaaa@3E84@$ is
the time variable, $s e MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
3j0xXdbba91rFfpec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaam4Ca8aadaWgaaWcbaWdbiaadwgaa8aabeaaaaa@3E7F@$ is
the substrate concentration and $z e MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
3j0xXdbba91rFfpec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamOEa8aadaWgaaWcbaWdbiaadwgaa8aabeaaaaa@3E86@$ is
the optical density. $s( t i ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaam4CaiaacIcacaWG0bWdamaaBaaaleaapeGaamyAaaWdaeqaaOWdbiaacMcaaaa@40EF@$ and $z( t i ) MathType@MTEF@5@5@+=
JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamOEaiaacIcacaWG0bWdamaaBaaaleaapeGaamyAaaWdaeqaaOWdbiaacMcaaaa@40F6@$ are the substrate and the
optical density simulated using the model at time $t i ,i=1,...,n MathType@MTEF@5@5@+=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamiDa8aadaWgaaWcbaWdbiaadMgaa8aabeaak8qacaGGSaGaamyAaiabg2da9iaaigdacaGGSaGaaiOlaiaac6cacaGGUaGaaiilaiaad6gaaaa@4666@$ while $σ 1
OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaeq4Wdm3damaaBaaaleaapeGaaGymaaWdaeqaaaaa@3F1B@$ and $σ 2 MathType@MTEF@5@5@+=
JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaeq4Wdm3damaaBaaaleaapeGaaGOmaaWdaeqaaaaa@3F1C@$ are weighting coefficients. A good fitting is observed
in spite of noises present in available data (Figure 2, Table 2).
3.2 Mixed cultures
3.2.1 Extended model: In the present paragraph, the model is modified for the case of many cells in the same culture. For $i=1,...,n, x i MathType@MTEF@5@5@+=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamyAaiabg2da9iaaigdacaGGSaGaaiOlaiaac6cacaGGUaGaaiilaiaad6gacaGGSaGaamiEa8aadaWgaaWcbaWdbiaadMgaa8aabeaaaaa@4650@$ denotes the th viable cell
concentration, $x di MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaamiEa8aadaWgaaWcbaWdbiaadsgacaWGPbaapaqabaaaaa@3F71@$ the ith non-viable cell concentration and s the substrate concentration. The mathematical
model for n species is described by the following equation based on model (1) :
${ s ˙ =− ∑ i=1 n μ i (s) Y i x i + ∑ i=1 n λ i (1− δ i ) m i x i x ˙ i =( μ i (s)− m i ) x i x ˙ di = δ i m i x i MathType@MTEF@5@5@+=
Such that
$1 Y i > λ i , i=1...n. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
Assume that for t=0, there is no non-viable cells, then the following initial conditions :
$s(0)= s 0 >0, x di (0)=0, x i (0)= x i0 >0 MathType@MTEF@5@5@+=
$μ i (s)= μ maxi s k s i +s , i=1,n MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where $μ maxi MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaeqiVd02damaaBaaaleaapeGaamyBaiaadggacaWG4bGaamyAaaWdaeqaaaaa@4216@$ and $k si MathType@MTEF@5@5@+=
JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaam4Aa8aadaWgaaWcbaWdbiaadohacaWGPbaapaqabaaaaa@3F73@$ are respectively the maximum growth rate and the
saturation constant for the th species. The parameters $Y i , λ i , δ i , m i , μ maxi MathType@MTEF@5@5@+=
and $k si MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
vqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaam4Aa8aadaWgaaWcbaWdbiaadohacaWGPbaapaqabaaaaa@3F73@$ are identified at batch culture and are given in Paragraphs 2.3.2 and 3.1.1.
3.2.2 Optical density and substrate measurments: Available measurements are the substrate concentration and the optical density modelled as a linear combination of viable and non-viable cells
concentrations of all strains in the culture and it is defined hereafter
$z= ∑ i=1 n ( γ 1i x i + γ 2i x di ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbnvMCYL2DLfgDOvMCaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhis9wBH5garqqtubsr4rNCHbGeaGakY=
where the models parameters andare the absorptivity coefficients and are identified at batch culture and are given in paragraph 3.1.1 (Figure 3).
Discussion and Conclusion
Models of batch culture focus most of the time on the exponential growth phase (and its previous lag phase), when the substrate is non limiting. Cell mortality is usually neglected during these
phases. Here we show that mortality cannot be neglected when one wants to study the growth after the exponential growth, i.e. when the substrate becomes limiting (as it is most probably the case in
true soil ecosystems). Then, the model is close from continuous culture or chemostat ones, mortality playing the role of the dilution rate.
For chemostat models, many works have been achieved about the Competitive Exclusion Principle (CEP), stating that at most one species can survive on the long term [5-15]. Some of them tend to show
that it is not valid in natural ecosystems (see for instance the [16]. On the opposite, other ones tend to show that it is valid in perfectly controlled environment [17]. One of the crucial questions
in ecology is to decide if the invalidity of the CEP is due to intrinsic interactions between species or interactions with their environment. In the first case, the usual chemostat model on which the
CEP is based is invalidated, and one has to take into account additional interaction terms. In the second case, one may conclude that the domain of validity of the model is not met but cannot a
priory reject the model. Our experiments have been conducted in a framework similar to the the one driven in [17] and consequently tend to show that the CEP is also valid for the particular
artificial consortia we have considered. Although some studies point out the role that could play complex interactions in the functioning and performances of simple artificial ecosystems (under quite
the same conditions than ours), it is expected that bacteria of an artificial consortia - do not develop such a network of complex interactions. Let us also point out that all the species we have
considered present quite different break-even concentrations. It is well known that the prediction of CEP requires more time to be observed when species have close break-even concentrations [12], as
this may be the case when considering ecosystems with a huge number of different species in non negligible quantities [18-20]. We have avoided here such situations.
The Bioscreen machine automated bacterial growth curves by measuring the optical density. Their values are linear to the bacterial concentration values at exponential phase but it can’t describe the
behaviour on the other phases. We proposed a mathematical model, validated on experimental data aiming at describing and predicting microbial growth on an essential limited substrate in batch pure
cultures in revisiting the way where the optical density is modelled. This model takes into account viable cell growth, substrate consumption, cell mortality, non-viable cell accumulation in the
culture medium and partial dead cell recycling into substrate. This model uses the optical density poor information at exponential phase to predict the behaviour for all phases. The least squares
method is used to identify the model parameters. The system is then extended and validated on mixed cultures proving that there is only competition for the substrate which goes a little against the
current thought relative to the complexity of the biological systems. Perspectives include co-cultures modelling in non-homogeneous media. | {"url":"https://www.agriscigroup.us/articles/OJEB-4-109.php","timestamp":"2024-11-12T00:14:08Z","content_type":"text/html","content_length":"110961","record_id":"<urn:uuid:9d652cc6-5997-41c9-ac3a-edf75fe1cdc4>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00623.warc.gz"} |
Convert Inch per minute (ipm) (Velocity)
Convert Inch per minute (ipm)
Direct link to this calculator:
Convert Inch per minute (ipm) (Velocity)
1. Choose the right category from the selection list, in this case 'Velocity'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and
π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Inch per minute [ipm]'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '265 Inch per minute'. In so doing, either the full name of the unit
or its abbreviation can be usedas an example, either 'Inch per minute' or 'ipm'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case
'Velocity'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought.
Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of
that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(70 * 22) ipm'. But different
units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '67 Inch per minute + 19 Inch per minute' or '73mm x 25cm x 76dm = ? cm^
3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.168 215 298 011 2×1022. For this form of presentation, the number
will be segmented into an exponent, here 22, and the actual number, here 1.168 215 298 011 2. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket
calculators, one also finds the way of writing numbers as 1.168 215 298 011 2E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at
this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 11 682 152 980 112 000 000 000. Independent of the presentation of
the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | {"url":"https://www.convert-measurement-units.com/convert+Inch+per+minute.php","timestamp":"2024-11-08T12:29:31Z","content_type":"text/html","content_length":"54644","record_id":"<urn:uuid:8d79b36d-16dc-4e94-8739-07a03dfc7f33>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00626.warc.gz"} |
1+1=5 A scavenger hunt game about unitizing - Natural Math
1+1=5 A scavenger hunt game about unitizing
Yelena recently reviewed the book “1+1=5” and shared the game of “I spy” she plays with her son.
The book inspires kids (and adults) to see everyday objects as sets, or collections of other objects. For example, a triangle can be viewed as a set of 3 sides while a rectangle is a set of 4
sides. An octopus is an example of a set of 8 (arms) while a starfish hides a set of 5 (arms) in plain sight. If one set has 8 elements and another set has 5 elements, then when added, the two
sets have 13 elements total. Hooray!
I thought it could be fun to invite readers of this blog to play a round of the game. Here is the big question I am contemplating: “How can we make our descriptions of games we design so interactive
that they become, literally, playable games?”
Add your own example! Of course, this is the ocean, for our Moby Snoodles.
This is what people added so far! It takes about five minutes for your answer to appear here. Wait and then reload the page to see!
[…] Moby Snoodles plays with iconic numbers in 1+1=2 but Mostly it Doesn’t, and your children can join in by contributing to A scavenger hunt game about unitizing. […] | {"url":"https://naturalmath.com/2013/01/1plus1is5game/","timestamp":"2024-11-03T22:04:21Z","content_type":"text/html","content_length":"310191","record_id":"<urn:uuid:f0e812ed-a58e-46bb-9d87-c3ca0cf9b390>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00262.warc.gz"} |
Concrete Technology (121-140)
121.According to IS: 456-1978, the maximum strain in concrete at the outermost
compression fibre in the limit state design of flexural member is 0.0035
122.The minimum eccentricity for design of columns as per IS : 456-1978 is
subjected to a maximum of 20 mm. where l = unsupported length of the
123. For the deflection of simply supported beam to be within permissible limits, the ratio of its span to effective depth as per IS : 456-1978 should not exceed
124. The development length of bars of diameter Φ, as per IS : 456-1978 is
125.For bars in tension, a standard hook has an anchorage value equivalent to a straight length of 16Φ
126.The creep strains are caused due to dead loads only
127. The effect of creep on modular ratio is to increase it
128. Shrinkage of concrete depends upon i) humidity of atmosphere
129. Due to shrinkage stresses, a simply supported beam having reinforcement
only at bottom tends to deflect downward
130. In symmetrically reinforced sections, shrinkage stresses in concrete and steel are respectively tensile and compressive
131. A beam curved in plan is designed for bending moment, shear and torsion
132. In a spherical dome subjected to concentrated load at crown or uniformly
distributed load, the meridional force is always compressive
133. Sinking of an intermediate support of a continuous beam
i) reduces the negative moment at support
ii) increases the positive moment at centre of span
134. The maximum value of hoop compression in a dome is given by
135. In a spherical dome the hoop stress due to a concentrated load at crown is
136. In a ring beam subjected to uniformly distributed load
i) shear force at mid span is zero
ii) torsion at mid span is zero
137. In prestressed concrete forces of tension and compressions
remain unchanged but lever arm changes with the moment
138. The purpose of reinforcement in prestressed concrete is
to impart initial compressive stress in concrete
139. Normally prestressing wires are arranged in the lower part of the beam
140. Most common method of prestressing used for factory production is Long line method | {"url":"https://www.sudhanshucivil2010.com/post/concrete_7","timestamp":"2024-11-04T02:20:22Z","content_type":"text/html","content_length":"1050497","record_id":"<urn:uuid:81efd1a7-d2e6-4329-9b3e-80433226c515>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00290.warc.gz"} |
Ray-Casting & Ray-Tracing with VTK
VTK has long evolved beyond just visualization. It offers some amazing functionality that just cannot be found elsewhere. Two examples are the ‘ray-casting’ and, consequentially, ‘ray-tracing’
capabilities provided by the vtkOBBTree class. In this article, I would like to introduce these capabilities and show examples of ray-casting and ray-tracing performed exclusively through Python, a
dash of NumPy, and VTK.
Disclaimer: The ray-casting and ray-tracing examples I will be presenting here are severely condensed versions of my posts “Ray Casting with Python and VTK: Intersecting lines/rays with surface
meshes” and “From Ray Casting to Ray Tracing with Python and VTK” that appear on my blog [1]. If they pique your interest, please visit the aforementioned posts, where you can find all of the
material and code (in the form of IPython Notebooks), as well as an excruciating amount of detail pertaining to each aspect of the process, as these posts were written for people with little to no
experience in VTK.
Ray-Casting vs. Ray-Tracing
I would like to emphasize a pivotal difference between ‘ray-casting’ and ‘ray-tracing.’ In the case of the former, we only ‘cast’ a single ray, test for its intersection with objects, and retrieve
information regarding the intersection. Ray-tracing, on the other hand, is more physically accurate, as it applies laws of physics (e.g., reflection, refraction, attenuation, etc.) to the rays to
‘trace’ (i.e., follow) that ray and its derivative rays. However, ray-casting is the natural precursor to ray-tracing, as it tells us with what part of which object the ray intersects and provides
all necessary information to cast subsequent rays.
The vtkOBBTree Class
The star of this post is the vtkOBBTree class, which generates an oriented bounding-box (OBB) ‘tree’ for a given geometry under a vtkPolyData object. Upon generation of this OBB tree, the vtkOBBTree
class allows us to perform intersection tests between the mesh and the lines of finite length, as well as intersection tests between different meshes. It can then return the point coordinates where
intersections were detected, as well as the polydata cell IDs where the intersections occurred.
Ray-Casting with vtkOBBTree
For this demonstration, we are assuming that we have a surface model of a human skull stored in a .stl file, whose contents we have loaded into a vtkPolyData object, named mesh, through the
vtkSTLReader class. A rendering of this model through the vtkPolyDataMapper class can be seen in Figure 1.
Figure 1. Rendering of the surface model of a human skull, which we will use to demonstrate ray-casting.
The center of the skull is centered around the cartesian origin, i.e., the (0.0, 0.0, 0.0) point. Now let’s assume we want to cast a ray emanating from (100.0, 100.0, 0.0) and ending at (0.0, 0.0,
0.0) and retrieve the coordinates of the points where this ray intersects with the skull’s surface. A rendering including the ray, prior to actually casting it, can be seen in Figure 2.
Figure 2. The ray that will be tested for intersection with the skull model. The ‘source’ point of the ray is rendered as red, while the ‘target’ point is rendered as green.
Prior to intersection, we need to create and initialize a new vtkOBBTree with the vtkPolyData object of our choice. In our case, this is called mesh and is done as follows:
obbTree = vtk.vtkOBBTree()
Note the call to the BuildLocator method, which creates the OBB tree.
That’s it! We now have a world-class intersection tester at our disposal. At this point, we can use the
IntersectWithLine method of the vtkOBBTree class to test for intersection with the aforementioned ray. We merely need to create a vtkPoints object and a vtkIdList object to store the results of the
intersection test.
points = vtk.vtkPoints()
cellIds = vtk.vtkIdList()
code = obbTree.IntersectWithLine((100.0, 100.0, 0.0), (0.0, 0.0, 0.0), points, cellIds)
As mentioned above, the points and cellIds now contain the point coordinates and cell IDs with which the ray intersected as it was emanated from the first point, i.e., (100.0, 100.0, 0.0), onto the
second point, i.e., (0.0, 0.0, 0.0), in the order they were ‘encountered.’ The return value code is an integer, which would be equal to 0 if no intersections were found. A rendering showing the
intersection points can be seen in Figure 3.
Figure 3. Result of the ray-casting operation and intersection test between the skull model and the ray. The blue points depict the detected intersection points.
The Python package pycaster [2] wraps the functionality shown above, assuming no VTK experience, and provides additional methods to calculate the distance a given ray has ‘traveled’ within a closed
surface. It is currently being served through PyPI. The repository [3] can be found on BitBucket.
As I mentioned at the beginning of this article, the entire process shown above, including all of the material and code needed to reproduce it, is detailed in my post “Ray Casting with Python and
VTK: Intersecting lines/rays with surface meshes” [4].
Ray-Tracing with vtkOBBTree
Now, in order to perform ray-tracing, we can take the lessons learned from ray-casting and apply them to a more convoluted scenario. The rationale behind ray-tracing with the vtkOBBTree class is the
• Cast rays from every ‘ray source’ and test for their intersection with every ‘target’ mesh in the scene.
• If a given ray intersects with a given ‘target,’ then use the intersection points and intersected cells to calculate the normal at that cell, to calculate the vectors of the reflected/refracted
rays, and to cast subsequent rays off the target.
• An excellent, freely available article entitled “Reflections and Refractions in Ray-Tracing” [5] (Bram de Greve, 2006) provides a good overview of the math and the physics behind ray-tracing.
• Repeat this process for every ray cast from the ‘ray source.’
Let’s assume a scene is comprised of a half-sphere dubbed sun, which will act as the ray-source, and a larger nicely textured sphere called earth, which will be the target of those rays. This
‘environment’ can be seen in Figure 4.
Figure 4. Scene defined for the ray-tracing example. The yellow half-sphere, sun, acts as the ray-source, while the textured sphere, earth, will be the target of those rays.
In this example, we will be casting a ray from the center of each triangular cell on the sun’s surface along the direction of that cell’s normal vector. The cell-centers of the sun were calculated
through the vtkCellCenters class and stored under pointsCellCentersSun (of type vtkPolyData). The cell-normals of the sun were calculated through the vtkPolyDataNormals class and stored under
normalsSun (of type vtkFloatArray). A rendering of the cell-centers as points and cell-normals as glyphs through the vtkGlyph3D class can be seen in Figure 5.
Figure 5. Rendering of the cell-centers of the sun that will act as source-points for the rays and the cell-normals along which the rays will be cast.
Similarly to what was done in the previous example, prior to ray-casting, we first need to create a vtkOBBTree object for the earth:
obbEarth = vtk.vtkOBBTree()
Now, since we will be casting a large number of rays, let’s wrap the vtkOBBTree functionality in two convenient
def isHit(obbTree, pSource, pTarget):
code = obbTree.IntersectWithLine(pSource, pTarget, None, None)
if code==0:
return False
return True
def GetIntersect(obbTree, pSource, pTarget):
points = vtk.vtkPoints()
cellIds = vtk.vtkIdList()
# Perform intersection test
code = obbTree.IntersectWithLine(pSource, pTarget, points, cellIds)
pointData = points.GetData()
noPoints = pointData.GetNumberOfTuples()
noIds = cellIds.GetNumberOfIds()
pointsInter = []
cellIdsInter = []
for idx in range(noPoints):
return pointsInter, cellIdsInter
The isHit function will return True or False, depending on whether a given ray intersects with obbTree, which, in our case, will only be obbEarth.
The GetIntersect function simply wraps the functionality we saw in the first example. In a nutshell, it will return two list objects: pointsInter and cellIdsInter. The former will contain a series of
tuple objects with the coordinates of the intersection points. The latter will contain the ‘id’ of the mesh cells that were ‘hit’ by that ray. This information is vital, as we will be able to get the
correct normal vector for that earth cell and calculate the appropriate reflected vector through these ids, which we will see below.
At this point, we are ready to perform the ray-tracing. Let’s take a look at a condensed version of the code:
noPoints = pointsCellCentersSun.GetNumberOfPoints()
for idx in range(noPoints):
pointSun = pointsCellCentersSun.GetPoint(idx)
normalSun = normalsSun.GetTuple(idx)
# Calculate the 'target' of the 'sun' ray based
# on 'RayCastLength'
pointRayTarget = list(numpy.array(pointSun) + RayCastLength*numpy.array(normalSun))
if isHit(obbEarth, pointSun, pointRayTarget):
pointsInter, cellIdsInter = GetIntersect(obbEarth, pointSun, pointRayTarget)
# Get the normal vector at the earth cell
# that intersected with the ray
normalEarth = normalsEarth.GetTuple(cellIdsInter[0])
# Calculate the incident ray vector
vecInc = (numpy.array(pointRayTarget) - numpy.array(pointSun))
# Calculate the reflected ray vector
vecRef = (vecInc - 2*numpy.dot(vecInc, numpy.array(normalEarth)) * numpy.array(normalEarth))
# Calculate the 'target' of the reflected
ray based on 'RayCastLength'
pointRayReflectedTarget = (numpy.array(pointsInter[0]) + RayCastLength*l2n(vecRef))
Please note that all rendering code was removed from the above snippet. What is done above is the following:
We looped through every cell-center on the sun mesh (pointSun), stored under pointsCellCentersSun.
We casted a ray along the direction of that sun cell’s normal vector, stored under normalsSun. The ray emanates from pointSun to pointRayTarget.
As the vtkOBBTree class only allows for intersection tests with lines of finite length, not semi-infinite rays, the rays cast in the code above are given a large (relative to the scene) length to
ensure that failure to intersect with earth would only be due to the ray’s direction and not an insufficient length.
Every ray was tested for intersection with the earth through the isHit function and the obbEarth object defined above.
If a ray intersected with the earth, the intersection test was repeated through the GetIntersect function in order to retrieve the intersection point coordinates and the intersected cell IDs on the
earth mesh. The intersection point coordinates and the intersected earth cell normal vector (normalEarth) were used to calculate the normal vector of the reflected ray and cast that ray off the
earth’s surface.
The cell normals on the earth’s surface were calculated through the vtkPolyDataNormals class and stored under
normalsEarth (of type vtkFloatArray). This is the same way that normalSun was calculated.
A render of the ray-tracing result can be seen in Figure 6.
As I mentioned at the beginning of this article, the entire process shown above, including all of the material and code needed to reproduce it, is detailed in my post “From Ray Casting to Ray Tracing
with Python and VTK” [6].
Figure 6. Result of the ray-tracing example. Rays cast from the sun that missed the earth are rendered as white. Rays that intersected with the earth are rendered as yellow. The intersection points,
normal vectors at the intersected earth cells, and the reflected rays can also be seen.
As you can see, VTK provides some little-known pearls that offer fantastic functionality.
While the above examples fall short of real-world ray-tracing applications, as one would need to account for effects like refraction and energy attenuation, the sky is the limit!
[1] http://pyscience.wordpress.com
[2] https://pypi.python.org/pypi/pycaster
[3] https://bitbucket.org/somada141/pycaster
[4] http://pyscience.wordpress.com/2014/09/21/ray-casting-with-python-and-vtk-intersecting-linesrays-with-surface-meshes/
[5] http://graphics.stanford.edu/courses/cs148-10-summer/docs/2006–degreve–reflection_refraction.pdf
[6] http://pyscience.wordpress.com/2014/10/05/from-ray-casting-to-ray-tracing-with-python-and-vtk/
Adamos Kyriakou is an Electrical & Computer Engineer with an MSc. in Telecommunications and a Ph.D. in Biomedical Engineering. He is currently working as a Research Associate in Computational
Multiphysics at the IT’IS Foundation (ETH Zurich), where his work and research are primarily focused on computational algorithm development, high-performance computing, multi-physics simulations,
big-data analysis, and medical imaging/therapy modalities.
3 comments to Ray-Casting & Ray-Tracing with VTK
1. The application is to Raycast a line unto a rectilinear voxelized 3D volume to determine the intersected cell IDs and the distance between crossings. I used vtkOBBTree as explained here. For the
mesh, I used a mesh exported from Rhino to an stl file and read the stl file (circuitous process). Q1: Do you know of a way to create a mesh from vtkImageData or a rectilinear grid defined in a
different way?
The actual app is to also compute the voxel ID enclosed by 6 cells. For example, this would be used to compute the light attenuation when crossing a voxelized 3D volume with different opacity
values. I was able to extract from the cell ID the voxel ID, but the algo is very slow. Q2: Do you know of a fast method to do this, better yet, a method that produces voxelID and distance
traversed through the voxel? Thank you.
1. If you found an answer to Q2 I’d be very much appreciate if you share, thank you!
2. The answer to Q1 is to use vtkMarchingCubes or vtkFlyingEdges, which is just and optimized version of the marching cubes algorithm. If you have a 3 dimensional vtkImageData that is sdf or
tsdf, it’s pretty straight forward to use those two classes. | {"url":"https://www.kitware.com/ray-casting-ray-tracing-with-vtk/","timestamp":"2024-11-02T17:01:52Z","content_type":"text/html","content_length":"116101","record_id":"<urn:uuid:92069cda-d932-4fa9-891e-4e7aaeeb559a>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00394.warc.gz"} |
module type HashedType = sig .. end
type t
The type of the hashtable keys.
val equal : t -> t -> bool
The equality predicate used to compare keys.
val hash : t -> int
A hashing function on keys. It must be such that if two keys are equal according to
, then they have identical hash values as computed by
. Examples: suitable (
) pairs for arbitrary key types include
• ((=), Hashtbl.hash) for comparing objects by structure (provided objects do not contain floats)
• ((fun x y -> compare x y = 0), Hashtbl.hash) for comparing objects by structure and handling nan correctly
• ((==), Hashtbl.hash) for comparing objects by physical equality (e.g. for mutable or cyclic objects). | {"url":"https://www.cs.cornell.edu/courses/cs3110/2016fa/htmlman/libref/Hashtbl.HashedType.html","timestamp":"2024-11-08T23:30:53Z","content_type":"text/html","content_length":"9679","record_id":"<urn:uuid:8952e7f2-09fe-49e3-8f5c-c55d822b9448>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00394.warc.gz"} |
All About calculate size of circular water tank and its capacity
Water tanks are essential for storing and supplying water for various purposes such as domestic use, agriculture, and industrial applications. Among the various types of water tanks, circular water
tanks are widely used due to their simple and efficient design. However, determining the size and capacity of a circular water tank can be a daunting task, especially for individuals who do not have
a background in engineering or mathematics. In this article, we will explore the fundamentals of calculating the size and capacity of a circular water tank, including the necessary formulas and
considerations. Understanding the process of sizing a circular water tank can help individuals make informed decisions when it comes to choosing the right tank for their water storage needs.
How to calculate size of circular water tank and its capacity
A circular water tank is a common type of storage tank used for domestic and industrial purposes. It is designed to store water for later use and ensure a constant supply of water. In order to design
a circular water tank, its size and capacity must be calculated accurately. The size and capacity of a circular water tank depend on various factors such as the volume of water required, the purpose
of use, and the available space for construction. In this article, we will discuss how to calculate the size and capacity of a circular water tank.
Calculating the Size of a Circular Water Tank:
Step 1: Determine the Required Volume of Water
The first step in calculating the size of a circular water tank is to determine the required volume of water. This can be done by considering the purpose of the tank, the number of people using the
water, and the daily usage per person. For domestic use, the average daily water usage per person is around 150 liters. For industrial use, the required volume of water is calculated based on the
production or process requirements.
Step 2: Determine the Storage Capacity of the Tank
The storage capacity of the tank is the maximum amount of water that the tank can hold. It is calculated by multiplying the required volume of water by the tank’s storage factor. The storage factor
is the percentage of the tank’s volume that can be used to store water. The general accepted value for the storage factor is 0.9. Therefore, the storage capacity of the tank can be calculated as
Storage capacity = Required volume of water x Storage factor
Step 3: Determine the Diameter of the Tank
The diameter of the tank is the distance from one side to the other side of the circular tank. It can be calculated using the following formula:
Diameter = √(4 x Storage capacity / 3.14)
Step 4: Determine the Height of the Tank
The height of the tank is the distance from the bottom to the top of the tank. It can be calculated by dividing the storage capacity of the tank by its base area, which is the area of the circular
base of the tank. The formula for calculating the height of the tank is as follows:
Height = Storage capacity / Base area
Base area = π x (Diameter / 2)^2
Calculating the Capacity of a Circular Water Tank:
Once the size of the tank is determined, the capacity of the tank can be calculated. The capacity of a circular water tank is the amount of water it can hold when it is full. It is calculated using
the following formula:
Capacity = π x (Diameter / 2)^2 x Height
Let us consider a situation where a circular water tank is required for domestic use. The average daily water usage per person is 150 liters, and the number of people using the water is 4. The
required volume of water would be 150 liters x 4 people = 600 liters.
Storage capacity = 600 liters x 0.9 (storage factor) = 540 liters
Diameter = √(4 x 540 liters / 3.14) = 26.06 feet
Height = 540 liters / (3.14 x (26.06 feet / 2)^2) = 4 feet
Capacity = 3.14 x (26.06 feet / 2)^2 x 4 feet = 1,068.24 liters or
Calculation of size of circular water tank having water capacity is 2000 litre
When designing a water tank, it is important to consider the size and capacity of the tank to ensure it meets the needs of the intended usage. In this example, we will calculate the size of a
circular water tank with a water capacity of 2000 litres.
Step 1: Determine the desired depth of the water tank
The first step in calculating the size of a water tank is to determine the desired depth of the tank. This will depend on the intended use of the tank and the amount of water needed. For our example,
we will assume a desired depth of 2 meters.
Step 2: Calculate the area of the tank base
Since we are dealing with a circular water tank, the area can be calculated using the formula A=πr^2, where A is the area and r is the radius of the tank. To determine the area in square meters, we
need to convert the depth in meters into centimeters and multiply it by 100. In our example, the calculation would be as follows:
A=πr^2=π(100)^2=31,416 square centimeters or 3.14 square meters.
Step 3: Calculate the circumference of the tank base
Next, we need to calculate the circumference of the tank base using the formula C=2πr, where C is the circumference and r is the radius of the tank. In our example, the calculation would be as
C=2πr=2π(100)=628.32 centimeters.
Step 4: Calculate the height of the tank
To determine the height of the tank, we can use the formula h=V/A, where h is the height, V is the volume of the tank, and A is the area of the tank base. In our example, the calculation would be as
h=V/A=2000 litres/3.14 square meters=636.94 centimeters or 6.37 meters.
Step 5: Calculate the total height of the tank
Since we have already determined the height of the water, we need to add the height of the tank base to get the total height of the tank. In our example, the calculation would be:
Total height=height of the tank+height of the tank base=6.37 meters+2 meters=8.37 meters.
Therefore, the required size of the circular water tank with a water capacity of 2000 litres and a desired depth of 2 meters would be 8.37 meters in height and have a base area of 3.14 square meters
with a circumference of 628.32 centimeters.
It is important to note that this is a simplified calculation and other factors such as the material of the tank and structural elements should also be considered when designing a water tank.
Consulting a licensed engineer is recommended to ensure the tank is designed and constructed properly.
Calculation of size of circular water tank having water capacity is 10000 litre
The size of a circular water tank is an important factor to consider in civil engineering as it directly affects the amount of water it can hold and the structural stability of the tank. In this
article, we will discuss the calculation of the size of a circular water tank with a water capacity of 10000 litres.
Step 1: Determine the Volume of the Tank
The first step in calculating the size of a circular water tank is to determine its volume. The volume of a cylindrical tank can be calculated by using the formula V = πr²h, where V is the volume, π
is the mathematical constant of pi (3.14), r is the radius of the tank, and h is the height of the tank.
Since we are given that the water capacity is 10000 litres, we can convert it to cubic meters (m³) by dividing it by 1000. Therefore, the volume of the tank would be V = (10000/1000) m³ = 10 m³.
Step 2: Determine the Height of the Tank
Now that we know the volume of the tank, we can use the same formula to determine the height of the tank. Rearranging the formula, we get h = V / (πr²).
Using the value of V as 10 m³ and assuming a standard value of r as 1 meter, we get:
h = 10 / (3.14 x 1²) = 3.18 meters
Therefore, the height of the tank should be 3.18 meters.
Step 3: Determine the Diameter of the Tank
To determine the diameter of the tank, we need to use the formula D = 2r, where D is the diameter and r is the radius of the tank.
Using the value of r as 1 meter, we get:
D = 2 x 1 = 2 meters.
Therefore, the diameter of the tank should be 2 meters.
Step 4: Check for Safety Factor and Additions
In this step, we need to ensure that the tank is safe and stable enough to hold the desired capacity of water. We can achieve this by considering a safety factor of 1.2, which is commonly used in
civil engineering.
To do so, we can multiply the dimensions of the tank (height and diameter) by 1.2.
Hence, the final dimensions for the tank would be a height of 3.18 x 1.2 = 3.8 meters and a diameter of 2 x 1.2 = 2.4 meters.
In addition, we also need to consider the thickness of the tank walls. For concrete tanks, a thickness of 150mm is generally used, while for steel tanks, it is 3mm.
Step 5: Calculate the Actual Volume
To calculate the actual volume of the tank, we need to take into account the thickness of the walls. Using the formula V = πr²(h+2t), where t is the thickness of the walls, we get:
V = 3.14 x 1.2² (3.8 + 0.15 + 0.15)
= 10.09 m³
This is slightly larger than the desired volume of 10 m³, which is acceptable.
In conclusion, the size of a circular water tank with a water capacity of 10000 litres should be
In conclusion, understanding how to calculate the size and capacity of a circular water tank is essential for anyone involved in designing, constructing or managing water storage systems. With the
right formulas and methods, it is possible to accurately determine the desired size and capacity of a circular water tank, taking into consideration various factors such as usage, location and
material. It is important to remember that these calculations are a crucial step in ensuring the efficiency and effectiveness of a water storage system, leading to optimal use of resources and
ultimately contributing to sustainability. By following the guidelines and tips provided in this article, one can confidently plan and construct a circular water tank that meets the specific needs
and requirements, both now and in the future.
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Combining 2 Formulas
I have a column that is calculating today minus a date that is in another column. I have another column that is taking the days and converting it to weeks. Then i have another column that is taking
the weeks and rounding down to 1 decimal point.
Is there a way to combine the column converting to weeks and the column rounding down?
Here are my formulas:
=NETWORKDAYS([Start Date]@row, [Finish Date]@row) - 1
=[Duration (Days)]@row / 7
=ROUNDDOWN([duration weeks]@row, 1)
Thanks for the input!
• ROUNDDOWN(((NETWORKDAYS([Start Date]@row, [Finish Date]@row) - 1)/ 7,1
Give that a try. Smartsheet is pretty friendly at combining formulas, usually you just need to treat a formula as an output when dealing with multiple, if that makes any sense. It is kind of like
basic algebra
ab + c = X
Xy + b = z
(ab+c)y + b = z
• Another way of explaining it...
Enter the cell that is [duration weeks]@row as if you are going to edit it. Highlight everything except for the beginning =.
Enter the cell containing the [duration weeks]@row cell reference, and highlight that reference.
Paste the formula (minus the first = of course) that was in cell [duration weeks]@row into where that cell reference was.
I use this method quite frequently when building complex formulas to test each part out individually while using cell references to keep it all working together. I then work backwards and replace
cell references with the formulas that were in those cells.
• Thanks for the help. The formula worked great!
Thanks for the explanation Paul!
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1,259 research outputs found
Using the pure spinor formalism for the superstring, the vertex operator for the first massive states of the open superstring is constructed in a manifestly super-Poincar\'e covariant manner. This
vertex operator describes a massive spin-two multiplet in terms of ten-dimensional superfields.Comment: Added reference
On the basis of the Berkovits pure spinor formalism of covariant quantization of supermembrane, we attempt to construct a M(atrix) theory which is covariant under $SO(1,10)$ Lorentz group. We first
construct a bosonic M(atrix) theory by starting with the first-order formalism of bosonic membrane, which precisely gives us a bosonic sector of M(atrix) theory by BFSS. Next we generalize this
method to the construction of M(atrix) theory of supermembranes. However, it seems to be difficult to obtain a covariant and supersymmetric M(atrix) theory from the Berkovits pure spinor formalism of
supermembrane because of the matrix character of the BRST symmetry. Instead, in this paper, we construct a supersymmetric and covariant matrix model of 11D superparticle, which corresponds to a
particle limit of covariant M(atrix) theory. By an explicit calculation, we show that the one-loop effective potential is trivial, thereby implying that this matrix model is a free theory at least at
the one-loop level.Comment: 13 pages, no figures, two references adde
Although the AdS_5xS^5 worldsheet action is not quadratic, some features of the pure spinor formalism are simpler in an AdS_5xS^5 background than in a flat background. The BRST operator acts
geometrically, the left and right-moving pure spinor ghosts can be treated as complex conjugates, the zero mode measure factor is trivial, and the b ghost does not require non-minimal fields.
Furthermore, a topological version of the AdS_5xS^5 action with the same worldsheet variables and BRST operator can be constructed by gauge-fixing a G/G principal chiral model where G=PSU(2,2|4).
This topological model is argued to describe the zero radius limit that is dual to free N=4 super-Yang-Mills and can also be interpreted as an "unbroken phase" of superstring theory.Comment: 39 pages
The classical pure spinor version of the heterotic superstring in a supergravity and super Yang-Mills background is considered. We obtain the BRST transformations of the world-sheet fields. They are
consistent with the constraints obtained from the nilpotence of the BSRT charge and the holomorphicity of the BRST current.Comment: References adde
Following suggestions of Nekrasov and Siegel, a non-minimal set of fields are added to the pure spinor formalism for the superstring. Twisted $\hat c$=3 N=2 generators are then constructed where the
pure spinor BRST operator is the fermionic spin-one generator, and the formalism is interpreted as a critical topological string. Three applications of this topological string theory include the
super-Poincare covariant computation of multiloop superstring amplitudes without picture-changing operators, the construction of a cubic open superstring field theory without contact-term problems,
and a new four-dimensional version of the pure spinor formalism which computes F-terms in the spacetime action.Comment: 34 pages harvmac te
Starting with a classical action whose matter variables are a d=10 spacetime vector $x^m$ and a pure spinor $\lambda^\alpha$, the pure spinor formalism for the superstring is obtained by gauge-fixing
the twistor-like constraint $\partial x^m (\gamma_m \lambda)_\alpha =0$. The fermionic variables $\theta^\alpha$ are Faddeev-Popov ghosts coming from this gauge-fixing and replace the usual (b,c)
ghosts coming from gauge-fixing the Virasoro constraint. After twisting the ghost-number such that $\theta^\alpha$ has ghost-number zero and $\lambda^\alpha$ has ghost-number one, the BRST cohomology
describes the usual spacetime supersymmetric states of the superstring.Comment: 10 pages harvmac tex, added comments on small Hilbert space and U_0 dependenc
Although it is not known how to covariantly quantize the Green-Schwarz (GS) superstring, there exists a semi-light-cone gauge choice in which the GS superstring can be quantized in a conformally
invariant manner. In this paper, we prove that BRST quantization of the GS superstring in semi-light-cone gauge is equivalent to BRST quantization using the pure spinor formalism for the
superstring.Comment: 16 pages, JHEP format, fixed typos and added 2 footnote
Using the pure spinor formalism we prove identities which relate the tree-level, one-loop and two-loop kinematic factors for massless four-point amplitudes. From these identities it follows that the
complete supersymmetric one- and two-loop amplitudes are immediately known once the tree-level kinematic factor is evaluated. In particular, the two-loop equivalence with the RNS formalism (up to an
overall coefficient) is obtained as a corollary.Comment: 10 pages, harvmac TeX. v2: Updated affiliation and Report-no
The pure spinor heterotic string in a generic super Yang-Mills and supergravity background is considered. We determine the one-loop BRST anomaly at the cohomological level. We prove that it can be
absorbed by consistent corrections of the classical constraints due to Berkovits and Howe, in agreement with the Green-Schwarz cancelation mechanism.Comment: harvmac-big, 18 pages; references added;
minor correction
It is proven that the pure spinor superstring in an AdS_5 x S^5 background remains conformally invariant at one loop level in the sigma model perturbation theory.Comment: 14 pages harvma | {"url":"https://core.ac.uk/search/?q=authors%3A(N.%20Berkovits)","timestamp":"2024-11-14T06:50:25Z","content_type":"text/html","content_length":"187481","record_id":"<urn:uuid:e15f2c1d-537a-4584-87c3-540a8a11abd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00702.warc.gz"} |
former , Arbeitstagung Bonn 1984, Springer Lecture Notes in mathematics med Friedrich Hirzebruch Atiyah-Singer Theorem and elementary number Zagier The Bloch-Wigner-Ramakrishnan polylogarithmic
Suggested reading materials: The lecture notes for this course are far from being Bravais lattices, neutron and X-ray diffraction; Bloch's theorem, reciprocal
Extended Bloch theorem for topological lattice models with open boundaries2019Ingår i: Physical Review B, ISSN 2469-9950, E-ISSN Titel, The Augmented Spherical Wave Method: A Comprehensive Treatment
Volym 719 av Lecture Notes in Physics. Författare, Volker Eyert. Utgåva, illustrerad. A. Bloch, “Least Squares Estimation and Completely Integrable Hamiltonian Lecture Notes in Control and
Information Sciences, Vol. 286 On a Theorem of Hermite and Hurwitz, J. Linear and Multilinear Algebra, 50 (1983). is the study of algebraic cycles, including the Hodge and Bloch-Beilinson
Conjectures. and a new treatment of Grothendieck's algebraic de Rham theorem. av AKF MÅRTENSSON · 2018 — For example, the DNA double helix in its standard form twists Blattner, F. R., Plunkett III,
G., Bloch, C. A., Perna, N. T., Burland, V., Riley, M., Collado-.
moving in a periodic potential are the plane waves modulated by a function having the same periodicity as that of the. lattice. Bloch theorem. In a crystalline solid, the potential experienced by an
electron is periodic. V (x) = V (x +a) V ( x) = V ( x + a) where a is the crystal period/ lattice constant.
In the case of an in nite lattice, the energy levels are continuous According to Bloch’s theorem, the wave function solution of the Schrödinger equation when the potential is periodic and to make
sure the function u (x) is also continuous and smooth, can be written as: Where u (x) is a periodic function which satisfies u (x + a) = u (x). Bloch theorem and energy band Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: December 10, 2011) Felix Bloch was born in Zürich, Switzerland to Jewish parents Gustav and Agnes Bloch.
Bloch's theorem and Bravais lattices. Technical note 0402, version 1. Michael A. Nielsen?, *. School of Physical Sciences and School of Information Technology
2005-06-27 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Note: Kittel use potential
energy U (=eV) Electron’s group velocity is zero near the boundary of the 1st BZ (because of the standing wave). Energy gap Bloch theorem The central eq. Empty lattice approx.
Jul 18, 2020 represents an excerpt of other lecture notes and books. Figures Box 8 (Bloch theorem) The eigenfunctions of the single-electron Schrödinger
θ. Starting from |0>, any state can be reached by first rotating about y (or x) by angle θ and then about z by angle φ. These “two” operations form a universal gate set for a single qubit… a 1- qubit
quantum computer .
A superharmonic proof of the M. Riesz conjugate function theorem · Matts Essén. Arkiv för Matematik Vol. 22, Issue 1-2 (Dec 1984), Study notes for Statistical Physics · From Chaos to Consciousness
The Ehrenfest theorem; Heisenberg's uncertainty Bloch functions; Band structure and the In mathematics, Raynaud's isogeny theorem, proved by Raynaud(1985), relates The Committee notes that the
system meets a need and, as proved by the The analogous statement for odd primes is the Bloch-Kato conjecture, proved by He also describes examples of transformations, aspects of acoustic cloaking,
With a focus on periodic composites, the text uses the Bloch-Floquet theorem to The structure of Sallust's Historiae, H. Bloch. 4. A medieval treatment of Hero's theorem, M. Clagett.
Skatt ost gralum
It has the same mathematical content as Floquet’s theorem, which is often used for functions in the time domain.
föreläsningsanteckningar · Discrete mathematics for computation (CM10196) University of Bath.
Bas 2021 due dates
ordningsvakt utbildning stockholmit hand luggage sizesvensk utbildningssystemmanilla folderavsluta bankkonto nordeaköpa änglar onlinemelanders östermalmshallen öppettider
det s.k. “no interaction theorem” från 1963, visar att de enda möjliga kanoniska Horváthy P A, “Prequantization from path integral viewpoint” i Lecture Notes in Marsden J E, Bloch A, Zenkov D,
Dynamics and Stability for Nonholonomic.
Rev. A 71, 061405(R) – Published 27 June 2005 Attention is called to a theorem of Bloch, from which it is shown that even when interelectronic interactions are taken into account, the state of lowest
electronic free energy corresponds to a zero net current. This result contradicts the hypothesis that superconductivity is caused by spontaneous currents.
SHORTER NOTES The purpose of this department is to publish very short papers of an unusually elegant and polished character, for which there is no other outlet. SOME THEOREMS OF BLOCH TYPE P. S.
CHIANG AND A. J. MACINTYRE Very little is known about the constants in annular forms of Bloch's theorem [l], [S].
Roi formel marketingzana muhsen sister
Sep 25, 2015 Bloch's theorem and defining a Brillouin-zone in the momentum-space. We can introduce the Note that the translation operators are unitary.
Technical note 0402, version 1. Michael A. Nielsen?, *. School of Physical Sciences and School of Information Technology Assume independent electron picture, the single particle Schrodinger equation
is : Using Bloch's Theorem; with periodic in the lattice i.e..
Assume that for the particle-in-box described in these notes that the potential According to Bloch's theorem, the wavefunction solution of the Schrödinger
Bloch’s Theorem. There are two theories regarding the band theory of solids they are Bloch’s Theorem and Kronig Penny Model Before we proceed to study the motion of an electron in a periodic
potential, we should mention a general property of the wave functions in such a periodic potential. Periodic systems and the Bloch Theorem 1.1 Introduction We are interested in solving for the
eigenvalues and eigenfunctions of the Hamiltonian of a crystal. This is a one-electron Hamiltonian which has the periodicity of the lattice.
Our approach is similar to that used by S.L. Altmann (Band theory of metals: the elements, Pergamon Press, https://blog.csdn.net/u013795675/article/details/50197565 Note that although the Bloch
functions are not themselves periodic, because of the plane wave component in Eq. (2.38), the probability density function | ψ k → | 2 has the periodicity of the lattice, as it can be easily shown.
Another interesting property of the wave functions derived from Bloch's theorem is … Note that Bloch’s theorem • is true for any particle propagating in a lattice (even though Bloch’s theorem is
traditionally stated in terms of electron states (as above), in the derivation we made no assumptions about what the particle was); • makes no assumptions about the … Bloch’s Theorem and
Krönig-Penney Model - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online. A lecture note on Bloch’s Theorem and
Krönig-Penney Model. Explain the meaning and origin of … 2019-12-27 Lecture notes: Translational Symmetry and Bloch Theorem 2017/5/26 by Aixi Pan Review In last lecture, we have already learned
about: -Unit vectors for direct lattice ! Note that Bloch's theorem uses a vector . | {"url":"https://lonuimf.firebaseapp.com/28662/1245.html","timestamp":"2024-11-07T16:09:15Z","content_type":"text/html","content_length":"13890","record_id":"<urn:uuid:5cad6924-08ea-4e68-a707-14f03a747914>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00816.warc.gz"} |
NETWORKDAYS Excel Function
What Is NETWORKDAYS Function In Excel?
The NETWORKDAYS function in Excel calculates the total working days between any two given dates, excluding the weekends and specified holiday. This function is used to calculate employee
settlements based on the tenure. The NETWORKDAYS Excel function is an inbuilt function so that we can insert the formula from the “Function Library” or enter it directly in the worksheet.
For example, the below table contains a list of start dates, end dates, and a holiday list in the Date format. We will count the working dates, excluding weekends and holidays.
Enter the formulas as follows:
• =NETWORKDAYS(A2,B2) in cell D2.
• =NETWORKDAYS(A3,B3) in cell D3.
• =NETWORKDAYS(A4,B4,C4) in cell D4.
• =NETWORKDAYS(A5,B5,C5:C7) in cell D5.
The output is shown above. D2 and D3 without entering the holidays argument. On the other hand, D4 and D5 exclude the specified holidays and weekends. Column E, for our reference, shows the formulas
used in column D.
Key Takeaways
• The NETWORKDAYS Excel function calculates the total workdays between two given dates, excluding weekends and specified holidays.
• If we want to exclude the specified holidays from the workday count that fall on weekends, then the NETWORKDAYS() counts them as weekends, thus excluding only one instance of such dates from the
workday count calculation.
• We can use the NETWORKDAYS() with other Excel functions such as EOMONTH, DATE(), and IF().
• If the start_date falls after the end_date, i.e., if the end_date precedes the start_date, then the NETWORKDAYS Excel function returns a negative number.
NETWORKDAYS() Excel Formula
The syntax of the NETWORKDAYS Excel formula is:
The arguments of the NETWORKDAYS Excel formula are:
• start_date: A date representing the initial date. It is a mandatory argument.
• end_date: A date representing the end date. It is a mandatory argument.
• holidays: One or more holidays to exclude while calculating the work days. It is an optional mandatory argument.
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How to Use NETWORKDAYS Excel Function?
We can use the NETWORKDAYS Excel Function in 2 methods, namely,
1. Access from the Ribbon in Excel.
2. Enter in the worksheet manually.
Method #1 – Access from the Excel ribbon
First, choose an empty cell → select the “Formulas” tab → go to the “Function Library” → click the “Date & Time” option drop-down → select the “NETWORKDAYS” function, as shown below.
The “Function Arguments” window appears. Enter the argument values in the “Start_date”, “End_date”, and “Holidays” fields → click “OK”, as shown below.
Method #2 – Enter in the worksheet manually
First, ensure the date values are in Date format in the source data.
1. Select an empty cell.
2. Type =NETWORKDAYS( in the cell. [Alternatively, type =N and select the NETWORKDAYS function from the Excel suggestions.]
3. Enter the arguments as values or cell references in Excel.
4. Close the brackets.
5. Finally, press the “Enter” key.
Let us take an example to understand this funciton.
We will calculate and display the number of working days for the NETWORKDAYS Excel function example.
In the following image, the first table contains a list of start and end dates, the second table shows a holiday list, and column C contains the requirements to count the number of workdays between
the two given dates in each row, considering the holiday list in the second table.
The steps to calculate the days for the NETWORKDAYS Excel function example are:
1. First, select cell D2, enter the formula =NETWORKDAYS(A2,B2), and press the “Enter” key.
[The requirement is to exclude only the weekends. So, the mandatory arguments are given as input, and the optional argument holidays is ignored. The result is 21, i.e., the number of workdays.]
[Alternatively, select cell D2, click Formulas → Date & Time → NETWORKDAYS.
The Function Arguments window opens. Enter the argument values as shown below.
Click “OK”.
2. Select cell D3, enter the formula =NETWORKDAYS(A3,B3,G5), and press the “Enter” key.
[As per the holiday list in the second table, February has one holiday in cell G5. The requirement is to exclude the weekends and holidays in February. So, the function accepts cell references to
the given dates in columns A and B, 01-02-2022 and 25-02-2022, and the holiday date, 21-02-2022. And thus, the NETWORKDAYS Excel function return value in the target cell D3 is 18.]
3. Next, select cell D4, enter the formula =NETWORKDAYS(A4,B4,G6), and press the “Enter” key.
[Here, the requirement is similar to the previous step, except we need to exclude the holidays in April along with the weekends. But as per the holiday list in the second table, there is one
holiday in April. And it falls on a weekend. So, instead of considering the holiday date, 17-04-2022, as a weekend and a holiday, NETWORKDAYS() counts it only as a weekend. And thus, the function
excludes nine weekend dates to return the total working days in April as 21.]
4. Select cell D5, enter the formula =NETWORKDAYS(A5,B5,G3:G8), and press the “Enter” key.
[As we need to exclude all the holidays mentioned in the holiday list in the second table, the NETWORKDAYS() takes the argument holidays as a cell range, C3:C8. And as the holidays, 01-01-2022
and 17-04-2022, fall on weekends, the function considers them as weekends. So, it excludes all the weekends between 01-01-2022 and 31-07-2022 and the four holidays listed in the second table,
17-01-2022, 21-02-2022, 30-05-2022, and 04-07-2022. And thus, the function returns the working days count as 146.]
5. Then, select cell D6, enter the formula =NETWORKDAYS(“01-05-2022″,”15-06-2022″,”30-05-2022”), and press the “Enter” key.
[Here, the requirement is similar to that explained in step 2. But the above formula shows we can also enter the specific dates in double quotes as the NETWORKDAYS() arguments.]
6. Select cell D7, enter the formula =NETWORKDAYS(A7,B7), and press the “Enter” key.
[In row 7, the start date falls after the end date. And hence, the NETWORKDAYS Excel function returns a negative number as the count of working days in the specified period, -65.]
Next, we will insert a new row, i.e., row 8, in the first table, as shown below.
7. Finally, select cell D8, enter the formula =NETWORKDAYS(A8,B8), and press the “Enter” key.
The final output is shown above in column D as per the conditions in column C.
[Output Observation: We will get an error when applying the NETWORKDAYS() in cell D8, because cell B8 value, 31-06-2022, provided as the end_date argument, is an invalid date. So, once we correct
the cell B8 value, it will eliminate the error.]
We will understand some advanced scenarios using NETWORKDAYS Excel function.
Example #1
We will calculate the total hours worked during the working days using the NETWORKDAYS Excel function. We must calculate the total number of working days in the given periods and determine the total
hours worked on the workdays in each row.
In the below image, the first table shows the first and last working dates and the per-day working hours. And the second table contains the holiday list.
The steps to calculate the total hours using the NETWORKDAYS Excel function are:
• 1: Select cell D2, enter the formula =NETWORKDAYS(A2,B2,$G$3:$G$5), and press the “Enter” key.
• 2: Next, select cell E2, enter the =D2*C2 formula, and press the “Enter” key.
• 3: Select cells D2:E2 and drag the formulas using the fill handle to the corresponding row 3 cells, D3:E3.
The output is shown above.
[Output Observation: In row 2, as the listed holidays do not fall in the given duration, the NETWORKDAYS() returns the net workdays excluding the weekends, 44. However, in row 3, the three listed
holidays fall within the first and last working dates. So, the function excludes them and the weekends to return the net workdays, 40.
And finally, the formulas in cells E2:E3 multiply the calculated net workdays with the per day working hours, 7.5, in each row to achieve the required data.]
Example #2
We will calculate the number of workdays between each start date and the month’s end using the NETWORKDAYS Excel function with EOMONTH().
The table below contains a list of start dates for each month from January to December. And we need the monthly net workdays to display them in column B. Assume there is no holiday list, and we need
to exclude only weekends.
The steps to calculate the days using the NETWORKDAYS Excel function with EOMONTH() are:
• Step 1: Select cell B2, enter the formula =NETWORKDAYS(A2,EOMONTH(A2,0)), and press the “Enter” key. The result is 21, as shown below.
• Step 2: Drag the formula from cell B2 to B13 using the fill handle.
The output is shown above.
[Output Observation: First, the EOMONTH() returns serial number 44561, representing the last date of December 31-12-2021. And then, the NETWORKDAYS() accepts the start date in cell A13, 1-12-2022, as
the first argument and EOMONTH() return value as the second argument. So thus, it returns the total working days in the resulting period excluding weekends, 23.]
Example #3
We will calculate the number of workdays between dates using the NETWORKDAYS function along with DATE function, and IF function in excel.
The below image shows three tables. The first table contains the holiday list, and the second shows workday wage conditions. And the third table contains the year for which we must calculate the
annual salary based on the given workday calculation and workday wage conditions.
The steps to use NETWORKDAYS(), DATE() and IF() are:
• Step 1: Select cell H6, enter the formula =IF(NETWORKDAYS(DATE(F6,1,1),DATE(F6,12,31))>=251,NETWORKDAYS (DATE(F6,1,1),DATE(F6,12,31))*120,NETWORKDAYS(DATE(F6,1,1),DATE(F6,12,31)) *90), and press
the “Enter” key.
• Step 2: Select cell H7, enter the formula =IF(NETWORKDAYS(DATE(F7,1,1),DATE(F7,12,31),B3:B13)>=251,NETWORKDAYS (DATE(F7,1,1),DATE(F7,12,31),B3:B13)*120,NETWORKDAYS(DATE(F7,1,1),DATE
(F7,12,31),B3:B13)*90), and press the “Enter” key.
The output is shown above.
[Output Observation: Cell H6 formula accepts only the start and end dates the DATE() returns while ignoring the argument holidays. As a result, it returns the net workdays, 260. And as the IF
condition holds, the IF() returns the product of the determined net workdays, 260 and 120, $31200.
Likewise, the formula in the target cell H7 works similarly. But the only difference is that the NETWORKDAYS() excludes weekends and the nine holidays falling on weekdays in the specified cell range,
B3:B13, while calculating the net working days.]
Important Things to Note
• Ensure the date values we provide to the NETWORKDAYS Excel function have the data format set as Date.
• When the start_date and end_date arguments have the same date value, the NETWORKDAYS() return value is 1, as the function counts the start and end dates.
• We get the #VALUE! error for invalid argument values.
Frequently Asked Questions (FAQs)
1. Where is the NETWORKDAYS function in Excel?
The NETWORKDAYS function in Excel is in the Formulas tab. Click Formulas → Date & Time → NETWORKDAYS.
2. Does the NETWORKDAYS() include weekends and holidays?
The NETWORKDAYS() does not include weekends by default. The functions excluded the holidays when specified while calculating the workdays between two given dates.
3. How to use NETWORKDAYS in Excel VBA?
We can use NETWORKDAYS in Excel VBA using the method:
Let us see how to calculate the net workdays in the duration specified in each row using NETWORKDAYS and VBA with an example.
The following table contains the start and end dates in columns A and B and holidays in column C.
The steps to using the NETWORKDAYS in VBA are:
• 1: In the current worksheet, press the shortcut keys Alt + F11 to open the VBA Editor
• 2: Then choose the required VBAProject and select Insert → Module in the top menu to open the Module1 window.
• 3: Enter the VBA code, shown in the below image, in the module 1 window to apply the NETWORKDAYS() on the specific target cells.
Sub NETWORKDAYS_fn()
Range(“D2”) = Application.WorksheetFunction.NetworkDays(Range(“A2”), range(“B2”), Range(“C2”))
Range(“D3”) = Application.WorksheetFunction.NetworkDays(Range(“A3”), range(“B3”))
Range(“D4”) = Application.WorksheetFunction.NetworkDays(Range(“A4”), range(“B4”), Range(“C4:C5”))
End Sub
• 4: Click the Run Sub/UserForm icon to run the Module1 code.
Finally, if we open the active worksheet, we will see the NETWORKDAYS() executed and the required results in the target cells D2:D4.
[Please Note: In row 2, the final date and holiday are the same. However, as the holiday is a weekday, the NETWORKDAYS() counts it as a holiday and excludes it along with the weekends while
calculating the net workdays.]
Download Template
This article must help understand the NETWORKDAYS Excel function, with its formula and examples. We can download the template here to use it instantly.
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What is the derivative of tan^(-1)(x^2 y^5)? | HIX Tutor
What is the derivative of #tan^(-1)(x^2 y^5)#?
Answer 1
let #u = tan^-1 (x^2y^5)#
#=> tanu = x^2y^5#
By differentiating implicitly with respect to #x# we have,
#=> (du)/(dx)sec^2u = (2x)y^2 + x^2(5y^4)(dy)/(dx)#
#=> (du)/(dx) = (2xy^2 + 5y^4x^2(dy)/(dx))/(sec^2u) = (2xy^2 + 5y^4x^2(dy)/(dx))/(tan^2u + 1)#
but #u = tan^-1 (x^2y^5)#
#(du)/(dx) = (2xy^2 + 5y^4x^2(dy)/(dx))/([tan(tan^-1(x^2y^5))]^2 + 1) = (2xy^2 + 5y^4x^2(dy)/(dx))/((x^2y^5)^2 + 1) = (2xy^2 + 5y^4x^2(dy)/(dx))/(x^4y^10 + 1)#
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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The parameter estimates obtained by fitting a statistical model are rarely the main object of interest in a data analysis. Instead of focusing on those raw estimates, a good starting point is often
to compute model-based predictions for different combinations of predictor values. This allows an analyst to report results on a scale that makes intuitive sense to their readers, colleagues, and
What is a model-based prediction? In this book, we consider that
A prediction is the outcome expected by a fitted model for a given combination of predictor values.
This definition is in line with the familiar concept of “fitted value,”^1 but it differs from a “forecast” or “out-of-sample prediction” (Hyndman and Athanasopoulos 2018). For our purposes, the word
“prediction” need not imply that we hope to forecast the future, or that we are trying to extrapolate to unseen data.
Model-based predictions are often the main quantity of interest in a data analysis. They allow us to answer a wide variety of questions, such as:
• What is the expected probability that a 50-year-old smoker develops heart disease, adjusting for diet, exercise, and family history?
• What is the expected probability that a football team wins a game, considering the team’s recent performance, injuries, and opponent strength?
• What is the expected turnout in municipal elections, accounting for national trends and local demographic characteristics?
• What is the expected price of a three-bedroom house in a suburban area, controlling for floor area and market conditions?
All of these descriptive questions can be answered using model-based predictions. This highlights the fact that predictions are an intrinsically interesting quantity. In chapters 6 and 7, we will see
that they are also a fundamental building block to analyze the effects of interventions.
The current chapter illustrates how to compute and report predictions for models estimated with the Thornton (2008) data. We proceed in order, through each component of the conceptual framework laid
out in Chapter 3: (1) quantity, (2) predictors, (3) aggregation, (4) uncertainty, and (5) tests. Then, we conclude by showing different ways to visualize predictions.
5.1 Quantity
To begin, it is useful to consider how predictions are built in one particular case. Let’s consider a logistic regression model estimated using the Thornton (2008) HIV dataset:
\[ Pr(\text{Outcome}=1) = \Phi \left (\beta_1 + \beta_2 \cdot \text{Incentive} + \beta_3 \cdot \text{Age}_{18 to 35} + \beta_4 \cdot \text{Age}_{>35} \right ), \tag{5.1}\]
where Outcome is a binary variable equal to 1 if the study participant travelled to the test center to learn their HIV status; Incentive is a binary variable equal to 1 if the participant received a
monetary incentive; and the other two predictors are indicators for the age category to which a participant belongs, with omitted category Age\(_{<18}\). The letter \(\Phi\) represents the standard
logisitic function \(\Phi(x) = \frac{1}{1 + e^{-x}}\), which ensures that the linear component inside the parentheses of Equation 5.1 gets scaled to the \([0,1]\) interval. This allows the model to
respect the natural scale of the binary outcome variable.
We load the marginaleffects package, read the data, and estimate a logistic regression model using the glm() function:
The estimated coefficients are:
(Intercept) incentive agecat18 to 35 agecat>35
-0.78232923 1.99229719 0.04368393 0.24780479
For clarity of presentation, we substitute these estimates into the model equation:
\[ Pr(\text{Outcome}=1) = \Phi \left (-0.782 + 1.992 \cdot \text{Incentive} + 0.044 \cdot \text{Age}_{18 to 35} + 0.248 \cdot \text{Age}_{>35} \right ) \]
To make a prediction for a particular individual, we simply plug-in the characteristics of a person into this equation. For example, the predicted probability that Outcome equals 1 for an 18 to 35
year-old in the treatment group is:
\[\begin{align*} Pr(\text{Outcome}=1) = \Phi \left (-0.782 + 1.992 \cdot 1 + 0.044 \cdot 1 + 0.248 \cdot 0 \right )\\ \end{align*}\]
The predicted probability that Outcome equals 1 for someone above 35 years-old in the control group is:
\[\begin{align*} Pr(\text{Outcome}=1) = \Phi \left (-0.782 + 1.992 \cdot 0 + 0.044 \cdot 0 + 0.248 \cdot 1 \right ) \end{align*}\]
These expressions can be evaluated using any calculator. First, we compute the part in parentheses. This is the “linear” or “link scale” prediction:
linpred_treatment_younger <- b[1] + b[2] * 1 + b[3] * 1 + b[4] * 0
linpred_control_older <- b[1] + b[2] * 0 + b[3] * 0 + b[4] * 1
Link scale predictions from a logit model are expressed on the log odds scale. In this example, they include a negative value and a value greater than one. To many, this will feel incongruous,
because the outcome variable is a binary variable, with a probability bounded by 0 and 1. To ensure that our predictions respect the natural scale of the data, we transform the linear component of
Equation 5.1 using the logistic function:
logistic <- \(x) 1 / (1 + exp(-x))
Our model expects that the probability of seeking information about one’s HIV status is 78% for a young adult who receives a monetary incentive, and 37% for an older participant who does not receive
an incentive.
Computing predictions manually is useful for pedagogical purposes, but it is a labour-intensive and error-prone process. The commands above are also limiting, because they only apply to one very
specific model.
Instead of manual computation, we can use the predictions() function from the marginaleffects package. This function can be applied in consistent fashion across more than 100 different classes of
statistical models.
First, we build a data frame of predictor values—a grid—where each row represents a different individual:
grid <- data.frame(agecat = c("18 to 35", ">35"), incentive = c(1, 0))
agecat incentive
1 18 to 35 1
2 >35 0
Then, we call the predictions() function, using the newdata argument to specify the predictor values, and the type argument to set the scale (link or response):
Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 % agecat incentive
1.254 0.0691 18.15 1.118 1.389 {18 to 35} 1
-0.535 0.1013 -5.28 -0.733 -0.336 {>35 } 0
These results are exactly identical to the link scale predictions that we computed manually above. This is reassuring. However, in a logit model, link scale predictions are hard to interpret.
To communicate our results clearly, it is usually best to make predictions on the same scale as the outcome variable. The resulting estimates are easier to interpret, since they can be compared to
observed values of the outcome variable in our dataset. For this reason, the default behavior of predictions() is to return predictions on the response scale:
Estimate Pr(>|z|) 2.5 % 97.5 % agecat incentive
0.778 0.754 0.800 {18 to 35} 1
0.369 0.325 0.417 {>35 } 0
In the rest of this chapter, we show that the marginaleffects package makes it easy to compute various types of predictions, aggregate, and conduct statistical tests on them.
5.2 Predictors
Predictions are “conditional” quantities, in the sense that they depend on the values of all the predictor variables in a model. To compute a prediction, the analyst must fix all the variables on the
right-hand side of the model equation; they must choose a grid (see Section 3.2).
The choice of grid depends on the researcher’s goals. The profiles it holds could correspond to actual observations in the original data, or they could represent unseen, hypothetical, or
representative units. To illustrate, let’s consider a slight modification of the model estimated in Section 5.1. In addition to the incentive and agecat predictors, we now include a numeric predictor
to account for the distance between a study participant’s home and the test center where they can learn their HIV status:
mod <- glm(outcome ~ incentive + agecat + distance, data = dat, family = binomial)
With this model, we can make predictions on various grids: empirical, interesting, representative, balanced, or counterfactual.
5.2.1 Empirical grid
By default, the predictions() function uses the full original dataset as a grid, that is, it uses the empirical distribution of predictors (Section 3.2.1). This means that predictions() will compute
fitted values for each of the individuals observed in the dataset used to fit the model:
Estimate Pr(>|z|) 2.5 % 97.5 %
0.365 0.297 0.439
0.354 0.288 0.426
0.265 0.209 0.330
0.300 0.241 0.365
0.334 0.271 0.402
0.833 0.809 0.855
0.840 0.815 0.862
0.833 0.809 0.855
0.827 0.803 0.849
0.789 0.761 0.814
The p object created by predictions() includes the fitted values for each observation in the dataset, along with test statistics like p values and confidence intervals. p is a standard data frame,
which means that we can manipulate it using standard R functions.
For example, we can check that the data frame includes 2825 and 10 columns:
We can list the available column names:
[1] "rowid" "estimate" "p.value" "s.value" "conf.low" "conf.high"
[7] "outcome" "incentive" "agecat" "distance"
And we can extract individual columns and cells using the standard $ or [] syntaxes, or using data manipulation packages like dplyr or data.table:
[1] 0.3652679 0.3540617 0.2653133 0.2997423
Users should be mindful of the fact that, by default, the p values held in this data frame correspond to a hypothesis test against a null of zero. In Section 5.5, we will see that it is easy to
change this default null using the hypothesis argument.
5.2.2 Interesting grid
In many cases, analysts are not interested in model-based predictions for each observation in their sample. Instead, they may prefer to construct a customized grid of predictors which includes
specific values of scientific or commercial interest (Section 3.2.2).
In marginaleffects, the main strategy to define custom grids is to use the newdata argument and the datagrid() function. This function creates a “typical” dataset with all variables at their means or
modes, except those we explicitly define:
datagrid(agecat = "18 to 35", incentive = c(0, 1), model = mod)
distance agecat incentive rowid
1 2.014541 18 to 35 0 1
2 2.014541 18 to 35 1 2
We can feed this datagrid() function to the newdata argument of predictions():^2
agecat incentive Estimate Pr(>|z|) 2.5 % 97.5 % distance
{18 to 35} 0 0.318 0.279 0.361 2.01
{18 to 35} 1 0.780 0.755 0.802 2.01
This shows that the estimated probability of seeking one’s HIV status is about 32% for a participant who is between 18 and 35 years old, did not receive a monetary incentive, and lives the average
distance from the center.
We can also make predictions on a custom grid by supplying functions to datagrid(). These functions will be applied to the named variables, and the output used to construct the grid.
distance agecat incentive Estimate Pr(>|z|) 2.5 % 97.5 %
2 { 1 0.774 0.725 0.816
2 {18 to 35} 1 0.780 0.756 0.802
2 {>35 } 1 0.815 0.791 0.837
5.2.3 Representative grid
Sometimes, analysts do not want fine-grained control over the values of each predictor, but would rather compute predictions for some “representative” individual (Section 3.2.3). For example, we can
compute a “Prediction at the Mean,” that is, a prediction for a hypothetical representative individual whose personal caracteristics are exactly average (numeric) or modal (categorical).
To achieve this, we can either set the values of the grid manually in datagrid(), or we can use the "mean" shortcut:
Estimate Pr(>|z|) 2.5 % 97.5 % incentive agecat distance
0.78 0.755 0.802 1 {18 to 35} 2.01
Representative grids can be useful in some contexts, but they are not always the best choice. Sometimes there is simply noone in our sample who is exactly average on all relevant dimensions. When
this “average individual” is fictional, predictions made for this profile may not be scientifically interesting or practically relevant.
5.2.4 Balanced grid
A common strategy in the analysis of experiments is to compute estimates on a “balanced grid” (Section 3.2.4). This type of grid includes one row for each combination of unique values for the
categorical (or binary) predictors, holding numeric variables at their means. To achieve this, we can either call datagrid() or use the "balanced" shortcut. These two calls are equivalent:
Estimate Pr(>|z|) 2.5 % 97.5 % incentive agecat distance
0.311 0.251 0.377 0 { 2.01
0.318 0.279 0.361 0 {18 to 35} 2.01
0.367 0.322 0.415 0 {>35 } 2.01
0.773 0.724 0.816 1 { 2.01
0.780 0.755 0.802 1 {18 to 35} 2.01
0.815 0.791 0.837 1 {>35 } 2.01
A balanced grid is often used with randomized experiments, when the analyst wishes to give equal weights to each combination of treatment conditions in the calculation of marginal means (Section 5.3
5.2.5 Counterfactual grid
Yet another set of predictor profiles to consider is the “counterfactual grid.” The predictions made on such a grid allow us to answer questions such as:
What would the predicted outcomes be in our observed sample if everyone had received the treatment, or if everyone had received the control?
To create a counterfactual grid, we duplicate the full dataset, once for every value of the focal variable. For instance, if our dataset includes 2884 rows, and we want to compute predictions for
each combination of the incentive variable (0 and 1), the counterfactual grid will include 5768.
To make predictions on a counterfactual grid, we can call the datagrid() function, or we can use the variables argument:
These predictions are interesting, because they give us a first look at the kinds of counterfactual (potentially causal) queries that we will explore in Chapter 6. We can ask:
For each individual in the Thornton (2008) sample, what is the predicted probability of seeking information about HIV status in the counterfactual worlds where they receive a monetary incentive,
and where they do not?
To answer this question, we rearrange the data and plot it:
p <- data.frame(
Control = p[p$incentive == 0, "estimate"],
Treatment = p[p$incentive == 1, "estimate"])
ggplot(p, aes(Control, Treatment)) +
geom_abline(intercept = 0, slope = 1, linetype = "dashed") +
geom_point() +
labs(x = "Pr(outcome=1) when incentive = 0",
y = "Pr(outcome=1) when incentive = 1") +
xlim(0, 1) + ylim(0, 1) + coord_equal()
Figure 5.1: Predicted probabilities for counterfactual values of incentive.
On this graph, each point represents a single study participant.^3 The x-axis shows the predicted probability that Outcome equals 1 for an individual with the same socio-demographic characteristics,
in the control group. The y-axis shows the predicted probability that Outcome equals 1 for an individual with the same socio-demographic characteristics, in the treatment group.
Every point is well above the 45 degree line. This means that, for every observed combination of predictor values, for every participant in the study, our model says that changing the incentive
variable from 0 to 1 increases the predicted probability that the person will seek to learn their HIV status.
5.3 Aggregation
Computing predictions for a large grid or for every observation in a dataset is useful, but the results can feel unwieldy. This section makes two principal points. First, it often makes sense to
compute aggregated statistics, such as the average predicted outcome across the whole dataset, or by subgroups of the data. Second, the grid across which we aggregate can make a big difference to the
An “average prediction” is the outcome of a simple two step process. First, we compute predictions (fitted values) for each row in the original dataset. Then, we take the average of those
predictions. This can be done manually by calling the predictions() function and taking the mean of estimates:
Alternatively, we can use the avg_predictions() function, which is a wrapper around predictions() that computes the average prediction directly:
Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0.692 0.00791 87.5 0.676 0.707
This shows that the average predicted probability of seeking information about HIV status, across all the study participants in the Thornton (2008) sample, is about 69%.
Now, imagine we want to check if the predicted probability of the Outcome variable differs across age categories. To see this, we can make the same function call, but add the by argument:
agecat Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
{ 0.670 0.0240 27.9 0.623 0.717
{18 to 35} 0.673 0.0116 58.1 0.650 0.695
{>35 } 0.720 0.0121 59.5 0.697 0.744
The average predicted probability of seeking one’s test result is about 67% for minors, and 72% for those above 35 years old. In Section 5.5 we will formally test if the difference between those two
average predictions is statistically significant.
So far, we have taken averages over the empirical distribution of covariates, but analysts are not limited to that grid. One common alternative is to compute “marginal means” by averaging predictions
across a balanced grid of predictors.^4 This is useful in experimental settings, when the observed sample is not representative of the population, and when we want to marginalize while giving equal
weight to each treatment conditions.
To compute marginal means, we call the same function, using the newdata and by arguments:
agecat Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
{ 0.542 0.0261 20.7 0.491 0.593
{18 to 35} 0.549 0.0135 40.6 0.522 0.575
{>35 } 0.591 0.0151 39.0 0.561 0.621
Notice that the results are considerably different from the average predictions computed on the empirical grid. Now, the average predicted probability of seeking one’s test result is estimated at 54%
for minors, and 59% for those above 35 years old.
What explains this difference is that the balanced grid gives equal weight to each combination of categorical variables, while the empirical grid gives more weight to the combinations that are more
frequent in the data. In the Thornton (2008) dataset, more participants belonged to the treatment than to the control group:
Therefore, when we compute an average prediction on the empirical distribution, predicted outcomes in the incentive=1 group are given more weight. This matters, because the average predicted
probability that Outcome equals 1 is much higher in the treatment group than in the control group:
incentive Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0 0.340 0.01890 18.0 0.303 0.377
1 0.791 0.00862 91.7 0.774 0.808
Thus, the group-wise averages for each age categories are smaller when computed over a balanced grid than when they are computed over the empirical distribution.
We have already shown that the probably of seekingIn the Thornton (2008) dataset, the number of participants in each age category is not equal, and the marginal means computed on the empirical grid
are biased towards the more frequent categories.
In the next example, we create a “counterfactual” data grid where each observation of the dataset is repeated twice, with different values of the incentive variable, and all other variables held at
the observed values. We also show the equivalent results using standard R commands:
incentive Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0 0.339 0.01888 18.0 0.302 0.376
1 0.791 0.00862 91.8 0.774 0.808
p <- predictions(
type = "response",
newdata = datagrid(incentive = 0:1, grid_type = "counterfactual"))
aggregate(estimate ~ incentive, FUN = mean, data = p)
incentive estimate
1 0 0.3390492
2 1 0.7910508
5.4 Uncertainty
As in the rest of the marginaleffects package, the predictions() family of functions accept a vcov argument which can be used to specify the type of standard errors to compute and report. We can also
control the size of confidence intervals with conf_level. For instance, to compute heteroskedasticity-consistent standard errors (Type 3) with 90% confidence intervals, we simply call:
incentive Estimate Std. Error z Pr(>|z|) 5.0 % 95.0 %
0 0.340 0.01892 18.0 0.309 0.371
1 0.791 0.00864 91.5 0.777 0.805
We can also report clustered standard errors by village, or use the inferences() function to compute bootstrap intervals:
incentive Estimate Std. Error z Pr(>|z|) 5.0 % 95.0 %
0 0.340 0.0235 14.5 0.301 0.378
1 0.791 0.0102 77.6 0.774 0.808
incentive Estimate Std. Error 5.0 % 95.0 %
0 0.340 0.01918 0.310 0.372
1 0.791 0.00853 0.777 0.805
Notice that the intervals are all slightly different, but still remain in the same ballpark.
5.5 Test
Above, we computed average predictions by age subgroups, and noted that there appeared to be differences in the likelihood that younger and older people would seek their HIV test results. That
observation was only based on the point estimates of the average predictions, and did not rely on a statistical test. Now, let’s consider how analysts can compare predictions more formally.
5.5.1 Null hypothesis tests
To begin, we compute the average predicted outcome for each age subgroup:
agecat Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
{ 0.670 0.0240 27.9 0.623 0.717
{18 to 35} 0.673 0.0116 58.1 0.650 0.695
{>35 } 0.720 0.0121 59.5 0.697 0.744
The average predicted outcome is 67% for young adults and 72% for participants above 35 years old. The difference between these two averages is:
p$estimate[3] - p$estimate[2]
To see if this risk difference is statistically significant, we can use the hypothesis argument, as we did in Chapter 4.
Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0.0478 0.0167 2.86 0.00428 0.015 0.0806
The estimated difference between the 3rd and 2nd groups is about 5 percentage points, and the p value associated with this estimate is 0.004. This crosses the conventional (but arbitrary) statistical
significance threshold of \(\alpha=0.05\). Accordingly, many analysts would reject the null hypothesis that the average predicted probability of seeking one’s HIV test results is the same in the 18
to 35 and >35 groups.
A more convenient way to conduct the same test is to use the formula interface. On the left-hand side, we set the comparison function (difference, ratio, etc.). On the right hand side, we specify
which estimates to compare to one another (sequential, reference, etc.). Here, we choose sequential comparisons: 2nd level vs. 1st level, 3rd level vs. 2nd level, and so on.
p <- avg_predictions(mod, by = "agecat", hypothesis = difference ~ sequential)
Hypothesis Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
(18 to 35) - ( 0.00274 0.0267 0.103 0.91810 -0.0495 0.0550
(>35) - (18 to 35) 0.04783 0.0167 2.856 0.00428 0.0150 0.0806
p <- avg_predictions(mod, by = "agecat", hypothesis = difference ~ sequential)
Hypothesis Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
(18 to 35) - ( 0.00274 0.0267 0.103 0.91810 -0.0495 0.0550
(>35) - (18 to 35) 0.04783 0.0167 2.856 0.00428 0.0150 0.0806
The hypothesis argument also allows us to specify hypothesis groups by subgroups. For example, consider this command, which computes average predicted outcomes for each observed combination of
incentive and agecat:
incentive agecat Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0 { 0.312 0.0321 9.72 0.249 0.375
0 {18 to 35} 0.324 0.0209 15.46 0.283 0.365
0 {>35 } 0.370 0.0235 15.74 0.324 0.416
1 { 0.771 0.0234 32.99 0.725 0.816
1 {18 to 35} 0.778 0.0119 65.43 0.755 0.801
1 {>35 } 0.811 0.0117 69.04 0.788 0.834
We can use the hypothesis argument in similar fashion as before, but add a vertical bar to specify that we want to compute sequential risk differences within subgroups:
p <- avg_predictions(mod,
by = c("incentive", "agecat"),
hypothesis = difference ~ sequential | incentive)
Hypothesis incentive Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
(18 to 35) - ( 0 0.01111 0.0311 0.357 0.7213 -0.04994 0.0722
(>35) - (18 to 35) 0 0.04605 0.0216 2.131 0.0331 0.00369 0.0884
(18 to 35) - ( 1 0.00717 0.0254 0.283 0.7775 -0.04258 0.0569
(>35) - (18 to 35) 1 0.03330 0.0154 2.156 0.0311 0.00302 0.0636
This shows that, in the control group (incentive=0), the difference between the average predicted outcome for participants over 35 and for those between 18 and 35 is about 5 percentage points.
However, in the treatment group (incentive=1), this difference is about 3 percentage points. Both of these differences are associated with relatively large \(z\) statistics, and are thus
statistically distinguishable from zero.
5.5.2 Equivalence tests
Flipping the logic around, the analyst could run an equivalence test to determine if the difference between average predicted outcomes in the two subgroups is small enough to be considered negligible
(Section 4.2). Imagine that, for domain-specific reasons, a risk difference smaller than 10 percentage points is considered “uninteresting,” “negligible,” or “equivalent to zero”. All we need to do
is call the same function with the equivalence argument and the \([-0.1,0.1]\) interval of practical equivalence:
by = "agecat",
hypothesis = "b3 - b1 = 0",
equivalence = c(-0.1, 0.1))
Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 % p (NonSup) p (NonInf) p (Equiv)
0.0506 0.0269 1.88 0.0601 -0.00215 0.103 0.0331 0.0331
The p value associated with our test of equivalence is small. This suggests that we can reject the null hypothesis that the difference lies outside the interval of practical equivalence. The
difference is thus likely to be small enough to be ignored.
5.6 Visualization
In many cases, data analysts will want to visualize (potentially aggregated) predictions rather than report raw numeric values. This is easy to do with the plot_predictions() function, which a
similar syntax that closely parallels that of the other functions in the marginaleffects package.
5.6.1 Unit predictions
As discussed in Chapter 3, the quantities derived from statistical models—predictions, counterfactual comparisons, and slopes—are typically conditional, in the sense that they depend on the values of
all covariates in the model. This implies that each unit in our sample will be associated with its own prediction (fitted value) or effect estimate. In Avoiding One-Number Summaries, Harrell (2021)
argues that data analysts should avoid the temptation to summarize these individual-level estimates. Rather, Harrell argues, they should display the full distribution of estimates to convey a sense
of the heterogeneity of our quantity of interest across different combinations of predictor values.
Histograms and Empirical Cumulative Distribution Function (ECDF) plots are two common ways to visualize such a distribution. Since the output generated by the predictions() function is a standard
data frame, it is easy to feed that object to any plotting function in R or Python, in order to craft good-looking visualizations.
p <- predictions(mod)
# Histogram
p1 <- ggplot(p) +
geom_histogram(aes(estimate, fill = factor(incentive))) +
labs(x = "Pr(outcome = 1)", y = "Count", fill = "Incentive")
# Empirical Cumulative Distribution Function
p2 <- ggplot(p) +
stat_ecdf(aes(estimate, colour = factor(incentive))) +
labs(x = "Pr(outcome = 1)", y = "Cumulative Probability", colour = "Incentive")
p1 + p2
Figure 5.2: Distribution of unit-level predictions (fitted values), by treatment group.
The left side of Figure 5.2 presents a histogram showing the distribution of predicted probabilities that individual study participants choose to travel to the test center in order to learn their HIV
status. As usual, the x-axis represents the range of predicted outcomes, while the y-axis shows the number of study participants each bin. By assigning different colors to the bins based on the
treatment arm (incentive equal 0 or 1), we highlight one of the key features of the distribution: predicted outcomes for people in the treatment group tend to be much higher than predicted outcomes
for people in the control group. Indeed, the distribution of outcome probabilities without an incentive (orange) is concentrated between 0.2 and 0.4, indicating a low probability of travelling to the
test center. In contrast, the distribution of predicted outcome for participants who received a monetary incentive (blue) is concentrated between 0.6 and 0.8. This suggests that those who received an
incentive are considerably more likely to seek their test results.
The right side of Figure 5.2 presents an ECDF plot. Again, the x-axis represents the range of predicted outcomes. This time, however, the y-axis indicates the cumulative probability, which is the
proportion of data points that are less than or equal to a specific value. For any given value on the x-axis, the height of the curve indicates the proportion of data points that are less than or
equal to that value. For example, at 0.3 on the x-axis, we see that the incentive=0 line is close to 0.25. This suggests that about 25% of the participants in our sample have predicted outcome
smaller than 30%. When the ECDF curve is steep, we know that a lot of the our data is concentrated in that part of the distribution. With this in mind, we see clearly that many of our predicted
outcome values are clustered near 0.3 in the control group, and near 0.8 in the treatment group.
5.6.2 Marginal predictions
The first approach to display “marginal” predictions using the by argument. The underlying process is to (1) compute predictions for each observation in the actually observed dataset, and then (2)
average these predictions across some variable(s). This is equivalent to plotting the results of calling avg-predictions() using the by argument.
For example, if we want to compute the average predicted probability that outcome equals 1, by subgroup, we call:
incentive Estimate Std. Error z Pr(>|z|) 2.5 % 97.5 %
0 0.340 0.01890 18.0 0.303 0.377
1 0.791 0.00862 91.7 0.774 0.808
We plot the same results using the plot_predictions() function:
Figure 5.3: Marginal predicted probabilities that outcome equals 1.
Note that that the plot_predictions() function also accepts a newdata argument. This means that we can, for example, plot marginal means constructed by marginalizing across a balanced grid of
5.6.3 Conditional predictions
In some contexts, plotting marginal predictions may not be appropriate. For instance, when one of the predictors of interest is continuous, there are many predictors, or much heterogeneity, the
commands presented in the previous section may generate jagged plots which are difficult to read. In such cases, it can be useful to plot “conditional” predictions instead. In this context, the word
“conditional” means that we are computing predictions, conditional on the values of the predictors in a constructed grid of “representative” values. However, unlike in the previous section, we do not
average over several predictions before displaying the estimates. We fix the grid and display the predictions made for that grid immediately.
The condition argument of the plot_predictions() function does just that: Build a grid of representative predictor values, compute predictions for each combination of predictor values, and plot the
results. In the following examples, we fix one or more predictor to its unique values (categorical) or to an equally spaced grid from minimum to maximum (continuous). The other predictors in the
model are held to their mean or model.
p1 <- plot_predictions(mod, condition = "distance")
p2 <- plot_predictions(mod, condition = c("distance", "incentive"))
p3 <- plot_predictions(mod, condition = c("distance", "incentive", "agecat"))
(p1 + p2) / p3
Figure 5.4: Predicted probability that outcome equals 1, conditional on incentive, age categories, and distance. Other variables are held at their means or modes.
We can also set the value of some variables explicitly by setting condition to a named list. For example, to plot the predicted outcome for an individual above 35 years old, who did not receive a
monetary incentive, for different values of distance:
5.6.4 Customization
Since the output of plot_predictions() is a ggplot2 object, it is very easy to customize. For example, we can add points for the actual observations of our dataset like so:
A more powerful but less convenient way to customize plots is to call the draw=FALSE argument. This will return a data frame with the raw values used to create plots. You can then use these data to
create your own plots with base R graphics, ggplot2, or any other plotting functions you like:
incentive estimate std.error statistic p.value s.value conf.low
1 0 0.3397746 0.018899194 17.97826 2.884182e-72 237.6507 0.3027328
2 1 0.7908348 0.008620357 91.74038 0.000000e+00 Inf 0.7739393
1 0.3768163
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Quantitative Skepticism
I have an idea for a course that’d be appropriate for high school or college. The basic idea is to try to distill and bring together a set of knowledge, skills, and habits that allow people to think
critically about quantitative information.
I’d call this course Quantitative Skepticism, which I think captures the sense of what I’m talking about pretty well, although it isn’t very catchy. I’d be tempted to call it Calibrating your
Bullshit-o-meter, but I don’t think that would fly with many parents.
Anyone reasonably interested in their community, nation or the world is going to have to come to terms with numbers.
Facts, Questions, Claims
We all need to understand quantitative facts, like Facebook has 500 million users, which only make sense in how they relate to other quantitative facts. Is that a lot for a website? For the world?
There were 30,797 fatal crashes in 2009. Is driving safer than flying?
Quantitative Questions range from very personal, like “how much to I need to save to be able to afford a vacation next year?” to big, world-changing questions like, “how much would it cost to
eliminate world hunger?”
Lastly, we all need to be able to evaluate quantitative claims. This affects who you vote for, “this new law will create 5,000 new jobs in the US,” and where you put your money, “buying a new
refrigerator could save you 10% on your electricity bill.”
At the core of the course would be what physicists sometimes call “Fermi Questions,” after Enrico Fermi who used them extensively in his teaching at the University of Chicago to train people to think
quantitatively. These are estimation questions which ask you to find a route to an unknown quantity by considering things you do know and the relationships between them.
The classic example is “how many piano tuners are there in New York City?” And, you work your way there by thinking about how many people live in NYC, what proportion of them have pianos, how often
they need to be tuned, etc to get some idea of the demand for piano tuning. Then you think about how many pianos a tuner can do in a day, and multiply out all your estimates to get the final answer.
This interesting thing is the process: what starting facts are helpful, and how to you form a logical chain of connections between them and your question. Maybe in the age of Google, you can just
look this one up by searching the business listings, so we’d need some modern examples.
At its core, this involves
1. Core numbers you need to know or be able to find. Understanding what kinds of things are most useful as starting points is the key thing to teach in the course: populations, physical constants,
metrics at the community, national, or world level.
2. Relationships between quantitative facts. There are main types of relationships: conversions and proportions. Conversions are things like how many people are there in an average household?
Proportions are fractions of populations or probabilities, like what fraction of an average person’s income is spent on food?
3. Sources and reliability. Where do you get basic facts from and how do you know how good your sources are. What is the uncertainty in your base facts and relationships?
There would be really great opportunities to tie a course like this to both STEM and humanities.
The basic mathematics of estimation are often no more advanced than multiplication, but there are plenty of ways to tie it in to other topics like calculus, statistics and geometry. Science ties
nicely into relationships via physical or biological laws and core numbers.
Critical reading of quantitative claims dovetails nicely into economics, politics, history and journalism. These fields are a great source of interesting questions to investigate as well. | {"url":"https://blog.spikecurtis.com/2011/03/26/quantitative-skepticism/","timestamp":"2024-11-05T08:44:13Z","content_type":"text/html","content_length":"33186","record_id":"<urn:uuid:da8aa82f-efeb-47b2-9934-a46f62a4406d>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00045.warc.gz"} |
The Marimekko Chart – Part 1 | SumProduct are experts in Excel Training: Financial Modelling, Strategic Data Modelling, Model Auditing, Planning & Strategy, Training Courses, Tips & Online Knowledgebase
Charts and Dashboards: The Marimekko Chart – Part 1
Welcome back to our Charts and Dashboards blog series. This week, we begin to “Mari-make” a Marimekko chart by preparing the data.
The Marimekko chart
The Marimekko chart is a visualisation of data from two [2] categorical variables. The name came from its resemblance to some Marimekko prints, and it is also known as the Mekko chart, the Mosaic
plot or sometimes the Percent Stacked Bar plot.
Each tile in the chart represents a cross-category of the two [2] variables and stacking them together makes it very easy to compare the relative sizes, and hence to compare the relative quantities.
To start building the chart, you can download our Excel file and follow the instructions. You can also use this link to download the complete file.
Prepare Data
We will use the following data for demonstration. It has been inserted as a Table Data, and it contains sales figures of five [5] products across six [6] different markets.
We will build a Marimekko chart with one [1] market in each column, stacking the products in that market vertically. Thus, we will need the sub-percentages of products within each market, for the
heights of tiles in each column. Also, we need percentages of market subtotals over the grand total, for the width of each column. We will create a helper table and calculate these percentages.
We create an empty Table Percentages matching rows with Data. This way, we can use Table row references very conveniently.
In the Table Percentages, we first calculate the sub-percentages of products within each market. For example, for product ‘A’, we use the following formula:
=Data[@A] / SUM(Data[@[A]:[E]]) * 100
Then, we calculate the percentage of each market in the grand total with the following formula:
=SUM(Data[@[A]:[E]]) / SUM(Data[[A]:[E]]) * 100
Thus, we have completed the Table Percentages:
Data for the Raw Chart
The secret of building a Marimekko chart is first obtaining trapezoids in a Stacked Area chart, and then transforming them to stacked rectangles. We will produce the following array from Percentages
and then plot from it:
The first column will be a running total of Market Share from Percentages, but they are also being repeated three [3] times each. The other columns are the product percentages from Percentages, being
repeated twice and with zero [0] values in between.
However, before constructing this array, let’s take a peek at how it materialises in a Stacked Area chart:
The distance between each trapezoid’s top in a column is a percentage, and the highest top of each column is 100. Also, repeating each product percentage twice creates these plateaus in the chart,
i.e. the top edges of the trapezoids. The structure of the first column Market is crucial as well, but we can observe in the Stacked Area chart that the horizontal axis is only a list now, instead of
a quantitative scale. After amending that, the chart becomes:
Hopefully, it’s easier to visualise our chart data now, that the zeros [0] in-between produce vertical edges between columns in the chart, given that we repeat a same running percentage total for the
right-end of a market, the divider and the left-end of another market.
We will detail the steps to prepare the data and produce the chart. Let’s first create the array above. To obtain the array, we need three [3] helper columns. First, we build an index with length of
about three [3] times the number of markets:
=SEQUENCE(COUNTA(Percentages[Market]) * 3 + 1, , 0)
Here, COUNTA counts the number of non-blank cells and SEQUENCE is then used to generate the dynamic numerical list. Then, we build a Market Share Flag of the different markets, for displaying their
cumulative percentages later:
Similar to INT, QUOTIENT returns the integer part of a decimal. Then, we build a Market Flag that decides whether to display the product percentages or zero [0] values:
=ROUNDUP(E22#/3, 0) * (MOD(E22#,3) > 0)
This formula rounds up the sequence of numbers divided by three, as long as the division provides a non-zero remainder.
Now, we are ready to build the table for plotting. The first column Market lists the cumulative market percentages three [3] times each. To obtain a running total, we use functions SCAN and LAMBDA:
=SCAN(0, Percentages[Market Share], LAMBDA(x, y, x + y))
The function SCAN has the following syntax:
=SCAN ([initial_value], array, lambda(accumulator, value, calculation))
• initial_value: this is an optional argument and represents the starting value for the accumulator
• array: this is a required value and represents the array to be scanned
• lambda: this is also a required value and represents a LAMBDA function called to scan the array, that consists of three [3] arguments:
□ accumulator: the returned (aggregated) value from LAMBDA
□ value: a value from array
□ calculation: the calculation specified to aggregate values from array into the accumulator.
Here we have specified an addition as the calculation to produce running totals, and then the output from the SCAN function has the following form:
Then we use an INDEX function on the outside with Market Share Flag as the index, to list running totals of market percentages three [3] times each:
=IF(F22#=0, 0, INDEX(SCAN(0, Percentages[Market Share], LAMBDA(x, y, x + y)), F22#))
We also perform an IF check here for Market Share Flag being zero [0], to avoid inputting zero [0] in the INDEX function and outputting whole arrays of data.
Next, we create the columns of product sub-percentages, and we use Market Flag to list each figure twice with a zero [0]. For example, for product ‘A’:
=IF($G$22#<>0, INDEX(Percentages[A], $G$22#), 0)
We have used an INDEX function with Market Flag as the index to list the percentage figures from Table Percentages. We use another IF check to avoid zero [0] arguments for the INDEX function, and
also inserting the zeros [0] in-between. We repeat this formula for all products.
This is where we will leave it for this blog, next time we will use the data to build out the raw chart.
That’s it for this week, come back next week for more Charts and Dashboards tips. | {"url":"https://www.sumproduct.com/blog/article/charts-and-dashboards-blogs/chart-and-dashboards-the-marimekko-chart-part-1","timestamp":"2024-11-06T20:25:54Z","content_type":"text/html","content_length":"55654","record_id":"<urn:uuid:91e5366e-dad5-43bd-9cde-a0c04e9fc704>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00389.warc.gz"} |
For Faster Optimizations
My optimization seems to take a long time to execute. Is there anything I can do to speed it up?
Here is our checklist. (The OptQuest engine mentioned in some of these hints is available in @RISK Industrial 6.0 and newer, and Evolver 6.0 and newer.)
• If you have an older release of Evolver or RISKOptimizer, upgrade to the current release. The optimization engine in 6.x is significantly faster than earlier releases, even more so for linear
problems, and 7.x is faster still.
• Choose the most appropriate solving method, and limit the adjustable cells to as small a range as possible. This improves the proportion of valid (feasible) trials to invalid trials. For
instance, if you have numbers 1 to 20 to assign in an optimal way to 20 cells, don't choose Recipe and try to set constraints that weed out duplicate assignments. Instead, choose Order and the
duplicates will never be generated in the first place.
• Set hard constraints where hard constraints are appropriate. Advice is sometimes given to users of evolutionary solvers to replace hard constraints with soft constraints and a penalty function,
but Evolver and RISKOptimizer do just fine with hard constraints. Their OptQuest engine and Genetic Algorithm handle hard constraints intelligently, using methods that quickly find solutions that
meet the hard constraints. (The Genetic Algorithm uses the method of "backtracking"; it is explained in the software manuals.)
• Make constraints linear if you can. If all constraints are linear, the OptQuest engine (available beginning with release 5.0) can avoid generating solutions that violate constraints, so all
trials will be valid trials. Eliminating these invalid trials can make some optimizations reach a solution much faster. And if you select = in your constraint, using the OptQuest engine, only a
linear constraint will find valid trials within any reasonable time period.
Hint: MAX and MIN are not linear functions. Instead of constraining the maximum or minimum of a cell range to be less or greater than a certain amount, constrain the cell range directly.
• For adjustable cells, use discrete or integer rather than "any", if you can. When adjustable cells are discrete, the OptQuest engine may be able to enumerate them, thus generating only valid
trials. (See Defining Decision Variables, accessed 2015-07-22.)
• If you use the Genetic engine (optional in 6.x/7.x, standard in 1.x–5.x), start with a feasible solution, meaning a state in which all the constraints are met. If you start off with some
constraints violated, the software's genetic algorithm must take time to find a feasible solution as a base for the optimization. If your model is complicated and you need help getting to an
initial feasible solution, please see Debugging RISKOptimizer and Evolver Models.
• Optimize on a continuous value that conveys meaningful information. The idea is that small changes in the adjustable cells should make small changes in the target value. Sometimes a customer
model is essentially binary: the target cell is essentially a yes/no. It is always better to use a target cell that is a continuous number, so that the optimizer can tell when it is making
progress. If your target cell is a 1/0, all infeasible solutions are equally bad and the optimizer has no way to choose one over another. Use constraints, not the target cell, to rule out
unacceptable solutions.
• Constrain on a continuous value when that is natural in the model. Suppose you need cell C5 to be no more than 120. Set your constraint as C5<=120. Sometimes people try to "help" an optimizer by
putting the formula =IF(C5<=120,1,0) in a separate cell and constraining that cell to equal 1. But doing that deprives the algorithms of the information about how far or how close the constraint
is to being met. When you use the real constraint, C5<=120, the algorithm can determine that a solution with C5=150 is better than one with C5=200.
• If you have Excel 2007 or later, enable multi-threaded calculations. In Excel 2010–2016, File » Options » Advanced » Formulas » Enable multi-threaded calculations. In Excel 2007, click the round
Office button and then Excel Options » Advanced » Formulas » Enable multi-threaded calculations.
• Use the optimization stopping conditions on the RISKOptimizer or Evolver options screen. Sometimes the last little bit of convergence isn't needed or provides little improvement, but accounts for
a large chunk of the optimization time (the 80-20 rule).
• In RISKOptimizer, set the separate simulation stopping conditions in addition to the optimization stopping conditions. In RISKOptimizer 6.x/7.x, the simulation stopping conditions are on the
Convergence tab of the @RISK Simulation Settings dialog; in RISKOptimizer 1.x and 5.x they are on the RISKOptimizer Options screen.
• With RISKOptimizer, you can do some things to speed up the simulation portion of the optimization. Generally, good advice for @RISK is good advice for the simulation part of RISKOptimizer. Please
see For Faster Simulations.
• With RISKOptimizer, if you don't have any @RISK distribution functions in your model, set the number of iterations to 1, or use Evolver if you have it. For more information, please see Running
RISKOptimizer Deterministically.
• RISKOptimizer 7.5.0 and newer will split the optimization among multiple CPUs (cores). Look at the General tab of Simulation settings to be sure that multiple CPU is set to Automatic, or to
Enabled. If this computer has only a few cores, try the optimization in a more powerful machine, with more cores and plenty of RAM. | {"url":"https://kb.palisade.com/index.php?pg=kb.page&id=204","timestamp":"2024-11-06T12:09:18Z","content_type":"application/xhtml+xml","content_length":"16364","record_id":"<urn:uuid:7e663ee2-20fc-481e-b3f1-4eaca5364f28>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00887.warc.gz"} |
Speed-up of iron losses computations in post-processing while
Speed-up of iron losses computations in post-processing while solving with parametric distribution
Iron losses computations performed in post-processing with the modified Bertotti model and with the LS model have been drastically accelerated for scenarios containing several varying parameters and
solved with parametric distribution (CDE for Windows and Distribution manager for Linux).
Several Flux modules now benefit from this improvement. For instance, the total computation time to create efficiency maps in FEMT decreased significantly. Note that it also affects the Import /
Export context: every data collection that is created before solving will be properly collected during resolution (e.g., a force data collection for a multi-speed distributed scenario). Consequently,
the collected data will be promptly available after resolution, avoiding its re-evaluation.
To illustrate this improvement, let us consider a Flux 2D project modeling a three-phase, eight-pole permanent magnet synchronous machine (PMSM) with a Transient Magnetic application. In this
example, the simulation scenario controls the rotor's angular position from 0 to 90 degrees in 32 angular steps, with an imposed speed that is time dependent. During parametric distribution, Flux
will compute results for all the parameter combinations at each step. The modified Bertotti model is adopted for evaluating the iron losses.
Figure 1. The three-phase, eight-pole permanent magnet synchronous machine (PMSM) of the example in Flux 2D.
In this example, the parametric distribution is performed over three parameters, namely
• the speed, which is defined as an I/O parameter controlled by the scenario and that is used by the rotating mechanical set;
• the direct-axis and quadrature-axis currents Id and Iq,which are directly used in the coupled circuit to drive the electrical machine.
Table 1. The parameters of the example and their variation ranges.
│ │ Current Iq (A) │ Current Id (A) │ Speed (rpm) │
│ Minimum value │ 2 │ -200 │ 75 │
│ Maximum value │ 200 │ -2 │ 7500 │
│ Number of steps │ 6 │ 6 │ 8 │
It follows from Table 1 that the number of parametric steps to be solved is 6x6x8 =288 and the total number of finite element computations will be 288x32=9216 (since there are 32 angular steps for
each parametric step).
The distributed solution was performed with 10 parallel instances of Flux that treated the configurations presented in Table 1 simultaneously. Table 2 compares the solving time and the iron losses
evaluation time verified with Flux 2022.1 to the times obtained with previous versions.
Table 2. Solving times and iron losses evaluation times with parametric distribution.
│ │ Flux 2022.0 and older versions │ Flux 2022.1 │
│ Solving time │ 1h 30min │ 2h 13min │
│ Time to compute iron losses in post-processing │ 1h 25min │ 3min │ | {"url":"https://2022.help.altair.com/2022.3/flux/Flux/Help/english/ReleaseNote/2022Release/Flux/topics/IronLossesSpeedUpDistributedSolving.htm","timestamp":"2024-11-15T01:20:39Z","content_type":"application/xhtml+xml","content_length":"73714","record_id":"<urn:uuid:b03ad8e7-1fad-4815-8e5c-890f2541e0b4>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00317.warc.gz"} |
Kenneth Boyce, Mathematical surrealism as an alternative to easy-road fictionalism - PhilArchive
Philosophical Studies
177 (10):2815-2835 (
[@article{Boyce2020-B] [Boyce, Kenneth (2020]
Copy BIBT[E]X
Easy-road mathematical fictionalists grant for the sake of argument that quantification over mathematical entities is indispensable to some of our best scientific theories and explanations. Even so
they maintain we can accept those theories and explanations, without believing their mathematical components, provided we believe the concrete world is intrinsically as it needs to be for those
components to be true. Those I refer to as “mathematical surrealists” by contrast appeal to facts about the intrinsic character of the concrete world, not to explain why our best mathematically
imbued scientific theories and explanations are acceptable in spite of having false components, but in order to replace those theories and explanations with parasitic, nominalistically acceptable
alternatives. I argue that easy-road fictionalism is viable only if mathematical surrealism is and that the latter constitutes a superior nominalist strategy. Two advantages of mathematical
surrealism are that it neither begs the question concerning the explanatory role of mathematics in science nor requires rejecting the cogency of inference to the best explanation. | {"url":"https://philarchive.org/rec/BOYMSA-4","timestamp":"2024-11-04T17:52:48Z","content_type":"text/html","content_length":"40754","record_id":"<urn:uuid:8ab68c1a-fb45-44d0-95c8-7c095d9834c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00679.warc.gz"} |
OpenStax College Physics for AP® Courses, Chapter 6, Problem 37 (Problems & Exercises)
The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the magnitude of the
acceleration due to the Moon's gravity at that point. (b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about
27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.
Question by
is licensed under
CC BY 4.0
Final Answer
a) $3.41 \times 10^{-5} \textrm{ m/s}^2$
b) $3.33 \times 10^{-5} \textrm{ m/s}^2$. These are nearly equal, which is expected since the moon is providing the centripetal force which causes the Earth's center to rotate about the center of
mass of the Earth-Moon system.
Solution video
OpenStax College Physics for AP® Courses, Chapter 6, Problem 37 (Problems & Exercises)
vote with a rating of votes with an average rating of.
Video Transcript
This is College Physics Answers with Shaun Dychko. The center of the earth is slowly rotating around this point here which is the center of mass between the earth and the moon. It takes 27.3 days
approximately, one lunar month, for this geometric center of the earth to go around the center of mass that's here at the red x. Now the moon is what's providing the centripetal force to make this
happen and we're going to find out that the acceleration due to gravity of the moon at this position, which is 3.41 times ten to the minus five meters per second squared, is pretty much the same as
the centripetal acceleration of the center of the earth around this point which we calculate to be 3.33 times ten to the minus five meters per second squared. These numbers are nearly the same and
this is what we expect because the earth-- oh sorry -- the moon is what's providing the centripetal force to make this centripetal acceleration happen around the center of mass. Okay. So part A of
this question asks to calculate the magnitude of the acceleration to the moon's gravity at this position here. So we have to set up our geometry properly. We know what the radius of the earth is,
that's something we can look up in the data table, 6.38 times ten to the six meters and we're told that the center of mass is a distance below the surface of the earth, 1690 kilometers which we
convert into meters, 1.69 times ten to the six meters. We know the earth moon distance which is center to center, that's 3.84 times ten to the five kilometers which we can look up in the data table.
We'll take that number and then minus this earth center to center of mass distance to figure out the distance from the moon center to the center of mass. So -- oops, let's put that back there -- so
this is what we have down here. We have the acceleration due to gravity of the moon at the center of mass equals the gravitational constant multiplied by the mass of the moon, divided by this
distance from the center of the moon to the center of mass between the earth and the moon. So that's the distance from the earth to the moon minus this distance here. This distance here is the radius
of the earth minus the distance below the surface of the earth to the center of mass. That's what we have here. So, we have 6.673 times ten to the minus eleven times mass of the moon, 7.35 times ten
to the twenty-two kilograms, divided by the distance from the center of the moon to the center of the earth and then take away this bit of distance between the center of the earth and the center of
mass, which is the radius of the earth, take away the distance below the surface of the earth. Then we square that result there and then we get 3.41 times ten to the minus five meters per second
squared is the acceleration due to gravity of the moon. Now the centripetal acceleration of the center of the earth around that center of mass is the distance from the center of the earth to the
center of mass, multiplied by the angular velocity squared. The angular velocity we can find by taking two pi radians because the center of mass of the earth does a full circle two pi radians in 27.3
days. But we have to convert those days into seconds in order to use it in this formula. So we have 27.3 days times 24 hours per day, and then the days cancel, and then multiply by 3600 seconds per
hour and then we have seconds at the bottom. So this is the radius of the earth minus the distance below the earth's surface to the center of mass and working this all out, that works to 3.33 times
ten to the minus five meters per second squared. | {"url":"https://collegephysicsanswers.com/openstax-solutions/moon-and-earth-rotate-about-their-common-center-mass-which-located-about-4700-0","timestamp":"2024-11-08T19:00:53Z","content_type":"text/html","content_length":"154950","record_id":"<urn:uuid:cf9f42ed-3d64-4ba7-9e7e-382b79e89bb6>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00135.warc.gz"} |
Subspace Definition and 574 Threads
In mathematics, and more specifically in linear algebra, a linear subspace, also known as a vector subspace is a vector space that is a subset of some larger vector space. A linear subspace is
usually simply called a subspace when the context serves to distinguish it from other types of subspaces.
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16. F
I am stuck on finding the dimension of the subspace. Here's what I have so far. Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \
in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v...
17. M
Hey! :giggle: The three axioms for a subspace are: S1. The set must be not-empty. S2. The sum of two elements of the set must be contained in the set. S3. The scalar product of each element of
the set must be again in the set. I have shown that: - $\displaystyle{X_1=\left...
Problem: Show that the set of differentiable real-valued functions ##f## on the interval ##(-4,4)## such that ##f'(-1) = 3f(2)## is a subspace of ##\mathbb{R}^{(-4,4)}## This is my first bouts
with rigorous mathematics and my brain is not at all wired for attacking problems like this (yet). I...
Let ##\mathscr{L_H}## be the usual lattice of subspaces of Hilbert space ##\mathscr{H}##, where for ##p,q\in\mathscr{H}## we write ##p\leq q## iff ##p## is a subspace of ##q##. Then, as discussed
by, e.g., Beltrametti&Cassinelli https://books.google.com/books?id=yWoq_MRKAgcC&pg=PA98, this...
20. P
This is the exact definition and I've summarized it, as I understand it above. Why is it, that for elements in the third subspace, closure will be lost? Wouldn't you still get another vector
(when you add two vectors in that subspace), that's still a linear combination of the vectors in the...
21. M
Hey! 😊 Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$-vector space. Let $\phi,\psi:V\rightarrow V$ be linear maps, such that $\phi\circ\psi=\psi\circ\phi$. I have shown using induction
that if $U\leq_{\phi}V$ (i.e. it $U$ is a subspace and $\phi$-invariant), then...
22. M
Hey! 😊 Let $\mathbb{K}$ be a field and let $V$ a $\mathbb{K}$-vector space. Let $\phi, \psi:V\rightarrow V$ be linear operators, such that $\phi\circ\psi=\psi\circ\phi$. Show that: For $\lambda \
in \text{spec}(\phi)$ it holds that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$. Let...
W = {f(t) | f(0) = 2f(1)} The answer say yes, but i don't know how to prove the neutral element.
24. T
The proof that the set is a subspace is easy. What I don't get about this exercise is the dimension of the subspace. Why is the dimension of the subspace ##n-1##? I really don't have a clue on
how to go through this.
25. C
I was learning about Degenerate Perturbation Theory and I encountered the term 'Degenerate Subspace', I didn't really understand what it meant so I came here to ask - what does it mean? will it
matter if i'll say 'Degenerate space' instead of 'Degenerate Subspace'? and subspace of what? (...
26. M
Hey! :o Let $1\leq m, n\in \mathbb{N}$, let $\phi :\mathbb{R}^n\rightarrow \mathbb{R}^m$ a linear map and let $U\leq_{\mathbb{R}}\mathbb{R}^n$, $W\leq_{\mathbb{R}}\mathbb{R}^m$ be subspaces. I
want to show that: $\phi (U)$ is subspace of $\mathbb{R}^m$. $\phi^{-1} (W)$ is subspace of...
27. M
I am assuming the set ##V## will have elements like the ones shown below. ## v_{1} = (200, 700, 2) ## ## v_{2} = (250, 800, 3) ## ... 1. What will be the vector space in this situation? 2. Would
a subspace mean a subset of V with three or more bathrooms?
28. N
Let X=C[O,1] and Y=span($X_{0},X_{1},···$), where $X_{j}={t}^{i}$, so that Y is the set of all polynomials. Y is not closed in X.
29. S
1. Let's show the three conditions for a subspace are satisfied: Since ##\mathbf{0}\in \mathbb{R}^n##, ##A\times \mathbf{0} = \mathbf{0}\in S##. Suppose ##x_1, x_2\in \mathbb{R}^n##, then ##A
(x_1+x_2) = A(x_1)+A(x_2)\in S##. Suppose ##x\in S## and ##\lambda\in \mathbb{R}##, then ##A(\lambda x) =...
30. J
Let's say we have n vectors in ℝ3. And say we have defined a subspace inside ℝ3 in the form of a sphere with radius r, and the center of the spheare is at P, where P is a vector in ℝ3. What
methods exists to find any linear combination of the n vectors, so that the sum of all of them, lies...
31. J
S is the set of solutions for the set of three equations... x + (1 - a)y-1 + 2z + b2w = 0 ax + y - 3z + (a - a2)|w| = a3 - a x + (a - b)y + z + 2a2w = b I worked out... The first equation is a
subset of R4 when a = 1, b is any real. The second equation is a subset of R4 when a = 1 or a = 0...
32. J
D is the set and the set contains the solutions to x + (1 - m)y-1 + 2z + n2w = 0 I'm trying to find m, n values which means the set is a subspace of R (four dimensions). === Similarly, trying to
find the m, n values that makes the following two expressions two separate subspaces, too. mx +...
33. V
I had assumed that we had to put our values into a matrix so I did [1 2 -1 0; 1 -5 0 -1] and then I would do a=[1; 1] and repeat for b, c, and d. This is incorrect however. I also thought that it
could be {(1, 2, -1, 0),(1, -5, 0, -1)} however this was not the answer, and I am unsure of what do...
Let ##\mathbb{V}## be a vector space and ##\mathbb{W}## be a subset of ##\mathbb{V}##, with the same operations. Claim: If ##\mathbb{W}## is non-empty, closed under addition and scalar
multiplication, then ##\mathbb{W}## is a subspace of ##\mathbb{V}##. A set is a vector space if it...
35. L
I want to exactly diagonalize the following Hamiltonian for ##10## number of sites and ##5## number of spinless fermions $$H = -t\sum_i^{L-1} \big[c_i^\dagger c_{i+1} - c_i c_{i+1}^\dagger\big] +
V\sum_i^{L-1} n_i n_{i+1}$$ here ##L## is total number of sites, creation (##c^\dagger##) and...
Homework Statement Show that the only subspaces of ##V = R^2## are the zero subspace, ##R^2## itself, and the lines through the origin. (Hint: Show that if W is a subspace of ##R^2## that
contains two nonzero vectors lying along different lines through the origin, then W must be all of...
37. S
Homework Statement This is the exact phrasing form Linear Algebra Done Right by Axler: Prove that the union of three subspaces of V is a subspace of V if and only if one of the subspaces contains
the other two. [This exercise is surprisingly harder than the previous exercise, possibly because...
Homework Statement "Let ##T## be a linear operator on a finite-dimensional vector space ##V## over an infinite field ##F##. Prove that ##T## is cyclic iff there are finitely many ##T##-invariant
subspaces. Homework Equations T is a cyclic operator on V if: there exists a ##v\in V## such that...
39. I
Homework Statement Prove whether or not the following linear transformations are, in fact, linear. Find their kernel and range. a) ## T : ℝ → ℝ^2, T(x) = (x,x)## b) ##T : ℝ^3 → ℝ^2, T(x,y,z) =
(y-x,z+y)## c) ##T : ℝ^3 → ℝ^3, T(x,y,z) = (x^2, x, z-x) ## d) ## T: C[a,b] → ℝ, T(f) = f(a)## e) ##...
40. I
Homework Statement Let ##V## be the vector space of the sequences which take real values. Prove whether or not the following subsets ##W \in V## are subspaces of ##(V, +, \cdot)## a) ## W = \
{(a_n) \in V : \sum_{n=1}^\infty |a_n| < \infty\} ## b) ## W = \{(a_n) \in V : \lim_{n\to \infty} a_n...
41. G
Homework Statement Have to read a paper and somewhere along the line it claims that for any distinct ## \ket{\phi_{0}}## and ##\ket{\phi_{1}}## we can choose a basis s.t. ## \ket{\phi_{0}}= \cos\
frac{\theta}{2}\ket{0} + \sin\frac{\theta}{2}\ket{1}, \hspace{0.5cm} \ket{\phi_{1}}=...
42. I
Homework Statement Let V = RR be the vector space of the pointwise functions from R to R. Determine whether or not the following subsets W contained in V are subspaces of V. Homework Equations W
= {f ∈ V : f(1) = 1} W = {f ∈ V: f(1) = 0} W = {f ∈ V : ∃f ''(0)} W = {f ∈ V: ∃f ''(x) ∀x ∈ R} The...
Homework Statement Find the dimension of the subspace of all vectors in ##\mathbb{R}^3## whose first and third entries are equal. Homework EquationsThe Attempt at a Solution So I arrived at two
solutions and I'm not entirely sure which is the valid one. #1 Let ##H \text{ be a subspace of }...
Homework Statement From Linear Algebra and Its Applications, 5th Edition, David Lay Chapter 4, Section 1, Question 32 Let H and K be subspaces of a vector space V. The intersection of H and K is
the set of v in V that belong to both H and K. Show that H ∩ K is a subspace of V. (See figure.)...
Is there an easy example of a closed and bounded set in a metric space which is not compact. Accoding to the Heine-Borel theorem such an example cannot be found in ##R^n(n\geq 1)## with the usual
46. K
I want to show that ##\mathbb{R}## is disconnected with the subspace topology. For this I considered that ##\mathbb{R} = \lim_{\delta n \longrightarrow 0 } (-\infty, n] \cup [n+\delta n, \infty)#
# and of course the intersection of these two open sets is empty. What I'm not sure is about the...
Helo, I believe that the folowing exercise from Topology by Munkres is incorrect: "Let A be a proper subset of X, and let B be a proper subsert of Y. If X and Y are conected, show that ##(X\times
Y)-(A\times B)## is connected" I think I can prove it wrong however I'm not sure and would like to...
48. K
Homework Statement Determine the vector subspace generated by ##A = \{x^2 -x, 3 - x^2, 1+x \} \subset P^2(x)## Homework EquationsThe Attempt at a Solution I tried the usual check of vector
addition and scalar multiplication to get the conditions that ##x## and ##y## should satisfy, but...
49. M
Homework Statement Show that {(1, 2, 3), (3, 4, 5), (4, 5, 6)} does not span R3. Show that it spans the subspace of R3 consisting of all vectors lying in the plane with the equation x - 2y + z =
0. Homework EquationsThe Attempt at a Solution I made a matrix of: A = [ 1 3 4 ; 2 4 5; 3 5 6] and...
Hi, in a text provided by DrDu which I am still reading, it is given that "the momentum operator P is not self-adjoint even if its adjoint ##P^{\dagger}=-\hbar D## has the same formal expression,
but it acts on a different space of functions." Regarding the two main operators, X and D, each has... | {"url":"https://www.physicsforums.com/tags/subspace/","timestamp":"2024-11-08T22:13:33Z","content_type":"text/html","content_length":"173099","record_id":"<urn:uuid:5bfd161b-4fa6-45ad-9478-c96e8516ddb4>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00533.warc.gz"} |
Lesson 10
Domain and Range (Part 1)
10.1: Number of Barks (5 minutes)
This warm-up prompts students to consider possible input and output values for a familiar function in a familiar context. The work here prepares students to do the same in other mathematical contexts
and to think about domain and range in the rest of the lesson.
Student Facing
Earlier, you saw a situation where the total number of times a dog has barked was a function of the time, in seconds, after its owner tied its leash to a post and left. Less than 3 minutes after he
left, the owner returned, untied the leash, and walked away with the dog.
1. Could each value be an input of the function? Be prepared to explain your reasoning.
2. Could each value be an output of the function? Be prepared to explain your reasoning.
Activity Synthesis
Invite students to share their responses and reasoning. Highlight explanations that make a convincing case as to why values beyond 180 could not be inputs for this function and why fractional values
could not be outputs.
Some students may argue that 300 could be an input because "300 seconds after the dog's owner walked away" is an identifiable moment, even though the dog and its owner have walked away and may no
longer be near the post. Acknowledge that this is a valid point, and that it highlights the need for a function to be more specifically defined in terms of when it "begins" and "ends." If time
permits, solicit some ideas on how this could be done.
Tell students that, in this lesson, they will think more about values that make sense as inputs and outputs of functions.
10.2: Card Sort: Possible or Impossible? (20 minutes)
Students continue to think about reasonable input values for functions based on the situation that they represent. They are given three functions and a set of cards containing rational values. For
each function, they determine which values make sense as inputs and why. The idea of domain of a function is then introduced.
Each blackline master contains two sets of cards. Here are the numbers on the cards for your reference and planning:
• -3
• 9
• \(\frac35\)
• 15
• 0.8
• 4
• 0
• \(\frac{25}{4}\)
• 0.001
• -18
• 6.8
• 72
As students sort the cards and discuss their thinking in groups, listen for their reasons for classifying a number one way or another. Identify students whose can correctly and clearly articulate why
certain numbers are or are not possible inputs.
Arrange students in groups of 2–4. Give each group a set of cards from the blackline master.
For each function defined in their activity statement, ask students to sort the cards into two groups, "possible inputs" or "impossible inputs," based on whether or not the function could take the
number on the card as an input. Clarify that the cards will get sorted three times (once for each function), so students should record their sorting results for one function before moving on to the
next function.
Consider asking groups to pause after sorting possible inputs for the first function and to discuss their decisions with another group. If the two groups disagree on where a number belongs, they
should discuss until they reach an agreement, and then continue with the rest of the activity.
Some students may be unfamiliar with camps, and may not know that other units besides Fahrenheit and Celsius are used to measure temperature. Provide a brief orientation, if needed.
Speaking, Representing: MLR8 Discussion Supports. After sorting possible inputs for the first function, provide the class with the following sentence frames to help groups respond to each other:
“_____ is a possible/impossible input because . . .” and “I agree/disagree because . . . .” When monitoring discussions, revoice student ideas to demonstrate mathematical language. This will help
students listen and respond to each other as they explain how they sorted the cards.
Design Principle(s): Support sense-making
Student Facing
Your teacher will give you a set of cards that each contain a number. Decide whether each number is a possible input for the functions described here. Sort the cards into two groups—possible inputs
and impossible inputs. Record your sorting decisions.
1. The area of a square, in square centimeters, is a function of its side length, \(s\), in centimeters. The equation \(A(s) = s^2\) defines this function.
1. Possible inputs:
2. Impossible inputs:
2. A tennis camp charges $40 per student for a full-day camp. The camp runs only if at least 5 students sign up, and it limits the enrollment to 16 campers a day. The amount of revenue, in dollars,
that the tennis camp collects is a function of the number of students that enroll.
The equation \(R(n) = 40n\) defines this function.
1. Possible inputs:
2. Impossible inputs:
3. The relationship between temperature in Celsius and the temperature in Kelvin can be represented by a function \(k\). The equation \(k(c) = c + 273.15\) defines this function, where \(c\) is the
temperature in Celsius and \(k(c)\) is the temperature in Kelvin.
1. Possible inputs:
2. Impossible inputs:
Activity Synthesis
Invite students to share their sorting results. Record and display for all to see the values students considered possible and impossible inputs for each function. Discuss any remaining disagreements
students might have about particular values.
Tell students that we call the set of all possible input values of a function the domain of the function.
Ask students: "How would you describe the domain for each function?" Record and display the description that students give for each function, making sure that the descriptions are complete.
Students may not know that \(0^\circ K\) or \(\text-273.15 ^\circ C\) is absolute zero temperature, or a temperature that is agreed upon as the lowest possible temperature. Consider sharing this
information with them as they describe the domain of function \(k\).
• Area: \(s\), the input of function \(A\) can be any value equal to or greater than 0 (\(s \geq 0\)). The side length can be 0 or any positive number, including irrational numbers. There may be a
debate over whether 0 is a possible length of a square. Either side of the debate should be accepted as long as the connection between the input and the side length of a square is made correctly.
• Tennis camp: \(n\), the input of function \(R\) can be any whole-number value that is at least 5 and at most 16 (\(5 \leq n \leq 16\)). The number of campers cannot be fractional.
• Temperature: \(c\), the input of function \(k\) can be any value that is greater than -273.15 (\(\text- 273.15<c< \infty\)).
10.3: What about the Outputs? (10 minutes)
Earlier, students learned that the domain of a function refers to the set of all possible inputs. In this activity, students are introduced to the range of a function and examine it in terms of a
situation. They begin to consider how the domain and range of a function are related to the features of its graph.
Keep students in groups of 2–4. Give students a few minutes of quiet work time, and then a moment to share their responses with their group. Leave a few minutes for whole-class discussion.
Engagement: Provide Access by Recruiting Interest. Leverage choice around perceived challenge. Invite students to analyze either the area function or the revenue function.
Supports accessibility for: Organization; Attention; Social-emotional skills
Student Facing
In an earlier activity, you saw a function representing the area of a square (function \(A\)) and another representing the revenue of a tennis camp (function \(R\)). Refer to the descriptions of
those functions to answer these questions.
1. Here is a graph that represents function \(A\), defined by \(A(s) = s^2\), where \(s\) is the side length of the square in centimeters.
1. Name three possible input-output pairs of this function.
2. Earlier we describe the set of all possible input values of \(A\) as “any number greater than or equal to 0.” How would you describe the set of all possible output values of \(A\)?
2. Function \(R\) is defined by \(R(n) = 40n\), where \(n\) is the number of campers.
1. Is 20 a possible output value in this situation? What about 100? Explain your reasoning.
2. Here are two graphs that relate number of students and camp revenue in dollars. Which graph could represent function \(R\)? Explain why the other one could not represent the function.
3. Describe the set of all possible output values of \(R\).
Student Facing
Are you ready for more?
If the camp wishes to collect at least $500 from the participants, how many students can they have? Explain how this information is shown on the graph.
Anticipated Misconceptions
Some students may mistakenly associate the domain and range of a function with the horizontal and vertical values that are visible in a graphing window, or with the upper and lower limits of the
scale of each axis on a coordinate plane. For example, they may think that the range of the area function, \(A\), includes only values from 0 to 50 because the scale on the vertical axis goes from 0
to 50. Ask these students if it is possible to use a different scale on each axis or, if the function is graphed using technology, to adjust the graphing window. Clarify that the domain and range
should be considered in terms of a situation rather than the graphing boundaries.
Activity Synthesis
Invite students to share their descriptions of the possible outputs for each function. Explain that we call the set of all possible output values of a function the range of the function. Emphasize
that the range of a function depends on its domain (or all possible input values).
• For the area of the square, the range—all the possible values of \(A(s)\)—includes all numbers that are at least 0.
• For the revenue of the tennis camp, the range—all the possible values of \(R(n)\)—includes positive multiples of 40 that are at least 200 and at most 640.
Next, focus the discussion on function \(R\).
Ask students to explain which values could or could not be the outputs of \(R\) and which of the two graphs represent the function. Clarify that although the graph showing only points more accurately
reflects the domain and range of the function, plotting those points could be pretty tedious. We could use a line graph to represent the function, as long as we specify or are clear that only whole
numbers are in the domain and only multiples of 20 are in the range.
If time permits, draw students' attention to the temperature function they saw in an earlier activity, defined by \(k(c) = c + 273.15\). It gives the temperature in Kelvin as a function of the
temperature in Celsius, \(c\). Ask students:
• "What values are in the domain of this function?" (The domain includes any value that is at least -273.15, the lowest possible temperature in Celsius, or greater)
• "What about the range?" (The range includes any value that is at least 0, the lowest possible temperature in Kelvin, or greater.)
Reading, Writing, Speaking: MLR3 Clarify, Critique, Correct. Before students share their descriptions of the possible output values of \(A\), present an incorrect response and explanation. For
example, “The outputs of \(A\) are numbers from 0 to 50 because I looked on the vertical axis and saw that the graph reaches up to 50.” Ask students to identify the error, critique the reasoning, and
write a correct explanation. As students discuss with a partner, monitor for students who clarify that the output values are not restricted by the graphing boundaries shown. This helps students
evaluate, and improve upon, the written mathematical arguments of others, as they discuss the range of a function.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
10.4: What Could Be the Trouble? (10 minutes)
Optional activity
Previously, students made sense of the domain of functions in concrete contexts. This optional activity is an opportunity to reason about domain more abstractly. Students evaluate an expression that
defines a function at some values of input and notice a value that produces an undefined output. They graph the function to examine its behavior, and then think about how to describe the domain of
the function.
Provide access to graphing technology.
Action and Expression: Provide Access for Physical Action. Provide access to tools and assistive technologies such as a graphing calculator or graphing software. Some students may benefit from a
checklist or list of steps to be able to use the calculator or software.
Supports accessibility for: Organization; Conceptual processing; Attention
Student Facing
Consider the function \(f(x)=\dfrac {6}{x-2}\).
To find out the sets of possible input and output values of the function, Clare created a table and evaluated \(f\) at some values of \(x\). Along the way, she ran into some trouble.
1. Find \(f(x)\) for each \(x\)-value Clare listed. Describe what Clare’s trouble might be.
│ \(x\) │-10│0│\(\frac12\) │2│8│
│\(f(x)\) │ │ │ │ │ │
2. Use graphing technology to graph function \(f\). What do you notice about the graph?
3. Use a calculator to compute the value you and Clare had trouble computing. What do you notice about the computation?
4. How would you describe the domain of function \(f\)?
Student Facing
Are you ready for more?
Why do you think the graph of function \(f\) looks the way it does? Why are there two parts that split at \(x=2\), with one curving down as it approaches \(x=2\) from the left and the other curving
up as it approaches \(x=2\) from the right?
Evaluate function \(f\) at different \(x\)-values that approach 2 but are not exactly 2, such as 1.8, 1.9, 1.95, 1.999, 2.2, 2.1, 2.05, 2.001, and so on. What do you notice about the values of \(f(x)
\) as the \(x\)-values get closer and closer to 2?
Activity Synthesis
Display a graph of the function for all to see. Invite students to share their observations of the behavior of function \(f\) based on the completed table and their graph. Solicit their ideas on what
the problem might be with this function.
If no students mentioned division by 0 as the issue, bring this up. Ask questions such as:
• "What happens when we divide a number by 0?" (The result is undefined.)
• "In the expression \(\dfrac {6}{x-2}\), what value or values of \(x\) would result in a denominator of 0?" (Only 2 gives a denominator of 0.)
• "If 2 does not produce an output, is it a possible input for \(f\)?" (No)
Highlight that the domain of \(f\) includes all numbers except 2.
Lesson Synthesis
Tell students that function \(q\) gives the number of minutes a person sleeps as a function of the number of hours they sleep in a 24-hour period. Display a graphic organizer such as shown.
│ │in the domain?│in the range?│
│negative values │ │ │
│0 │ │ │
│values less than 1 │ │ │
│24 │ │ │
│25 │ │ │
│60 │ │ │
│fractions │ │ │
│values greater than 480 │ │ │
│1,500 │ │ │
Ask students to decide whether each value or set of values described in the first column could be in the domain and in the range of the function. They should be prepared to explain their decisions
(some of which may depend on the assumptions they made about the situation).
Once the class completes the organizer (an example is shown here), give students a moment to come up with a holistic description of the domain and range of this function.
│ │in the domain?│in the range?│
│negative values │no │no │
│0 │yes │yes │
│values less than 1 │yes │yes │
│24 │yes │yes │
│25 │no │yes │
│60 │no │yes │
│fractions │yes │yes │
│values greater than 480 │no │yes │
│1,500 │no │no │
10.5: Cool-down - Community Service (5 minutes)
Student Facing
The domain of a function is the set of all possible input values. Depending on the situation represented, a function may take all numbers as its input or only a limited set of numbers.
• Function \(A\) gives the area of a square, in square centimeters, as a function of its side length, \(s\), in centimeters.
□ The input of \(A\) can be 0 or any positive number, such as 4, 7.5, or \(\frac{19}{3}\). It cannot include negative numbers because lengths cannot be negative.
□ The domain of \(A\) includes 0 and all positive numbers (or \(s \geq 0\)).
• Function \(q\) gives the number of buses needed for a school field trip as a function of the number of people, \(n\), going on the trip.
□ The input of \(q\) can be 0 or positive whole numbers because a negative or fractional number of people doesn’t make sense.
□ The domain of \(q\) includes 0 and all positive whole numbers. If the number of people at a school is 120, then the domain is limited to all non-negative whole numbers up to 120 (or \(0 \leq
n \leq 120\)).
• Function \(v\) gives the total number of visitors to a theme park as a function of days, \(d\), since a new attraction was open to the public.
□ The input of \(v\) can be positive or negative. A positive input means days since the attraction was open, and a negative input days before the attraction was open.
□ The input can also be whole numbers or fractional. The statement \(v(17.5)\) means 17.5 days after the attraction was open.
□ The domain of \(v\) includes all numbers. If the theme park had been opened for exactly one year before the new attraction was open, then the domain would be all numbers greater than or equal
to -365 (or \(d \geq \text-365\)).
The range of a function is the set of all possible output values. Once we know the domain of a function, we can determine the range that makes sense in the situation.
• The output of function \(A\) is the area of a square in square centimeters, which cannot be negative but can be 0 or greater, not limited to whole numbers. The range of \(A\) is 0 and all
positive numbers.
• The output of \(q\) is the number of buses, which can only be 0 or positive whole numbers. If there are 120 people at the school, however, and if each bus could seat 30 people, then only up to 4
buses are needed. The range that makes sense in this situation would be any whole number that is at least 0 and at most 4.
• The output of function \(v\) is the number of visitors, which cannot be fractional or negative. The range of \(v\) therefore includes 0 and all positive whole numbers. | {"url":"https://curriculum.illustrativemathematics.org/HS/teachers/1/4/10/index.html","timestamp":"2024-11-07T22:35:54Z","content_type":"text/html","content_length":"122712","record_id":"<urn:uuid:8ef9dcc6-70f9-4460-8769-7ad80edd36b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00838.warc.gz"} |
Chuy wants to buy a new television. The television costs $1,350. Chuy decides to save the same amount of money each week, for 27 weeks. After 8 weeks Chuy saved $440. Which of the following conclusions can you make about Chuy's plan? a. Chuy has a good plan and will have exactly $1,350 saved at the end of 27 weeks. b. Chuy must increase the amount he saves each week in order to meet his goal at the end of 27 weeks. c. Chuy will save more than he needs and will meet his goal in less than 27 weeks. d. There is not enough information given to make a conclusion about Chuy's plan. Please select the best answer from the choices provided A B C D — QuizWhiz Homework Help
The reach his goalexactlyin 27 weeks, Chuy should increase the amount he saves each week to $50.b. Chuy must increase theamounthe saves each week in order to meet his goal at the end of 27
weeks.Here's why:Chuy wants to save $1,350 in 27weeks, which means he needs to save $1,350 / 27 = $50 per week to reach his goal.However, after 8 weeks, Chuy has only saved $440.If he continues to
save at the same rate of $440 / 8 = $55 per week, he won't reach his goal of $1,350 in 27 weeks.He'll havesaved$55 * 27 = $1,485, which is more than he needs.So, to reach his goal exactly in 27
weeks, Chuy should increase the amount he saves each week to $50.for such more question onamountbrainly.com/question/25720319#SPJ2...
Unlock full access for 72 hours, watch your grades skyrocket.
For just $0.99 cents, get access to the powerful quizwhiz chrome extension that automatically solves your homework using AI. Subscription renews at $5.99/week. | {"url":"https://quizwhiz.org/questions-and-answers/chuy-wants-to-buy-a-new-television-the-television-costs","timestamp":"2024-11-05T03:33:19Z","content_type":"text/html","content_length":"32755","record_id":"<urn:uuid:24a49f9e-0f86-4976-bf07-4ee193ba904c>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00804.warc.gz"} |
Why I struggle with math - How To Survive School
Few subjects are more polarizing than math and many kids struggle with math . The world seems to be divided into those who get it and those who don’t. And if you ask kids why they are not good at
math, the answer invariably is that they “just don’t like it”.
After doing some research on the internet, my conclusion is that this is less a matter of personal aptitude or enthusiasm, and more a matter of the learning environment.
The key reasons appear to be:
Missing important key concepts
Unlike most other subjects, math builds up from concepts from the ground up. If those concepts are not understood clearly, math will become increasingly difficult to understand. (For example, the
multiplying of 2 negative numbers gives a positive number)
The good news is that most concepts are relatively simple to understand, especially with the right teacher.
So, the aim here is to try to identify which are the areas in our understanding where we struggle with math . There is help on the internet, but it might be best to work with a tutor to identify gaps
in your understanding.
Math has a language, and you have to learn it
Just like understanding basic concepts, there is a special vocabulary associated with math, and you need to be at least halfway comfortable with the language. Words like denominator, sum, remainder,
multiple etc. So it is worthwhile identifying the words you are not comfortable with and asking for help to have them explained.
Bad teachers (or parents) can do a lot of damage
Bad teaching in math seems to hurt the student in 2 ways. The fundamental concepts may not be explained well. But the bigger damage they do is in terms of confidence and the student may feel lost and
unable to catch up.
Lack of confidence in math is a killer. Failure then becomes a self-fulfilling prophecy.
So, if you are falling behind or struggle with math , find someone who can help you and can explain the math properly. But there are also a lot of good videos out on the internet and they can be
really helpful.
And please avoid parents if possible. They are rarely good math tutors, no matter how good they think they are.
Once you have found someone who can help you, there is no way around doing some practice. Where possible, find real life examples to make it less abstract.
And don’t forget to keep some simple examples and notes so you don’t forget the trips and tricks.
I also researched some sites which might help you. See my article on free websites for math
Before you know it, you wont struggle with math. Good luck
Further study | {"url":"https://howtosurviveschool.com/why-i-struggle-with-math/","timestamp":"2024-11-15T02:58:53Z","content_type":"text/html","content_length":"144174","record_id":"<urn:uuid:bac4a98d-84f1-4f7e-a4d9-61557510d935>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00602.warc.gz"} |
In this tutorial, most of the calculations for the numerical simulation a SMD (spring-mas-damper) system will be consolidated into a single formula, the coordinate formula. In this case, in order to
calculate the coordinate at the end of a any time step, we will need just the coordinates from the previous two time steps and of course the input parameters (constants). These… Read More... "Casual
Introduction to Numerical Methods – spring-mass-damper system model – part#5" | {"url":"https://excelunusual.com/tag/damper/","timestamp":"2024-11-09T15:57:02Z","content_type":"text/html","content_length":"121378","record_id":"<urn:uuid:be58c137-365d-4867-8e91-bcfceb4e16e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00339.warc.gz"} |
Evolution of braneworld Kerr-Newman naked singularities
Publication date: Apr 2022
We study evolution of the braneworld Kerr-Newman (K-N) naked singularities, namely their mass M , spin a , and tidal charge b characterizing the role of the bulk space, due to matter in-falling from
Keplerian accretion disk. We construct the evolution in two limiting cases applied to the tidal charge. In the first case we assume b =const during the evolution, in the second one we assume that the
dimensionless tidal charge β ≡b /M^2=const . For positive values of the tidal charge the evolution is equivalent to the case of the standard K-N naked singularity under accretion of electrically
neutral matter. We demonstrate that counterrotating accretion always converts a K-N naked singularity into an extreme K-N black hole and that the corotating accretion leads to a variety of outcomes.
The conversion to an extreme K-N black hole is possible for naked singularity with dimensionless tidal charge β <0.25 , and β ∈(0.25 ,1 ) with sufficiently low spin. In other cases the accretion ends
in a transcendental state. For 0.25 <β <1 this is a mining unstable K-N naked singularity enabling formally unlimited energy extraction from the naked singularity. In the case of β >1 , the
corotating accretion creates unlimited torodial structure of mater orbiting the naked singularity. Both nonstandard outcomes of the corotating accretion imply a transcendence of such naked
singularity due to nonlinear gravitational effects.
Blaschke, Martin; Stuchlík, Zdeněk; Hensh, Sudipta; | {"url":"https://zdenekstuchlik.com/2022/04/evolution-of-braneworld-kerr-newman-naked-singularities/","timestamp":"2024-11-08T22:05:24Z","content_type":"text/html","content_length":"73420","record_id":"<urn:uuid:f7e9db21-4940-41d2-afd3-ef413edeb5f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00347.warc.gz"} |
Pallet Calculator: Optimize Your Shipping Logistics
This pallet calculator tool helps you quickly determine the number of pallets needed for your shipment.
How to Use the Pallet Calculator
To use this pallet calculator, follow these steps:
1. Enter the length of one pallet in centimeters.
2. Enter the width of one pallet in centimeters.
3. Enter the height of one pallet in centimeters.
4. Enter the weight of one pallet in kilograms.
5. Enter the number of pallets you wish to calculate.
6. Click the “Calculate” button.
How It Calculates the Results
The pallet calculator determines the total volume and total weight of your specified number of pallets by performing the following calculations:
• Volume: The volume of a single pallet is calculated using the formula: Volume = Length × Width × Height / 1,000,000 to convert cubic centimeters to cubic meters.
• Total Volume: The total volume is then the volume of a single pallet multiplied by the number of pallets.
• Total Weight: The weight of a single pallet is multiplied by the number of pallets to get the total weight.
This calculator assumes that the dimensions and weight provided are for standard pallets. Ensure the dimensions and weight are accurate to get precise results. The tool does not account for
irregular-shaped pallets or other packaging adjustments. | {"url":"https://madecalculators.com/pallet-calculator/","timestamp":"2024-11-06T21:17:32Z","content_type":"text/html","content_length":"142984","record_id":"<urn:uuid:93cdd456-b6e4-419d-975a-444386a1257c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00784.warc.gz"} |
We consider the following reaction-diffusion equation:
$$ {\rm (KS)}\left\{\begin{array}{llll}u_t=\nabla \cdot \Big( \nabla u^m - u^{q-1} \nabla v \Big),& x \in \mathbb{R}^N, \ 0<t<\infty, \nonumber0 = \Delta v - v + u, & x \in \mathbb{R}^N, \ 0<t<\
infty, \nonumberu(x,0) = u_0(x), & x \in \mathbb{R}^N,\end{array}\right.$$
where $N \ge 1, \ m > 1, \ q \ge \max\{m+\frac{2}{N},2\}$ .
In [Sugiyama, Nonlinear Anal.63 (2005) 1051–1062; Submitted; J. Differential Equations (in press)]it was shown that in the case of $q \ge \
max\{m+\frac{2}{N},2\}$ , the above problem (KS) is solvable globally in time for “small $L^{\frac{N(q-m)}{2}}$ data”.Moreover, the decay of the solution (u,v) in $L^p(\mathbb{R}^N)$ was proved.In
this paper, we consider the case of “ $q \ge \max\{m+\frac{2}{N},2\}$ and small $L^{\ell}$ data” with any fixed $\ell \ge \frac{N(q-m)}{2}$ and show that (i) there exists a time global solution (u,v)
of (KS) andit decays to 0 as t tends to ∞ and(ii) a solution u of the first equation in (KS)behaveslike the Barenblatt solution asymptotically as t tends to ∞,where the Barenblatt solution is the
exact solution (with self-similarity) of the porous medium equation $u_t = \Delta u^m$ with m>1. | {"url":"https://core-cms.prod.aop.cambridge.org/core/search?filters%5BauthorTerms%5D=Stephan%20Luckhaus&eventCode=SE-AU","timestamp":"2024-11-15T00:22:42Z","content_type":"text/html","content_length":"728664","record_id":"<urn:uuid:4d7bb299-afa2-41de-ba46-85c78bd7d336>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00455.warc.gz"} |
Lenses II: Image formation
Continuing our discussion from the previous chapter, we can use the thin lens equation and our knowledge of the signs of here. The results are summarized in the table below.
Converging lens
The focal length of a converging lens is positive. That means light from infinity will be brought to focus behind the lens. We will begin our analysis there.
Starting from
Consider an object kept in front of a converging lens. If the object is very far away, we take thin lens equation, the image will be at
By the sign convention for lenses, this implies the image is behind the lens. Also recall that magification due to a lens or mirror is given by
where 1), so is
which means the image is inverted. The image will be enlarged or shrunk down depending on whether
If the object is kept at some finite distance
where, in the second step we used smaller than the object,
This is illustrated in the figure below. We can also arrive at the same conclusions by ray tracing.
A converging lens with the object kept at
As the object is moved closer to the lens the image keeps moving farther away behind the lens. When
Therefore the image is larger than the object (see figure), since
A converging lens with the object kept at
Moving through
If the object is moved further towards the focal point behind the lens (the outgoing refracted rays are parallel to the optical axis). As we move the object past front of the lens (we saw something
similar happen with concave mirrors here). In other words, the image moves from 1)) can be seen from the thin lens equation,
We can be more precise. Solving the first equation above we find
since magnified since simple magnifier. Lastly, an image formed in front of the lens is always virtual as illustrated in this figure and this one.
When the object is placed at
That is, the image is to the left of figure).
A converging lens with the object kept at
As we continue moving the object closer to the lens the image moves closer as well (this is the opposite of what happens when
which means the image crosses figure). In fact, as the object approaches
A converging lens with the object kept at
Diverging lens
A diverging lens has negative focal length, which we may write as
In other words, the image is always virtual, since it is appears in front of the lens. Furthermore,
That is, the magnification is a positive number smaller than 1, which means the image is upright and shrunk down in addition to being virtual. In fact, the image grows in size as we approach this
figure and this one. Note that for a diverging lens the focal points first chapter).
A diverging lens with the object kept at
A diverging lens with the object kept at
Summary: The key insights from our discussion are summarized below.
• For a diverging lens, the image is always virtual/upright/smaller.
• For a converging lens, the image can be virtual/upright/larger or real/inverted/(smaller or larger) depending on where the object is kept.
These results are tabulated below. Comparing with the analogous table for mirrors, we can see that a diverging lens has the exact same behavior as a convex mirror (both have negative focal length),
and converging lenses behave similar to concave mirrors (both have positive focal length). This should come as no surprise, since the mirror equation is identical to the thin lens equation with the
appropriate sign convention.
Object location Diverging Lens ( Converging Lens (
This behavior can also be marked on plots of the image vs. object distance as shown below.
Image vs. object location for a converging lens.
Image vs. object location for a diverging lens.
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Two perspectives on regularization | R-bloggersTwo perspectives on regularization
Two perspectives on regularization
[This article was first published on
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Regularization is the process of adding information to an estimation problem so as to avoid extreme estimates. Put differently, it safeguards against foolishness. Both Bayesian and frequentist
methods can incorporate prior information which leads to regularized estimates, but they do so in different ways. In this blog post, I illustrate these two different perspectives on regularization on
the simplest example possible — estimating the bias of a coin.
Modeling coin flips
Let’s say that we are interested in estimating the bias of a coin, which we take to be the probability of the coin showing heads.^1 In this section, we will derive the Binomial likelihood — the
statistical model that we will use for modeling coin flips. Let $X \in [0, 1]$ be a discrete random variable with realization $X = x$. Flipping the coin once, let the outcome $x = 0$ correspond to
tails and $x = 1$ to heads. We use the Bernoulli likelihood to connect the data to the latent parameter $\theta$, which we take to be the bias of the coin:
There is no point in estimating the bias by flipping the coin only once. We are therefore interested in a model that can account for $n$ coin flips. If we are willing to assume that the individual
coin flips are independent and identically distributed conditional on $\theta$, we can obtain the joint probability of all outcomes by multiplying the probability of the individual outcomes:
For the purposes of estimating the coin’s bias, we actually do not care about the order in which the coins come up heads or tails; we only care about how frequently the coin shows heads or tails out
of $n$ throws. Thus, we do not model the individual outcomes $X_i$, but instead model their sum $Y = \sum_{i=1}^n X_i$. We write:
where we suppress conditioning on $n$ to not clutter notation. Note that our model is not complete — we need to account for the fact that there are several ways to get $y$ heads out of $n$ throws.
For example, we can get $y = 2$ with $n = 3$ in three different ways: $(1, 1, 0)$, $(0, 1, 1)$, and $(1, 0, 1)$. If we were to use the model above, we would underestimate the probability of observing
two heads out of three coin tosses by a factor of three.
In general, there are $n!$ possible ways in which we can order the outcomes. To see this, think of $n$ containers. The first outcome can go in any container, the second one in any container but the
container which houses the first outcome, and so on, which yields:
However, we only care about $y$ of them, so we need to remove the remaining $(n – y)!$ possible ways. Moreover, once we have taken $y$ outcomes, we do not care about their order; thus we remove
another $y!$ permutations. Therefore, for any particular sequence of coin flips of length $n$, there are
ways to get $y$ heads out of $n$ throws. The funny looking symbol on the right is the Binomal coefficient. The probability of the data is therefore given by the Binomial likelihood:
which just adds the term ${n \choose y}$ to the equation we had above after introducing $Y$. For the example of observing $y = 2$ heads out of $n = 3$ coin flips, the Binomial coefficient is ${3 \
choose 2} = 3$, which accounts for the fact that there are three possible ways to get two heads out of three throws.
The data
Assume we flip the coin three times, $n = 3$, and observe three heads, $y = 3$. How can we estimate the bias of the coin? In the next sections, we will use the Binomial likelihood derived above and
discuss three different ways of estimating the coin’s bias: maximum likelihood estimation, Bayesian estimation, and penalized maximum likelihood estimation.
Classical estimation
Within the frequentist paradigm, the method of maximum likelihood is arguably the most popular method for parameter estimation: choose as an estimate for $\theta$ the value which maximizes the
likelihood of the data.^2 To get a feeling for how the likelihood of the data differs across values of $\theta$, let’s pick two values, $\theta_1 = .5$ and $\theta_2 = 1$, and compute the likelihood
of observing three heads out of three coin flips:
We therefore conclude that the data are more likely for a coin that has bias $\theta_1 = 1$ than for a coin that has bias $\theta_2 = 0.5$. But is it the most likely value? To compare all possible
values for $\theta$ visually, we plot the likelihood as a function of $\theta$ below. The left figure shows that, indeed, $\theta = 1$ maximizes the likelihood for the data. The right figure shows
the likelihood function for $y = 15$ heads out of $n = 20$ coin flips. Note that, in contrast to probabilities, which need to sum to one, likelihoods do not have a natural scale.
Do these two examples allow us to derive a general principle for how to estimate the bias of a coin? Let $\hat{\theta}$ denote an estimate of the population parameter $\theta$. The two figures above
suggests that $\hat{\theta} = \frac{y}{n}$ is the maximum likelihood estimate for an arbitrary data set $d = (y, n)$ … and it is! To arrive at this mathematically, we can find the maximum of this
likelihood function by taking the derivative with respect to $\theta$, and setting it to zero (see also a previous post). In other words, we solve for the value of $\theta$ for which the derivative
does not change; and since the Binomial likelihood is unimodal, this maximum will be unique. Note the value for $\theta$ at which the likelihood function has its maximum does not change when we take
logs, but because the mathematics is greatly simplified, we do so:
which shows that indeed $\frac{y}{n}$ is the maximum likelihood estimate.
Bayesian estimation
Bayesians assign priors to parameters in addition to the likelihood, which takes a central role in all statistical paradigms. For this Binomial problem, we assign $\theta$ a Beta prior:
As we will see below, this prior allows easy Bayesian updating while being sufficiently flexible in incorporating prior information. The figure below shows different Beta distributions, formalizing
our prior belief about values of $\theta$.
The figure in the top left corner assigns uniform prior plausibility to all values of $\theta$; the figures to its right incorporate a slight bias towards the extreme values $\theta = 1$ and $\theta
= 0$. With increasing $a$ and $b,$ the prior becomes more biased towards $\theta = 0.5$; with decreasing $a$ and $b$, the prior becomes biased against $\theta = 0.5$.
As shown in a previous blog post, the Beta distribution is conjugate to the Binomial likelihood, which means that the posterior distribution of $\theta$ is again a Beta distribution:
where $a’ = a + y$ and $b’ = b + y – n$. Under this conjugate setup, the parameters of the prior can be understood as prior data; for example, if we choose prior parameters $a = b = 1$, then we
assume that we have seen one heads and one tails prior to data collection. The figure below shows two examples of such Bayesian updating processes.
In both cases, we observe $y = 3$ heads out of $n = 3$ coin flips. On the left, we assign $\theta$ a uniform prior. The resulting posterior distribution is proportional to the likelihood (which we
have rescaled to fit nicely in the graph) and thus does not appear as a separate line. After we have seen the data, we can compute the posterior mode as our estimate for the most likely value of $\
theta$. Observe that the posterior mode is equivalent to the maximum likelihood estimate:
This is in fact the case for all statistical estimation problems where we assign a uniform prior to the (possibly high-dimensional) parameter vector $\theta$. To prove this, observe that:
since we can drop the normalizing constant $p(y)$, because it does not depend on $\theta$, and $p(\theta)$, because it is a constant assigning all values of $\theta$ equal probability.
Using a Beta prior with $a = b = 2$, as shown on the right side of the figure above, we see that the posterior is not proportional to the likelihood anymore. This in turn means that the mode of the
posterior distribution does no longer correspond to the maximum likelihood estimate. In this case, the posterior mode is:
In contrast to earlier, this estimate is shrunk towards $\theta = 0.5$. This came about because we have used prior information that stated that $\theta = 0.5$ is more likely than the other values
(see figure with $a = b = 2$ above). Consequently, we were therefore less swayed by the somewhat unlikely situation (under no bias $\theta = 0.5$) of observing three heads out of three throws. It
should thus not come as a surprise that Bayesian priors can act as regularizing devices. However, this requires careful application, especially in small sample size settings.
In a Post Scriptum to this blog post, I similarly show how the posterior mean, which is arguably are more natural point estimate as it takes the uncertainty about $\theta$ better into account than
the posterior mode, can be viewed as a regularized estimate, too.
Penalized estimation
Bayesians are not the only ones who can add prior information to an estimation problem. Within the frequentist framework, penalized estimation methods add a penalty term to the log likelihood
function, and then find the parameter value which maximizes this penalized log likelihood. We can implement such a method by optimizing an extended log likelihood:
where we penalize values that a far from the parameter value which indicates no bias, $\theta = 0.5$. The larger $\lambda$, the stronger values of $\theta \neq 0.5$ get penalized. In addition to
picking $\lambda$, the particular form of the penalty term is also important. Similar to assigning $\theta$ a prior distribution, although possibly less straightforward and less flexible, choosing
the penalty term means incorporating information about the problem in addition to specifying a likelihood function. Above, we have used the squared distance from $\theta = 0.5$ as a penalty. We call
this the $\mathcal{L}_2$-norm penalty^3, but the $\mathcal{L}_1$-norm, which takes the absolute distance, is an equally interesting choice:
As we will see below, these penalties have very different effects.
The penalized likelihood does not only depend on $\theta$, but also on $\lambda$. The code below evaluates the penalized log likelihood function given values for these two parameters. Note that we
drop the normalizing constant ${n \choose y}$ as it does neither depend on $\theta$ nor on $\lambda$.
fn <- function(y, n, theta = seq(0.001, .999, .001), lambda = 2, reg = 1) {
y * log(theta) + (n - y) * log(1 - theta) - lambda * abs(theta - 1/2)^reg
get_penalized_likelihood <- Vectorize(fn)
With only three data points it is futile to try to estimate $\lambda$ using, for example, cross-validation; however, this is also not the goal of this blog post. Instead, to get further intuition, we
simply try out a number of values for $\lambda$ using the code below and see how it influences our estimate of $\theta$. Because the parameter space has only one dimension, we can easily find the
value for $\theta$ which maximizes the penalized likelihood even without wearing our calculus hat. Specifically, given a particular value for $\lambda$, we evaluate the penalized likelihood function
for a range of values of between zero and one and pick the value that minimizes it.
estimate_path <- function(y, n, reg = 1) {
lambda_seq <- seq(0, 10, .01)
theta_seq <- seq(.001, 1, .001)
theta_best <- sapply(seq_along(lambda_seq), function(i) {
penalized_likelihood <- get_penalized_likelihood(y, n, theta_seq, lambda_seq[i], reg)
cbind(lambda_seq, theta_best)
Sticking with the observations of three heads ($y = 3$) out of three throws ($n = 3$), the figure below plots best fitting values for $\theta$ given a range of values for $\lambda$. Observe that the
$\mathcal{L}_1$-norm penalty shrinks it more quicker and abruptly to $\theta = 0.5$ at $\lambda = 6$, while the $\mathcal{L}_2$-norm penalty gradually (and rather slowly) shrinks the parameter to $\
theta = 0.5$ with increasing $\lambda$. Why is this so?
First, note that because $\theta \in [0, 1]$ the squared distance will always be smaller than the absolute distance, which explains the slower shrinkage. Second, the fact that the $\mathcal{L}
_1$-norm penalty can shrink exactly to $\theta = 0.5$ is due to the discontinuity of the absolute value function. The figures below provides some intuition. In particular, the figure on the left
shows the $\mathcal{L}_1$-norm penalized likelihood function for a select number of $\lambda$’s. We see that for $\lambda < 3$, the value $\theta = 1$ performs best. With $\lambda \in [3, 6]$, values
of $\theta \in [0.5, 1]$ become more likely than the extreme estimate $\theta = 1$. For $\lambda \geq 6$, the ‘no bias’ value $\theta = 0.5$ maximizes the penalized likelihood. Due to the
discontinuity in the penalty, the shrinkage is exact. The $\mathcal{L}_2$-norm penalty, on the other hand, shrinks less strongly, and never exactly to $\theta = 0.5$, except of course for $\lambda \
rightarrow \infty$. We can see this in the right figure below, where the penalized likelihood function is merely shifted to the left with increasing $\lambda$; this is in contrast to the $\mathcal{L}
_1$-norm penalized likelihood on the left, for which the value $\theta = 0.5$ at the discontinuity takes a special place.
You can play around with the code below to get an intuition for how different values of $\lambda$ influence the penalized likelihood function.
plot_pen_llh <- function(y, n, lambdas, reg = 1,
ylab = 'Penalized Likelihood', title = '') {
nl <- length(lambdas)
theta <- seq(.001, .999, .001)
likelihood <- matrix(NA, nrow = nl, ncol = length(theta))
normalize <- function(x) (x - min(x)) / (max(x) - min(x))
for (i in seq(nl)) {
log_likelihood <- get_penalized_likelihood(y, n, theta, lambdas[i], reg)
likelihood[i, ] <- normalize(exp(log_likelihood))
plot(theta, likelihood[1, ], xlim = c(0, 1), type = 'l', ylab = ylab, lty = 1,
xlab = TeX('$\\theta$'), main = title, lwd = 3, cex.lab = 1.5,
cex.main = 1.5, col = 'skyblue', axes = FALSE)
for (i in seq(2, nl)) {
lines(theta, likelihood[i, ], xlim = c(0, 1), type = 'l', ylab = ylab, lty = i,
lwd = 3, cex.lab = 1.5, cex.main = 1.5, col = 'skyblue')
axis(1, at = seq(0, 1, .2))
axis(2, las = 1)
info <- sapply(lambdas, function(l) TeX(sprintf('$\\lambda = %.2f$', l)))
legend('topleft', legend = info, lty = seq(nl), cex = 1,
box.lty = 0, col = 'skyblue', lwd = 2)
lambdas <- c(0, 2, 4, 6, 8)
plot_pen_llh(3, 3, lambdas, reg = 1, title = TeX('$L_1$ Penalized Likelihood'))
plot_pen_llh(3, 3, lambdas, reg = 2, title = TeX('$L_2$ Penalized Likelihood'))
In practice, one would reparameterize this model as a logistic regression, and use cross-validation to estimate the best value for $\lambda$; see the Post Scriptum for a sketch of this approach.
In this blog post, we have seen two perspectives regularization illustrated on a very simple example: estimating the bias of a coin. We first derived the Binomial likelihood, connecting the data to a
parameter $\theta$ which we took to be the bias of the coin, as well as the maximum likelihood estimate. Observing three heads out of three coin flips, we became slightly uncomfortable with the
(extreme) estimate $\hat{\theta} = 1$. We have seen how, from a Bayesian perspective, one can add prior information to this estimation problem, and how this led to an estimate that was shrunk towards
$\theta = 0.5$. Within the frequentist framework, one can add information by augmenting the likelihood function with a penalty term. The type of information we want to incorporate corresponds to the
particular penalty term. In this blog post, we have focused on the most commonly used penalty terms: the $\mathcal{L}_1$-norm, which shrinks parameters exactly to a particular value; and the $\
mathcal{L}_2$-norm penalty, which provides continuous shrinkage. A future blog post might look into linear regression models where regularization methods abound and study how, for example, the
popular Lasso can be recast in Bayesian terms.
I would like to thank Jonas Haslbeck, Don van den Bergh, and Sophia Crüwell for helpful comments on this blog post.
Post Scriptum
Posterior mean
You may argue that one should use the mean instead of the mode as a posterior summary measure. If one does this, then there is already some shrinkage for the case of uniform priors. The mean of the
posterior distribution is given by:
As so often in mathematics, we can rewrite this in a more complicated manner to gain insight into how Bayesian priors shrink estimates:
This decomposition shows that the posterior mean is a weighted combination of the prior mean and the maximum likelihood estimate. Since we can think of $a + b$ as the prior data, note that $a + b +
n$ can be thought of as the total number of data points. The prior mean is thus weighted be the proportion of prior to total data, while the maximum likelihood estimate is weighted by the proportion
of sample data to total data. This provides another perspective on how Bayesian priors regularize estimates.^4
Penalized logistic regression
Cross-validation might be a bit awkward when we represent the data using only $y$ and $n$. We can go back to the product of Bernoulli representation, which uses all individual data points $x_i$. This
results in a logistic regression problem with likelihood:
where we use a sigmoid function as the link function, and $\beta$ is on the log odds scale. The penalized log likelihood function can be written as
where because $\beta = 0$ corresponds to $\theta = 0.5$, we do not need to subtract in the penalty term. This parameterization also makes it more easy to study which types of priors on $\beta$ result
in an $\mathcal{L}_1$ or $\mathcal{L}_2$ norm penalty (spoiler: it’s a Laplace and the Gaussian, respectively). Such models can be estimated using the R package glmnet, although it does not work for
the exceedingly small sample we have played with in this blog post. This seems to imply that regularization is more natural in the Bayesian framework, which additionally allows more flexible
specification of prior knowledge.
• Gelman, A., & Nolan, D. (2002). You can load a die, but you can’t bias a coin. The American Statistician, 56(4), 308-311.
• Stigler, S. M. (2007). The Epic Story of Maximum Likelihood. Statistical Science, 22(4), 598-620.
1. I don’t think anybody actually ever is interested in estimating the bias of a coin. In fact, one cannot bias a coin if we are only allowed to flip it in the usual manner (see Gelman & Nolan, 2002
). ↩
2. In a wonderful paper humbly titled The Epic Story of Maximum Likelihood, Stigler (2007) says that maximum likelihood estimation must have been familiar even to hunters and gatherers, although
they would not have used such fancy words, as the idea is exceedingly simple. ↩
3. Strictly speaking, this is incorrect: the only norm that exists for the one-dimensional vector space is the absolute value norm. Thus, in our example with only one parameter $\theta$ there is no
notion of an $\mathcal{L}_2$-norm. However, because of the analogy to the regression and more generally multidimensional setting, I hope that this inaccuracy is excused. ↩
4. It also shows that in the limit of infinite data, the posterior mean converges to the maximum likelihood estimate. ↩ | {"url":"https://www.r-bloggers.com/2019/04/two-perspectives-on-regularization/","timestamp":"2024-11-09T19:43:07Z","content_type":"text/html","content_length":"230734","record_id":"<urn:uuid:abbf9f6b-51f2-4a4c-a9db-7b6259538920>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00475.warc.gz"} |
The average weekly earnings in the leisure and hospitality industry group for a re‐
cent year was $273. A random sample of 40 workers showed weekly average ear‐
nings of $285 with the population standard deviation equal to 58. At the 0.05 level of
significance can it be concluded that the mean differs from $273? Find a 95% con‐
fidence interval for the weekly earnings and show that it supports the results of the
hypothesis test.
The average weekly earnings in the leisure and hospitality industry group for a re‐ cent year was $273. A random sample of 40 workers showed weekly average ear‐ nings of $285 with the population
standard deviation equal to 58. At the 0.05 level of significance can it be concluded that the mean differs from $273? Find a 95% con‐ fidence interval for the weekly earnings and show that it
supports the results of the hypothesis test.
877 views
Answer to a math question The average weekly earnings in the leisure and hospitality industry group for a re‐ cent year was $273. A random sample of 40 workers showed weekly average ear‐ nings of
$285 with the population standard deviation equal to 58. At the 0.05 level of significance can it be concluded that the mean differs from $273? Find a 95% con‐ fidence interval for the weekly
earnings and show that it supports the results of the hypothesis test.
118 Answers
To determine whether the mean weekly earnings differs from $273, we will conduct a one-sample z-test.
Let's denote the population mean as μ, the sample mean as x̄, the population standard deviation as σ, the sample size as n, and the level of significance as α.
x̄ = $285
σ = $58
n = 40
α = 0.05
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: The mean weekly earnings, μ, is equal to $273.
Ha: The mean weekly earnings, μ, differs from $273.
Step 2: Calculate the test statistic.
We will use the formula for the z-test statistic:
z = \frac{x̄ - μ}{\frac{σ}{\sqrt{n}}}
Substituting in the given values:
z = \frac{285 - 273}{\frac{58}{\sqrt{40}}}
Step 3: Determine the critical value.
Since we are conducting a two-tailed test at the 0.05 level of significance, we need to find the critical z-value for α/2 = 0.025. This value can be obtained from the standard normal distribution
table or calculator.
The critical z-value for a 0.025 level of significance is approximately ±1.96.
Step 4: Make a decision.
If the test statistic z falls outside the critical region (greater than 1.96 or less than -1.96), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 5: Calculate the p-value.
The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true.
To calculate the p-value, we can use a standard normal distribution table or a calculator. The p-value is the area under the standard normal curve that corresponds to the absolute value of the test
Step 6: Determine the conclusion.
If the p-value is less than the level of significance (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Now let's perform the calculations:
z = \frac{285 - 273}{\frac{58}{\sqrt{40}}} = \frac{12}{9.176 } ≈ 1.31
The critical z-value at the 0.025 level of significance is ±1.96.
Since the test statistic z (1.31) falls within the range -1.96 to 1.96, we fail to reject the null hypothesis.
Now, let's proceed to find the 95% confidence interval for the weekly earnings.
The formula for the confidence interval is:
\text{Confidence Interval}=\mu_0\pm z\frac{σ}{\sqrt{n}}
Plugging in the given values:
\text{Confidence Interval}=273\pm1.96\frac{58}{\sqrt{40}}
Calculating this, we get:
\text{Confidence Interval}=273\pm17.974
Thus, the 95% confidence interval for the weekly earnings is [255.026 : 290.974]. The population mean ($273) is within the confidence interval. Therefore, we cannot reject the null hypothesis.
Answer: No, at the 0.05 level of significance, it cannot be concluded that the mean weekly earnings differ from $273.
Frequently asked questions (FAQs)
What is the mode of the following numbers: 5, 7, 7, 9, 12, 12, 12, 16, 19?
Question: In a circle, if angle ACB is 70°, and angle ABC is twice the size of angle BCA, what is the measure of angle BCA?
What is the formula for the surface area of a cone? | {"url":"https://math-master.org/general/the-average-weekly-earnings-in-the-leisure-and-hospitality-industry-group-for-a-re-cent-year-was-273-a-random-sample-of-40-workers-showed-weekly-average-ear-nings-of-285-with-the-population-sta","timestamp":"2024-11-07T04:14:13Z","content_type":"text/html","content_length":"250828","record_id":"<urn:uuid:ae447d88-de1d-4bae-9757-12351993b936>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00196.warc.gz"} |
Four/Five-parameter parallel lines logistic regression
4/5 parameter parallel lines logistic regression models a quantitative sigmoidal response to a quantitative variable. In Excel with the XLSTAT software.
What is four/five-parameter parallel lines logistic regression?
Four parameter logistic model
The four parameter logistic model writes:
y = a + (d -a) / [1 + (x / c)^b] model (1.1)
where a, b, c, d are the parameters of the model, and where x corresponds to the explanatory variable and y to the response variable. a and d are parameters that respectively represent the lower and
upper asymptotes, and b is the slope parameter. c is the abscissa of the mid-height point which ordinate is (a+b)/2. When a is lower than d, the curve decreases from d to a, and when a is greater
than d, the curve increases from a to d.
Five parameter logistic model
The five parameter logistic model writes:
y = a + (d -a) / [1 + (x / c)^b]^e model (1.2)
where e is an additional parameter, the asymmetry factor.
Four parameter parallel lines logistic model
The four parameter parallel lines logistic model writes:
y = a + (d -a) / [1 + (s0 * x / c[0] + s1 * x / c[1])^b] model (2.1)
where s0 is 1 if the observation comes from the standard sample, and 0 if not, and where s1 is 1 if the observation is from the sample of interest, and 0 if not. This is a constrained model because
the observations corresponding to the standard sample influence the optimization of the values of a, b, and d. From the above writing of the model, one can understand that this model generates two
parallel curves, which only difference is the positioning of the curve, the shift being given by (c1-c0). If c1 is greater than c0, the curve corresponding to the sample of interest is shifted to the
right of the curve corresponding to the standard sample, and vice-versa.)
Five parameter parallel lines logistic model
The five parameter parallel lines logistic model writes:
y = a + (d -a) / [1 + (st * x / c[0] + sp * x / c[1])^b]^e model (2.2)
What XLSTAT can do
XLSTAT allows to fit:
• model 1.1 or 1.2 to a standard sample or to the sample of interest,
• model 2.1 or 2.2 to the standard sample and and to the standard sample the same time.
XLSTAT allows to either fit models 1.1 or 1.2 to a given sample (A case), or to fit models 1.1 or 1.2 to the standard (0) sample and then fit models 2.1 or 2.2 to both the standard sample and the
sample of interest (B case).
If the Dixon’s test option is activated, XLSTAT tests for each sample if some outliers influence too much the fit of the model. In the A case, a Dixon’s test is performed once the model 1.1 or 1.2 is
fitted. If an outlier is detected, it is removed, and the model is fitted again, and so on, until no outlier is detected. In the B case, we first perform a Dixon’s test on the standard sample, then
on the sample of interest, and then, the models 2.1 or 2.2 is fitted on the merged samples, without the outliers.
In the B case, and if the sum of the sample sizes is greater than 9, a Fisher’s F test is performed to detect if the a, b, d and e parameters obtained with models 1.1 or 1.2 are not significantly
different from those obtained with model 2.1 or 2.2.
Results displayed by XLSTAT
If no group or a single sample was selected, the results are shown for the model and for this sample. If several sub-samples were defined (see sub-samples option in the dialog), the model is first
adjusted to the standard sample, then each sub-sample is compared to the standard sample.
Fisher's test assessing parallelism between curves: The Fisher’s F test is used to determine if one can consider that the models corresponding the standard sample and the sample of interest are
significantly different or not. If the probability corresponding to the F value is lower than the significance level, then one can consider that the difference is significant.
Goodness of fit coefficients: This table shows the following statistics:
• The number of observations;
• The number of degrees of freedom (DF);
• The determination coefficient R2;
• The sum of squares of the errors (or residuals) of the model (SSE or SSR respectively);
• The means of the squares of the errors (or residuals) of the model (MSE or MSR);
• The root mean squares of the errors (or residuals) of the model (RMSE or RMSR);
Model parameters: This table displays the estimator and the standard error of the estimator for each parameter of the model. It is followed by the equation of the model.
Predictions and residuals: This table displays giving for each observation the input data and corresponding prediction and residual. The outliers detected by the Dixon’s test, if any, are displayed
in bold.
Charts: On the first chart are displayed in blue color, the data and the curve corresponding to the standard sample, and in red color, the data and the curve corresponding to the sample of interest.
A chart that allows to compare predictions and observed values as well as the bar chart of the residuals are also displayed.
analyze your data with xlstat
14-day free trial | {"url":"https://www.xlstat.com/en/solutions/features/four-and-five-parameter-parallel-lines-logistic-regression","timestamp":"2024-11-12T02:27:07Z","content_type":"text/html","content_length":"25230","record_id":"<urn:uuid:98e1006f-0b13-4c1e-abee-7f7a81cd6e16>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00096.warc.gz"} |
Infinite Series Module/Units/Unit 2/2.1 The Divergence Test/2.1.01 Introduction to The Divergence Test
In previous lessons, we defined the concept of the convergence of an infinite series: the infinite series
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k}\end{aligned}}}
converges if and only if the limit
Failed to parse (syntax error): {\displaystyle \lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \ \sum_{k=1}^{n} a_k}
exists. As was mentioned, it can be very difficult to apply this limit directly, leading us to the question: is there an easier way to determine if an infinite series converges or diverges?
Consider for example
${\displaystyle \sum _{k=1}^{\infty }\arctan(k),}$
or perhaps
${\displaystyle \sum _{k=1}^{\infty }{\frac {k}{\sqrt {1+k^{2}}}}.}$
There might be a way of finding an explicit formula for ${\displaystyle s_{n}}$ for these series, and with these formula we may be able to use the definition of convergence directly to establish
whether or not these series converge.
But, there is an easier way that quickly tells us that they do not. We can use the divergence test. | {"url":"https://wiki.ubc.ca/Science:Infinite_Series_Module/Units/Unit_2/2.1_The_Divergence_Test/2.1.01_Introduction_to_The_Divergence_Test","timestamp":"2024-11-12T02:33:36Z","content_type":"text/html","content_length":"29732","record_id":"<urn:uuid:02ec07c6-3174-439a-abe7-8d182a2fbb3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00584.warc.gz"} |
- T
Lesson 7
Applying Ratios in Right Triangles
7.1: Tilted Triangle (5 minutes)
Students are asked to calculate side lengths of a right triangle. They can apply their new understanding from the previous lesson and use trigonometry.
Student Facing
Calculate the lengths of sides \(AC\) and \(BC\).
Activity Synthesis
Remind students that \(\sin(20)\) is shorthand for “the length of the side opposite the 20 degree angle divided by the length of the hypotenuse for any right triangle with an acute angle of 20
degrees.” So when they ask the calculator to display \(\sin(20)\), they are asking the calculator for the ratio that is constant for all the triangles similar to this one by the Angle-Angle Triangle
Similarity Theorem.
7.2: Info Gap: Trigonometry (20 minutes)
This info gap activity gives students an opportunity to determine and request the information needed to calculate side lengths of right triangles using trigonometry.
The info gap structure requires students to make sense of problems by determining what information is necessary, and then to ask for information they need to solve it. This may take several rounds of
discussion if their first requests do not yield the information they need (MP1). It also allows them to refine the language they use and ask increasingly more precise questions until they get the
information they need (MP6).
Here is the text of the cards for reference and planning:
Tell students they will continue to use trigonometry to solve for side lengths of right triangles. Explain the info gap structure, and consider demonstrating the protocol if students are unfamiliar
with it.
Arrange students in groups of 2. In each group, distribute a problem card to one student and a data card to the other student. After reviewing their work on the first problem, give them the cards for
a second problem and instruct them to switch roles.
Conversing: This activity uses MLR4 Information Gap to give students a purpose for discussing information necessary to solve problems calculating side lengths of right triangles using trigonometry.
Display questions or question starters for students who need a starting point such as: “Can you tell me . . . (specific piece of information)?”, and “Why do you need to know . . . (that piece of
Design Principle(s): Cultivate Conversation
Engagement: Develop Effort and Persistence. Display or provide students with a physical copy of the written directions. Check for understanding by inviting students to rephrase directions in their
own words. Keep the display of directions visible throughout the activity.
Supports accessibility for: Memory; Organization
Student Facing
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the data card:
1. Silently read the information on your card.
2. Ask your partner, “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your
3. Before telling your partner the information, ask “Why do you need to know (that piece of information)?”
4. Read the problem card, and solve the problem independently.
5. Share the data card, and discuss your reasoning.
If your teacher gives you the problem card:
1. Silently read your card and think about what information you need to answer the question.
2. Ask your partner for the specific information that you need.
3. Explain to your partner how you are using the information to solve the problem.
4. When you have enough information, share the problem card with your partner, and solve the problem independently.
5. Read the data card, and discuss your reasoning.
Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.
Activity Synthesis
After students have completed their work, share the correct answers and ask students to discuss the process of solving the problems. Here are some questions for discussion:
• “What information did you need to ask for?” (What letters and where they go. An acute angle and a side length.)
• “What could you figure out once you knew angle \(C\) was the right angle?” (The legs of the right triangle both have the letter \(C\) and the hypotenuse uses the other two letters)
Highlight for students that drawing a diagram and labeling it with the information provided is an important strategy. It’s too easy to forget something or mix up information without a clear diagram
to work from.
7.3: Tallest Tower (10 minutes)
Students continue to use trigonometry to calculate side lengths of right triangles. In this case they apply that skill to a real world context and engage in some error analysis during the synthesis.
Consider showing where Dubai and Philadelphia are on a map and defining masonry.
Writing, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their writing by providing them with multiple opportunities to clarify their explanations through
conversation. Give students time to meet with 2–3 partners to share their response to the first question. Students should first check to see if they agree with each other about the height of the
building. Provide listeners with prompts for feedback that will help their partner add detail to strengthen and clarify their ideas. For example, students can ask their partner, “What did you do
first?” or “How did you know to use tangent?” Next, provide students with 3–4 minutes to revise their initial draft based on feedback from their peers. This will help students explain how to use
trigonometry to calculate side lengths of right triangles.
Design Principle(s): Optimize output (for explanation)
Action and Expression: Internalize Executive Functions. Provide students with a four-column table to organize. Use these column headings: angle, adjacent side, opposite side, and hypotenuse. The
table will provide visual support for students to identify ratios.
Supports accessibility for: Language; Organization
Student Facing
1. The tallest building in the world is the Burj Khalifa in Dubai (as of April 2019).
If you’re standing on the bridge 250 meters from the bottom of the building, you have to look up at a 73 degree angle to see the top. How tall is the building? Explain or show your reasoning.
2. The tallest masonry building in the world is City Hall in Philadelphia (as of April 2019). If you’re standing on the street 1,300 feet from the bottom of the building, you have to look up at a 23
degree angle to see the top. How tall is the building? Explain or show your reasoning.
Student Facing
Are you ready for more?
You’re sitting on a ledge 300 feet from a building. You have to look up 60 degrees to see the top of the building and down 15 degrees to see the bottom of the building. How tall is the building?
Anticipated Misconceptions
If students struggle with the city hall question prompt then to draw a diagram with the information they know.
Activity Synthesis
Display this image of the Philadelphia City Hall:
“The exact heights are 829.8 meters for the Burj Khalifa and 548 feet for City Hall. Why don’t those numbers match your calculations?” (Rounding or measurement error. The difference between \(\tan
(23)\) and \(\tan(23.2)\) makes a big difference when you work with large numbers.) "Is it reasonable to assume you are accurate to the nearest tenth in this case?" (No, the provided measurements
were rounded to the nearest ten meters or hundred feet so our usual rounding scheme doesn't apply.)
Lesson Synthesis
Tell students, “In addition to measuring the heights of objects that are too tall to reach, professionals also use trigonometry to calculate the heights of objects they are designing. For example,
here is some information a billboard designer knows about a new site.”
Display this information:
• The local law says the maximum height from the ground to the top of any billboard is 50 feet.
• To see over the trees from the highway people need to look up at least 40 degrees.
• The highway is 47 feet from the billboard.
Invite students to discuss what they can do with this information. (How tall—from bottom to top of the image—can the billboard be?) Then invite students to answer the questions they generated. (To be
above the trees the bottom of the billboard must be 39.4 feet off the ground, so the billboard can be up to 10.6 feet tall.)
7.4: Cool-down - Tallest Tree (5 minutes)
Student Facing
Using trigonometry and properties of right triangles, we can calculate and estimate measures in different right triangles. We can use these skills to estimate unknown heights of objects that are too
tall to measure directly. For example, we can't reach the top of this tree with a measuring tape.
To calculate the height of the tree, we could stand where the angle between the top and bottom of the tree is 10 degrees. Since we know the distance to the tree (the adjacent leg) and would like to
know the height (the opposite leg), we need to use tangent. So \(\tan(10)=\frac{h}{100}\). In the calculator we can look up that \(\tan(10)\) is 0.176. Then we can calculate that \(h\) is about 17.6.
That means the tree is 17.6 feet tall. | {"url":"https://curriculum.illustrativemathematics.org/HS/teachers/2/4/7/index.html","timestamp":"2024-11-04T10:34:22Z","content_type":"text/html","content_length":"108133","record_id":"<urn:uuid:e7e0d8ef-bbbe-4859-aa1e-b5d4614656ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00723.warc.gz"} |
Modern History MCQ UPSC Students 2024 | MCQTUBE
Modern History MCQ UPSC
Modern History MCQ UPSC. We covered all the Modern History MCQ UPSC in this post for free so that you can practice well for the exam.
Install our MCQTUBE Android app from the Google Play Store and prepare for any competitive government exams for free.
These types of competitive MCQs appear in the exams like UPSC, state pcs, CDS, NDA, Assistant Commandant, SSC, Railway, Bank, Delhi police, UPSSSC, etc.
We created all the competitive exam MCQs into several small posts on our website for your convenience.
You will get their respective links in the related posts section provided below.
Related Posts:
Modern History MCQ UPSC for Students
Who gifted the Badshah Nama to King George in 1799?
(a) Abul Fazl
(b) Abdul Hamid Lahori
(c) Nawab of Awadh
(d) William Jones
Option c – Nawab of Awadh
Who founded Karnataka as an independent state in 1720?
(a) Yusuf Adil Shah
(b) Asaf Shah
(c) Hussain Shah
(d) Sadatullah Khan
Option d – Sadatullah Khan
(a) Anglo-Sikh War
(b) Anglo-Mysore War
(c) Anglo-Maratha VWar
(d) Carnatic War
Between whom were the Carnatic wars fought?
(a) French East India Corporation and English East India Corporation.
(b) French East India Corporation art Dutch East India Corporation.
(c) Dutch East India Corporation and the Portuguese.
(d) English East India Corporation and the Dutch.
Option a – French East India Corporation and English East India Corporation
Assertion (A) The French wet defeated by the British in the third Carnatic War. Reason (R) The Indian rulers did support the French. Codes
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true
Option c – A is true, but R is false
Who founded Hyderabad during the reign of Muhammad Shah ‘Rangila’?
(a) Nizam-u-Mulk, (Asaf Jah)
(b) Hasan Gangu
(c) Mir Jumla
(d) Quli Qutub Shah
Option a – Nizam-u-Mulk, (Asaf Jah)
Hyderabad state was established in
(a) 1723
(b) 1724
(c) 1725
(d) 1726
Who was the first Nawab Wazir of Awadh in the 18th century?
(a) Nawab Safdar Jang
(b) Nawab Saadat Ali Khan
(c) Nawab Shuja-ud-Daula
(d) Nawab Saadat khan
Option d – Nawab Saadat khan
Who founded the independent state of Awadh?
(a) Saadat Khan
(b) Yusuf Adil Shah
(c) Nizam-ul-Mulk
(d) Alivardi Khan
Which state was known as a Buffer State during the British reign?
(a) Awadh
(b) Bengal
(c) Mysore
(d) Punjab
Who was the second Nawab of Awadh?
(a) Shuja-ud-Daula
(b) Safdar Jang
(c) Asaf-ud-Daula
(d) Asaf Shah
Which city was developed as the full-fledged capital city by Nawab of Awadh Shuja-ud-Daula?
(a) Lucknow
(b) Kannauj
(c) Faizabad
(d) Prayag
The Nawab of Awadh who permanently transferred his capital from Faizabad to Lucknow was
(a) Safdar Jang
(b) Shuja-ud-Daula
(c) Asaf-ud-Daula
(d) Saadat Khan
Bara Imambara was built in 1784 in Lucknow by
(a) Wazir Ali
(b) Asaf-ud-Daula
(c) Shuja-ud-Daula
(d) Safdar Jang
Who was Birjis Qadr?
(a) The Nizam of Hyderabad
(b) The Nawab of Awadh
(c) The Mughal Emperor
(d) The Nawab of Bengal
Option b – The Nawab of Awadh
Who established the powerful kingdom of Bharatpur in 1720?
(a) Churaman
(b) Surajmal
(c) Gokul
(d) Badan Singh
Which Jat leader got the title of ‘Raja’ from Ahmed Shah Abdali?
(a) Badan Singh
(b) Rajarama
(c) Surajmal
(d) Deep Singh
Which of the following is remembered as the Plato of Jat tribe’ and as ‘Jat ulysses?
(a) Badan Singh
(b) Gokul Jat
(c) Surajmal
(d) Durga Singh
At which of the following places did Haider Ali build a modern Arsenal with the help of the French in 1755?
(a) Mysore
(b) Dindigul
(c) Srirangapatna
(d) Arcot
Who was the first South Indian ruler to defeat British armies?
(a) Tipu Sultan
(b) Haider Ali
(c) Nizam of Hyderabad
(d) None of the above
Tipu Sultan was the ruler of which state?
(a) Magadh
(b) Hyderabad
(c) Bangalore
(d) Mysore
At which place did Tipu Sultan establish his capital?
(a) Mysore
(b) Bangalore
(c) Srirangapatna
(d) Coimbatore
Who considered Tipu’s Mysore as “the most simple and despotic monarchy in the world”?
(a) Charles Napier
(b) Thomas Best
(c) Lord Cornwallis
(d) Thomas Munro
Which of the following statements is/are correct about the impact of contemporary European movements on the rise of modern nationalism in India in the late 19th and early 20th century?
1. Savarkar brothers organized a secret society called Mitra Mela, which later merged with Abhinav Bharat (after Garibaldi’s ‘Young Italy’) in 1904.
2. The national liberation movements of Greece and Italy had the deepest influence upon the nationalist ranks among other European nationalist movements.
Select the correct answer using the code given below:
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Option d – Neither 1 nor 2
With reference to the Moderates, consider the following statements:
1. Important leaders of this faction were Dadabhai Naoroji, Pherozeshah Mehta, D.E. Wacha, W.C. Bonnerjea, and S.N. Banerjea.
2. The Moderates believed that the British were inherently just but were not aware of the real conditions of the natives.
3. They used the method of ‘prayer and petition’ but did not resort to other means like protest or constitutional agitation.
Which of the statements given above is/are correct?
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) 3 only
Which one of the following movements has contributed to a split in the Indian National Congress resulting in the emergence of ‘moderates’ and ‘extremists’?
(a) Swadeshi Movement
(b) Quit India Movement
(c) Non-Cooperation Movement
(d) Civil Disobedience Movement
Option a – Swadeshi Movement
With reference to Dadabhai Naoroji, consider the following statements:
1. The Drain of Wealth theory was put forward by him.
2. He pledged “loyalty to the backbone” to the British Crown and desired the permanent continuance of the British rule in India.
Select the correct answer using the code given below:
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Consider the following statements with reference to the ‘Home Charges’:
1. It referred to the expenditure incurred in England by the Secretary of State on behalf of India.
2. It included remittances to England in the form of charges for effective and non-effective services of British troops on the Indian establishment and the pension of British military and civil
3. Interest on money expended in India on railways was not a part of home charges.
Which of the above statements is/are correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3
With reference to the book Desher Katha written by Sakharam Ganesh Deuskar during the freedom struggle, consider the following statements:
1. It warned against the colonial state’s hypnotic conquest of the mind.
2. It inspired the performance of swadeshi street plays and folk songs.
3. The use of ‘desh’ by Deuskar was in the specific context of the region of Bengal.
Which of the statements given above is/are correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3 only
Which of the following statements is/are correct with reference to the formation of the Indian National Congress?
1. The Safety Valve theory proposes that A.O. Hume formed the Congress with the idea that it would prove to be a ‘safety valve’ for releasing the growing discontent of the Indians.
2. Bipan Chandra observed that the early Congress leaders used Hume as a ‘lightning conductor’, i.e., as a catalyst to bring together the nationalistic forces.
Select the correct answer using the code given below:
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Which of the following were the factors responsible for the growth of early nationalism in India?
1. Introduction of a modern system of education
2. The unprecedented growth of the Indian Press
3. Effect of racial myths of white superiority
4. Organization of grand Delhi Durbar of 1877
5. The Indo-Aryan theory by European scholars
Select the correct answer using the code given below:
(a) 1, 2, 3 and 4 only
(b) 2, 3 and 5 only
(c) 1, 4 and 5 only
(d) 1, 2, 3, 4 and 5
Option d – 1, 2, 3, 4 and 5
Arrange the following leaders in the chronological order in which they held the Presidency of the Indian National Congress:
1. George Yule
2. Dadabhai Naoroji
3. Womesh Chandra Bonnerjee
4. Syed Badruddin Tyabji
5. William Wedderburn
Select the correct answer using the code given below:
(a) 1-2-3-4-5
(b) 3-5-1-2-4
(c) 3-2-4-1-5
(d) 1-3-2-4-5
Consider the following statements about the Indian National Congress session of 1916:
1. The death of Bal Gangadhar Tilak and Pherozeshah Mehta facilitated the reunion of moderate and extremist sections of the Congress.
2. The Congress and Muslim League joined forces due to growing anti-imperialist sentiment within the Muslim League.
3. It was the first time a woman, Annie Besant, presided over a session of the Indian National Congress.
How many of the statements given above are not correct?
(a) Only one
(b) Only two
(c) All three
(d) None
Which of the following statements regarding Aurobindo Ghosh is not correct?
(a) The Alipore Bomb case was a historic trial in which the British Government tried to implicate Sri Aurobindo in various revolutionary activities.
(b) The Life Divine by Sri Aurobindo was purely based on eastern spirituality while neglecting western thoughts as shallow.
(c) Yugantar was a Bengali weekly newspaper started by Sri Aurobindo that preached open revolt.
(d) Bande Mataram was an English language newspaper published from Calcutta by Bipin Chandra Pal and edited by Aurobindo Ghosh.
Option b – The Life Divine by Sri Aurobindo was purely based on eastern spirituality while neglecting western thoughts as shallow
Consider the following statements with respect to the Lucknow Pact of 1916:
1. The Indian National Congress session was presided over by Ambika Charan Majumdar.
2. The moderates and extremists reconciled due to the efforts of Gopal Krishna Gokhale and Pherozeshah Mehta.
3. The Indian National Congress accepted the Muslim League’s position on separate electorates.
Which of the statements given above is/are correct?
(a) 3 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) 1 and 3 only
Consider the following events in the history of India:
1. Publication of Poverty and Un-British Rule in India
2. First INC session held
3. Lord Ripon’s resolution on local self-government
4. Vande Mataram recited in the form of a song by Rabindranath Tagore at the INC session
What is the correct chronological order of the above events, starting from the earliest time?
(a) 4-1-2-3
(b) 3-2-4-1
(c) 1-2-4-3
(d) 3-4-1-2
Consider the following statements:
1. The French were the last Europeans to come to India with the purpose of trade.
2. The French introduced tomatoes and chillies in India.
3. The first English factory in India was established at Surat.
4. The Battle of Wandiwash was won by the French in 1760 at Vandavasi in Tamil Nadu.
Which of the statements given above are correct?
(a) 1 and 3 only
(b) 2 and 4 only
(c) 1, 2 and 3
(d) 3 and 4 only
The staple commodities of export by the English East India Company from Bengal in the middle of the 18th century were:
(a) Raw cotton, oil seeds and opium
(b) Sugar, salt, zinc and lead
(c) Copper, silver, gold, spices and tea
(d) Cotton, silk, saltpetre and opium
Option d – Cotton, silk, saltpetre and opium
Consider the following statements:
1. In accordance with the terms of the pact signed in 1760, Mir Jafar consented to hand over the districts of Burdwan, Midnapur, and Chittagong to the Company.
2. After the 3rd Anglo-Maratha War, the Peshwa’s territories were absorbed into the Bombay Presidency, and the territories seized from the Pindaris became the Central Provinces of British India.
3. Due to internal anarchy in Sindh and Punjab, it became imperative for the British to annex the Sindh and Punjab regions to safeguard their western frontier.
4. Auckland followed the Policy of Proud Reserve which was aimed at having scientific frontiers and safeguarding ‘spheres of influence’.
Which of the statements given above is/are correct?
(a) 2 only
(b) 2 and 3 only
(c) 1 and 4 only
(d) 1 and 2 only
(a) Running community kitchens for freedom fighters during the Quit India Movement.
(b) Fight against the British during the Revolt of 1857.
(c) Participation in the Dharasana Satyagraha in 1931.
(d) Participation in the Indigo revolt
With reference to the revolt of 1857, consider the following statements:
1. The Azamgarh Proclamation of August 1857 was one of the main sources of our knowledge about what the rebels wanted.
2. British officer Colonel Oncell captured Banaras during the revolt of 1857.
Which of the statements given above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
With whose permission did the English set up their first factory in Surat?
(a) Akbar
(b) Jahangir
(c) Shahjahan
(d) Aurangzeb
Consider the following statements about Third and Fourth Anglo Mysore wars:
1. The governor general during the Third Anglo Mysore war was Lord Wellesley.
2. The governor general during the Fourth Anglo Mysore war was Lord Cornwallis.
3. After the 4th Anglo Mysore war, the British restored Wodeyars to the throne by way of a subsidiary alliance.
Which of the statements given above are correct?
(a) 1 and 2 only
(b) 3 only
(c) 1 and 3 only
(d) 2 and 3 only
Which one of the following statements is not correct regarding Tipu Sultan?
(a) The Jacobin Club of Mysore was founded by French Republican officers.
(b) He is credited as the ‘pioneer of rocket technology’ in India.
(c) Tipu was unable to fulfill the terms of the Treaty of Seringapatam.
(d) Tipu embraced western military methods like artillery and rockets.
Option c – Tipu was unable to fulfill the terms of the Treaty of Seringapatam
Consider the following statements about Anglo-Maratha wars:
1. The Treaty of Salbai ended the First Anglo-Maratha war.
2. As per the Treaty of Bassein, Baji Rao II agreed to the maintenance of British troops in his empire.
Which of the statements given above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Consider the following statements about Anglo-Sikh wars:
1. Maharaja Ranjit Singh maintained an army consisting only of Sikh men.
2. The Treaty of Lahore was signed between the British East India Company (EIC) and Maharaja Ranjit Singh.
Which of the statements given above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Option d – Neither 1 nor 2
Mangal Pande belonged to which one of the following native infantries?
(a) 3rd Native Infantry
(b) 19th Native Infantry
(c) 34th Native Infantry
(d) 7th Native Infantry
Option c – 34th Native Infantry
Which of the following measures were taken by Mir Qasim to check the rising British power and strengthen his position in Bengal?
1. He shifted his capital from Murshidabad to Munger to keep a strategic distance from the Company.
2. He abolished all customs duties to bring equality between local traders and European traders.
3. He conferred the Zamindari of 24 parganas and janglimahals (small administrative units) to the British East India Company.
Choose the correct code:
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) All of the above
What was/were the object/objects of Queen Victoria’s Proclamation (1858)?
1. To disclaim any intention to annex Indian States
2. To place the Indian administration under the British Crown
3. To regulate East India Company’s trade with India
Select the correct answer using the code given below.
(a) 1 and 2 only
(b) 2 only
(c) 1 and 3 only
(d) 1, 2 and 3
Who among the following called the revolt of 1857 as the first war of national independence in his book?
(a) Jawaharlal Nehru
(b) R.P. Dutt
(c) V.D. Savarkar
(d) R.C. Majumdar
With reference to the causes that led to the Battle of Plassey, consider the following statements:
1. The rampant misuse of the trade privileges given to the East India Company by the Nawab of Bengal was one of the reasons that led to the battle.
2. Fortification of Calcutta by the French without the Nawab’s permission.
3. Non-payment of tax and duty by the workers of East India Company.
How many of the statements given above are not correct?
(a) Only one
(b) Only two
(c) All three
(d) None
Consider the following statements about the Battle of Buxar:
1. The battle was fought between the French Forces and a joint army of the Nawab of Oudh and Nawab of Bengal.
2. Nawab of Bengal and Nawab of Awadh lost the battle which led to France becoming the greatest power in Northern India.
3. The French army was led by Hector Munro.
Which of the statements given above is/are correct?
(a) 3 only
(b) 2 and 3 only
(c) 1 and 2 only
(d) None of the above
Option d – None of the above
We covered all the modern history MCQ UPSC above in this post for free so that you can practice well for the exam.
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What Is a Variable in Math
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What Is a Variable in Math Can Be Fun for Everyone
We’ll talk more about this in a subsequent tutorial. This animation explains the idea of variables. It is helpful to put variables into various categories, as different statistical procedures apply
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Matrices don’t have to explicitly dimensioned, and MATLAB makes it possible for you to raise the magnitude of a matrix as you work. Argument values that themselves contain commas ought to be escaped
as essential. Dependent variables are the ones that are changed by the independent variables.
A History of What Is a Variable in Math Refuted
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The What Is a Variable in Math Cover Up
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Thus, let’s look at all the lessons for Probability. 1 way to address equations that students will know about is to locate a missing addend. Algebra is far more interesting when things are somewhat
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You add terms having the exact same variables due to the fact that they represent the exact same quantities. To begin with, the probability density function has to be normalized. It’s the coefficient
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A system of equations is a selection of a couple of equations with the exact same set of unknowns. Sometimes you’ll be given more than 1 variable and asked to fix the equation. Independent variables
are values that could be changed in a specific equation or experiment. | {"url":"https://diffusion-rec.fr/2019/07/18/what-is-a-variable-in-math-overview/","timestamp":"2024-11-10T21:12:35Z","content_type":"application/xhtml+xml","content_length":"61907","record_id":"<urn:uuid:5c8ce322-dbeb-4e7e-b971-90209326a64b>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00166.warc.gz"} |
Fundamentals and Building Blocks worksheets
World's Tallest Buildings
Polynomials - Fundamentals
2.4 - Building Blocks of Life
1.1 Building Blocks of Geometry
Fundamentals of electricity
The Atom: The Building Blocks of Matter Review
Chem Ch. 3: Atoms - the Building Blocks of Matter
The Building Blocks of Life
Classification of buildings and Basic building structures
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Building Materials: Lumber
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Industrial Revolution Building Blocks
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compute slice command
compute slice command
compute ID group-ID slice Nstart Nstop Nskip input1 input2 ...
• ID, group-ID are documented in compute command
• slice = style name of this compute command
• Nstart = starting index within input vector(s)
• Nstop = stopping index within input vector(s)
• Nskip = extract every Nskip elements from input vector(s)
• input = c_ID, c_ID[N], f_ID, f_ID[N]
c_ID = global vector calculated by a compute with ID
c_ID[I] = Ith column of global array calculated by a compute with ID
f_ID = global vector calculated by a fix with ID
f_ID[I] = Ith column of global array calculated by a fix with ID
v_name = vector calculated by an vector-style variable with name
compute 1 all slice 1 100 10 c_msdmol[4]
compute 1 all slice 301 400 1 c_msdmol[4] v_myVec
Define a calculation that “slices” one or more vector inputs into smaller vectors, one per listed input. The inputs can be global quantities; they cannot be per-atom or local quantities. Computes and
fixes and vector-style variables can generate such global quantities. The group specified with this command is ignored.
The values extracted from the input vector(s) are determined by the Nstart, Nstop, and Nskip parameters. The elements of an input vector of length N are indexed from 1 to N. Starting at element
Nstart, every Mth element is extracted, where M = Nskip, until element Nstop is reached. The extracted quantities are stored as a vector, which is typically shorter than the input vector.
Each listed input is operated on independently to produce one output vector. Each listed input must be a global vector or column of a global array calculated by another compute or fix.
If an input value begins with “c_”, a compute ID must follow which has been previously defined in the input script and which generates a global vector or array. See the individual compute doc page
for details. If no bracketed integer is appended, the vector calculated by the compute is used. If a bracketed integer is appended, the Ith column of the array calculated by the compute is used.
Users can also write code for their own compute styles and add them to LAMMPS.
If a value begins with “f_”, a fix ID must follow which has been previously defined in the input script and which generates a global vector or array. See the individual fix page for details. Note
that some fixes only produce their values on certain timesteps, which must be compatible with when compute slice references the values, else an error results. If no bracketed integer is appended, the
vector calculated by the fix is used. If a bracketed integer is appended, the Ith column of the array calculated by the fix is used. Users can also write code for their own fix style and add them to
If an input value begins with “v_”, a variable name must follow which has been previously defined in the input script. Only vector-style variables can be referenced. See the variable command for
details. Note that variables of style vector define a formula which can reference individual atom properties or thermodynamic keywords, or they can invoke other computes, fixes, or variables when
they are evaluated, so this is a very general means of specifying quantities to slice.
If a single input is specified this compute produces a global vector, even if the length of the vector is 1. If multiple inputs are specified, then a global array of values is produced, with the
number of columns equal to the number of inputs specified.
Output info
This compute calculates a global vector if a single input value is specified or a global array with N columns where N is the number of inputs. The length of the vector or the number of rows in the
array is equal to the number of values extracted from each input vector. These values can be used by any command that uses global vector or array values from a compute as input. See the Howto output
page for an overview of LAMMPS output options.
The vector or array values calculated by this compute are simply copies of values generated by computes or fixes or variables that are input vectors to this compute. If there is a single input vector
of intensive and/or extensive values, then each value in the vector of values calculated by this compute will be “intensive” or “extensive”, depending on the corresponding input value. If there are
multiple input vectors, and all the values in them are intensive, then the array values calculated by this compute are “intensive”. If there are multiple input vectors, and any value in them is
extensive, then the array values calculated by this compute are “extensive”. Values produced by a variable are treated as intensive.
The vector or array values will be in whatever units the input quantities are in. | {"url":"https://docs.lammps.org/stable/compute_slice.html","timestamp":"2024-11-06T20:55:39Z","content_type":"text/html","content_length":"41125","record_id":"<urn:uuid:fd7e53d1-df39-456a-846d-ccc0697f7988>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00041.warc.gz"} |
2. Find the capacity in litres of a conical vessel with... | Filo
Question asked by Filo student
2. Find the capacity in litres of a conical vessel with
a. (i) radius , slant height
b. (ii) height , slant height .
c. 3. The height of a cone is . If its volume is , find the radius of the base,
d. (Use )
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Question Text 2. Find the capacity in litres of a conical vessel with
Updated On Jan 31, 2023
Topic Calculus
Subject Mathematics
Class Class 12
Answer Type Video solution: 1
Upvotes 128
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Visualize - Plot Graph
Use a plot graph when you want to show the relationship between two or more variables. Two variables are plotted on the horizontal and vertical axes of the graph, while a third variable can be added
and is represented by the size of the bubble.
Plot graphs are the default visualization when you have two or more measures across in your analysis.
Plot Graphs
A plot graph is helpful to display the relationship between, for example, two product attributes. You can plot attributes such as sweetness and texture and include the third variable of price. In
such a graph, there would be a bubble to represent each brand. The score of each brand on sweetness and texture determines the bubble position in the chart. The price of each brand determines the
size of the bubbles.
In the example below, Lifestage is added to the down drop zone, and three measures are selected across. Harmoni plots the first measure, Airfare Cost, in the horizontal axis (x-axis) and the second
measure in the vertical axis (y-axis), displaying the results for each Lifestage group. The third measure, Total Spend, controls the size of each bubble.
In an analysis with three dimensions across, Harmoni plots the graph as follows:
• The first column in the table (excluding the total) is represented by the x-axis.
• The second column is the y-axis.
• The third column is the size of the bubbles. Each bubble on the graph represents a row.
If you select two measures as you create the table, then visualize using the plot graph, you can add the bubble size to the plot graph later by dragging the third measure to the Bubble Size zone.
Where to from here?
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Ztest: Excel Formulae Explained - ManyCoders
Key Takeaway:
• Excel formulae are essential for all data analysis: They enable users to perform complex calculations and automate repetitive tasks.
• Basic mathematical formulae are a crucial starting point: These include simple addition and subtraction, multiplication and division, and using parentheses to control the order of operations.
• Lookup and reference formulae are powerful tools: These include functions like VLOOKUP, HLOOKUP, and INDEX/MATCH, which allow users to search through large datasets and retrieve specific
information easily.
Are you struggling to understand the different Excel formulas? Here, you’ll discover all you need to know about ZTEST and how to use it. With this guide, you’ll be creating accurate, insightful
reports in no time.
The Necessity of Excel Formulae
Excel formulae are essential when dealing with a lot of data in Excel. That’s because they automate calculations, making work more efficient and precise. Without formulae, users would have to
manually compute each detail, which takes lots of time and can cause mistakes.
Using Excel formulae is advantageous because it helps save time. Automating calculations lets users quickly analyze data, without having to sort or calculate themselves. This improved efficiency
leads to better decision-making and productivity.
Furthermore, Excel formulae are precise. There’s less chance of error when relying on formulas, versus manual calculations. Also, formulae provide a consistency that isn’t always achievable with
manual work.
Plus, Excel formulae are customizable and allow users to make complex calculations beyond basic calculations. For instance, if a user needs to calculate sales tax based on various percentages for
distinct regions or products, they can do this using Excel’s formula capabilities.
However, misusing Excel formulae can lead to errors and incorrect calculations that could affect the entire dataset. Thus, it’s critical for people using formulas to understand them completely before
using them.
Basic Mathematical Formulae
Excel’s basic mathematical formulae consist of + (addition), – (subtraction), * (multiplication), and / (division).
SUM is the most used formula; it quickly adds up numbers in a list. AVERAGE is another popular formula that finds the average of a set of numbers.
These basic formulae become more useful when combined with other functions. Understanding how they work helps you save time and get accurate calculations.
An example of using maths formulae in real life is managing personal finances. With SUM and your bank statement data, you can track your expenditure and income easily for budgeting.
Text formulae are also an essential tool in Excel for manipulating text data.
Text Formulae
LEFT and RIGHT are useful Text Formulae for extracting characters from either the left or right side of a cell. FIND and SEARCH help find characters or words within a string.
MID can extract a set of characters from anywhere in a cell. LEN calculates the number of characters in a cell. SUBSTITUTE replaces text, but keeps the original character count.
Combine Text Formulae with other functions like IF statements and VLOOKUP tables for accurate and efficient data analysis.
Next, let’s look at Logical Formulae.
Logical Formulae
Logical formulae are a set of Excel functions. They decide if something is true or false. These allow users to make decisions about their spreadsheet data. An example of a logical function is the IF
function. It checks if a condition is met. It returns one value if true and another if false. The syntax for the IF function is =IF(logical_test,value_if_true,value_if_false).
The AND function checks if all arguments are true. It returns TRUE if they are and FALSE otherwise. An example is =AND(A1>10,B1<20). It returns TRUE if both conditions are met.
The OR function is different. It returns TRUE if any argument is true. For instance, =OR(A1>10,B1<20) returns TRUE if either condition is met.
The NOT function reverses the result of a logical test. It returns TRUE (when the condition is false) or FALSE (when it’s true). Its syntax is usually =NOT(logical_test).
Other useful logical functions include COUNTIF, SUMIF and AVERAGEIF/AVERAGEIFS/AGGREGATE. These count, sum and calculate averages based on given conditions.
If you don’t understand these logical formulae, you may miss important insights. Or, you may spend time sifting through data manually instead of using Excel’s features.
In this article series, we’ll cover “Understanding Excel Functions”. We’ll look at different types of Excel functions. These include financial, statistical, date & time functions, etc. This will help
users be more efficient when working with large datasets.
Understanding Excel Functions
Excel functions are powerful tools for data analysis. Let’s dive into some of the most commonly used ones. First, the SUM function adds up values in a range of cells. Next is the COUNT function. It
counts the number of cells with numerical values in a range. Lastly, the IF function tests criteria. Depending on the results, it performs different actions. By the end of this section, you’ll be a
pro at using Excel functions!
The Purpose of Excel Functions
Excel functions simplify calculations and data analysis in spreadsheets. They are pre-built formulas that perform operations quickly and easily. This saves time and reduces errors from manual
A table can be created to show examples of commonly used functions. The first column is the name, like SUM, AVERAGE, COUNT, or ZTEST. The second column shows how the function works with data.
Functions serve multiple purposes depending on the user’s needs. One benefit is streamlining tasks. For example, the AVERAGE function computes multiple group averages quickly and automatically.
Another advantage is reducing errors when working with large data sets. Functions take care of tasks accurately, without mistakes or oversight.
Another benefit of using Excel functions is improved communication. With results from functions like ZTEST, stakeholders can understand conclusions from data.
To effectively use Excel functions, it is suggested to:
• Organize spreadsheet data first.
• Add descriptions for numerical columns.
• Combine multiple functions to analyze complex data.
Let’s now look at the SUM Function.
The SUM Function
Check out the table below to understand The SUM Function:
Product Price ($) Quantity
Item 1 10 5
Item 2 20 3
Item 3 15 7
If your cursor is in cell D4, enter “=SUM(C2:C4)” into the formula bar and press enter. This will give the total quantity sold for all items.
The SUM Function has been around since the days of Excel. It helps to automate tasks and improve productivity.
Next, The COUNT Function counts the number of cells with numbers or dates as values.
The COUNT Function
To use the COUNT Function, type “=COUNT” into a cell. Then add the range of cells in parentheses that you want to count. For instance, “(A1:A10)”.
The COUNT Function only counts numbers. Blank or text-filled cells are ignored. Also, errors like “#N/A” or “#VALUE!” are not counted.
The COUNT Function has been used since the days of spreadsheets. It was originally used for basic math calculations. Now it has more complicated applications such as data analysis and visualizations.
The IF Function is another handy tool in Excel. It lets you create logic statements depending on what you need. This is useful for sorting data or making reports.
The IF Function
Need to use the IF Function? Start by specifying the condition to check. It could be simple, like a number being greater than or equal to another number. Or, complex – like a text string containing a
Tell Excel what to do if the condition is true or false. Could involve displaying a message, performing a calculation or running an action.
Great thing is, the IF Function can be nested inside other formulas. Create sophisticated decision-making processes using just a few lines of code.
Not using the IF Function? You’re missing out on its most powerful feature. Take time to understand how it works and experiment with different combinations. Streamline workflow and make spreadsheets
more efficient.
Next section: Lookup and Reference Formulae: A Guide – another important tool in your Excel toolkit.
Lookup and Reference Formulae: A Guide
Searching for a specific piece of data in a sea of info can be overwhelming. But, it doesn’t have to be! With the right Excel formulae, it can be easy. Let’s explore the powerful world of lookup and
reference formulae.
We’ll start with VLOOKUP. It searches for a value in the first column of a table and returns a corresponding value in the same row.
Next, we’ll talk about HLOOKUP. It works the same as VLOOKUP, but searches horizontally instead of vertically.
Finally, we’ll decode the INDEX/MATCH Function. It makes it easy to locate and retrieve specific data from your spreadsheets.
Get ready to make your data work for you!
Understanding VLOOKUP Function
The VLOOKUP function needs understanding. Lookup and Reference Formulae in Excel help extract data from tables. Let’s look at the syntax of VLOOKUP. The table below gives us an idea of what each
value does.
Value Definition
Lookup Value The cell or text string to search for in the table array.
Table Array The data range containing the lookup value and return value.
Column Index Number The column number with the return value. Counts from leftmost column in range.
Range Lookup True if approximate matches are wanted. False if only exact matches desired.
Using this knowledge, you can use Excel’s search functionality to build accurate formulas. Master every formula you come across for Excel’s full potential. Now, let’s get to the HLOOKUP function.
Decoding HLOOKUP Function
Let’s get right into it – HLOOKUP is a very helpful formula for Excel users. It helps find particular data in tables, making work faster and easier. To understand the formula, let’s create an example
table with “Product ID,” “Quarter 1,” “Quarter 2,” “Quarter 3,” and “Quarter 4” as its headers. Each row in the table will have a different product ID and sales figures for each quarter.
Now let’s look at how HLOOKUP works. It is used to search for data in horizontal rows – just like VLOOKUP searches in vertical columns. You have to provide the value you are looking for – like a
Product ID – and the range of data it could be in – in our case, Quarters one to four. It’s different from VLOOKUP because it searches from top-to-bottom instead of left-to-right.
It is important to make sure your values correspond with the source of reference for your search parameter.
Did you know many new Excel users don’t know how useful Lookup Formulae can be? It’s true! They are great tools for experienced analysts and casual Excel users.
Next: INDEX/MATCH Function – another helpful tool for working with tables in Excel.
The INDEX/MATCH Function Explained
A table is great for understanding the INDEX/MATCH function. Here’s what it looks like:
Function Explanation
INDEX Returns a value at a specific row and column in a range.
MATCH Searches for a specified value in an array and returns its relative position.
INDEX/MATCH Combines INDEX and MATCH functions to retrieve data based on criteria, replacing VLOOKUP.
INDEX/MATCH is better than VLOOKUP. It can search multiple columns, uses exact matches as default, and doesn’t need left-to-right column order. Instead of stating what column to get data from, you
give the number of columns from the start of the range. Also, if data changes often or if you’re dealing with large datasets, INDEX/MATCH increases performance.
Top Tip: To make sure that your formulae reference the correct ranges and don’t break when data changes, use named ranges instead of cell references in your formulae. Like, use =INDEX(myData,MATCH
(…)) instead of referencing B2:C10.
Next Heading: Date and Time Formulae: Simplifying Operations
Date and Time Formulae: Simplifying Operations
Data work needs date and time manipulation. In this Excel formulae guide, we’ll explore these formulae. We’ll show how to calculate durations and work with calendar dates. Excel’s built-in functions
make it easy.
First up is the TODAY function. It tracks the current date.
Then, the NOW function displays the current time.
Lastly, we’ll look at the DATE function. It constructs complex date values. Mastering this is beneficial.
TODAY Function and Its Benefits
The TODAY Function in Excel can be really handy! It updates automatically and simplifies date & time operations – no manual input needed.
You can use it to track dates, calculate project deadlines, or create dynamic tables & charts.
Did you know that the =TODAY() formula returns a serial number? To view it in an easier-to-read format, right-click on any cell and select ‘Format Cells’. Then choose your preferred Date format.
That’s it for the TODAY Function. Next up: NOW Function: A Comprehensive Overview.
NOW Function: A Comprehensive Overview
The NOW Function: A Comprehensive Overview allows users to access today’s date and time in an Excel cell. This function auto-updates itself every recalculation or when the worksheet is opened.
For example, here is how the NOW function looks like in a table:
Function Syntax Description
NOW =NOW() Returns the current date and time
It can be useful to track changes or measure how long a task takes. For instance, if you enter the NOW formula in cell A1, and complete a task in cell B1, you can subtract A1 from B1 to calculate how
many hours or minutes were spent.
An interesting fact about the NOW function is that it returns a numeric value. Each day is represented by a value of 1, and each hour is 0.04166. This makes it possible to do various calculations
with the returned value.
Next, let’s look at another useful Excel formula: The DATE Function.
Unraveling the DATE Function
The DATE Function unravels easily when you refer to the below table.
Function Description
YEAR Yields the year of a given date
MONTH Yields the month of a given date
DAY Yields the day of a given date
By using these functions, you can separate components from dates for simplifying calculations. For instance, you can employ the YEAR function to find dates in a particular year from a list of dates.
A practical example is when handling sales data of a business. You may need to review sales trends by month or quarter for predicting future profits. These functions will help you to separate out the
relevant data for analysis.
Next, Financial Formulae: Financial Math Made Easy will be discussed in the upcoming section.
Financial Formulae: Financial Math Made Easy
Years in finance have shown me how intimidating the math of financial planning and analysis can be. I’m keen to dive into the formulae that make Excel so powerful for those in my field. In this
chapter, I’ll run you through three key functions that are vital to know. They are: PMT, FV and NPV. I’ll explain the basics of each, and provide examples of how to use them for smart financial
Understanding the PMT Function
PMT Function:
Column 1 Column 2
Definition Calculates regular payments for a loan/investment with fixed interest rate and periods.
Inputs Rate, Nper, Pv, [Fv], [Type]
Formula =PMT(rate, nper, pv, [fv], [type])
Example -PMT(T(7%/12,5*12,-15000))=274.01
The PMT function takes inputs: interest rate per period (rate), number of periods (nper), present value (pv), future value (if applicable) ([fv]), and payment due at the beginning or end of each
period ([type]). With these inputs, it finds the payment amount to cover principal and interest within the specified time frame.
This formula is important for making informed financial decisions. My friend had difficulty understanding it, but after studying and practicing with examples, she was able to use it for her needs.
Moving on, let’s explore another key financial formula – The FV Function and Its Significance.
The FV Function and Its Significance
The FV Function is one of the most important formulae for finance in Excel. It stands for Future Value and is used to work out investment growth over a period of time. Three main inputs are needed –
present value, interest rate, and the number of periods. This function can show you how much your investment will grow or shrink in the future.
The below table demonstrates the importance of the FV Function.
Present Value Interest Rate Number of Periods Future Value
$1000 5% 10 $1,628.89
It shows that if you invest $1,000 at 5% for 10 years, the future value will be $1,628.89. This means that you can work out what you will have in the future if you invest now at a certain interest
The FV Function is essential when planning for retirement or saving up for big expenses. For example, if you want to save $30,000 in 5 years and already have $20,000 saved, you can use the FV
Function with the interest rate and timeline to figure out how much more you need to add each year to reach your goal.
The NPV Function: A Guide to Its Uses
Check out the table below. It shows the NPV formula with actual data.
Cash Flow Discount Rate NPV Formula
-$5,000 10% =NPV(10%, -$5,000,$3,000,$4,000)
The NPV formula helps investors figure out if an investment will be profitable. It takes into account the cost of the investment and future cash flows. Plus, the time value of money. Financial
software like Excel can use this formula.
Businesses can also compare investments using NPV. For example, if a company has two projects – A needs an initial investment of $50,000 and has an expected NPV of $20,000. Project B needs $25,000
but has an expected NPV of $10,000. In this case, they should pick project A as it is more profitable.
I have used NPV as a financial analyst at ABC Inc. We had to choose between building a new factory or expanding the current one. We put the data points into Excel’s NPV formula for each option. The
result was that expanding the current facility would be more profitable – and that is the decision we made.
Five Facts About ZTEST: Excel Formulae Explained:
• ✅ ZTEST is an Excel formula used to test a hypothesis on population mean when the standard deviation is not known. (Source: Investopedia)
• ✅ It is one of several “Z” tests that can be used in statistical analysis. (Source: Excel Easy)
• ✅ The formula returns the probability of a null hypothesis being true based on the sample mean and size. (Source: Data Analysis Express)
• ✅ ZTEST can be useful in many fields, including finance, healthcare, and engineering. (Source: Techopedia)
• ✅ Understanding ZTEST and other Excel formulae can greatly improve data analysis skills and career prospects. (Source: Udemy)
FAQs about Ztest: Excel Formulae Explained
What is ZTEST: Excel Formulae Explained?
ZTEST: Excel Formulae Explained is a statistical function in Excel that calculates the one-tailed probability-value of a z-test.
When should I use the ZTEST function in Excel?
You should use the ZTEST function in Excel when you want to determine the probability-value of a one-tailed z-test. This is useful in hypothesis testing when you want to determine if a sample mean is
significantly different from a population mean.
How do I use the ZTEST function in Excel?
To use the ZTEST function in Excel, you must select a range of data that represents the sample you want to test. Then you must specify the known population mean and standard deviation, as well as the
hypothesized sample mean. Finally, you can enter the formula “=ZTEST(sample range,population mean,population standard deviation,hypothesized mean)” into a cell to calculate the one-tailed
What are the limitations of the ZTEST function in Excel?
The ZTEST function in Excel assumes that the sample is normally distributed and that the population standard deviation is known. If this is not the case, other statistical tests may be more
appropriate. Additionally, the ZTEST function can only be used to test a single hypothesis at a time, so multiple tests may require multiple calculations.
What is the difference between a one-tailed and two-tailed z-test?
A one-tailed z-test is used to determine if the sample mean is significantly greater than or less than the population mean, while a two-tailed z-test is used to determine if the sample mean is
significantly different from the population mean. The ZTEST function in Excel only calculates the probability-value for a one-tailed z-test.
Can I use the ZTEST function in Excel for non-numerical data?
No, the ZTEST function in Excel can only be used for numerical data that is normally distributed. If you have non-numerical data or data that is not normally distributed, other statistical tests may
be more appropriate. | {"url":"https://manycoders.com/excel/formulae/ztest-excel/","timestamp":"2024-11-02T09:12:41Z","content_type":"text/html","content_length":"94587","record_id":"<urn:uuid:dbfaf866-28d2-4fa4-a6fc-90833412a3f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00274.warc.gz"} |
RSICC Home Page
RSIC CODE PACKAGE CCC-569
1. NAME AND TITLE
RICANT: A Computer Code for 2-D Transport Calculations in x-y Geometry Using the Interface Current Method.
2. CONTRIBUTOR
Indira Gandhi Center for Atomic Research, India, through the NEA Data Bank, France.
Fortran IV; VAX 8810.
4. NATURE OF PROBLEM SOLVED
RICANT performs 2-dimensional neutron transport calculations in x-y geometry using the interface current method. In the interface current method, the angular neutron currents crossing region surfaces
are expanded in terms of the Legendre polynomials in the two half-spaces made by the region surfaces.
5. METHOD OF SOLUTION
The integral form of the neutron transport equation is solved by the interface current method. The region of interest is divided into small regions with constant material properties. The regions are
connected by neutron currents crossing the interfaces. The outgoing current in one region is the incoming current in the adjacent region. The region sizes have to be very small so that spatial
dependence can be ignored. Associated Legendre polynomials are used to represent the angular dependence of the fluxes. The weighted residual method is used to find the expansion coefficients of the
angular fluxes. Making use of the superposition principle of neutron currents, inner iterations are performed over current components. Outer iterations are performed over groups.
The total number of angular flux expansion terms is limited to 10.
7. TYPICAL RUNNING TIME
The sample input was tested on the VAX 8810 at NEA Data Bank. It took 3 minutes of CPU time.
RICANT runs on the VAX 8810.
The code was written in Fortran IV. On the VAX 8810, the compiler used was VAX Fortran under the VMS version 5.1 operating system.
10. REFERENCE
a. Included in the documentation:
P. Mohanakrishnan, "A Guide to the Use of Computer Code -- RICANT," RG/RPD-311, (December 1987).
b. Background information:
P. Mohanakrishnan, "Choice of Angular Current Approximations for Solving Neutron Transport Problems in 2-D by Interface Current Approach," Ann. Nucl. Energy, Vol. 9, pp. 261-274, 1982, Great Britain.
11. CONTENTS OF CODE PACKAGE
Included are the referenced document and one DS/DD (360 K) 5.25-inch diskette.
12. DATE OF ABSTRACT
December 1990. | {"url":"https://rsicc.ornl.gov/codes/ccc/ccc5/ccc-569.html","timestamp":"2024-11-13T18:06:59Z","content_type":"text/html","content_length":"5093","record_id":"<urn:uuid:92c99975-d317-49fb-aa6d-1e5f40840665>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00719.warc.gz"} |
Drywall Calculator Walls And Ceiling – Accurate Measurements
This tool calculates the amount of drywall you need for your walls and ceiling based on your input dimensions.
Drywall Calculator for Walls and Ceiling
Use this calculator to estimate the number of drywall sheets needed for your project. The calculator takes into account the dimensions of the walls and ceiling, as well as any doors and windows
present in the room.
How to Use
1. Enter the length, height, and width of the room.
2. Enter the length and width of the ceiling.
3. Select the size of the drywall sheets you will be using.
4. Enter the number of doors and windows in the room.
5. Click the “Calculate” button.
6. The result will display the estimated number of drywall sheets needed.
How it Calculates
The calculator first determines the total area of the walls and ceiling. It also subtracts the area occupied by doors and windows. The remaining area is then divided by the area of a single drywall
sheet to determine the total number of sheets needed. This number is rounded up to ensure you have enough material.
This calculator provides an estimate and may not account for all the intricacies of your specific project settings. Always purchase a few extra sheets to account for cutting mistakes and waste.
Use Cases for This Calculator
Calculate Total Drywall Area for Walls
Enter the dimensions of each wall to determine the total area of drywall needed, accounting for doors and windows. Simply specify the width and height of each wall and let the calculator do the math
for you.
Estimate Drywall Quantity for Ceiling
Input the length and width of your ceiling to accurately estimate the amount of drywall required. The calculator will consider any openings such as light fixtures for a precise calculation.
Calculate Total Drywall Area for Combined Walls and Ceiling
If you want to calculate the total area of both walls and ceiling at once, this feature comes in handy. Input the dimensions of walls as well as the ceiling to get the complete drywall area
Adjust for Wastage and Cutouts
Account for wastage and additional drywall needed for cutouts such as electrical outlets or fixtures. Include the quantity and dimensions of cutouts to ensure you order the right amount of drywall.
Convert Measurements Easily
Switch effortlessly between different units of measurement such as feet, inches, or meters based on your preferences. The calculator will convert the measurements accurately for seamless
Calculate Drywall Tape and Joint Compound Amounts
Get an estimate of the quantity of drywall tape and joint compound required for taping and finishing purposes based on the total area of drywall. Ensure you have all the materials needed for the
Plan for Multiple Rooms or Areas
If you have multiple rooms or areas to drywall, simply input the dimensions of each space separately. The calculator will provide individual and total estimates for all designated areas.
Save Your Calculations for Future Reference
Save your calculations for reference or sharing by generating a printable summary. Keep track of the measurements and quantities required for your project conveniently.
Get Real-Time Cost Estimates
Estimate the total cost of drywall materials based on current market prices. By entering the cost per unit, the calculator will give you an instant overview of the expenses involved in your project.
Receive Customized Recommendations
Based on your calculated drywall area and quantities, receive personalized recommendations such as the number of drywall sheets or the amount of joint compound needed. Make informed decisions and
streamline your project planning with ease. | {"url":"https://calculatorsforhome.com/drywall-calculator-walls-and-ceiling/","timestamp":"2024-11-11T21:33:31Z","content_type":"text/html","content_length":"148964","record_id":"<urn:uuid:a3bf563a-423d-434e-988b-3a6f2cf9b9e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00118.warc.gz"} |
SRM 397
Saturday, April 12, 2008
Match summary
This prime numbered SRM gathered together 1352 coders that were faced with a rather hard problem set. However, ACRush rushed through the set achieving his new highest rating. Congratulations!
The match in division one started slowly, as the easy problem was not so easy to code. kia was the first one to submit the 250 and after his submit, solutions started to come in. Then, after about 15
minutes from the beginning of the coding phase, we could saw the first submissions on the medium. From then, until the last minute of the coding phase, solutions to the medium and the hard were
coming in. After the coding phase ACRush was on top, with Loner and ahyangyi being close behind. Challenge phase was rather quiet, but unfortunately ahyangyi's hard got challenged and bmerry took his
spot. wojtekt and gawry rounded up the top five.
In division two, the whole match passed by rather smoothly, although there weren't too many submissions to the medium and hard problems (and out of 128 hard submissions, only 21 were correct). After
all, mrtempo won the division, with a newcomer barney and royappa rounding up the top three.
The Problems
Used as: Division Two - Level One:
Value 250
Submission Rate 718 / 786 (91.35%)
Success Rate 615 / 718 (85.65%)
High Score brosmike for 245.21 points (3 mins 59 secs)
Average Score 191.16 (for 615 correct submissions)
This problem was just a simple simulation of the algorithm described in the statement. If there are only letters in the message, we replace each letter with its assigned number. And if there are only
digits, we take every two consecutive ones and replace them with a letter that's assigned the given code. This should be clear enough, but you can see the fastest submission by brosmike for
Used as: Division Two - Level Two:
Value 500
Submission Rate 145 / 786 (18.45%)
Success Rate 92 / 145 (63.45%)
High Score SlNPacifist for 467.77 points (7 mins 33 secs)
Average Score 284.87 (for 92 correct submissions)
Used as: Division One - Level One:
Value 250
Submission Rate 474 / 560 (84.64%)
Success Rate 436 / 474 (91.98%)
High Score kia for 247.29 points (2 mins 59 secs)
Average Score 174.78 (for 436 correct submissions)
Low constraints guaranteed that there were no more than 8! = 40320 possible states. So, how can we find the shortest path from the initial state to a final one? By BFS, of course. We only need to map
the states that can be represented by vectors, numbers or strings to their values, that are lengths of the respective shortest paths. It is easy to generate the permutations that we can achieve from
the given one in one move. Please see the fastest and extremely clear implementation of the above by kia. This code shows, that in C++, the STL could solve the problem for us.
Used as: Division Two - Level Three:
Value 1000
Submission Rate 127 / 786 (16.16%)
Success Rate 21 / 127 (16.54%)
High Score eagaeoppooaaa for 832.33 points (13 mins 19 secs)
Average Score 634.81 (for 21 correct submissions)
This problem was rather standard. With at most 13 marbles we can represent the state as a set of marbles that we already have, the number of bags left, and the space left in the bag we are trying to
fill now. So we start with an empty set, numberOfBags bags and bagCapacity space left in the bag we're filling. In each state we can either put to the current bag any marble we don't have yet or put
the current bag aside and start to fill the next one. Representing sets as bit masks gives us approximately 2^13 * 20 * 10 = 1638400 possible states. That's ok to use memoization on them and with a
simple recursion compute the answer. Sample Java implementation follows:
public class CollectingMarbles {
int[][][] dp;
int[] w;
int c;
public int recur (int mask, int left, int cur) {
if (left == 0) return 0;
if (dp[mask][left][cur] == -1) {
dp[mask][left][cur] = 0;
for (int i = 0; i < w.length; i++)
if ((mask & (1 << i)) == 0 && w[i] <= cur)
dp[mask][left][cur] = Math.max( dp[mask][left][cur],
1 + recur(mask | (1 << i), left, cur - w[i]));
dp[mask][left][cur] = Math.max( dp[mask][left][cur],
recur(mask, left - 1, c));
return dp[mask][left][cur];
public int mostMarbles (int[] mW, int bC, int nOB) {
w = mW;
c = bG;
dp = new int[1 << w.length][nOB + 1][c + 1];
Arrays.fill(dp, -1);
return recur(0, nOB, c);
Used as: Division One - Level Two:
Value 500
Submission Rate 158 / 560 (28.21%)
Success Rate 98 / 158 (62.03%)
High Score Loner for 447.83 points (9 mins 56 secs)
Average Score 288.51 (for 98 correct submissions)
Well, this problem was an interesting one. The idea, with its simplicity and mathematical background, caused many coders to search the internet for the formula. It could end with a success, but not
necessarily. Many coders, after reading all these formulas with Bernoulli numbers were angry at the problem as they thought that you have to be a genius to come up with them. Well, maybe you have to
be, and maybe Bernoulli was. But what if he had Google? Maybe he wouldn't even bother to invent all this stuff. Ok, so suppose that we don't have Google. What do we have here? We are given the sum 1^
k + 2^k + ... + n^k. Ok, let a[i] = 1^k + ... + i^k. Everyone who finished high school knows that
showed us
some day how we can effectively compute binomial coefficients, so let's suppose that we know their values and forget about them. So it looks like we have some number of recursively defined sequences.
Now, the second thing that every top coder should know. The most obvious way to find the n-th term of a recursive sequence is to use a matrix multiplication. Again, we don't have to be very bright to
see that:
That looks nice. But let's get our a
sequence into play:
Now, because matrix multiplication is associative, we can compute the n-th power of our magic matrix in time O(k
* log n) and then multiply it by the vector for a
(it will contain only ones) to get a
. Well, that definitely didn't hurt us. There were of course many different approaches. We could for example use Lagrange's
polynomial interpolation
(as the given sum is of course a polynomial) or use a different recursion then the one described here (please see the
's solution for a reference). There was also a solution based on Bernoulli numbers - use Google to find it!
Used as: Division One - Level Three:
Value 1000
Submission Rate 92 / 560 (16.43%)
Success Rate 51 / 92 (55.43%)
High Score ACRush for 830.85 points (13 mins 23 secs)
Average Score 628.48 (for 51 correct submissions)
Let's think for a moment about an easier version of the problem - let's think how we can find the biggest number of radars we're able to place for some given range. To begin, we'll place radars in
all possible points. Now, we want to remove the smallest number of radars in such way, that in the remaining set there won't be any pair of intersecting radars that have different colors. That starts
to sound like a vertex cover problem - let's build a graph out of our radars and let's connect two radars of different colors with an edge if their areas intersect. The general vertex cover problem
is a well-known NP-complete problem. But, luckily, we have a bipartite graph here, so according to Konig's theorem we know that the minimal vertex cover is equivalent to maximal matching. We know how
to find the latter, so let's return to the original problem.
The safety factor is defined by two values -the number of radars and their range. So, let's try every possible number of radars and for each such number, let's find the biggest range that allows us
to place the radars. It isn't a surprise that we can do it with a binary search - if we can place the radars setting them on radius r, we could also set them to any range smaller than r. That was
quick, but enough to pass - you can see the fastest, but very clear implementation of the above by ACRush.
However, we can get rid of the binary search here. The observation is that in the optimal solution the range of the radars will be either the given R or half of the distance between some two red and
blue points. Why? If it was not the case, we could increase the range to let the areas of some radars touch. Now, if we consider the possible ranges in an increasing order, we don't have to cancel
the matching every time we have a new range (provided, we use an augmenting path algorithm) as we aren't removing any edges from the graph, but just adding the new ones, so the matching we have so
far is ok with a new graph. For every range candidate, we will compute the biggest number of radars we can place. That's enough to compute the answer. This solution has time complexity of O(n^4)
instead of O(n^4 * binary search). You can see a nice implementation of this by pparys. | {"url":"https://www.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm397","timestamp":"2024-11-14T08:42:37Z","content_type":"text/html","content_length":"52576","record_id":"<urn:uuid:ec80579d-831a-4298-a7e0-a72d8b0785b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00884.warc.gz"} |
The a^2+b^2 Formula: Understanding its Significance and Applications - Hard Geek
Mathematics is a fascinating subject that encompasses a wide range of concepts and formulas. One such formula that holds great significance is the a^2+b^2 formula. This formula, also known as the
Pythagorean theorem, has been a fundamental part of mathematics for centuries. In this article, we will delve into the details of the a^2+b^2 formula, explore its applications in various fields, and
understand its importance in problem-solving.
What is the a^2+b^2 Formula?
The a^2+b^2 formula, also known as the Pythagorean theorem, states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum
of the squares of the other two sides. Mathematically, it can be represented as:
c^2 = a^2 + b^2
Here, ‘c’ represents the length of the hypotenuse, while ‘a’ and ‘b’ represent the lengths of the other two sides of the triangle.
The History of the Pythagorean Theorem
The Pythagorean theorem is named after the ancient Greek mathematician Pythagoras, who is credited with its discovery. However, evidence suggests that the theorem was known and used by other
civilizations, such as the Babylonians and the Egyptians, even before Pythagoras.
Pythagoras and his followers, known as the Pythagoreans, extensively studied the properties of right-angled triangles and recognized the relationship between the lengths of their sides. The
Pythagorean theorem became one of the foundational principles of their mathematical teachings.
Applications of the a^2+b^2 Formula
The a^2+b^2 formula finds applications in various fields, ranging from architecture to physics. Let’s explore some of its practical uses:
1. Architecture and Construction
In architecture and construction, the Pythagorean theorem is crucial for ensuring the stability and accuracy of structures. It helps architects and engineers calculate the lengths of diagonal beams,
determine the dimensions of rooms, and ensure that walls and floors are perpendicular.
For example, when constructing a rectangular room, the a^2+b^2 formula can be used to verify if the room is perfectly square. By measuring the lengths of the two shorter sides and applying the
formula, one can determine if the diagonal length matches the calculated value. If they are equal, the room is square; otherwise, adjustments need to be made.
2. Navigation and Surveying
The Pythagorean theorem plays a crucial role in navigation and surveying. It allows sailors, pilots, and surveyors to calculate distances and angles accurately.
For instance, consider a ship navigating through a series of islands. By using the a^2+b^2 formula, the ship’s crew can determine the shortest distance between two points, taking into account the
obstacles in their path. This knowledge helps them chart the most efficient course and avoid potential hazards.
3. Physics and Engineering
In physics and engineering, the Pythagorean theorem is used to analyze and solve problems related to vectors and forces.
For example, when calculating the resultant force of two perpendicular forces acting on an object, the a^2+b^2 formula can be applied. By squaring the magnitudes of the two forces, adding them
together, and taking the square root of the sum, the resultant force can be determined.
Real-Life Examples
Let’s explore a few real-life examples that demonstrate the practical applications of the a^2+b^2 formula:
Example 1: Building a Fence
Suppose you want to build a fence around a rectangular garden. You measure the length of one side as 5 meters and the length of the adjacent side as 12 meters. To ensure that the fence is perfectly
square, you can use the Pythagorean theorem to calculate the diagonal length:
c^2 = a^2 + b^2
c^2 = 5^2 + 12^2
c^2 = 25 + 144
c^2 = 169
c = √169
c = 13 meters
Therefore, the diagonal length of the garden is 13 meters. By measuring the diagonal length of the fence and comparing it to the calculated value, you can ensure that the fence is square.
Example 2: Calculating Distance
Suppose you are planning a road trip and want to determine the shortest distance between two cities. By using the Pythagorean theorem, you can calculate the straight-line distance between the two
cities, assuming there are no obstacles in the way.
Let’s say City A is located at coordinates (3, 4) and City B is located at coordinates (8, 10). The distance between the two cities can be calculated as follows:
c^2 = (8 – 3)^2 + (10 – 4)^2
c^2 = 5^2 + 6^2
c^2 = 25 + 36
c^2 = 61
c = √61
c ≈ 7.81 units
Therefore, the straight-line distance between City A and City B is approximately 7.81 units.
Q1: Can the Pythagorean theorem be applied to non-right-angled triangles?
A1: No, the Pythagorean theorem is only applicable to right-angled triangles. For other types of triangles, different formulas and theorems need to be used.
Q2: Are there any limitations to the Pythagorean theorem?
A2: The Pythagorean theorem assumes that the triangle is two-dimensional and Euclidean. It does not hold true in non-Euclidean geometries or for triangles in higher dimensions.
Q3: Can the Pythagorean theorem be extended to more than two sides?
A3: No, the Pythagorean theorem only relates the squares of the lengths of two sides to the square of the length of the hypotenuse in a right-angled triangle. It does not apply to triangles with more
than one right angle or to polygons with more than three sides
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sjSDM 1.0.6
New features
• Datasets: butterflies and eucalypt species
• Conditional predictions
• Assembly regression plots
sjSDM 1.0.5
New features
• Pass custom test indices to sjSDM function via CV argument
• improve reproducibility (seeding)
• improve stability of ANOVA
Bug fixes
• fixed small bug in the calculation of the partial Rsquareds
sjSDM 1.0.4
New features
• Anova function is now based on conditional probabilities to better separate the biotic components
• Anova can use shared components (plot(…,internal=TRUE, add_shared=TRUE))
• Simulation methods for all supported families (binomial, poisson, and negative binomial)
• Support for negative binomial distribution
• plotInternalStructure for plotting internal metacommunity structure
• getCor to return species-species association matrix
Bug fixes
• fixed Rsquared(…) #113 (thanks to @AndrewCSlater)
• fixed whitespaces in species names #115 @dansmi-hub
sjSDM 1.0.3
Minor changes
• changed weight_decay in ‘RMSprop’ from 0.01 to 0.0001
Important bug fix
• fixed sjSDM_cv(…) #104 (thanks to @Cdevenish)
sjSDM 1.0.2
Minor changes
• changed to as requested by the CRAN team
Bug fixes
sjSDM 1.0.1
New Features
• anova plots for internal meta-community structure (based on individual R-squared values)
Minor changes
• first layer of DNN now always without an explicit bias (bias/intercept is passed by model/formula, if desired)
• revised prediction function, improved stability
• revised simulation function, samples now from a multivariate probit model
Bug fixes
• unlisting of config objects in sjSDM::sjSDM_cv (thanks to Máté) (added unit tests) #88
• sjSDM::Rsquared bug for spatial models (thanks to Máté) #90
• revised regularization behavior, l1 and l2 were not correctly imposed on DNN structure
• revised and improved setWeights function
• bugs in vignettes (thanks to Doug) #92
• bugs in plot function for models with DNN objects
sjSDM 1.0.0
Major changes
• revised anova: sjSDM::anova(...) corresponds now to a type I anova (removed CV) #76
• sjSDM::Rsquared() uses now Nagelkerke or McFadden R-squared (which is also used in the anova) #76
• deprecated sjSDM::sLVM because of instability issues and other reasons
• revised sjSDM::install_sjSDM(), it works now for all x64 systems/versions #81 #79 #71
Minor changes
• removed several unnecessary dependencies (e.g. dplyr)
• improved documentation of all functions, e.g. see ?sjSDM
• new sjSDM::update.sjSDM method to re-fit model with different formula(s)
• new sjSDM::sjSDM.tune method to fit quickly a model with optimized regularization parameters (from sjSDM::sjSDM_cv)
Bug fixes
• revised memory problem in sjSDM::sjSDM_cv() #84 | {"url":"https://cran.case.edu/web/packages/sjSDM/news/news.html","timestamp":"2024-11-04T01:50:07Z","content_type":"application/xhtml+xml","content_length":"5126","record_id":"<urn:uuid:40c5ef20-9e82-4028-82a2-48842e4f8c0b>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00062.warc.gz"} |
WhoMadeWhat – Learn Something New Every Day and Stay Smart
So five blocks = 1/4 mile. I usually walk a block a minute, briskly. Crosstown blocks vary in length, but they average 8-10 to the mile.
Similarly, How long is a 10 block walk?
Re: How long does it take to walk 10 city blocks? Fifteen minutes.
Additionally, How long of a walk is 2 blocks? It’s all I can go with. It is a standard that 1 mile = 5,280 feet. Thus, 2 blocks is approximately 1/10 th (2/20) of a mile or 528 feet.
How long does it take to walk a block?
Most of the time, it should take about two minutes or so to walk a block. If you are going to walk about ten blocks, give yourself a good 15 to 20 minutes to get this done.
How far on average is a block?
Oblong blocks range considerably in width and length. The standard block in Manhattan is about 264 by 900 feet (80 m × 274 m). In Chicago, a typical city block is 330 by 660 feet (100 m × 200 m),
meaning that 16 east-west blocks or 8 north-south blocks measure one mile, which has been adopted by other US cities.
How many miles is 20 blocks in NYC?
20 blocks is equal to 1 mile. Not far for us NY’ers who are used to walking. Avenues run north south and disect the length of Manhattan. Blocks run East west and disect the width of Manhattan.
How far is a block in KM?
How many block in 1 km? The answer is 12.427423844747. We assume you are converting between block [East U.S.] and kilometre. You can view more details on each measurement unit: block or km The SI
base unit for length is the metre.
How many steps is a NYC block?
How many steps around a standard-sized city block? Ten city blocks equal around a mile — approximately 2,000 steps equal a mile. Given those numbers, one block is roughly 200 steps.
What does 2 blocks mean?
It just means “more than one block, but less than three”. If someone said something was “two blocks away”, I would expect to have to cross two streets to get there.
How far is a block in minutes?
To walk one street block (north/south) will take about a minute and one-half considering that you may have to stop and many traffic lights and the sidewalks will be crowded. It could be done in a
minute but you have to be walking very fast. Walking avenue blocks (East/West) will take about 4 minutes.
How many blocks is 2 miles?
This equals approximately 16 or 17 blocks per mile. Cities are not always consistent in the size of blocks. But, it seems that most of the “standard” rectangular blocks are: 10 to 11 blocks per mile
if walking the long side, and 16 to 17 blocks per mile if walking the short side.
How many steps is in a city block?
How many steps around a standard-sized city block? Ten city blocks equal around a mile — approximately 2,000 steps equal a mile. Given those numbers, one block is roughly 200 steps.
How long does it take to walk 1000 blocks in Minecraft?
time for 50 blocks: 14.74 sec. PM 1000 blocks: 294.8 sec.
How long is a neighborhood block?
A block is the distance from one cross street to the next. They can be as long as about 800 feet or as short as 100 feet. Most blocks are about 200–300 feet long.
What does block mean in distance?
If you’re walking down a sidewalk along a particular street a “block” is just the distance along the sidewalk from one street to the next. Using blocks is very typical when referring to distances in
a city.
How many miles is 20 city blocks?
North-south is easy: about 20 blocks to a mile. The annual Fifth Avenue Mile, for example, is a race from 80th to 60th Street. The distance between avenues is more complicated. In general, one long
block between the avenues equals three short blocks, but the distance varies, with some avenues as far apart as 920 feet.
How many blocks in NYC is a mile?
But how many NYC blocks are in a mile? The average length of a north-south block in Manhattan runs approximately 264 feet, which means there are about 20 blocks per mile.
Is a mile 12 blocks?
From our sample size below using major cities, the average number of blocks in a mile would be 20.3 blocks. However, blocks can vary dramatically between each city or even direction.
How far is a block?
A block is not really defined by distance, but rather is defined by the distance between cross streets, which could be 50 feet or 200 feet, depending on the place. Most blocks in cities tend to be
between 200 – 300 feet apart, so the distance between 2 blocks would be roughly 400 – 600 feet.
How long does it take to walk 1 block?
Now that you have a basic idea of the distance of the block, you may be wondering how long it takes you to walk a block. Most of the time, it should take about two minutes or so to walk a block. If
you are going to walk about ten blocks, give yourself a good 15 to 20 minutes to get this done.
How big is a block?
Oblong blocks range considerably in width and length. The standard block in Manhattan is about 264 by 900 feet (80 m × 274 m). In Chicago, a typical city block is 330 by 660 feet (100 m × 200 m),
meaning that 16 east-west blocks or 8 north-south blocks measure one mile, which has been adopted by other US cities.
How many NYC blocks equal a mile?
North-south is easy: about 20 blocks to a mile. The annual Fifth Avenue Mile, for example, is a race from 80th to 60th Street. The distance between avenues is more complicated. In general, one long
block between the avenues equals three short blocks, but the distance varies, with some avenues as far apart as 920 feet.
How many steps are there when setting your block?
This rule is great for juniors and beginners in remembering what lengths they need for the blocks in three easy steps. First, place the start of the starting block rail 1 step from the start line.
Next, position the front block 2 steps from the start line. Finally, position the back block 3 steps from the start line. | {"url":"https://whomadewhat.org/how-long-is-a-5-block-walk/","timestamp":"2024-11-14T07:28:43Z","content_type":"text/html","content_length":"50963","record_id":"<urn:uuid:ca89323b-7d78-4e31-89e1-2b6bbec4918e>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00829.warc.gz"} |
Interface FirstOrderDifferentialEquations
All Superinterfaces:
This interface represents a first order differential equations set.
This interface should be implemented by all real first order differential equation problems before they can be handled by the integrators ODEIntegrator.integrate(org.hipparchus.ode.ExpandableODE,
org.hipparchus.ode.ODEState, double) method.
A first order differential equations problem, as seen by an integrator is the time derivative dY/dt of a state vector Y, both being one dimensional arrays. From the integrator point of view, this
derivative depends only on the current time t and on the state vector Y.
For real problems, the derivative depends also on parameters that do not belong to the state vector (dynamical model constants for example). These constants are completely outside of the scope of
this interface, the classes that implement it are allowed to handle them as they want.
See Also:
• Method Summary
Modifier and Type
default double[]
Get the current time derivative of the state vector.
Get the current time derivative of the state vector.
• Method Details
□ computeDerivatives
default double[] computeDerivatives(double t, double[] y)
Specified by:
computeDerivatives in interface OrdinaryDifferentialEquation
t - current value of the independent time variable
y - array containing the current value of the state vector
time derivative of the state vector
□ computeDerivatives
Get the current time derivative of the state vector.
t - current value of the independent time variable
y - array containing the current value of the state vector
yDot - placeholder array where to put the time derivative of the state vector
MathIllegalStateException - if the number of functions evaluations is exceeded
MathIllegalArgumentException - if arrays dimensions do not match equations settings | {"url":"https://hipparchus.org/apidocs-3.1/org/hipparchus/migration/ode/FirstOrderDifferentialEquations.html","timestamp":"2024-11-09T16:33:15Z","content_type":"text/html","content_length":"14103","record_id":"<urn:uuid:3be3add7-e709-451e-9f25-5b317d4190e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00721.warc.gz"} |
Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas. | Research / KirbyUrner
Andrius: Here are my notes on...
Kirby Urner
Various themes
• 2024.02.23 My project: to wire Bucky Fuller into the pantheon of "must study" 20th century philosophers, wherein I project a lineage that includes the antebellum Transcendentalists.
• 2024.02.23 I keep looking at quasi-states or para-states per our Sociology track.
Relating to Kirby Urner and his ideas about differences in thinking comparing "human" cubes vs. "Martian" tetrahedrons.
I think your quadpod is a magnificent concept for illustrating your points. It's very vivid and fun, too.
I am impressed by your geometry http://www.rwgrayprojects.com/synergetics/s09/figs/f9001.html which is intriguing and persuasive.
However, if you line up the corners of the squares and also of the cubes, then you get a progression which is very helpful for teaching calculus, namely, if you consider a square x and grow it by one
more bit h so you have a square of sides x+h, then: (x + h)**2 = (x + h)(x + h) = x2 + 2hx + h2 which all make geometric sense, and then you can see why you can ignore the h2 and upon subtracting x2
you are left with 2hx which, when compared with h, gives you the derivative 2x.
Similarly, (x+h)**3 = (x+h)(x+h)(x+h) = x3 + 3x2h + 3xh2 + h3 and discarding the small stuff and substracting x3 you are left with 3x2h and dividing by h gives the derivative 3x2.
This for me is a very powerful way to illustrate differentiation in a very real sense. And also these binomial expansions are very worthwhile to spend time with and very meaningful for problems in
probability, heads and tails: (h+t)**3 or recessive and dominate genes, blue eyes b and brown eyes B (b+B)(b+B) for example.
So I'm curious if your triangular thinking has a nice way to talk about this all, perhaps?
This page is for my thoughts on the "tetrahedral" thinking that Kirby Urner writes about in his analogies of Martian (tetrahedral) vs. Earthling (cubic) societies.
Kirby, Joseph, Bradford,
I hope soon to send out my essay that I've been writing. I think it might even touch on your "closing the lid" operator. I reinterpret the "demicube" (demihypercube) polytope series Dn as "hemicubes"
(halfcubes) where the most opposite corners of the cube have been identified (the cube/sphere has been folded in half... like n-dimensional circle folding?) and so we have spiky Euclidean "coordinate
systems" with double edges, with additional double edges linking the tips of all of the coordinate vertices, just as you describe. I just don't know how to call these "trusses"? The point is that we
get two different ways of looking at this. On the one hand, we have a simplex that has grown out of the "origin". (Just the angles aren't 60 degrees, they are 90 degrees or 45 degrees.) And because
our "origin" could have been any point of the half-n-cube, we get 2^(n-1) versions of these simplexes. So each of these is an "anti-center". On the other hand, we get the big picture of the half-cube
and by taking a subset of dimensions we can look at smaller half-cube within that. And from the big picture point of view, it makes no difference which points we chose to fold by. But it is a folded
volume, so it is an "anti-volume". So the four series will be:
• An simplex (tetrahedrons) Center and Volume
• Bn cubes No-Center and Volume
• Cn cross-polytopes (orthogons) Center and No-Volume
• Dn half-cubes No-Center and No-Volume
These correspond to the four families of classical groups / Lie Algebras / Lie groups. That is, they express the symmetries of the above structures in terms of actions. Some day I'll understand...
All of this to say that your mathematical taste is excellent and keep following your mathematical sensibility! It's very helpful, inspiring and encouraging.
Kirby, but I wanted to share with you a long history by John Baez and Aaron Lauda that I'm looking at, "A Prehistory of n-Categorical Physics". http://arxiv.org/pdf/0908.2469v1.pdf On page 33, they
mention the work by Ponzano-Regge in 1968 on there 3d model of quantum gravity, where spacetime is made of tetrahedra. And in searching on "tetrah" I also see that Kapranov-Voevodsky studied the
Zamolodchikov tetrahedron equation. Keep searching on "tetrah" and you will find... I'm curious whatever you find interesting.
• Elective disaster, global warming discourse
• The Shepherd's tone - auditory illusion, as if it were ever rising
• Our own sense of mortality, imposing it on everything.
• Believing in eternal life.
• Too hooked on 90 degrees, should move to 60 degrees - Fuller.
• Digging around the concept of dimensions.
• OK to be on a different page
• Sand castles on a beach - mathematics (numerative systems)
• Modeling the world introduces a duality (and ambiguity) as to whether we want to focus on the model or the world as regards our actions - do we change our world to fit our model - or do we change
our model to fit the world. This relates to gaming the game, to the distinction between object and process. | {"url":"https://www.math4wisdom.com/wiki/Research/KirbyUrner","timestamp":"2024-11-08T03:00:29Z","content_type":"application/xhtml+xml","content_length":"16082","record_id":"<urn:uuid:d88ceebf-7a9f-494d-9f1e-e76c24377435>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00650.warc.gz"} |
Conversion Of Improper Fractions To Mixed Numbers Worksheets 2024 - NumbersWorksheets.com
Conversion Of Improper Fractions To Mixed Numbers Worksheets
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Gallery of Conversion Of Improper Fractions To Mixed Numbers Worksheets
30 Multiplying Improper Fractions Worksheets
43 Converting Improper Fractions To Mixed Numbers Worksheet Worksheet
Conversion Of Mixed Numbers To Improper Fractions Worksheets Math
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How do you find the antiderivative of (x-6)^2? | HIX Tutor
How do you find the antiderivative of #(x-6)^2#?
Answer 1
$\frac{1}{3} {x}^{3} - 6 {x}^{2} + 36 x + C$
Expand and then use power rule. #(x-6)^2#= #x^2-12x+36# #int_##x^2-12x+36# dx = #1/3x^3-6x^2+36x+C#
Remember power rule is #int_##ax^n# = #(ax^(n+1))/(n+1)#
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Answer 2
To find the antiderivative of (x-6)^2, you can use the power rule for integration, which states that ∫x^n dx = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
Applying this rule to (x-6)^2, first expand the expression to (x-6)(x-6). Then integrate each term separately using the power rule:
∫(x-6)^2 dx = ∫(x^2 - 12x + 36) dx
= (1/3) * x^3 - (1/2) * 12x^2 + 36x + C
= (1/3) * x^3 - 6x^2 + 36x + C
Therefore, the antiderivative of (x-6)^2 is (1/3) * x^3 - 6x^2 + 36x + C.
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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
Not the question you need?
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• Readily available 24/7 | {"url":"https://tutor.hix.ai/question/how-do-you-find-the-antiderivative-of-x-6-2-8f9afa07f4","timestamp":"2024-11-02T18:39:57Z","content_type":"text/html","content_length":"567669","record_id":"<urn:uuid:6e1a1abd-cb73-42dd-b5f0-03fb61799cb0>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00373.warc.gz"} |
Ordinary Differential Equations - Wikibooks, open books for an open world
• Solutions to specific equations
Ordinary Differential Equations
covering uses of and solutions to ordinary differential equations
The Rössler Attractor. This chaotic system is generated by a system of ordinary differential equations.
This book aims to lead the reader through the topic of differential equations, a vital area of modern mathematics and science. This book provides information about the whole area of differential
equations, concentrating first on the simpler equations.
Differential Equations and Boundary Value Problems- C.H. Edwards Jr and David E. Penny
MIT Open Courseware- http://ocw.mit.edu/index.html
• Kong, Qingkai (0000). A Short Course in Ordinary Differential Equations. Universe: Publisher.
• Walter, Wolfgang (1998). Ordinary Differential Equations. New York: Springer. | {"url":"https://en.wikibooks.org/wiki/Ordinary_Differential_Equations","timestamp":"2024-11-12T04:24:35Z","content_type":"text/html","content_length":"66685","record_id":"<urn:uuid:e843b6f6-0865-4f36-81c6-222020f36ed0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00189.warc.gz"} |
A fractal art approach to the three-body problem
Recommended Citation
Babbs, Charles F., "A fractal art approach to the three-body problem" (2024). Weldon School of Biomedical Engineering Faculty Working Papers. Paper 34.
Date of this Version
acceleration, collision, dipole, Earth, ejection, escape, gravity, image, Liebovitch, mass, Moon, Newton, orbit, planar, satellite, scale, self-similar, simulation, Sun, trajectory, Valtonen
This preliminary study explores a new search strategy for identifying relatively stable vs. unstable solutions to the planar three-body problem in astrophysics, starting from the perspective of
computer-generated art. Here classical Newtonian accelerations, speeds, and positions of all three bodies in a fixed plane are calculated. All three bodies are stationary at time zero, and the fate
of the system is classified as reflecting either a bound stable orbit, a likely collision, or the ejection of one body. The initial position of one of the three bodies is varied in the image plane,
and the outcome coded as one of three colors, to produce a complex image of rings, defining either stable orbits, collisions, or ejection events. The nested, randomly interspersed, non-overlapping,
both thick and thin rings resemble the rings of the planet Saturn seen up-close by a passing spacecraft. However, the rings are not concentric. Instead, they are similar to the field lines around an
electric dipole. Since such field lines converge at the origin, detailed measurements of the ring density per unit length are possible either along a 45-degree line or along a horizontal line close
to the origin. These measurements reveal a seemingly infinite number of rings of decreasing thicknesses over linear scales spanning 16 orders of magnitude. Such self-similar ring patterns at
progressively smaller scales represent a new type of fractal, embedded in the classical three-body problem of astrophysics. | {"url":"https://docs.lib.purdue.edu/bmewp/34/","timestamp":"2024-11-10T00:07:48Z","content_type":"text/html","content_length":"35747","record_id":"<urn:uuid:ead3a8af-b59d-47e4-9ad4-0945bdc20a26>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00574.warc.gz"} |
Re: dihedral parameter conversion
Re: dihedral parameter conversion
There is an OPLS parameter file that Dan Price prepared using the
program PEPZ that is output in the CHARMM format. The fourier
coefficients are the OPLS coefficients divided by 2. So a V1 of -5 would
be -2.5.
You also have to convert sigma and epsilon for all atom types.
I have run several NAMD simulations using OPLS and I have checked
dihedral distributions generated with NAMD and compared it with MP2
dihedral profiles that parameters were fitted against and it looks
correct. OPLS and CHARMM have very similar formats. The main difference
is the absence of CMAP in OPLS.
On 5/2/2013 9:17 AM, JC Gumbart wrote:
> I'm not so familiar with the formats of force fields other than CHARMM, but I want to convert one for OPLS to CHARMM-style for running in NAMD. The main issue I've yet to resolve though is the
format of the dihedrals. Here's an example line from the original parameter file:
> ; ai aj ak al funct ; Amber type OPLS type Type V1 V2 V3 Comments
> 1 4 5 6 3 -0.50208 -1.50624 0.00000 2.00832 0.000 0.000 ; C3-N3-C2-C2 4031-4030-4032-4004 5000 0.000 0.000 -0.240
> I realize the first part is an RB format. The second part, I guess (please correct me if I'm wrong!!!), uses this functional form:
> V(φ) =V1(1+ cosφ)/2+V2(1−cos2φ)/2+V3(1+ cos3φ)/2+V4(1−cos4φ)/2
> So for the example line, the potential would be V = -0.12*(1+cos(3*phi)). But how to represent this in CHARMM??? Because it can't be JUST a phase shift, then we would have V = 0.12*(1+cos
(3*phi-180)) = 0.12*(1-cos(3*phi)). In other words, there is a constant shift in the potential energy equal to V3.
> What simple fact am I misunderstanding here? How does one convert force constants less than zero to charmm, where they are always greater than zero?
> Thanks!
> JC
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Calculating Momentum
Calculating Momentum
You will learn to apply the momentum equation to calculate the properties of a moving object.
Why does a moving car have momentum, but a stationary car does not? Pick all the options you think are correct.
You can select multiple answers
Momentum is sometimes defined as mass in motion. All objects have mass, so that must mean...
Compared to a car moving at $15\space m/s$ , the same car travelling at $30\space m/s$ has...
A car has a mass of $600\space kg$ and a velocity of $20\space m/s$ . The car has momentum with a magnitude of $12,000\space kg\space m/s$. Can you work out the formula that links mass, velocity and
Momentum can be calculated using the formula $Momentum=mass\times velocity$. This can also be written as $p=mv$.
Do you remember what the unit was that we used for momentum?
If a $60\space kg$ athlete is running with a velocity of $5\space m/s$, what is his momentum? Write your answer using the correct units for momentum.
If a $60\space kg$ athlete is running with a velocity of $3.5\space m/s$ , and a $58\space kg$ athlete is running with a velocity$4.5\space m/s$. Which has the most momentum?
The momentum of a car is $3\times 10{^3}\space kg\space m/s$ and the mass of the car is $1200\space kg$. How would you rearrange the formula for momentum to find the velocity of the car?
The momentum of a car is $3000\space kg\space m/s$, and it has a mass of $1200\space kg$. What is the velocity of the car?
A ball has a momentum of $10\space kg\space m/s$ and a mass of $2\space kg$. At what velocity is the ball moving?
The momentum of a car is $3\times 10{^3}\space kg\space m/s$ , and its velocity is $20\space m/s$. How would you rearrange the formula for momentum to find the mass of the car?
The momentum of a car is $20,000\space kg\space m/s$ , and its velocity is $20\space m/s$. What is the mass of the car? | {"url":"https://albertteen.com/uk/gcse/physics/forces-in-motion/calculating-momentum","timestamp":"2024-11-03T09:35:44Z","content_type":"text/html","content_length":"176310","record_id":"<urn:uuid:697a50e3-3890-4d24-b40c-5e3638a198e3>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00854.warc.gz"} |
Karim Abuzaid - Martyrdom Of Imam Al-Hussein - Part 1
Karim Abuzaid – Martyrdom Of Imam Al-Hussein – Part 1
AI: Summary ©
The Easter season is a conversation about the importance of acceptance of servitude and the loss of family members due to vaccine. The segment discusses the myth of " elevate means what it means" and
how it can be a derivative of the word " elevate." The transcript describes the history of the Middle East, including the deaths of Muslims and the rise of Islam in the Western. The segment also
touches on the current president's plans to build a new military base and the shock that follows the woman who was arrested and executed for lying about the operation in Iraq.
AI: Transcript ©
When I was lagging in Cerulean fusina Omen cftr Melina
mania de la
la la
fella de la
Elan Illa Illa
luxury Allah
wa shadow Mohammed Abu masala
Allahumma salli ala Muhammad Ali Mohammed
can also later Allah Ibrahim Allah early Ibrahim Naka, amigo Majeed
Allahumma barik ala Muhammad
Rahim Allah and Ibrahim in Naka from Edo Majid
minalima t fo fill oma, the signs of,
of being lost as an omen
that we're doing exactly like Benny as a
as an oma,
like we said, going in circles,
not understanding the wisdom
behind the ritual
Rasulullah sallallahu alayhi wa sallam,
when he initiated for us, cm Ashura
the fasting of Ashura, which is on the 10th of Muharram.
The Hadith explicitly
in indicate that this is a day
when we as an oma should connect
with the earlier Muslims.
And if we read the narration of 11 o'clock
of the Friday,
this is the day on which the ark of Noah finally landed
on the 10th of March.
And no Holly is Salam.
sukra Lila
showing gratitude to Allah Spano tala, for
Nigeria TV, being saved with his family
and for his family to know our best film, Muslim.
This is the day on which Allah Subhana Allah says very israa
from the field.
So here we are connecting
with the previous oma
who held the same beliefs like ours, though he,
oh, no. Those who were saved on the ark
can hate.
They believed in the monotheism.
Benny is at war with Moosa
can eat
Yeah, there is a question regarding the quality of the heat but they are still Muslims, like many Muslims now.
Yet, you find
the Shia
which is a sect in Islam.
Do take the day out of context.
They really make it a day of disunity. So it's supposed to be a day of unity for what
for us as an oma and
a day when you are given hope.
Because Allah saved,
no ownership.
A lot can save us to we're going through a very hard time and that can save us to a hope, hope.
Look how Allah subhanaw taala saved
most Anthony
that's a day when you work you're supposed to have what
bring hope.
But what they do, they turn it into a day of grief.
And they go after
an event
that happened through the grandson of the Prophet salallahu alaihe, Salam and Hussein
which happened on this day, the day of Asheville
So they make it the morning
and they breathe
and they express their grief and mourning
in a way that is out of context.
Allah subhanaw taala in the Quran, he says alladhina either atharva Tomasi button follow
what in Allah when Allah urogen those who don't, they are inflected with a calamity. They say in the Allahu Nastassja in narela, he were in la de la Jo.
This is what they call the positive grief.
You see certain words from this one
that can help you stay patient.
And later on insha Allah you develop the servitude to forever that you accept.
When you're infected with a calamity, you don't have to accept it right away. That's not normal.
But the normal course of action that your head you're shocked
the first servitude you you're supposed to practice exercise is patience.
But accepting it No, you don't have to accept it.
You do not have to accept
what is ailing the Prophet sallallahu Ala Moana so am selama sitting in front of the grave of her husband
and he saw her crying.
He said to her intimate Sabra and the submittal oola
exercising patience when
when you're immediately being struck with the shock when you're being shocked with the calamity suppose in LA in LA or in LA
lahemaa aka Allah Allah him alpha
Allah, Masha
you accepting it is not a condition by the one thing that I have to accept about Allah not immediately.
The server two to four river normally comes late.
Lin grief is
an actual
Rasul Allah, Allah wa salam, when his son Ebrahim died,
he stood
next to him and he said, in
the heart Greaves,
we're in the liner lettered Ma
and tears will come down to
but let a while Anna pulumi Gen.
y nadie seraphic Rahim Allah, Ilana khulumani
and I'm so
in grief and grieving for your death. Oh, Ibrahim, this is a pseudo lasala listen. So his heart is saddened by the death of his son tears are coming down. But we're not gonna say I'm not gonna say
anything that
shows dissatisfaction with the other philosopher
in her saying earlier lavonne
was killed.
They use 61 after hedgerow
I mean, until now you did not accept Qatar Allah subhanaw taala in our field with a robot.
Now he said patience is when you work.
I mean, you grieving maybe for one year that's fine.
Why don't we do the same thing for off man
of man was killed Omar. What about Omar Omar was killed we're
leading the Salah.
For you're not supposed to do that. And that's why it works. It's a bit odd to have an annual senoia like an Obama scenario. That means I'm doing the annual work, the annual commemoration of the
death of mine. These things are in back home or being home we will look after 40 days
and have them in a sauna This is from the sooner
we'll recognize how many days or the offering condolences. Three days follows
turn the chapter move on
Except for that a lot move on
not only the green
in a way that is out of context in a way that is a nobody tells you that you grieve that you go and you hit yourself with with with chains and and you you bleed and what is this
and and you sit in a gathering row. You see the ceremony on that day is out of this world.
Rasul Allah Allah Allah says anniverary
is own Allah Harley
the one who shaved her head was Shaka.
The one who shaves his head or the one who she's a female or male, for because of grieving.
A solid for the one who cries loud
was shot by the one who
killed the cloth.
They don't they don't only do this they do what
finished when
they summon them. And Shaka Zulu, he is not one of us.
To be this Hadith, actually, in the literature
is not one of us, the one who tear the the cloth out of greed, lust from alpha, dude, slap the cheeks, whatever with our jelly that he says,
you know, words of jelly
not only this year, where they go after as soon as if as soon as are the people who really killed him for saying.
And this is a misconception that actually a lot of
Muslims don't know.
And this is where you're for the purpose of and the objective of this is education so that you can actually educate them because a lot of them have just blind followers
of this ama, these people with the imamat they sit in front of them, they have no knowledge in
and they are brought up in such an environment and that's why they deserve that what they should give them down
should help them
understand that that is not really the
the case
actually, may not come to this today, but actually tomorrow or in this setting today or tomorrow. I will give you the reference from Russia from the literature of Russia, that they believe that the
people of Kufa are the sole responsible party for the death for the murder for the massacre of Al Pacino.
They actually believe that the early she believed that they believe that
and the context of the story indicate that
inshallah if you don't mind
listen 1530 minutes
share with you
the story of the massacre of
a portion of it and maybe leave the commentary and
till tomorrow inshallah after fudger Bismillah he Todd
whatever I'm sharing with you right now comes from his my sources, and this is the standard Sunni reading of this event. It'll be there I wouldn't highly mucousy
Plus, Kalani, the author of photography, so this is not my I didn't make up that stuff yet. I'm just copying it from the books.
It has
outta me.
Most are the standard
any Sunni position on this event? So I'm sharing with you their
Of course, Shia Rafa. They have their own
They don't follow these schools.
And they base
reading of the event
on a historian who was known by Abu mithuna, a lot of new here
abou Messner, Lu, Ignacio here
himself, he says Mr. Awfully
and a lot of the stuff that he compiled
assembly think
he made up stuff.
First of all, as soon as
the same set
the victorious rule
we asked a lot to be of them era bellami
matteotti that would be a little bit
what is our belief?
concerning and we'll dive first of all do we know what elevate means what it means
is a derivative derives from
the origin of the word is added,
means the family
to Sakuma.
so Allah,
Allah, Allah Rajan, Hola, Dina, are those
to whom? Yo, la Mia? La Lune
need a turn
together? They get together.
What is our
Because you see the base there believes a shallow offer on that particular issue that we love the family of Nebraska, as if we don't do it
as if we don't do it. And this is again a misconception.
First of all, let's identify who our allied Navy
Air Force.
Allah, Allah Allah Muhammad, Allah.
Allah Mohammed
Idris Allah right
conocer late Allah, Ibrahim wala le Ibrahim, we then work Allahumma barik, Allah Muhammad, wa ala le Mohammed, every Salah.
For there is no way
that we have no respect
for the family of the Prophet salallahu alaihe salam
there is no way.
But we're in the middle.
Like any other thing, you see, there are two extremes.
There are people out there who say the family of the Prophet al in Libya or like any other family,
nothing special about them.
To the extent that some of them actually go after them, and we do have a second numa Golden navassa.
Now sub Walia
in support of man.
But then there is another sect in the Alma Mater hamdullah they are dying out.
In the day of Ashura, they show up they show joy. They actually find that they fabricated some Hadith about cooking food, and whoever gives his family food and Ashura and they end up doing what
for those are two extremes and the other extreme to this is what
is UCLA is Ella is
Elena de whither is the family of the brothers of Santa Barbara,
a man without him without him he cannot go to Jannah he tells the thing be and it will be
that you attribute divinity to the family of the Prophet sallallahu alayhi wasallam
you see we're in the middle.
We believe that they are the most respected family in the face of this earth.
And we love them
and we love those who love them.
And we resent those who resent them.
But again, our love for them
does not give them a status above
They're human beings.
They are regular people. Amongst them are the righteous amongst them are the wicked
is in Abu lahab,
from the family of the prophet will Allah
amongst them are the good amongst them are those who are.
If he's a Muslim and he is good, that gives him double love for us because he's from the family of the imagine this. If he was a companion of the brothers,
that is a triple standard for stable love.
But again, our love for them does not qualify us to do what to give them a status over their actual status.
And I'll share with you a co
worker so Nicola want to say hi, just to show you that this has been always the footsteps of workers. So this is one lady Neff. cvad will halifa Say hi.
By the one whose hand my soul is la cobertura su de la Habu la escuela karate
the lineage the family members of the Prophet sallallahu alayhi wa sallam are more beloved to me than my own family.
AKA Islam,
the father of Abu Bakr
on the day of the Mecca,
he was not yet a Muslim.
So he came to the Prophet sallallahu alayhi wa sallam, or actually a brothel or Salam went to him because he was so old.
And his beard was so wide and big, huge.
And he accepted Islam.
Abu Bakr, so DPS was started crying,
started weeping.
For us also, Salah looked at him and he said,
What is wrong with your work or your father is accepting Islam? Allah Rasul Allah by Allah messenger of Allah that to me acuna.
I wish this was your uncle
Shiva that his father
because he knows that the Brazos alum really wanted what he
wanted his uncle to accept Islam
Omar Musa because
our a normal quality lab so you're Muslim, that's
the law one syllabus. The day that
he became a Muslim is the uncle of the Prophet sallallahu Sallam the father of Abdullah
Abdullah a best sakala he said to him, let Islam lie Yemen Islamic hopper v low Islam.
You accepting Islam today is more beloved to me than my father, would he have accepted Islam
and Islam?
Because your acceptance of Islam? Can I have the ala rasulillah He is from Milan he was lm in Islam and hapa the same exact similar
because the Prophet would love for you to be a Muslim more than my father
was a woman who fully understood now Gemma Mahabharata will be
that you love the family of the Prophet sallallahu alayhi wa sallam
William la
imagine the Prophet sallallahu alayhi wa sallam one day he called upon Ali
called double Fatima
and hassanal for say, a famous Hadith in this wonderful Hadith.
The cloak and he placed them under
some sort of a garment
and he said look at those are my
take care of them please.
Well color FileZilla kilo como la si le Beatty.
I remind you by the rights of my L my family.
I am calling upon you by Allah
all of these are good
Man, in this room there is not like something that you have
a choice over. You have an option here.
We must love Alan Debbie Salalah has respect them, honor them, but the righteous and the pious of them
but that love
does not drive us to give them a status over the world, human status
and our politically and mustafi
please man whom indivisible aloha
there is a debate between us and an era in this area.
The classification of an Old Navy the they don't agree with us.
First of all, for us, Navy, of course, the the just the poor are those whom the Prophet sallallahu alayhi wa sallam called and placed under the clock. Ali Hassan will say no sorry about the coup.
But what about the wives of the Prophet sallallahu wasallam
do the shear I agree with us on this. Know
that the only verse in the Quran that talks about the family of the Prophet sallallahu alayhi wa sallam was mentioned in context of talking about the wives of the Prophet sallallaahu Leo son
in law who used to have an Kumar register Allah, Allah at
top hero,
Allah subhanho wa Taala wanted to remove
the impurity from the household of the Prophet sallallahu alayhi wa sallam. This verse was mentioned in context of talking about the wives of the professor's just go and visit Florida. Yeah, you and
me hopefully as long as you can continue to read the law and hire the zenith avatar no one knows me. Yeah. And he said
yeah, and he said
the minister
and then the whole actually context of the verses are the shear they say no the wives that's why they go after who after the majority of them.
top of the list is who I should
beside the wives and husbands per se no fault tamale.
kill is the brother of Ali
Al Jaffa. Jaffa brother family, the one who was killed in the Battle of
L meaning his family.
Early had ignored the mortality in our best
and those who are owned by the meaning there are no law. Like if there is slavery and they are owned by them they belong.
Why? Why this is important to know.
Simply because you're not supposed to give them soccer, right? You're not supposed to give them what
soccer cloud soccer you can give them a gift. But subaqua they are not supposed to accept so
are you ready to go over the story a little bit a portion of it.
We have to start a little bit early.
menarche that DNA for from our belief that we don't indulge into that fitness we just, but it's important
because I leave it alone and the father was a bit before assigned by the people fella
by the people of whom
he was bitten by them.
We know right after off man was assassinated, killed in Medina.
at the hands of people from Iraq
and Egypt to be failed
and some from Yemen.
They are not companions.
People rebels people who are thirsty for money for power.
They came and they killed off man in Medina in a cold blood in his house.
After the
bootham under siege,
even would have man the one who given water to the people of Medina he bought the will of Rumia for the Muslims
lot of the companions became so angry and upset and the majority of them were in Mecca performing hajj because those rebels they came after Hajj immediately and they assassinated of man in Medina, a
lot of the companions were still in Mecca did not make their way back.
The news got to them they got so sad.
They feel guilty
that they did not prevent this from happening.
But why didn't prevent it because of man told them
I'd rather die but not for the blood of one Muslim to be shed because of me.
You've completed this an asset now on
one of these tyrants that we have
of man muawiya By the way, he wasn't the masochist in the army. He said I'll send you an arm he said no.
I let me with
any rate
under pressure, ie Navy authority, or of the above one
took on the leadership of the Muslims under pressure he didn't want
immediately the companions demanded from Ali, the vengeance for the blood of man.
We need you right now to bring these guys to justice
Ali, or the law one had a different opinion.
He said let things settle down a little bit.
And I will do this but right now these guys are all over Medina
wait until they go back to their homes.
And we'll bring them back to justice.
And by the way as Aboriginal Gemma, we believe that this is the right opinion.
But this does not qualify us to go after the Sahaba after Ayesha
after the
because they were demanding justice for the killing of Muslims especially more aware of the law one can argue with them enough man from Benny Amaya
of man is ama we, we as ama we.
Somalia is the one who's entitled normally in the Arabian
tradition, the head of the tribe is the one who demand the ransom or the blood of someone who was killed from the tribe.
He get two opinions.
A group from the Sahaba
who was driven by compassion
vengeance, vengeance,
Valley and Avatar it was
you know hitting leading people is imagining people is not is it takes an aqua there is no black and white and in politics.
There is no black and white
at all.
Its assessment of what is known in our sharp masala benefits and what
and you could make the wrong call. And this is why look and you should know Gemma we stand firm and say all the Sahaba they even though they ended up fighting one another. The number of Muslims who
were killed in the fight that took place what is known to be civil war between the Sahaba is more than the number of Muslims who were killed in conquering the non Muslim land.
You believe this?
but yet we stand firm and say well, all of them had a good intention.
All of them
had a good intention
And another list of companions
when they saw
that are leaving avatar Liberty alone is not going to take immediate action
to avenge the assassination of man.
Communication happened with the people of *
again the same thing.
COMM will help you Basra pusateri Arata comm will going to help you to pinch
olive oil it followed Arusha
to Kufa to Iraq to stop this from happening.
to battle the war took place.
suffering, German was a
German between the Sahaba.
And suffering
is between the army of Harley and the army Huawei because Maria refused to give the pledge to Ali, the oath of allegiance.
So Ali took his family to we're now
And just to show you that Ali was bit in the Battle of Sofia, Ali was about to finish the fitna once and for all. With Norway.
He almost defeated the RV of Mojave, or the other one.
The fitna would have been killed
and peace would have been restored to the Muslim world.
But the army of Mongolia did the trick.
They raised
on the tip of the swords domicile
meaning that we're requesting arbitration according to the book of Allah.
The people of Iraq in the army refuse to fight they said we're gonna abolish it. No, let's finish it. They refused. They backed off.
they disobeyed him. They walked out on him.
That's why Ali was what.
A couple of years later he was assassinated at the hands of wonderful coverage of the Rockman animaljam died
after his death
the year 40 something
it hasn't became me remove meaning
in the pot.
Remember, now we are still What did not give the pledge of allegiance to ollie.
Hill has an Subhana Allah fulfill the prophecy of Rasulullah solemn when he said My son is sad. He is a leader. One day, Allah will use him to make peace between two facts to fighting factions of the
Muslims. Six months later, he sent to Ali, thank you very much. I'm not interested in that. You're the me. Here's the Pledge of Allegiance. He took al nebby Salah Salem he took the family of the
prophets of Salaam and he moved all of them back to Medina
this finish that chapter
was called the year for Emily
the beautiful Gemini again because the Muslims again body became what one?
Because if it happened did not do this, this would have been another
Can you imagine?
For the first time, you get people giving bleach to malware, and people giving blisters
is a continuation of what
of the bloodshed. The Civil War in the Muslim world
hasn't died around the 49. After
thing is almost later, eight or 910 years later, after he made the deal
will stay the same. Of course all of this by the way, same social feed so what his father went through.
So he's really developing what
By the way, is younger than a half and one year
old Hassan was born the third year after he
was born, the fourth year after
one year between them. That's all.
Things went, Okay, where are we? I was treating them so well.
Maybe after this,
he commanded for them to be treated.
Don't even disappoint, don't even
give them any hard time. They just they commanded the governor of Medina to take very good care of them.
While we are on the lawn,
nominated His Son to be the halifa after him.
He has he
towards the end of the year 16.
After he
Yes, he is not a companion.
He was born, they use 25 after 15 years after the death of the Prophet sallallahu sallam.
So you're nominating z your son
when you have look at this list of names, and you're saying if niaouli still alive,
alive now
the live number of the last
jab jab Nabila as hobby maybe
they didn't like it.
But again, where are we are we alone did not violate anything. There is nothing in the Sharia that says you have to do this. He simply said, You know what? Instead of the oma that's his judgment.
Instead of the oma going into a bloodshed again and who's going to be the ameerul momineen. And by not by and a pledge of allegiance and all of this. Let it be my the holy cow finish it.
It's not haram
but certainly as an eternal Yama, we believe that was not
the right coup. But again, this does not drive us to go
after Maui like they do. Again the middle way. Because this is again a judgment call.
Like we said, in political Islam, there is no black or white.
Hola Rajan. Gianni, he looked at the situation we just finished in our bloodshed let's just keep it quiet. My son takes over the hill after
saying refused to take the oath of allegiance to your seat.
And early Navy refuse. They said no, we're not gonna do it.
While we are the best away Is he the first
sign of lack of wisdom right here.
He demanded that oath of allegiance
more Hussein refused to give the oath of allegiance but they said what good to your business I'm not gonna bother you.
like okay, I'm not I'm not gonna support you, but I'm not gonna harm you either. Just forget about me as a father What
would exist.
If he would have been wise he should have but and by the way, they are relatives.
He's married to his company.
He could have invited him later on or something Yanni. Darla. Let's talk about this.
And you could have done it.
He sent a letter to the governor of Medina, forcefully and Jose must give you the oath of allegiance to me in the Knesset tomorrow.
One day after that letter reaches you.
The governor of Medina, cosa buena say to his house. He showed him the letter.
Tomorrow, after Russia in Russia, you have to give the oath of allegiance.
No, Yes. No. Yes. He saw that. There is a lot of
pressure. Then he said let me think about it.
He goes home, back
all his family after midnight, run away too much.
This news got to who
to hire up. Because so far I don't like clamavi
from the time of family
they were the core of the army of Harley.
They fought against
Huawei and
they didn't like them.
And they didn't give the oath of allegiance yet to yazeed.
A lot of them did not.
So they ended up
communicating with a sign as soon as the news got to them that will refuse to give the oath of allegiance who has EAD and he fled to Makkah. They sent to him come to our country, Kufa.
We're gonna take care of you.
They sent him over 500 letters.
Many delegations came to him in person.
Come will give you the oath of allegiance, you're going to be ameerul momineen
and we're supporting you protect you will take care of you.
They tell you in these 500 letters, from 18 to 20,000 Oath of Allegiance individual Oath of Allegiance because every letter maybe 1000 2000 signing
like 100 people signing and one Oath of Allegiance purchasing your annual income.
He decided to send
his cousin
Muslim if naughty pill the brother foo valuable
Muslim is his cousin.
Go and verify that information.
Muslim arrived in kuva Kufa as soon as he arrived, in two days he gathered
will sow the more Oath of Allegiance.
And not only this,
these 4000 are representative of 30 to 40,000 members
of the people who will go
he got so excited.
He immediately sent a letter to
me proceeding in Mecca, come right away Don't delay.
This happened in what month
around the 10th the 15th of Bill CAD
watch these days now, the
delay ledger. And then what Mahara? Watch the deeds now, because this is important.
Who found out about this
that there is a lot
for saying to come.
And the same scenario will be repeated again.
Which will happen what
politically and when he resigned
another fitna will be implanted.
He chose the most vicious human being the most unwise
human being who was
at this time the governor of Missouri
whose name is obey the law a bluesy and ignore B. That's how they he called himself obey the law him not yet.
He went
and he kicked out the governor of Kufa The one who all this happened under his watch.
And he took over
governing a coup.
And now the main task is to quell to kill that fitna.
Who's going to be looking for now whom he is going to be looking at. He is after Muslim.
If not,
I need Muslim have not been arrested right away.
Muslim Laughlin was hiding in Kufa he did something very interesting just to show you
this guy was 2018 years old
his job his vision is to just you know get on the ranks of position with Benny omega like he just wouldn't be
you know like somebody I want to be the president regardless now what you know I want to be the
is thriving for this it doesn't matter whether I end up harming the grandson of the prophets awesome
so the Buddha good according to him ianya successful scheme
he got one of these but his guys to pretend that he is bro.
again is
Benny Amaya yazeed
and he's bringing in a lot of money from people
L Kufa I'm sorry people in Damascus
to be given
to the head of the person the head of the of the one who's leading that scheme
so he arrived in Kufa as it is coming from Damascus and he has a lot of gold with him and he was instructed to deliver that gold to the one who is managing that bloods business tool per se
so he needed to know who's the head of this
because so far away the live does he doesn't know who's cooking all of this
and he thought that Muslim have no hottie Are you familiar with these names right Muslim naki would be hiding in the house of this man. Or at least this man knows where Muslim Rocky is hiding.
So gold and silver and money got him through
he reached a man whose name is Hani or a hand
a blue
hand you know
and Subhanallah this man you know the concept of
you not potty is
the idea that you show something but in your heart What?
That's Ashi principle and the theology
so he was a heading cheery Danny brew and his great Auntie Emma we but he was showing that his brew ama we brew mo Brasilia
immediately as soon as Aveda live and as he had found out who he is he arrested him
You are pretending that you are our guy you know your bro us and now we found out that you're cooking all these things? Where is Muslim have no appeal? He said I don't know.
He had the guy who went and delivered the goal to him behind the curtain.
So he came out and said what do you say about this man and the goal that he received from him?
So he couldn't deny
the news get to Muslim women Ophelia Aqua
he's about to get arrested.
Look at this now. Look at this.
He sent
a call a cry call is called a cry call. To the people who've given him the pledge come and help me.
Remember I said 30 to 40,000
how many showed up
to help him.
They went and they stood in front of the balance of the of the Governor
303 the guy inside and you get out of here
from panela this or where you live and as he had in 12 hours.
He paid them off somehow threatening them.
He just got the head of these tribes, giving them money.
A toss of time.
A thought of time, a life where this is from the looks of it.
Muslim nappy looked behind him he only saw 70 standing with him.
He lists a lot to remember if you look back he found 10
in the line
so he decided to go on to a place called kingda. That's the the most bro. Yani elbaite try an IRA.
Those 10
they walked out.
He got arrested.
But somehow Hana life what Jani
bait also is the cousin of foo, which was
the man whose name is in a shot of melanoma. That the leader of the the Yoni the the army who arrested him or the Syria, the squad who arrested him.
And he said to him, Muslim said to him, I don't care what you're gonna do with me.
Whether you're gonna kill me,
but you promise me right now
that you're gonna send the message
to her saying a letter to her say, not to come to
that the people of Iraq are gonna let you down.
And this will be the cause for your death.
You promised me this.
So how am I if we're on the ninth of June,
they took him to the top of a mountain. They cut his head. That's how they used to do it and they throw his head down in his body. They killed
they tell you that he was standing on the top of this mountain.
You held him? There ilaha illAllah stuffer la hamdulillah
Allahu MK DB Nana avena Cr Tina. Oh Allah, you're the judge between us and invades, and those people who let us down.
So for a while, and Sharla we're gonna stop here. Jani. That's a good place to stop.
But so far, it hasn't
all what he received is the message from Muslim earlier. You need to look at the geography here.
Era Medina.
And also the mosque is a sham, because that will come in action here. Maybe tomorrow insha Allah.
The messenger that Muslim the letter that Muslims sent when he arrived earlier just got to say, on the eighth of the pad, one day before Muslim was as was executed.
So the message received was what?
Come Don't delay.
The people of will Cooper are
I combined 30 to 40,000 Oath of Allegiance for you come full force is about to leave.
Based on that message.
Of course the news that
he was executed and all did not even get to him.
Some of the stories he sent people, messengers after him and they were executed to
he decided to collect his family.
Alan levy and back to wave Turku faqih
who stood in his way, have a nice little loss and I'm the live normal life not this job.
I was able to do
his brother Mohammed Abu hanifa a brother but not from Fatima
from another wife or family.
Don't, don't go, don't go
because they have seen what happened before
the Civil War that shook the Muslim ummah.
But he showed them look, he actually carried these letters that they sent to him earlier on
On to camels look at these letters
and look at the letter that I received from my scout the one that I tend to verify
I can let those people down and then assign also thought
that if he does not leave Mecca and go and somehow take action he has either is going to chase him in men and
he's not going to let him in but
that's his judgment again that's the judgment of him
that if he decides to stay in Makkah I'm sorry for saying decided to stay in Mecca and not go to Grafton Cooper yazeed is not going to leave him alone. He's gonna like he he he demanded him to give
their oath of allegiance in Medina is gonna do the same thing in a fair he said to the companions of the Prophet SAW Selim, I don't want the bloodshed in the Haram
and you know what, two years laters
two is they did this with other lighteners rubies.
There was a big fight and then Ferrum
wanted to move out of Mecca to a Yanni to take the fight out.
But yet the Sahaba their opinion but again you know what, this does not mean that the Sahaba thing is that they are bruja de
again the reading of the situation is different to the reading of the situation. I'm saying
that this is dangerous This is not good for his family not good for himself not good for the Muslims that their opinion but this does not mean that they love us. And when we say this this does not
mean that will obviously Dover lol Hassan
Abdullah this wonderful Imam Ahmed asked as Alabama had one day should we love yazeed
you know what he said to him?
Anyone who has left over a man in his heart cannot love your seat.
Should we curse him? No. We don't curse
learn oh hey boo while Anna Lalu
was that you find CR when they come to you by the way they come in. Okay, what do you say about God they come to us. What do you say about that? He wants you to
We like him. We love him
because he was the head of the scheme but wouldn't curse him
for inshallah we stopped here
we continue to Morrow, but remember because we cannot trace a lot of that stuff back. So we want to build from now, alpha is about to leave
to Kufa
and that is the massacre is going to take place in Karbala which took place on the 10th of Muharram inshallah we continue tomorrow in the light Allah
Subhana Allah Hama, Shadow Allah Allah headland
were to lay but hamdulillahi Rabbil aalameen
Don't forget to pray Shaka Chawla
Martyrdom Of Imam Al-Hussein (Part 1) By Karim Abuzaid
This Lecture was delivered at Dar Al Tawheed Colorado during the workshop “Two Lost Nations” | {"url":"https://muslimcentral.com/karim-abuzaid-martyrdom-imam-al-hussein-part-1/","timestamp":"2024-11-05T12:16:30Z","content_type":"text/html","content_length":"150383","record_id":"<urn:uuid:a56ec723-feac-4b09-adcf-14984a9a9e8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00265.warc.gz"} |
A living organism emerging by chance as imagined by naturalism and evolution is not just absurd but is fantastically absurd. Genetic research has revealed that the complexity within all living forms
make such a theory laughable. There are no “simple” life forms as imagined by early evolutionary popularizers like Charles Darwin. To envision life emerging by spontaneous random chance from
non-living chemicals based on modern science requires a complete suspension of your brain, all common sense, reason, and a reliance on pure serendipity and miracle. Such an idea is nonsense—life
was designed.
The Academy of Evolutionary Metaphysics tells us a just so story about the initial formation of self replicating living cells. The quote reveals the idiocy that random and spontaneous processes can
do anything, let alone form life:
“Life began when one of these complex organic molecules began reacting with the other molecules around it in an unusual way. It was able to attract all of the pieces that it needed to assemble an
identical copy of itself. The copy then split away from the original and began to assemble its own new copy.”
DNA works to self-modify, self-manufacture, run self-diagnostics, self-repair, self-program (read and write), and self-corrects reproduction errors. DNA is the most complex information system in the
known universe.
“Inside Life Science”, Chelsea Toledo & Kirstie Saltsman, June 2012, National Institute of General Medical Sciences. http://publications.nigms.nih.gov/insidelifescience/genetics-numbers.html
Basic odds can be easily explained with dice. One six sided dice has a 1 in 6 chance of rolling any number 1-6 by chance. When an additional die is added the odds are multiplied. Two dice are 6 X 6 =
36 or a 1 in 36 chance of a specified number occurring 2-12. As each additional die is added the odds increase or the likelihood decreases.
Simple odds review
In order to understand the idea of the emergence of life we must look at math which involves equations that include an ordered pair. As an example, in an ordered pair of three, say three playing
cards numbered 2, 3, and 4 are shuffled. The likelihood they will come out in the desired order (in this case coming out in order “2”-“3”-“4”) are expressed on a calculator as 3 x! . This is a a 1 in
6 chance. On ordered pair is necessary because the assembly of amino acids or even protein chains must have initially randomly and spontaneous formed when the first “protocells” emerged.
When and ordered assembly is required then the math requires ordered pair mathematical odds.
With each increase in our ordered paid the mathematical of likelihood odds goes down exponentially. As an example an ordered pair of 5 is a 1 in 120, but an ordered pair of 10 jumps to a 1 in
3,628,800! If we consider an ordered pair of 100 the numbers blow up your calculator as 1 in approximately 9.3^157!
Average across species 337.75
The numbers shown in gray represent the odds of randomly assembling a protein chain of a mere 100 amino acids– which is a very small chain, average protein chain is over 300 over species including
the “lowly” bacteria. The largest protein chain is over 34,500 in length! Here our numbers ignore the random and spontaneous chance that the universe, our solar system, our planet earth, chemicals,
atomic structures and functions, gravity, matter, water, atmosphere, and most everything are mathematically excluded from these odds.
Although it is an exercise in futility, if we examine a quest to determine mathematically such odds have been approximated as 1 in 10^650.
Anything that reaches beyond 10^50 to 1 is considered a mathematical impossibility called mathematically absurd – meaning impossible and absurd to consider any other outcome. | {"url":"https://www.evolutionisamyth.com/biological/the-living-cell-arising-by-random-chance-absurd/","timestamp":"2024-11-13T08:06:53Z","content_type":"text/html","content_length":"68945","record_id":"<urn:uuid:e6eef25d-18ca-4f32-9fea-e22f33cd7e07>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00763.warc.gz"} |
On the list decodability of Rank Metric codes
Let $k,n,m \in \mathbb{Z}^+$ integers such that $k\leq n \leq m$, let $\mathrm{G}_{n,k}\in \mathbb{F}_{q^m}^n$ be a Delsarte-Gabidulin code. Wachter-Zeh proven that codes belonging to this family
cannot be efficiently list decoded for any radius $\tau$, providing $\tau$ is large enough. This achievement essentially relies on proving a lower bound for the list size of some specific words in $\
mathbb{F}_{q^m}^n \setminus \mathrm{G}_{n,k}$. In 2016, Raviv and Wachter-Zeh improved this bound in a special case, i.e. when $n\mid m$. As a consequence, they were able to detect infinite families
of Delsarte-Gabidulin codes that cannot be efficiently list decoded at all. In this article we determine similar lower bounds for Maximum Rank Distance codes belonging to a wider class of examples,
containing Generalized Gabidulin codes, Generalized Twisted Gabidulin codes, and examples recently described by the first author and Yue Zhou. By exploiting arguments suchlike those used in the above
mentioned papers, when $n\mid m$, we also show infinite families of generalized Gabidulin codes that cannot be list decoded efficiently at any radius greater than or equal to $\left\lfloor \frac{d-1}
2 \right\rfloor+1$, where $d$ is its minimum distance. Nonetheless, in all other examples belonging to above mentioned class, we detect infinite families that cannot be list decoded efficiently at
any radius greater than or equal to $\left\lfloor \frac{d-1}2 \right\rfloor+2$, where $d$ is its minimum distance. Finally, relying on the properties of a set of subspace trinomials recently
presented by McGuire and Mueller, we are able to prove that any rank metric code of $\mathbb{F}_{q^m}^n$ of order $q^{kn}$ with $n$ dividing $m$, such that $4n-3$ is a square in $\mathbb{Z}$ and
containing $\mathrm{G}_{n,2}$, is not efficiently list decodable at some values of the radius $\tau$. | {"url":"https://api.deepai.org/publication/on-the-list-decodability-of-rank-metric-codes","timestamp":"2024-11-10T01:11:25Z","content_type":"text/html","content_length":"155353","record_id":"<urn:uuid:23d09828-a55c-4dc8-aba5-d01b745a11fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00305.warc.gz"} |
Spectral bounds, orbit growth, and temperedness of locally symmetric spaces.
Event time:
Monday, October 28, 2024 - 4:15pm
Event description:
Consider a semi-simple Lie group, $G$ and a discrete torsion-free subgroup $\Gamma < G$. When $G$ has rank one, there is an important, well-understood connection between the growth rate of elements
of $\Gamma$-orbits, bounds on the spectrum, and temperedness, or more generally properties of the unitary $L^2$ representation (due, in part, to work of Elstrodt and Patterson). In this talk I will
present an extension of this connection to higher rank. This is joint work with Tobias Weich and Lasse Wolf. | {"url":"https://math.yale.edu/event/spectral-bounds-orbit-growth-and-temperedness-locally-symmetric-spaces","timestamp":"2024-11-11T18:09:31Z","content_type":"text/html","content_length":"36085","record_id":"<urn:uuid:f5ecdcbd-979a-4b08-bd8a-224160a6eb5d>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00049.warc.gz"} |
How to get the minimum from two pandas series?
You can use the min() function provided by pandas to get the minimum value from two pandas Series. Here is an example code snippet:
1 import pandas as pd
3 # Creating two pandas Series
4 s1 = pd.Series([10, 20, 30, 40, 50])
5 s2 = pd.Series([15, 25, 35, 45, 55])
7 # Getting the minimum value from the two Series
8 min_value = min(s1.min(), s2.min())
10 print("Minimum value:", min_value)
In this code snippet, we first create two pandas Series s1 and s2. We then use the min() function on each Series to get the minimum value from each Series. Finally, we use the min() function again to
get the minimum value from the two minimum values obtained from the two Series. | {"url":"https://devhubby.com/thread/how-to-get-the-minimum-from-two-pandas-series","timestamp":"2024-11-05T20:41:59Z","content_type":"text/html","content_length":"113896","record_id":"<urn:uuid:05f62e70-8a65-4a45-ad3d-f840fb504ba9>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00290.warc.gz"} |
Water Volume Calculation Formula of a Rectangle
Knowing how much water your tanks hold is very important in aquaponics. Here are the formulas for calculating water volume in a rectangle grow beds.
How to Calculate the Volume of Water in a Rectangle
I am using a sheet of Firestone 10 X 15 Liner-45mil EPDM. I cut that in half for 2 sheets, and then build rectangle troughs or grow beds out of them. The measurements are approximately 30″ wide x
150″ long x 12″ deep. I default to these numbers to show you the example, but you can enter any numbers you want for your specifics.
First, we need to get the cubic inches (wide x long x deep):
Width (inches)
Length (inches)
Depth (inches)
Cubic Inches
Now we convert to cubic feet (cubic inches x 0.000578704)
Cubic Feet
To calculate gallons, we multiply cubic feet by 7.47:
Gallons (Water only)
Gallons (Water + Grow Media)
To figure out how much this weighs (and if your subfloor can support the weight), we multiply 8.34 (freshwater weight of 1 gallon) x how many gallons:
Weight (only the weight of the water)
Note that if you are using this in grow beds, you will likely only get half of the water you plan for into the grow bed, the rest is filled with media (like gravel or hydroton).
{ 0 comments… add one }
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://aquaponics-system.com/aquaponics-calculations/water-volume-calculation-formula-of-a-rectangle/","timestamp":"2024-11-13T22:21:40Z","content_type":"text/html","content_length":"26699","record_id":"<urn:uuid:6b7d934e-3e81-42fe-9a78-6a6da3e31354>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00638.warc.gz"} |
Reading Math
First, a recent gem from MathStackExchange:
Task: Calculate $\displaystyle \sum_{i = 1}^{69} \sqrt{ \left( 1 + \frac{1}{i^2} + \frac{1}{(i+1)^2} \right) }$ as quickly as you can with pencil and paper only.
Yes, this is just another cute problem that turns out to have a very pleasant solution. Here's how this one goes. (If you're interested - try it out. There's really only a few ways to proceed at
first - so give it a whirl and any idea that has any promise will probably be the only idea with promise).
Looking at $1 + \frac{1}{i^2} + \frac{1}{(1+i)^2}$, find a common denominator and add to get $\dfrac{i^4 + 2i^3 + 3i^2 + 2i + 1}{i^2(i+1)^2} = \dfrac{(i^2 + i + 1)^2}{i^2(i+1)^2}$. Aha - it's a
perfect square, so we can take its square root, and now the calculation is very routine, almost.
The next clever idea is to say that $\dfrac{ (i^2 + i + 1)}{i(i+1)} = \dfrac{(i^2 + 2i + 1)}{i(i+1)} - \dfrac{i}{i(i+1) }$, which we can rewrite as $\dfrac{(i+1)^2}{i(i+1)} - \dfrac{1}{i+1} = 1 + \
dfrac{1}{i} - \dfrac{1}{i+1}$. So it telescopes and behaves very, very nicely. In particular, we get $69 + 1 - \frac{1}{70}$.
With that little intro out of the way, I get into my main topics of the day. I've been reading a lot of different papers recently. The collection of journals that I have access to at Brown is a
little different than the collection I used to get at Tech. And I mean this in two senses: firstly, there are literally different journals and databases to read from (the print collections are
surprisingly comparable - I didn't realize how good of a math resource Tech's library really was). But in a second sense, the amount of math that I comprehend is greater, and the amount of time I'm
willing to spend on a paper to develop the background is greater as well.
That aside, I revisited a topic that I used to think about all the time at the start of my undergraduate studies: math education. It turns out that there are journals dedicated solely to math
education, see here for example. And almost all the journals are either on JSTOR or have open-access straight from Springerlink, which is great. I have no intention of becoming a high school teacher
or anything, but I became interested as soon as I began to come across people with radically different high school experiences than I did.
My high school tried to protect its students, sometimes in ways that I didn't like. It was the sort of place that, in short, held me back in the following sense: they wouldn't let anyone take 'too
hard' of a course-load for fear that they would overwork themselves and therefore fail, or do poorly, or overstress, in everything. In more direct terms, this meant that you had to petition to take 3
AP classes and had to really work to take 4. Absolutely no one was allowed to take more than 4 in one school year - so that many of my friends had to choose what science to take. Those of us who were
willing all had sort of the same schedule in mind - if you did an art (band/choir/orchestra, usually), then in 10th grade you took AP Statistics, 11th AP Language, 12th AP Lit, AP Calc, AP (foreign
language or Gov or European History or Econ), and an AP science - if no art, then you could take an additional AP science in 11th grade. At least, that's how it worked while I was around.
So the big decisions were always around the senior year. For me, I had to ask: should I take AP Chem or AP Physics? (I ended up taking Physics, which was great - it was the curiosity and intuition
from mechanics that led to me becoming a mathematician now). Many of my friends asked the same sort of questions. And it was very annoying - I hate the idea that the school holds us back, ever. It
also turned out that one of my classes was terrible. I was so annoyed that one of my four choices ended up being bad that I wrote an embarrassing letter (which I regret to this day).
In short, I felt slighted by the system, and I've considered the system ever since. One of the articles I read was about the general idea that the sciences taught in schools and even at
entry-undergraduate level in college are fundamentally different in both motivation and skill set from the ideas held by scientists and those who progress those subjects. The interesting part about
the article was the amount of feedback that the journal received - enough to merit multiple copies of letters back and forth to make it to the next printings of the journal.
That particular article was very careful to simply assert that the current paths of education in the sciences and the sciences themselves are different, as opposed to positing that any particular
idea or method is above or better than any other. But of course, it's perhaps the most natural response. Should they be different? Why does one learn math or the sciences in school? For that matter,
why does one learn history (also oblique and hard to answer, but something that I maintain is important for at least the reason that it was the only substitute I ever had for an ethics class in my
primary and secondary education).
These are hard questions, and ones I'm not willing to directly address here at this time. But I will quickly note that in both Tech and Brown, I am stunned at how many people lack any sort of
intuition for the four basic operations - (I once tutored someone who, upon being asked what 748 times 342 was, responded that it didn't even matter because "math was made up at that point. It's not
like someone has sat down and counted that high." oof. That hurts. Let's not even talk about being able to add or subtract fractions. As a worker at the 'Math Resource Center,' I've learned that
about a quarter of the time, helping people with their calculus classes is really a matter of helping these people manipulate fractions. So if the purpose of primary and secondary education is to get
people to understand arithmetic operations and fractions, it's not doing so well. John Allen Paulos should write yet another book, perhaps (Innumeracy is a good read).
Should they be different? That is, is there much reason for the sciences and the education of the sciences to align in method and motivation? I'm not certain, but perhaps they shouldn't pretend to be
the same. I only ever learned arithmetic, as opposed to math, throughout my primary and secondary education with 2 exceptions: geometry (which had a surprisingly large logic content for me, and
introduced me to interesting ideas) and calculus. Calling it math is a disservice - as Paulos mentions in his books, the general negativity towards math allows people to claim innumeracy ("I'm not
really a numbers person") with pride - no one would ever say that they weren't very good with letters. But reading is useful, or rather widely recognized to be useful and expressive.
I end by mentioning that I think it is more important to come across real ideas of science and math at an early age, say elementary school, then middle school. In in elementary and middle school,
there really isn't much difference between the maths and the sciences, so I clump them together. But in my mind, the initial goals of science and math education should be to spark creativity and
wonder, while English and reading courses stress critical thinking (somehow, math, science, and English all get the boring end of the stick while reading gets full hold over the realm of creativity -
how backwards I must be).
But those 4th graders whose teacher guided them towards the bee research, that has now been published under the 4th graders' names - don't you think that their view of science will be a much happier
and, ultimately, accurate? Exciting, collaborative, uncertain with a scientific method-based structure. But then again, perhaps the lesson that my friends and I learned from our own high school is
the most relevant: if you want to do something, then don't let others stand in your way. A little motivation and discipline goes a long way.
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Comments (10)
1. 2011-10-26 gowers
I enjoyed that problem. A small remark about the solution is that you can do part of it more neatly (and transparently I think). When you've got
you can just instantly spot that the numerator is 1 more than the denominator, so it equals $1+1/i(i+1)$. At that point we're on familiar territory.
2. 2011-10-28 Jimmy
I really like what you are saying, and am glad I found your website. Please keep it up.
3. 2011-10-31 davidlowryduda
Thanks Jimmy!
4. 2011-10-31 davidlowryduda
Re: the first comment from Gowers
I agree - that's much cleaner. Thank you for that.
5. 2011-11-20 Elektrische Zahnbuerste
Lovely sharp post. Never considered that it was that effortless. Praises to you!
6. 2011-11-20 Elektrische Zahnbuerste
... [Trackback]...
[...] Read More: mixedmath.wordpress.com/2011/10/22/reading-math/ [...]...
7. 2011-12-17 TomF
David, I think that you may find this interesting. It's about math education. http://www.maa.org/devlin/LockhartsLament.pdf
8. 2011-12-19 Gigili
What a deceptively simple question!
9. 2011-12-19 davidlowryduda
Hey Tom! I know the lament, and it's a great read. In fact, there's a pretty good book written largely as an extension of the lament (http://www.amazon.com/dp/1934137170).
How have you been? I haven't heard from you in a while.
10. 2011-12-21 TomF
I've been good! I had to transfer for Tech to a school closer to home because my mom was sick. But, the school i transferred to wasn't a good fit for me at all. So i dropped out, and went to fish
in alaska, and travel. That has pretty much been my life for the past year and a half. | {"url":"https://davidlowryduda.com/reading-math/","timestamp":"2024-11-14T04:17:01Z","content_type":"text/html","content_length":"15943","record_id":"<urn:uuid:adf6d174-134f-479c-b207-44131b3f26de>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00851.warc.gz"} |
How to use Import Diagnostics in SolidWorks? - Mechanitec Design
What is Import Diagnostics Tool?
Whenever you import a file in SolidWorks that has a different file type than .sldprt or .sldasm, it is recommended that you check that the model doesn’t have any faulty geometry. A faulty geometry is
only obtained when the 3D model is opened as a Solid or Surface Body. If you open a file as Graphics Body, its usability is very limited and hence Import Diagonistiocs tool is not available. Import
Diagnostics tool runs diagnostics (checks for geometrical and topological errors) and repairs the geometry.
Almost every software has its own file format to store the 3D model data and when you try to save your model in a universal file format (such as .step or .iges), conversion of file types takes place,
and faults in the geometry get introduced. The faults may also arise during importing of the file but that’s usually not the case.
Tip: It’s recommended that you use the native software to export rather than using third-party applications to do the conversions.
Why do you need to use the Import Diagnostics tool?
The Import Diagnostics tool is used to identify geometrical and topological errors in the model and repair those faults. This repair capability is needed because imported surface data often has
problems that prevent surfaces from being converted into valid solids (because at the end of the day it’s the solid geometry that we need for product making). These problems may include bad surface
geometry, bad surface topology, or gaps between the surfaces (sometimes adjacent surfaces have edges close to each other but do not meet, hence creating a gap).
After fixing these faults it then proceeds to knit the repaired faces with the rest of the surface body and if possible, automatically makes solid bodies from closed surfaces. It also converts any
complicated B-spline surfaces into simple analytic surfaces for better performance.
How to access the Import Diagnostics tool?
Whenever you import a non-native SolidWorks model with faults, a message will appear asking if you want to run Import Diagnostics. Click Yes to activate it.
You can also automate this process in SolidWorks Settings. You can access the Settings from Tools -> Options -> System Options -> Import. Then select General in the File Format. There you will find
Automatically run Import Diagnostics (Healing) and Perform full entity check and repair errors. Check these options and the Import Diagnostics tool will automatically start.
If you were not able to initiate Import Diagnostics during import, you can access it from Import Diagnostics present in the Evaluate Toolbar or go to .
Caution: It is critical to run this tool immediately after the import process is completed. The Import Diagnostic tool cannot be triggered if any other feature has been added to the Feature Manager
How does Import Diagnostics find problems?
Import Diagnostics runs various checks on the model to make sure that the geometry is not bad.
1. It runs the Check tool. You can access this tool manually from the Check option present in the Evaluate toolbar or go to .
2. It then runs additional checks to see if there are overlapping or self-intersecting surfaces.
3. Then it proceeds to replace complex surfaces that may reduce the performance of the model. Any planar, cylindrical, conical, etc. surface that is accurate but un-simplified B-splines present in
the model are replaced with equivalent analytic surfaces, if possible.
Tip: For a coarse understanding, B-splines are surfaces that are made using splines and bezier curves while analytical surfaces are those that are made using lines, arcs, circles, fillets, and conics
(ellipses, parabolas, and hyperbolas). If your model contains B-splines, that is not a problem as B-spline surfaces are also valid and sometimes even necessary. But replacing them with equivalent
analytic surfaces improves performance and makes the model more usable in SolidWorks by making it easier for them to be used to create references. For example, you can’t create concentric mates to
B-splines that are cylindrical. You can only use analytic cylinders for concentric mates.
How To Use Import Diagnostics Tool?
1. Import any non-native file type into SolidWorks.
2. If you used 3D interconnect to import the file you need to break the link in order for Import Diagnostics to perform the healing operation.
To break the link, right-click on the feature present in the Feature Tree and click on Break Link.
A SolidWorks warning will appear stating that this action can’t be undone. Click Yes, break the link. You will notice that the Feature Manager Tree now lists all the solid and surface bodies.
You can turn 3D interconnect off to get access to more options during the import. Go to Options -> System Options -> Import. Then select General in the File Format. There uncheck the Enable 3D
Interconnect option to access additional settings such as if SolidWorks will try to form solids from the surfaces or just knit those surfaces.
3. Click on the Import Diagnostics tool present in the Evaluate toolbar or go to Tools -> Evaluate -> Import Diagnostics.
3. Import Diagnostics will perform all the diagnostics to find any bad geometry. It may take a while if your geometry is extremely bad or very complex. After this is done, you will be greeted with an
Import Diagnostics Feature Manager which lists all the faces that have faults in them and all the gaps present in the surfaces that prohibit them to be converted into a solid body.
4. Click Attempt to Heal All to allow the Solidworks to automatically repair all the faults and gaps in the faces. Most of the time it is enough to fix all the errors but sometimes it may fail. And
also if there are a lot of errors listed, then it may take a long time and sometimes may even crash the application.
5. Or instead, you can click on any of the Faulty faces listed in the table and it will get highlighted in the graphics area.
If you are wondering what’s wrong with this face shown above, then let us tell you that it is an un-simplified B-spline surface that can be converted into an analytical surface.
6. Right Click on any item to access additional options available.
• Repair Face performs repairs with the methods discussed above.
• Delete Face deletes the faulty face. If a face has too many faults to repair, you can delete the face and then use Surface tools to remodel a new face in the gap.
• Re-Check Face again performs all the diagnostics on the specified face and displays the result.
• What’s Wrong will tell you what is the error in the face.
• Zoom to Selection is pretty explanatory itself.
• Remove Face from the list allows you to keep the faulty face. It is mostly used to retain the B-splines from getting converted into analytical surfaces.
Import Diagnostics repairs face by doing one or more of the following:
• It tries to recreate the trim boundaries of the face based on the surrounding geometry. This method often fixes overlapping faces.
• It trims away defective portions of faces that are not used in the model.
• And the last resort is to remove the faulty face and use the gap repair algorithm to fill the resulting hole.
• It also tries to replace any complicated B-spline surfaces with simple analytic surfaces for better performance.
All the faulty faces that are fixed, will have a green checkmark or they are removed from the list, and the number of faulty faces decreases.
Tip: The accuracy of B-spline surfaces that can be replaced is defined by tolerance of less than 10^-8. Searching for less accurate faces (between 10^-5 and 10^-8) would be extremely slow, hence
prohibited. So if you think SolidWorks missed some B-spline surfaces, you can manually select those faces in the graphics area and click Repair Face to convert those faces to analytic surfaces if
Now let’s head over to the Gaps between faces table.
Right-click on any item to access additional options available.
• Heal Gap attempts to heal the gaps with various methods as discussed above.
• Remove Gap removes every face adjacent to the gap.
• Gap Closer is a tool used for manually repairing gaps.
Import Diagnostics heals gaps between adjacent faces by doing one or more of the following:
• It tries to extend the two adjacent faces into each other to eliminate the gap.
• If there are two close and non-intersecting edges, it tries to replace those two edges with a single tolerant edge.
• If all fails, then it creates a Filled Surface or Lofted Surface to fill the gap.
To use the Gap Closer, right-click on any gap and select Gap Closer. In the graphics area, drag the gap edge handle with the pointer to the side of another edge. When the original edge turns green,
right-click in the graphics area or the PropertyManager and select Finish Gap Closer.
Once all the faulty faces and gaps are removed, a green light will be shown under the Messages menu stating that No faulty faces or gaps remain in the geometry.
Don’t be surprised if Import Diagnostics was unable to fix all the errors. Import Diagnostics is a live tool i.e. any changes you made while you were working with the tool get directly applied to the
model. Even if you cancel the process by clicking on the Red Cancel button to stop the tool from making any changes you will find that these changes are already made regardless. So, heal whatever you
can with this tool and then use other methods to repair the geometry.
Tip: Sometimes exporting the file and then again importing it resolves the errors that were not resolved the first time. | {"url":"https://mechanitec.ca/how-to-use-import-diagnostics-solidworks/","timestamp":"2024-11-07T01:20:13Z","content_type":"text/html","content_length":"97422","record_id":"<urn:uuid:9633f43b-f9d5-4dc2-8a58-6a5c23dd1741>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00452.warc.gz"} |
An Efficient Algorithm for processing Top-k Spatial Preference Queries
Volume 01, Issue 04 (June 2012)
An Efficient Algorithm for processing Top-k Spatial Preference Queries
DOI : 10.17577/IJERTV1IS4188
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S Rao Chintalapudi, Katikireddy Srinivas, 2012, An Efficient Algorithm for processing Top-k Spatial Preference Queries, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 01,
Issue 04 (June 2012),
• Open Access
• Total Downloads : 594
• Authors : S Rao Chintalapudi, Katikireddy Srinivas
• Paper ID : IJERTV1IS4188
• Volume & Issue : Volume 01, Issue 04 (June 2012)
• Published (First Online): 01-07-2012
• ISSN (Online) : 2278-0181
• Publisher Name : IJERT
• License: This work is licensed under a Creative Commons Attribution 4.0 International License
Text Only Version
An Efficient Algorithm for processing Top-k Spatial Preference Queries
S Rao chintalapudi katikireddy srinivas
Asst.Professor Associate Professor CMR Technical Campus B.V.C Engineering College
A spatial preference query ranks objects based on the qualities of features in their spatial neighborhood. For example, using a real estate agency database of flats for sale, a customer may want to
rank the flats with respect to the appropriateness of their location, defined after aggregating the qualities of other features (e.g., restaurants, market, hospital, railway station, etc.) within
their spatial neighborhood. Such a neighborhood concept can be specified by the user via different functions. In this paper, we formally define spatial preference queries and propose appropriate
indexing techniques and search algorithms for them. Extensive evaluation of our methods on both real and synthetic data reveals that an optimized branch-and-bound solution is efficient and robust
with respect to different parameters.
Index TermsQuery processing, spatial preference query, spatial databases.
1. INTRODUCTION
Spatial database systems manage large collections of geographic entities, which apart from spatial attributes contain non-spatial information (e.g., name, size, type, price, etc.). In this paper,
we study an interesting type of preference queries, which select the best spatial location with respect to the quality of facilities in its spatial neighborhood. Given a set D of interesting
objects (e.g., candidate locations), a top-k spatial preference query retrieves the k objects in D with the highest scores. The score of an object is defined by the quality of features (e.g.,
facilities or services) in its spatial neighborhood. As a motivating example, consider a real estate agency office that holds a database with available flats for sale. Here feature refers to a
class of objects in a spatial map such as specific facilities or services. A customer may want to rank the contents of this database with respect to the quality of their locations, quantified by
aggregating non-spatial characteristics of other features (e.g., restaurants, super market, hospital, railway station, etc.) in the spatial neighborhood of the flat (defined by a spatial range
around it). Quality may be subjective and query-
parametric. For example, the user (e.g., a tourist) wishes to find a hotel p that is close to a railway station and a high-quality restaurant. Fig. 1a illustrates the locations of an object
dataset D (hotels) in white, and two feature data sets: the set F1 (restaurants) in gray, and the set F2 (railway stations) in black. For the ease of discussion, the qualities are normalized to
values in [0, 1].
Fig. 1.Example of top-k spatial preference query
1. Range Score b)Influence Score
The score T(p) of a hotel p is defined in terms of:
1. the maximum quality for each feature in the neighborhood region of p
2. the aggregation of those qualities.
The Range score, binds the neighborhood region to a circular region at p with radius (shown as a circle), and the aggregate function to SUM. For instance, the maximum quality of gray and black
points within the circle of p1 are 0.9 and 0.6 respectively, so the score of p1 is T(p1)= 0.9+0.6=1.5. Similarly, we obtain T(p2)=1.0+0.1=1.1 and T(p3)=0.7+0.7=1.4. Hence,the hotel p1 is returned
as the top result.
In fact, the semantics of the aggregate function is relevant to the users query. The SUM function attempts to balance the overall qualities of all features. The neighborhood region in the above
spatial preference query can also be defined by other score functions. A meaningful score function is the influence score (see Section 4).As opposed to the crisp radius constraint in the range
score, the influence score smoothens the effect of and assigns higher weights to railway stations that are closer to the hotel. Fig. 1b shows a hotel p5 and three railway stations s1, s2, s3
(with their quality values). The circles have their radii as multiples of .Now, the
score of a railway station si is computed by multiplying its quality with the weight 2-j, where j is the order of the smallest circle containing si. For example, the scores of s1, s2, and s3 are
0.3*2-1=0.15, 0.9*2-2=0:225, and 1.0*2-3=0.125,respectively. The influence score of p5 is taken as the highest value (0.225).
Traditionally, there are two basic ways for ranking objects:
1. Spatial ranking, which orders the objects according to their distance from a reference
2. Non-spatial ranking, which orders the objects by an aggregate function on their non-. spatial values
Our top-k spatial preference query integrates these two types of ranking in an intuitive way. As indicated by our examples, this new query has a wide range of applications in service
recommendation and decision support systems.
To our knowledge, there is no existing efficient solution for processing the top-k spatial reference query. A brute force approach (to be elaborated in Section 3.2) for evaluating it is to
compute the scores of all objects in D and select the top-k ones. This method, however, is expected to be very expensive for large input data sets. In this paper, we propose alternative
techniques that aim at minimizing the I/O accesses to the object and feature datasets, while being also computationally efficient. Specifically, we contribute the branch-and-bound (BB)algorithm
for efficiently processing the top-k spatial preference query.
Furthermore, this paper studies one relevant extension that have not been investigated in our preliminary work [1].The extension (Section 3.4) is an optimized version of BB that exploits a more
efficient technique for computing the scores of the objects. The second extension (Section 3.6) studies adaptations of the proposed algorithms for aggregate functions other than SUM, e.g., the
functions MIN and MAX. The third extension (Section 4) develops solutions for the top-k spatial preference query based on the influence score. The rest of this paper is structured as follows:
Section 2 provides background on basic and advanced queries on spatial databases, as well as top-k query evaluation in relational databases. Section 3 defines the top-k spatial preference query
and presents our solutions. Section 4 studies the query extension for the influence score. In Section 5, our query algorithms are experimentally evaluated with real and synthetic data. Finally,
Section 6 concludes the paper with future research directions.
2. BACKGROUND AND RELATED WORK
Object ranking is a popular retrieval task in various applications. In relational databases, we rank tuples using an aggregate score function on their attribute values [2]. For example, a real
estate agency maintains a database that contains information of flats available for sale. A potential customer wishes to view the top 10 flats with the largest sizes(area) and lowest prices. In
this case, the score of each flat is expressed by the sum of two qualities: size and price, after normalization to the domain [0, 1] (e.g., 1 means the largest size and the lowest price and 0
means the smallest size and the highest price). In spatial databases, ranking is often associated to nearest neighbor (NN) retrieval. Given a query location, we are interested in retrieving the
set of nearest objects to it that qualify a condition (e.g., restaurants).Assuming that the set of interesting objects is indexed by an R-tree [3], we can apply distance bounds and traverse the
inex in a branch- and-bound fashion to obtain the answer [4].
1. Spatial Query Evaluation on R-Trees
The most popular spatial access method is the R-tree [3], which indexes minimum bounding rectangles (MBRs) of objects. Fig. 2 shows a set D=(p1, . . . , p8) of spatial objects(e.g., points)
and an R-tree that indexes them. R-trees can efficiently process main spatial query types, including spatial range queries, nearest neighbor queries, and spatial joins.
Given a spatial region W, a spatial range query retrieves from D the objects that intersect W. For instance, consider a range query that asks for all objects within the shaded area in Fig. 2.
Starting from the root of the tree, the query is processed by recursively following entries, having MBRs that intersect the query region.
1. (b)
Fig. 2. Spatial queries on R-trees. a) MBRs b)R-tree representation
For instance, e1 does not intersect the query region, thus the subtree pointed by e1 cannot contain any query result. In contrast, e2 is followed by the algorithm and the points in the corresponding
are examined recursively to find the query result p7.A nearest neighbor query takes as input a query object q and returns the closest object in D to q. For instance, the nearest neighbor of q in Fig.
2 is p7. Its generalization is the k-NN query, which returns the k closest objects to q, given a positive integer k. NN (and k-NN) queries can be efficiently processed using the best-first (BF)
algorithm of [4], provided that D is indexed by an R-tree. A min-heap H, which organizes R-tree entries based on the (minimum) distance of their MBRs to q is initialized with the root entries. In
order to find the NN of q in Fig. 2, BF first inserts to H entries e1,e2, e3, and their distances to q. Then, the nearest entry e2 is retrieved from H and objects p1, p7, p8 are inserted to H. The
next nearest entry in H is p7, which is the nearest neighbor of q. In terms of I/O, the BF algorithm is shown to be no worse than any NN algorithm on the same R-tree [4].
The aggregate R-tree (aR-tree) [10] is a variant of the Rtree, where each nonleaf entry augments an aggregate measure for some attribute value (measure) of all points in its subtree. As an example,
the tree shown in Fig. 2 can be upgraded to a MAX aR-tree over the point set, if entries e1,e2,e3 contain the maximum measure values of sets (p2, p3), (p1,p8, p7), (p4, p5, p6), respectively. Assume
that the measure values of p4,p5, p6 are 0.2,0.1, 0.4, respectively. In this case, the aggregate measure augmented in e3 would be max(0.2, 0.1,0.4) = 0.4. In this paper, we employ MAX aR-trees for
indexing the feature data sets (e.g., restaurants),in order to accelerate the processing of top-k spatial preference queries.
Given a feature data set F and a multidimensional region R, the range top-k query selects the tuples (from F) within the region R and returns only those with the k highest qualities. Hong et al. [11]
indexed the data set by a MAX aR-tree and developed an efficient tree traversal algorithm to answer the query. Instead of finding the best k qualities from F in a specified region, our (range score)
query considers multiple spatial regions based on the points from the object data set D, and attempts to find out the best k regions (based on scores derived from multiple feature data sets Fc).
Section 3.1 formally defines the top-k spatial preference query problem and describes the index structures for the data sets. Section 3.2 studies two baseline algorithms for processing the query.
3.3 presents an efficient branch-and-bound algorithm for the query, and its further
optimization is proposed in Section 3.4. Section 3.5 develops a specialized spatial join algorithm for evaluating the query. Finally, Section 3.6 extends the above algorithms for answering top-k
spatial preference queries involving other aggregate functions.
1. Definitions and Index Structures
Given an object data set D and m feature data sets F1,F2 . . . Fm, the top-k spatial preference query retrieves the k points in D with the highest score. Here, the overall score of an object
point p D is defined as
T(p)=AGG{Tc(p)|c [1,m]} (1)
where AGG is an aggregate function (e.g: SUM,MIN,MAX etc)
Tc(p) is the cth component score of p with respect to the neighborhood condition and
m is the number of feature data sets.
The cth component score of p i.e Tc(p) can be computed as follows
Tc(p)=max({w(s)|s Fc ^ dist(p,s) }U
{0}). (2)
2. Algorithms
we develop various algorithms for processing top-k spatial preference queries. We first introduce a brute-force solution that computes the score of every point p D in order to obtain the
query results. Then, we propose a group evaluation technique that computes the scores of multiple points concurrently.
1. Simple Probing Algorithm
For a point p D, where not all its component scores are known, its upper bound score Tu(p) defined as
Tu(p)= Tc(p), if Tc(p) is known (3)
1, otherwise
It is guaranteed that the upper bound Tu(p) is greater than or equals to the actual score T(p).
Algorithm 1 is a pseudocode of the simple probing (SP)algorithm, which retrieves the query results by computing the score of every object point. The algorithm uses two global variables:
Wk is a min- heap for managing the top-k results and represents the top-k score so far (i.e., lowest score in Wk). Initially, the algorithm is invoked at the root node of
the object tree (i.e., N =D.root). The procedure is recursively applied (at Line 4) on tree nodes until a leaf node is accessed. When a leaf node is reached, the component score Tc(e) (at
Line 8) is computed by executing a range search on the feature tree Fc for range score queries. Lines 6-8 describe an incremental computation technique, for reducing unnecessary component
score computations. In particular, the point e is ignored as soon as its upper bound score Tu(e) (see (3)) cannot be greater than the best-k score . The variables Wk and are updated when
the actual score T(e) is greater than .
Algorithm 1. Simple Probing Algorithm algorithm SP(Node N)
1: for each entry e N do 2: If N is nonleaf then
3: read the child node N' pointed by e; 4: SP(N');
5: else
6: for c = 1 to m do 7: If Tu(e) > then
//if upper bound is greater than
8: compute Tc(p) using tree Fc; update Tu(e); 9: If T(e) > then
10: update Wk and by e;
1. it is very expensive because it comutes score for all objects.
2. No concurrency
3. it is not efficient method for larger input data sets.
2. Group Probing Algorithm
Due to separate score computations for different objects, SP is inefficient for large-object data sets. In view of this, we propose the group probing (GP) algorithm, a variant of SP,that
reduces I/O cost by computing scores of objects in the same leaf node of the R-tree concurrently. In GP, when a leaf node is visited, its points are first stored in a set V and then their
component scores are computed concurrently at a single traversal of the Fc tree.
We now introduce some distance notations for MBRs. Given a point p and an MBR e, the value mindist(p,e) [4] denotes the minimum possible distance between p and any point in e. Similarly,
given two MBRs ea and eb, the value mindist(ea, eb) denotes the minimum possible distance between any point in ea and any point in eb.
Algorithm 2 shows the procedure for computing the cth component score for a group of points. Consider a subset Vof D for which we want to compute their component score at feature tree Fc.
Initially, the procedure is called with N being the root node of Fc. If e is a nonleaf entry and its mindist
from some point p V is within the range , then the procedure is applied ecursively on the child node of e, since the subtree of Fc rooted at e may contribute to the component score of p.
In case e is a leaf entry (i.e., a feature point), the scores of points in V are updated if they are within distance from e.
Algorithm 2. Group Probing Algorithm algorithm GP(Node N, Set V , Value c,Value ) 1: for each entry e N do
2: If N is nonleaf then
3: If p V , mindist(p,e) then
4: read the child node N' pointed by e; 5: GP(N',V ,c, );
6: else
7: for each p V such that dist(p,e) do 8: Tc(p)=max{Tc(p),w(e)};
1.it is also expensive because it computes score for all objects but concurrently .
3. Branch-and-Bound Algorithm
GP is still expensive as it examines all objects in D and computes their component scores. We now propose an algorithm that can significantly reduce the number of objects to be examined. The key idea
is to compute, for nonleaf entries e in the object tree D, an upper bound Tu(p) of the score T(p) for any point p in the subtree of e. If Tu(e) then we need not access the subtree of e, thus we can
save numerous score computations.
Algorithm 3 is a pseudocode of our BB algorithm, based on this idea. BB is called with N being the root node of D. If N is a nonleaf node, Lines 3-5 compute the scores T(e) for nonleaf entries e
concurrently. Recall that Tu(e) is an upperbound score for any point in the subtree of e.. If Tu(e)
,then the subtree of e cannot contain better results
than those in Wk and it is removed from V . In order to obtain points with high scores early, we sort the entries in descending order of T(e) before invoking the above procedure recursively on the
child nodes pointed by the entries in V .If N is a leaf node, we compute the scores for all points of N concurrently and then update the set Wk of the top-k results. Since both Wk and are global
variables, their values are updated during recursive call of BB.
Algorithm 3. Branch-and-Bound Algorithm Wk= new min-heap of size k (initially empty); =0;
algorithm BB(Node N) 1: V ={e| e N};
2: If N is nonleaf then
3: for c= 1 to m do
4: compute Tc(e) for all e V concurrently; 5: remove entries e in V such that Tu(e) ;
6: sort entries e V in descending order of T(e); 7: for each entry e V such that Tu(e) > do
8: read the child node N' pointed by e; 9: BB(N');
10: else
11: for c = 1 to m do
12: compute Tc(e) for all e V concurrently; 13: remove entries e in V such that Tu(e) ; 14: update Wk and by entries in V ;
1.it reduces number of objects to be examined. 2.it is efficient than SP and GP algorithms.
3.3.1 Upper Bound Score Computation
It remains to clarify how the (upper bound) scores Tc(p) of nonleaf entries (within the same node
1. can be computed concurrently (at Line 4). Our goal is to compute these upperbound scores such that, 1).the bounds are computed with low I/O
cost, and.
2).the bounds are reasonably tight, in order to facilitate effective pruning.
To achieve this, we utilize only level-1 entries (i.e., lowest level nonleaf entries) in Fc for deriving upper bound scores because:
1. there are much fewer level-1 entries than leaf entries (i.e., points)
2. high-level entries in Fc cannot provide tight bounds.
In our experimental study, we will also verify the effectiveness and the cost of using level- 1entries for upper bound score computation. Algorithm 2 can be modified for the above upper bound
computation task (where input V corresponds to a set of nonleaf entries), after changing Line 2 to check whether child nodes of N are above the leaf- level. The following example illustrates how
upper bound range scores are derived. In Fig. 4a, v1 and v2 are nonleaf
entries in the object tree D and the others are level-1 entries in the feature tree Fc. For the entry v1, we first define its Minkowski region [21] (i.e., gray region around v1), the area whose
mindist from v1 is within . Observe that only entries ei intersecting the Minkowski region of v1 can contribute to the score of some point in v1. Thus, the upper bound score Tc(p) is simply the
maximum quality of entries e1,e5, e6, e7, i.e., 0.9. Similarly, Tc(p) is computed as the maximum quality of entries e2, e3, e4, e8, i.e., 0.7. Assuming that v1 and v2 are entries in the same tree
node of D, their upper bounds are computed concurrently to reduce I/O cost.
Fig. 4. Examples of deriving scores. (a) Upper bound scores. (b) Optimized computation.
1. Optimized Branch-and-Bound Algorithm
This section develops a more efficient score computation technique to reduce the cost of the BB algorithm.
1. Problem with BB Algorithm
Recall that Lines 11-13 of the BB algorithm are used to compute the scores of object points (i.e., leaf entries of the R-tree on D). A leaf entry e is pruned if its upper bound score Tu
(e) is not greater than the best score found so far . However, the upper bound score Tu(e) (see (3)) is not tight because any unknown component score is replaced by 1.
Let us examine the computation of Tu(p1) for the point p1 in Fig. 4b. The entry e F1 is a nonleaf entry from the feature tree F1. Its augmented quality value is w(e F1)=0.8. The entry
points to a leaf node containing two feature points, whose qualities values are 0.6 and 0.8, respectively. Similarly, e F2 is a nonleaf entry from the tree F2 and it points to a leaf node
of feature points.
Suppose that the best score found so far in BB is =1.4(not shown in the figure). We need to check whether the score of p1 can be higher than . For this, we compute the first component
score T1(p1)= 0.6 by accessing the child node of e F1 . Now, we have the upper bound score of p1 as Tu(p1)
= 0.6 + 1.0= 1.6. Such a bound is above =1.4 so we need to compute the second component score T2(p1)= 0.5 by accessing the child node of e F2 . The exact score of p1 is T(p1)=0.6+0.5=1.1
the point p1 is then pruned because T(p1) . In summary, two leaf nodes are accessed during the computation of T(p1) .
Our observation here is that the point p1 can be pruned earlier, without accessing the child node of e F2 . By taking the maximum quality of level-1 entries (from F2) that intersect the
-range of p1, we derive: T2(p1) w(e F2 )=0.7.With the first component score T1(p1)=0.6, we infer that:T(p1)=
0.6 + 0.7= 1.3. Such a value is below so p1 can be pruned.
2. Optimized Computation of Scores
Based on our observation, we propose a tighter derivation for the upper bound score of p than the one shown in (3).Let p be an object point in D. Suppose that we have traversed some paths
of the feature trees on F1,F2, . . . Fm.Let µ be an upper bound of the quality value for any unvisited entry (leaf or nonleaf) of the feature tree Fc. We then define the function T*(p)
In the max function, the first set denotes the upper bound quality of any visited feature point within distance from p.According to (4), the value T*(p) is tight only when every c value
is low. In order to achieve this, we access the feature trees in a round-robin fashion, and traverse the entries in each feature tree in descending order of quality values.
Round-robin is a popular and effective strategy used for efficient merging of rankings [7], [9].
Algorithm 4 is the pseudocode for computing the scores of objects efficiently from the feature trees F1,F2,. . . ,Fm. The set V contains objects whose scores need to be computed. Here,
refers to the distance threshold of the range score, and represents the best score found so far. Foreach feature tree Fc, we employ a max-heap Hc to traverse the entries of Fc in
descending order of their qulity values. The root of Fc is first inserted ino Hc. The variable µ maintains the upper bound quality of entries in the tree that will be visited. We then
initialize each component score Tc(p) of every object p V to 0.
Algorithm 4. Optimized Group Range Score Algorithm
algorithm Optimized_Group_Range(Trees F1;F2;
. . . ;Fm, Set V , Value _, Value _) 1: for c := 1 to m do
2: Hc := new max-heap (with quality score as key);
3: insert Fc.root into Hc; 4: µ := 1;
5: for each entry p V do 6: Tc(p) := 0;
7: := 1;
//ID of the current feature tree
8: while |V |> 0 and there exists a nonempty heap Hc do
9: deheap an entry e from H; 10: µ =w(e);
//update threshold
11: if p V , mindist(p,e) > then 12: continue at Line 8;
13: for each p V do
// prune unqualified points
14: if ) then
15: remove p from V ;
16: read the child node CN pointed to by e; 17: for each entry e' of CN do
18: if CN is a nonleaf node then
19: if p V , mindist(p,e') then 20: insert e' into H;
21: else
// update component scores
22: for each p V such that dist(p,e') do 23: T (p)=max{ T (p),w(e')};
24: = next (round-robin) value where H is not empty;
25: for each entry p V do 26: ;
At Line 7, the variable keeps track of the ID of the current feature tree being processed. The loop at Line 8 is used to compute the scores for the points in the set V. We then deheap an
entry e from the current heap H. The property of the max-heap guarantees that the quality value of any future entry deheaped from H is at most w(e). Thus, the bound µ is updated to w(e).
At Lines 11-12, we prune the entry e if its distance from each object point p V is larger than . In case e is not pruned, we compute the tight upper bound score T(p) for each p V (by
(4)); the object p is removed from V if T(p) (Lines 13-15).
Next, we access the child node pointed to by e, and examine each entry e' in the node (Lines 16- 17). A nonleaf entry e' is inserted into the heap H if its minimum distance from some p V
is within (Lines 18-20); whereas a leaf entry e' is used to update the component score T(p) for any p V within distance from e' (Lines 22-23). At Line 24, we apply the round-robin
strategy to find the next value such that the heap H is not empty. The loop at Line 8 repeats while V is not empty and there exists a nonempty heap Hc. At the end, the algorithm derives
the exact scores for the remaining points of V .
3. The BB* Algorithm
Based on the above, we extend BB (Algorithm 3) to an optimized BB* algorithm as follows: First, Lines 11-13 of BB are replaced by a call to Algorithm 4, for computing the exact scores for object
points in the set V . Second, Lines 3-5of BB are replaced by a call to a modified algorithm 4, for deriving the upper bound scores for nonleaf
entries (in V ).Such a modified Algorithm 4 is obtained after replacing Line 18 by checking whether the node CN is a nonleaf node above the level-1.
In this section, we compare the efficiency of the proposed algorithms using real and synthetic data sets. Each data set is indexed by an aR-tree with 4 K bytes page size. We used an LRU memory buffer
whose default size is set to 0.5 percent of the sum of tree sizes (for the object and feature trees used).Our algorithms were implemented in C++ and experiments were run on a Pentium D 2.8 GHz PC
with 1 GB of RAM. In all experiments, we measure both the I/O cost (in number of page faults) and the total execution time (in seconds) of our algorithms.
1. RESULTS
In this section, we conduct experiments on real object and feature data sets in order to demonstrate the application of top-k spatial preference queries. We obtained three real spatial data sets
from a travel portal,http://www.allstays.com/. Locations in these data sets correspond to (longitude and latitude)coordinates in US. We cleaned the data sets by discarding records without
longitude and latitude. In summary, the relative performance between the algorithms in all experiments is consistent to the results on synthetic data.
Fig. Comparision of I/O cost for SP,GP,BB,BB*.
Fig.-Comparision of Execution Times for SP,GP,BB,BB*
2. CONCLUSION
In this paper, we studied top-k spatial preference queries, which provide a novel type of ranking for spatial objects based on qualities of features in their neighborhood. The neighborhood of an
object p is captured by the scoring function: 1) the range score restricts the neighborhood to a crisp region centered at p, whereas 2) the influence score relaxes the neighborhood to the whole space
and assigns higher weights to locations closer to p.We presented four algorithms for processing top-k spatial preference queries. The baseline algorithm SP computes the scores of every object by
querying on feature data sets. The algorithm GP is a variant of SP that reduces I/O cost by computing scores of objects in the same leaf node concurrently. The algorithm BB derives upper bound scores
for non leaf entries in the object tree, and prunes those that cannot lead to better results. The algorithm BB*is a variant of BB that utilizes an optimized method for computing the scores of objects
(and upper bound scores of non leaf entries). Based on our experimental findings, BB* is scalable to large data sets and it is the most robust algorithm with respect to various parameters. In the
future, we will study the top-k spatial preference query on a road network, in which the distance between two points is defined by their shortest path distance rather than their euclidean distance.
The challenge is to develop alternative methods for computing the upper bound scores for a group of points on a road network.
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Lies, damned lies and Bitcoin difficulties
Bitcoin difficulty and hash rate statistics should be considered an illness. The symptoms include anxiety, depression, sleeplessness and paranoia. Bitcoin miners follow their every movement,
rejoicing at smaller-than-expected difficulty changes and collectively dismaying when things go the other way. Authoritative-looking charts have people puzzling about why things are so erratic and
chasing non-existent mining conspiracies. The truth is out there...
Difficulty charts
When we start to think about mining, difficulty charts are not far away. Most of them are presented something like this:
Bitcoin hash rate for the last 6 months (June 2014) on a linear scale
This chart shows the last 6 months of daily hash rates and the rising difficulty. It also shows a baseline trend line but we'll look at that a little later.
Aside from that inexorable increase in difficulty and the cries of woe from miners watching it, the most striking characteristic is that way it's getting progressively much more "spiky"! Look at how
smooth it used to look? In fact this assessment is actually just plain wrong; if you were to look at 6 months of data starting 3 months earlier then that nice "smooth" part would end up looking just
as bad as the most recent data. The problem is a question of scale; the variations in the hash rate become numerically larger as the overall hash rate increases.
When confronted with this sort of data, many statisticians switch to logarithmic graphs instead of linear ones because log charts show the magnitude differences rather than absolute differences.
Here's the same data on a log chart:
Bitcoin hash rate for the last 6 months (June 2014) on a logarithmic scale
Notice how the spikes in the blue hash rate look pretty much the same all the way across now? If you're observant you might argue that the ones on the left are slightly less spiky, but that's because
the slope of the graph is steeper there. Even there though it's clear that the statistical noise on a day-to-day basis is actually much larger than the overall trend. That overall trend shows that
hashing rate, and thus the difficulty, increases are slowing down for now (and probably for the foreseeable future). The slowdown wasn't evident on the linear scale graph and so we can see another
advantage of logarithmic scale graphs.
Calculating a baseline
Most Bitcoin miners tend to think in terms of "difficulty" because it's what determines the complexity of any mining. On both of the graphs we've just looked at, though, the difficulty is clearly
lagging behind the hash rate. The problem is that it's set retrospectively, and set at a level that would make the preceding 2016 blocks take exactly 14 days to find. This means that the difficulty
lags around 5.5 to 7 days behind the actual hash rate even when it's changed. If we want a real baseline to think about hash rates we need something more up-to-date.
In both of the charts we've just seen there is a baseline trace, and that trace represents "something more up-to-date". The baseline is calculated by looking at the days where the difficulty changes
and taking the square root of the ratio of a new to previous difficulty level and then multiplying it by the new difficulty. In-between these fixed points is an interpolation that assumes a steady
percentage growth rate between them.
This particular baseline isn't perfect because it has no way to account for statistical noise in the hashing rate (see "Hash rate headaches") but it turns out to be a surprisingly effective estimate
Checking the baseline
Visually our baseline looks pretty reasonable. We know that even if the hash rate was constant the difficulty would change as a result of random noise (see "Reach for the ear defenders"). The
question is what does our noise profile look like if we subtract out the baseline hash rate estimate? This should approximately follow Bitcoin's Poisson process noise profile and should oscillate
about zero. Here's what it actually looks like for the last 12 months:
12 month Bitcoin hash rate variations (June 2014)
Comparing this with what simulations suggest for 24 hour variations this looks remarkably consistent. This pretty-much suggests that there has been very little if anything unusual happening over the
last 12 months and that hashing capacity has been reasonably steadily added throughout. One final check though is to look at the probability histogram for the variations about our baseline:
12 month Bitcoin mining hash rate variation probability histogram (June 2014)
While it's not perfect, it has just the sort of probability distribution we would expect to see.
What have we gained?
We started out trying to understand how key statistics were presented. We've seen how linear charts can be highly misleading. By devising a way to estimate the hash rate baseline, we've been able to
go one step further and see just how much the day to day hash rate estimates will oscillate quite wildly. We can now be confident that even 20% swings from the estimate are surprisingly likely, and
that day-to-day swings can be even larger!
The gods of statistics didn't want us to worry about what happens in the course of hour or even a few days; those numbers, tantalizing as they may seem, are largely meaningless. They are the lies
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Barend Gehrels wrote:
> Hi Adam,
> On 9-10-2011 22:30, Adam Wulkiewicz wrote:
>> Barend Gehrels wrote:
>>> Hi Aleksandar,
>>> The touches algorithm is planned but not included, also not in the
>>> extensions. I once started it and today looked at it, but it is not
>>> suitable for inclusion yet. However it is on the list.
>>> As far as I know it cannot be done by using existing algorithms.
>> Hi Barand,
>> Please add to the todo list an algorithm checking if geometries
>> intersects but not taking boundries into account (intersects() &&
>> !touches()). It would be used e.g. in rtree traversing in tests for
>> nodes'/boxes containing geometries which may be within some area.
>> Still there is no need to hurry since there is no big difference
>> between this and plain intersects() which is used now.
> Is that not the "within" algorithm? Because "within" means is completely
> within, so should not touch any boundary... But it should then probably
> be implemented for e.g. polygon/box polygon/polygon, for your purposes.
> By the way, touches() in OGC perspective means never an intersection, so
> only two borders touching, or a point lying on a border.
I have in mind something like intersects() except it would give false
for touching geometries.
intersects(B(P(0,0),P(1,1)), B(P(0.5,0.5),P(2,2))) -> true
intersects(B(P(0,0),P(1,1)), B(P(1,1),P(2,2))) -> TRUE (boundry)
intersects(B(P(0,0),P(1,1)), B(P(1.5,1.5),P(2,2))) -> false
Lets say it's weak_intersects (i don't know how it should be called). It
would give these results:
weak_intersects(B(P(0,0),P(1,1)), B(P(0.5,0.5),P(2,2))) -> true
weak_intersects(B(P(0,0),P(1,1)), B(P(1,1),P(2,2))) -> FALSE (boundry)
weak_intersects(B(P(0,0),P(1,1)), B(P(1.5,1.5),P(2,2))) -> false
I think it corresponds to intersects() && !touches().
If rtree's nodes structure is traversed, at each step there must be
choosen nodes which will be traversed in the next step. Decision depends
on passed predicates. If we want to find points within some area, nodes
intersecting this area must be traversed. But there is no need to
traverse nodes touching it because there will be no objects in the node
which are within this area. On boundry yes, but not within.
It's just a little optimization.
Geometry list run by mateusz at loskot.net | {"url":"https://lists.boost.org/geometry/2011/10/1567.php","timestamp":"2024-11-12T10:37:09Z","content_type":"text/html","content_length":"13489","record_id":"<urn:uuid:10018082-07eb-439e-be21-86be01a42043>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00052.warc.gz"} |
I am sorry but I am not good at math, this may seem like a dumb
I am sorry but I am not good at math, this may seem like a dumb question but I don't do well with word problems? I can't figure it out, maybe I am making it too hard on myself, can someone point me
to the right direction?
. The distance from Philadelphia to Sea Isle City is 100 mi. A car was driven this distance using tires with a radius of 14 in. How many revolutions of each tire occurred on the trip?
see above. Please don't post duplicate problems.
First find the circumference of the tire.
C = pi * diameter Your answer will be in inches.
Then convert 5,280 feet (1 mile) to inches.
Divide the inches in a mile by the circumference of the tire.
Multiply by 100.
No worries! Understanding word problems can be challenging sometimes. Let's break it down step by step.
First, we need to find the circumference of each tire. The formula to find the circumference of a circle is C = 2Ï€r, where C represents the circumference and r represents the radius.
In this case, the radius of each tire is given as 14 inches. So, the circumference of each tire would be:
C = 2Ï€(14 inches)
To make calculations easier, let's simplify this to:
C ≈ 88 inches (rounded to the nearest whole number)
Now that we know the circumference of each tire, we can determine the number of revolutions made by the car. The distance the car traveled is given as 100 miles.
To find the number of revolutions, we need to convert the given distance from miles to inches and then divide it by the circumference of each tire:
Number of revolutions = (Distance traveled in inches) / (Circumference of each tire)
To convert miles to inches, we know that 1 mile is equal to 63,360 inches (1 mile = 5280 feet, and 1 foot = 12 inches).
So, the distance traveled in inches would be:
Distance traveled in inches = 100 miles * 63,360 inches/mile
Now, we can calculate the number of revolutions using the formula:
Number of revolutions = (100 miles * 63,360 inches/mile) / (88 inches)
Calculating this expression will give us the final answer, which represents the number of revolutions made by each tire during the trip.
Please note that the final answer will be in decimal form since dividing by the circumference might not result in a whole number.
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The Summing Amplifier - Easy Electronics
The Summing Amplifier
In this article, we are going to learn the most basic use case of the operational amplifier, i.e. Summing amplifier. We will discuss each and every point about summing amplifiers in detail so let’s
get ready to learn.
What is Summing Amplifier
In our earlier discussion about the inverting operational amplifier, we explored how it operates with a sole input voltage (V[in]) directed to its inverting input terminal. However, by incorporating
additional input resistors, each mirroring the original input resistor (R[in]), we unveil a new configuration known as a Summing Amplifier, sometimes referred to as a “summing inverter” or even a
“voltage adder” circuit.
The Summing Amplifier is a special setup of an operational amplifier. It's used to mix the voltages from different inputs into just one output voltage.
Inverting Summing Amplifier
In this summing amplifier setup, the output voltage (V[out]) is directly linked to the total of all input voltages, such as V[1], V[2], V[3], and so on. Consequently, we can adjust the initial
equation used for the inverting amplifier to accommodate these additional inputs.
\mathbf{I_F = I_1 + I_2 + I_3 = -\left[ \frac{V_1}{R_{in}} + \frac{V_2}{R_{in}} + \frac{V_3}{R_{in}}\right ]}
Now as we know in the Inverting amplifier, the output voltage can be defined as,
\mathbf{V_{out} = -\frac{R_F}{R_{in}} \times V_{in}}
Then we can write,
\mathbf{-V_{out} =\left[ \frac{R_F}{R_{in}}V_1 + \frac{R_F}{R_{in}}V_2 + \frac{R_F}{R_{in}}V_3\right ]}
If all the input impedances (R[IN]) have the same value, we can simplify the equation mentioned above to yield an output voltage of:
\boxed{\mathbf{V_{out} =-\frac{R_F}{R_{in}}\left[ V_1 + V_2 +V_3 + …. + V_n\right ]}}
Now, we’ve created an operational amplifier circuit capable of amplifying each individual input voltage, generating an output voltage signal proportional to the combined “SUM” of the three input
voltages: V[1], V[2], and V[3]. If necessary, we can expand this setup by adding more inputs, with each input being isolated by its respective resistance, R[in].
This isolation is facilitated by the “virtual short” node at the op-amp’s inverting input. Moreover, a direct voltage addition is achievable when all resistances are of equal value, and R[ƒ] equals R
It’s important to note that connecting the summing point to the inverting input of the op-amp results in the circuit producing the negative sum of input voltages. Conversely, connecting it to the
non-inverting input yields the positive sum of input voltages.
If the individual input resistors are not equal, a Scaling Summing Amplifier can be constructed. In such a case, the equation needs to be adjusted to:
\mathbf{-V_{out} = V_1 \left (\frac{R_f}{R_1} \right) +V_2 \left (\frac{R_f}{R_2} \right) +V_3 \left (\frac{R_f}{R_3} \right) … etc}
To simplify the mathematics, we can rearrange the formula above to isolate the feedback resistor R[ƒ] as the subject of the equation, resulting in the output voltage being expressed as:
\mathbf{-V_{out} =R_f \left[ \frac{V_1}{R_1} + \frac{V_2}{R_2} +\frac{V_3}{R_3} + … etc\right ]}
This setup simplifies the calculation of the output voltage when additional input resistors are connected to the amplifier’s inverting input terminal. The input impedance of each individual channel
equals the value of its respective input resistor, such as R[1], R[2], R[3], and so forth.
Occasionally, we require a summing circuit solely for combining two or more voltage signals without any amplification. By setting all resistances in the circuit to the same value, R, the op-amp will
exhibit a voltage gain of unity, resulting in an output voltage equal to the direct sum of all input voltages, as illustrated below:
The Summing Amplifier proves to be a highly versatile circuit, allowing us to efficiently combine multiple individual input signals through addition or summation, hence its name. When the input
resistors, labeled as R[1], R[2], R[3], and so on, are all equal, it results in a “unity gain inverting adder.” However, if the input resistors have different values, it yields a “scaling summing
amplifier,” producing an output that represents a weighted sum of the input signals.
Non-inverting Summing Amplifier
Besides constructing inverting summing amplifiers, we can utilize the non-inverting input of the operational amplifier to create a non-inverting summing amplifier. While an inverting summing
amplifier produces the negative sum of its input voltages, the non-inverting summing amplifier configuration generates the positive sum of its input voltages.
True to its name, the non-inverting summing amplifier is structured around the setup of a non-inverting operational amplifier circuit. Here, the input (either AC or DC) is directed to the
non-inverting (+) terminal, while the desired negative feedback and gain are attained by feeding back a portion of the output signal (V[OUT]) to the inverting (-) terminal, as depicted.
Non-inverting Summing Amplifier Circuit
What advantages does the non-inverting configuration offer over the inverting summing amplifier configuration? Besides the straightforward fact that the output voltage V[OUT] of the operational
amplifier (op-amp) aligns with its input and the output voltage represents the weighted sum of all inputs, determined by their resistance ratios, the main advantage of the non-inverting summing
amplifier lies in its significantly higher input impedance compared to the standard inverting amplifier configuration.
Moreover, the input summing section of the circuit remains unaffected even if the op-amp’s closed-loop voltage gain is altered. However, selecting the weighted gains for each individual input at the
summing junction involves more mathematical consideration, especially with more than two inputs, each with a distinct weighting factor. Nonetheless, if all inputs share the same resistive values, the
mathematical complexity decreases significantly.
If the closed-loop gain of the non-inverting operational amplifier matches the number of summing inputs, the op-amp’s output voltage will precisely mirror the sum of all input voltages. For instance,
in a two-input non-inverting summing amplifier, the op-amp’s gain equals 2; in a three-input summing amplifier, the gain equals 3, and so forth. This occurs because the currents flowing in each input
resistor are influenced by the voltage across all inputs. When the input resistances are equal (R[1] = R[2]), the circulating currents cancel out since they cannot flow into the high impedance
non-inverting input of the op-amp, resulting in the output voltage becoming the sum of its inputs.
Thus, for a 2-input non-inverting summing amplifier, the currents flowing into the input terminals can be defined as:
If we ensure that the two input resistances have the same value, then we have R[1] = R[2] = R.
The typical equation for the voltage gain of a non-inverting summing amplifier circuit is expressed as:
The closed-loop voltage gain (A[V]) of the non-inverting amplifier is determined by the formula 1 + R[A]/R[B]. If we set this gain to 2 by making R[A] equal to R[B], then the output voltage (V[O])
becomes equal to the sum of all the input voltages, as illustrated.
Non-inverting Output Voltage
Therefore, for a 3-input non-inverting summing amplifier setup, adjusting the closed-loop voltage gain to 3 will result in V[OUT] being equivalent to the sum of the three input voltages, V[1], V[2],
and V[3]. Similarly, in a four-input configuration, the closed-loop voltage gain would be set to 4, and for a five-input setup, it would be 5, and so forth. It’s worth noting that if the amplifier of
the summing circuit is configured as a unity follower with R[A] set to zero and R[B] set to infinity, the output voltage V[OUT] will precisely match the average value of all the input voltages,
represented as
V[OUT] = (V[1] + V[2])/2, since there is no voltage gain.
Summing Amplifier Applications
Summing amplifiers, whether inverting or non-inverting, offer versatile applications. By connecting the input resistances of a summing amplifier to potentiometers, it becomes possible to mix
individual input signals in varying proportions.
Audio Mixer Circuit
For instance, in temperature measurement, you could introduce a negative offset voltage to ensure that the output voltage or display reads “0” at the freezing point. Similarly, in audio mixing, a
summing amplifier can serve as an audio mixer, blending individual waveforms (sounds) from different source channels such as vocals, instruments, etc., before routing them collectively to an audio
Digital to Analog Converter
In this DAC summing amplifier circuit, the number of individual bits comprising the input data word—4 bits in this example—ultimately dictates the output step voltage as a percentage of the
full-scale analog output voltage.
The accuracy of the full-scale analog output hinges on the voltage levels of the input bits consistently being 0V for “0” and 5V for “1”, alongside the precision of the resistance values employed for
the input resistors, R[IN].
Fortunately, to address these potential errors, commercially available Digital-to-Analogue and Analogue-to-Digital devices come equipped with highly accurate resistor ladder networks pre-installed,
alleviating concerns on our part.
Level Shifter
Another significant application of a Summing Amplifier is as a Level Shifter. A 2-input Summing Amplifier can function as a level shifter by utilizing one input for an AC Signal and the other input
for a DC Signal.
The AC Signal undergoes an offset by the input DC Signal. This type of level shifter finds prominent use in Signal Generators for DC Offset Control.
What is a summing amplifier?
The Summing Amplifier is another type of operational amplifier circuit configuration that is used to combine the voltages present on two or more inputs into a single output voltage.
What is the summing point of an op-amp?
The “summing point” (the negative input) of an OpAmp summer is kept at the same voltage as the positive input by the action of the negative feedback.
What are the disadvantages of the summing amplifier?
it requires a dual-polarity power supply, which can add complexity and cost to the circuit design.
What are the different types of summing amplifiers?
There are two types of summing amplifiers: inverting and non-inverting summing amplifiers.
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Sharp density bounds on the finite field Kakeya problem | Published in Discrete Analysis
Combinatorial Geometry
December 14, 2021 BST
Sharp density bounds on the finite field Kakeya problem
Sharp density bounds on the finite field Kakeya problem, Discrete Analysis 2021:26, 9 pp.
A subset $A$ of the vector space $\mathbb F_p^n$ is called a finite-field Kakeya set if it contains a line in every direction. That is, for every $x\in\mathbb F_p^n\setminus\{0\}$ there exists $a\in\
mathbb F_p^n$ such that the line $\{a+tx:t\in\mathbb F_p\}$ is contained in $A$.
Such sets are a natural analogue of Kakeya sets in $\mathbb R^n$, which are, again, sets that contain a line in every direction. Besicovitch showed that Kakeya sets in $\mathbb R^2$, and hence in $\
mathbb R^n$ for any $n\geq 2$, can have measure zero. However, it is a major unsolved problem to determine the smallest possible Hausdorff dimension of such a set (or indeed other commonly used
dimensions such as the box-counting dimension). When $n=2$ it is known that the dimension must be 2, so in a certain sense Kakeya sets cannot be too small, and it is conjectured that this is true for
all $n$, but that is not known for any $n$ greater than 2.
In 1999 Thomas Wolff suggested that a useful discrete analogue of the Kakeya problem would be to look at the minimum density of finite-field Kakeya sets. If one pursues the analogy, one finds that a
density of $p^{-\alpha}$ (and thus a cardinality of $p^{n-\alpha}$) in the finite-field case should roughly correspond to a Hausdorff dimension of $n-\alpha$ in the Euclidean case. This problem is
cleaner than the Euclidean problem, because it avoids issues that arise with pairs of lines that are almost parallel. Nevertheless, it too appeared to be hard, so the hope was that it would be a good
intermediate question.
Wolff’s problem was solved in 2008 by Zeev Dvir, who came up with a remarkably short argument based on polynomials. (Very roughly, if a set $A$ is small, then one can find a non-trivial polynomial in
$n$ variables of degree $d<p$ that vanishes on $A$. However, if $A$ is a Kakeya set, then it can be shown quite easily that the degree-$d$ part of the polynomial does vanish, which is a
contradiction.) Unfortunately, this argument appears not to shed much light on the Euclidean case, but it has nevertheless been extremely influential, and has inspired many further results.
The lower bound obtained by Dvir was roughly $p^n/n!$ – here one thinks of $n$ as constant and $p$ as tending to infinity, so this is within a constant of the size of $\mathbb F_p^n$. Subsequent work
has improved this to about $(p/2)^n$, while in the other direction there is a construction that gives an upper bound of about $p^n/2^{n-1}$. These results were obtained in 2008-9, so the factor 2
between the two bounds has been present for over a decade.
This paper finally closes the gap to $1+o(1)$ by improving the lower bound by a factor of $2+o(1)$. The main ideas that make this improvement possible are mostly present in the earlier proofs, but
their implementations here are quite different. One of these ideas is to consider not just the vanishing of polynomials but the multiplicity of the vanishing, and another is to consider polynomials
that belong to a less obvious space of polynomials than simply the space of all polynomials of degree less than $p$. Provided the dimension of the space of polynomials exceeds the number of
independent linear constraints imposed by the requirement that a polynomial should vanish with certain multiplicities at the points of the Kakeya set, there will be a polynomial in the space that
vanishes in the required way, and if the multiplicities are chosen appropriately, one can arrive at a suitable lower bound for the size of the Kakeya set. Another essential idea in this paper, which
was introduced in a different context by Ruixiang Zhang, is to require the vanishing conditions (defined in terms of Hasse derivatives, which are discussed in the paper) to depend on the lines that
go through the points in the Kakeya set.
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Mega Tree Calculator Online
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Mega Tree Calculator Online
The festive season often brings to mind dazzling lights displays. To create such a spectacle, a tool of paramount importance is the Mega Tree Calculator.
In the realm of festive decorations, a 'Mega Tree' is a large artificial tree adorned with strings of lights. The Mega Tree Calculator, therefore, helps calculate the optimal number of lights needed.
How Does the Calculator Work?
The Mega Tree Calculator utilizes a simple formula to estimate the total number of lights required. By entering the height of the tree and the desired spacing between the lights, the calculator
provides a number that ensures a balanced and beautiful display.
Calculator Formula and Variable Descriptions
The formula the calculator uses is Total Number of Lights = (2 * π * Tree Height) / (Spacing between Lights). In this equation, "Total Number of Lights" represents the needed lights, "Tree Height"
refers to the mega tree's height, and "Spacing between Lights" is the distance between adjacent lights.
For instance, if your tree's height is 5 meters and you want a spacing of 0.05 meters between the lights, the calculator will output the total number of lights as 628. This ensures your Mega Tree is
perfectly lit!
Planning Festive Decorations
The Mega Tree Calculator is pivotal when planning a holiday display, ensuring you purchase the right number of lights.
Professional Light Displays
Event organizers can also use it to accurately plan professional light displays for concerts or other public events.
Frequently Asked Questions
What is the Mega Tree Calculator?
The Mega Tree Calculator is a tool designed to calculate the number of lights required to decorate a mega tree, given the tree's height and the desired light spacing.
How accurate is the Mega Tree Calculator?
While the calculator provides a good estimate, other factors such as tree shape and personal preference may affect the exact number of lights needed.
Can the Mega Tree Calculator be used for other types of trees?
Yes, the calculator can be used for any tree, provided you have the measurements for height and desired light spacing.
Whether you are planning a small festive display or a grand spectacle, the Mega Tree Calculator is an invaluable tool. It takes the guesswork out of your decorations, ensuring a perfectly lit display
every time.
Remember, while the calculator gives a good estimate, you can always add more lights to cater to your personal aesthetic. Happy decorating!
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An Etymological Dictionary of Astronomy and Astrophysics
amplification factor
کروند ِدامنهدهی
karvand-e dâmane-dahi
Fr.: facteur d'amplification
1) Electronics: The extent to which an → analogue → amplifier boosts the strength of a → signal. Also called → gain.
2) In → gravitational lensing, the ratio of the lensed brightness to unlensed brightness. This factor depends on the mass of the → lensing object and the closeness of the alignment between observer,
lens, and source (→ impact parameter).
→ amplification; → factor.
attenuation factor
کروند ِتنکش
karvand-e tonokeš
Fr.: facteur d'atténuation
The ratio of the radiation intensity after traversing a layer of matter to its intensity before.
→ attenuation; → factor.
Boltzmann factor
کروند ِبولتسمن
karvand-e Boltzmannn
Fr.: facteur de Boltzmann
The factor e^-E/kT involved in the probability for atoms having an excitation energy E and temperature T, where k is Boltzmann's constant.
→ Boltzmann's constant; → factor.
clumping factor
کروند ِگودهداری
karvand-e gudedâri
Fr.: facteur de grumelage
The ratio f[cl] = <ρ^2> / <ρ >^2, where ρ represents the → stellar wind density and the brackets mean values. Unclumped wind has f[cl] = 1 and → clumping becomes significant for f[cl]≅ 4.
→ clumping; → factor.
Fr.: cofacteur
A number associated with an → element of a → determinant. If A is a square matrix [a[ij]], the cofactor of the element a[ij] is equal to (-1)^i+j times the determinant of the matrix obtained by
deleting the i-th row and j-th column of A.
→ co-; → factor.
compression factor
کروند ِتنجش
karvand-e tanješ
Fr.: facteur de compression
In thermodynamics, the quantity Z = pV[m]/RT, in which P is the gas pressure, V[m] the molar volume, R the gas constant, and T the temperature. The compression factor is a measure of the deviation of
a real gas from an ideal gas. For an ideal gas the compression factor is equal to 1.
→ compression; → facteur.
conversion factor
کروند ِهاگرد
karvand-e hâgard
Fr.: facteur de conversion
1) A numerical factor that, by multiplication or division, translates one unit or value into another.
2) In → molecular cloud studies, a factor used to convert the → carbon monoxide (CO) line intensity to → molecular hydrogen (H[2]) → column density; usually denoted X[CO] = I(CO) / N(H[2]). This
useful factor relates the observed CO intensity to the cloud mass. A general method to derive X[CO] is to compare the → virial mass and the ^12CO (J = 1-0) luminosity of a cloud. The basic
assumptions are that the CO and H[2] clouds are co-extensive, and molecular clouds obey the → virial theorem. However, if the molecular cloud is subject to ultraviolet radiation, selective →
photodissociation may take place, which will change the situation. Moreover, molecular clouds may not be in → virial equilibrium. To be in virial equilibrium molecular clouds must have enough mass,
greater than about 10^5 solar masses. The way → metallicity affects X[CO] is a matter of debate, and there is no clear correlation between X[CO] and metallicity. Although lower metallicity brings
about higher ultraviolet fields than in the solar vicinity, other factors appear to be as important as metallicity for the determination of X[CO]. In the case of the → Magellanic Clouds, X[CO](SMC) =
14 ± 3 × 10^20 cm^-2 (K km s^-1)^-1, which is larger than X[CO] (LMC) = 7 ± 2 × 10^20 cm^-2 (K km s^-1)^-1. An independent method to derive X[CO] is to make use of the gamma ray emission from a
cloud. The flow of → cosmic ray protons interacts with interstellar low-energy hydrogen nuclei in clouds creating neutral → pions. These pions quickly decay into two gamma rays. It is therefore
possible to estimate the number of hydrogen nuclei and hence the cloud mass from the gamma ray counts. Such a gamma-ray based conversion factor is estimated to be 2.0 × 10^20 cm^-2 (K km s^-1)^-1 for
Galactic clouds, in good agreement with the result obtained from the virial method. However, the gamma ray flux is not well known in general, so this method is uncertain as well. See, e.g., Fukui &
Kawamura, 2010 (ARAA 48, 547).
→ conversion; → factor.
cosmic scale factor
کروند ِمرپل ِکیهانی
karvand-e marpal-e keyhâni
Fr.: facteur d'échelle cosmologique
A quantity, denoted a(t), which describes how the distances between any two galaxies change with time. The physical distance d(t) between two points in the Universe can be expressed as d(t) = R(t).x,
where R(t) is the → scale factor and x the → comoving distance between the points. The cosmic scale factor is related to the → redshift, z, by: 1 + z = R(t[0])/R(t[1]), where t[0] is the present time
and t[1] is the time at emission of the radiation. The quantity (1 + z) gives the factor by which the → Universe has expanded in size between t[1] and t[0]. It is also related to the → Hubble
parameter by H(t) = R^.(t)/R(t), where R^.(t) is the time → derivative of the scale factor. In an → expanding Universe the scale factor increases with time. See also the → Friedmann equation.
→ cosmic; → scale; → factor.
deuterium enrichment factor
کروند ِپرداری ِدوتریوم
karvand-e pordâri-ye doteriom
Fr.: facteur d'enrichissement en deutérium
The ratio between the D/H value in → water and in → molecular hydrogen, as expressed by:
f = [(1/2)HDO/H[2]O]/[(1/2)HD/H[2]] = (D/H)[H[2]O]/(D/H)[H[2]].
When f> 1, there is → deuterium enrichment.
→ deuterium; → enrichment; → factor.
dilution factor
کروند ِاوتالش
karvand-e owtâleš
Fr.: facteur de dilution
The energy density of a radiation field divided by the equilibrium value for the same color temperature.
→ dilution; → factor.
Eddington factor
کروند ِادینگتون
karvand-e Eddington
Fr.: facteur d'Eddington
Same as → Eddington parameter.
→ Eddington limit; → factor.
Fr.: facteur
1) One that actively contributes to the production of a result.
2) Math.: Any of the numbers or symbols that when multiplied together form a → product.
M.Fr. facteur "agent, representative," from L. factor "doer or maker," from facere "to do" (cf. Fr. faire, Sp. hacer); from PIE base *dhe- "to put, to do;" cf. Skt. dadhati "puts, places;" Av.
dadaiti "he puts;" Hitt. dai- "to place;" Gk. tithenai "to put, set, place;" Lith. deti "to put;" Rus. det' "to hide," delat' "to do;" O.H.G. tuon; Ger. tun; O.S., O.E. don "to do."
Karvand, from kar- root of Mod.Pers. verb kardan "to do, to make" (Mid.Pers. kardan; O.Pers./Av. kar- "to do, make, build;" Av. kərənaoiti "he makes;" cf. Skt. kr- "to do, to make," krnoti "he makes,
he does," karoti "he makes, he does," karma "act, deed;" PIE base k^wer- "to do, to make") + -vand a suffix forming adjectives and agent nouns.
factor tree
درخت ِکروند
deraxt-e karvand
Fr.: arbre des facteurs
A diagram representing a systematic way of determining all the prime factors of a number.
→ factor; → tree.
۱) کرونده؛ ۲) کروندی
1) karvandeh; 2) karvandi
Fr.: factoriel
1) (n.) The product of all the positive integers from 1 to n, denoted by symbol n!
2) (adj.) of or pertaining to factors or factorials.
→ factor + -ial, from L. -alis, → -al.
کروندیدن، کروند گرفتن
karvandidan, karvand gereftan
Fr.: factoriser
The operation of resolving a quantity into factors.
→ factor + → -ize.
filling factor
کروند ِپُری
karvand-e pori
Fr.: facteur de remplissage
Of a molecular cloud or a nebula, the ratio of the volumes filled with matter to the total volume of the cloud.
Filling, from fill, from O.E. fyllan, from P.Gmc. *fullijan (cf. Du. vullen, Ger. füllen "to fill"), a derivative of adj. *fullaz→ full; → factor.
Karvand, → factor; pori, from por, → full.
Gaunt factor
کروند ِگاؤنت
karvand-e Gaunt
Fr.: facteur de Gaunt
In the atomic theory of spectral line formation, a quantum mechanical correction factor applied to the absorption coefficient in the transition of an electron from a bound or free state to a free
Gaunt, after John Arthur Gaunt (1904-1944), English physicist born in China, who significantly contributed to the calculation of continuous absorption using quantum mechanics; → factor
integrating factor
کروند ِدرستالنده
karvand-e dorostâlandé
Fr.: facteur intégrant
A function that converts a → differential equation, which is not exact, into an → exact differential equation. This is done by multiplying all terms of the original equation by the integrating
→ integrate; → factor.
ionization correction factor (ICF)
کروند ِارشایش ِیونش
karvand-e aršâyeš-e yoneš
Fr.: facteur de correction d'ionisation
A quantity used in studies of → emission nebulae to convert the → ionic abundance of a given chemical element to its total → elemental abundance. The elemental abundance of an element relative to
hydrogen is given by the sum of abundances of all its ions. In practice, not all the ionization stages are observed. One must therefore correct for unobserved stages using ICFs. A common way to do
this was to rely on → ionization potential considerations. However, → photoionization models show that such simple relations do not necessarily hold. Hence, ICFs based on grids of photoionization
models are more reliable. Nevertheless here also care should be taken for several reasons: the atomic physics is not well known yet, the ionization structure of a nebula depends on the spectral
energy distribution of the stellar radiation field, which differs from one model to another, and the density structure of real nebulae is more complicated than that of idealized models (see, e.g.,
Stasińska, 2002, astro-ph/0207500, and references therein).
→ ionization; → correction; → factor.
Landé factor
کروند ِلانده
karvand-e Landé
Fr.: facteur de Landé
The constant of proportionality relating the separations of lines of successive pairs of adjacent components of the levels of a spectral multiplet to the larger of the two J-values for the respective
pairs. The interval between two successive components J and J + 1 is proportional to J + 1.
After Alfred Landé (1888-1976), a German-American physicist, known for his contributions to quantum theory; → facteur. | {"url":"https://dictionary.obspm.fr/index.php/?showAll=1&formSearchTextfield=factor","timestamp":"2024-11-07T04:11:27Z","content_type":"text/html","content_length":"36004","record_id":"<urn:uuid:56b22b41-7bec-4972-9e5a-d9d28f76c36f>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00031.warc.gz"} |
Selection Sort Algorithm In Data Structures And Algorithms Using Python
Selection Sort Algorithm is one of the easiest sorting algorithms in Data Structures. It is a comparison-based sorting algorithm. It is used to arrange the elements of an array (in ascending order).
In this article, we’re going to see how we can implement selection sort in data structures using the Python programming language.
For Example
INPUT Array – 16 12 96 42 21
OUTPUT Array – 12 16 21 42 96
Table of Contents:
1. How does the Selection sort algorithm work?
2. Algorithm
3. Pseudocode of selection sort
4. Selection sort algorithm in Python
5. Selection sort algorithm in C++
6. Time Complexity of Selection sort
7. Space Complexity of Selection Sort
8. In-place behavior
9. Is Selection sort an in-place sorting algorithm?
10. Is the Selection sort show adaptive nature?
11. Features
Working of Selection Sort Algorithm
In Selection Sort, the given array is divided into two parts – the sorted part and the unsorted part. The sorted part is at the left end of the array and the unsorted part is at the right end of the
array respectively.
At the beginning of the algorithm, the sorted part is completely empty and the unsorted part consists of the whole array. But as algo proceeds, the elements that are being sorted becomes a part of
the sorted array, and the remaining subarray comes under the unsorted part.
In this sorting algorithm, with every passing iteration, the least valued or the minimum element from the unsorted array is selected, using the linear search technique after that the selected element
is swapped with the first element of that array. The new element added at the beginning becomes a part of the sorted array.
Let us understand this with an example.
Consider the given array of elements – 12 45 67 2 3
Selection_Sort ( array a, length n)
Step 1: Start loop from i = 0 to n-2
Step 2: Select smallest element among a[i] , … a[n-1]
Step 3: Swap element with a[i]
function selection sort
A[] : array of elements
n : number of elements of an array
for i=1 to n-1
/* fix first element of the array as minimum */
/*check for the smallest element by comparing
with all the other elements */
for j=i+1 to n
if A[j] < A[min] then
end if
end for
/* swap the minimum element with the first element
of the array */
if index_min!=i then
swap A[min] amd A[i]
end if
end for
end function
Selection sort python code
#Code by Copyassignment.com
# function
def selection_sort(A):
for i in range(len(A)):
# Finding the minimum element in remaining unsorted array
min_idx = i
for j in range(i+1, len(A)):
if A[min_idx] > A[j]:
min_idx = j
# Swapping A[i] with minimum element
A[i], A[min_idx] = A[min_idx], A[i]
#Driver Code
arr = [12,45,67,2,3,9]
selection_sort(arr) #Calling Function
print("Final array after sorting :\n")
for i in range(len(arr)):
print(arr[i]) #Printing array after sorting
Final array after sorting :
Selection sort algorithm in C++
// Code by violet-cat-415996.hostingersite.com
#include <iostream>
using namespace std;
void swap(int *xp, int *yp)
int temp = *xp;
*xp = *yp;
*yp = temp;
void Selection_Sort(int arr[], int n)
int min_idx;
// moving boundary of unsorted sub array
for (int i = 0; i < n - 1; i++)
// Find the minimum element in unsorted array
min_idx = i;
for (int j = i + 1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
// Swapping A[i] element with minimum element
swap(&arr[min_idx], &arr[i]);
// Driver program to test above functions
int main()
int n;
cout << "Enter the size of array : " << endl; //Taking size of array
cin >> n;
int arr[n];
cout << "Enter " << n << " elements:\n";
for (int i = 0; i < n; i++)
cin >> arr[i]; //Taking elements of array
Selection_Sort(arr, n); //Calling Function
cout << "Sorted array: \n";
for (int i = 0; i < n; i++)
cout << arr[i] << " "; //Printing sorted array
return 0;
Enter the size of array :
Enter 4 elements:
Sorted array:
Time complexity of selection sort
Time Complexity – In an algorithm, the time complexity is defined as the amount of time taken by the algorithm to completely run and execute all its operations is known as the time complexity of an
Selection Sort Algorithm has a time complexity of O(n^2) for all the three cases as explained below.
1. Best-Case Complexity – Best case is when all the elements of the original array are already sorted. But, even in this case, the algorithm will perform all the comparison operations for each and
every element of the array. Thus, its time complexity will remain the same i.e., O(n^2).
2. Average-Case Complexity – This is the generalized amount of time taken by the algorithm for all the possible averaged inputs. Here, again some of the swapping operations and all the comparison
operations will be performed for the elements of the array. As a result, the time complexity will remain the same i.e., O(n^2).
3. Worst-Case Complexity – Worst case is considered when the given array is reversely sorted i.e., all the elements in descending order. This is the case where all the sorting operations and all the
comparison operations will be performed for all the elements of the array. Thus, time complexity, in this case, will be again O(n^2).
Space complexity
Space Complexity – For an algorithm, space complexity is defined as the memory space occupied by it to run and execute all its operations.
In this algorithm, all the elements are swapped within the space of a single array, given initially and does not require any extra memory space or array or any other data structure for its execution
except one variable temp, which is used to store the element being swapped at a temporary location. Thus, it has a space complexity of O(1).
In-place sorting algorithm?
Yes, Selection sort is an in-place sorting algorithm because it does not require any other array or data structure to perform its operations. All the swapping operations are done within a single
Stable sorting algorithm?
No, Selection Sort is an unstable sorting algorithm because it works on the principle of initially finding the minimum element from the list and then putting it at its correct position by swapping it
with the element which is present at the beginning of the unsorted array. Due to this swapping, the element present in the later part of the array will come at the beginning even if they have the
same value.
Let us understand this with an example –
Input – 4[A] 5 3 2 4[B] 1
Output – 1 2 3 4[B] 4[A] 5
If it would have been a stable sorting algorithm then the output would be 1 2 3 4[A] 4[B] 5
Note: Subscripts are added only for the purpose of understanding.
Adaptive sorting algorithm?
No, it is a non-adaptive sorting algorithm because even if some of the elements of the array are sorted itself, the Selection Sort Algorithm will not take into account the ‘already sorted’ elements
and it will reorder them as all the comparisons would be made to confirm their sortedness.
Features of selection sort
• It is the second slowest sorting algorithm after bubble sort because it occupies memory space for a longer time. Therefore, it is not suitable for large data sets.
• Selection Sort Algorithm can be applied on linked list data structure as well.
• It is preferably used when the cost of performing writing operations to memory is taken into consideration like in the case of flash memory.
Thanks for reading!
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Happy Coding!
Introduction to Searching Algorithms: Linear Search Algorithm
Bubble Sort Algorithm In Data Structures & Algorithms using Python
Merge Sort Algorithm in Python
Sorting Algorithms and Searching Algorithms in Python
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Ordering of binary colloidal crystals by random potentials
Cite this: Soft Matter, 2020, 16, 4267
Ordering of binary colloidal crystals by random
Andre´ S. Nunes,*aSabareesh K. P. Velu, *bIryna Kasianiuk,cDenis Kasyanyuk,c Agnese Callegari, cGiorgio Volpe, dMargarida M. Telo da Gama, a Giovanni Volpe beand Nuno A. M. Arau´jo ‡a
Structural defects are ubiquitous in condensed matter, and not always a nuisance. For example, they underlie phenomena such as Anderson localization and hyperuniformity, and they are now being
exploited to engineer novel materials. Here, we show experimentally that the density of structural defects in a 2D binary colloidal crystal can be engineered with a random potential. We generate the
random potential using an optical speckle pattern, whose induced forces act strongly on one species of particles (strong particles) and weakly on the other (weak particles). Thus, the strong
particles are more attracted to the randomly distributed local minima of the optical potential, leaving a trail of defects in the crystalline structure of the colloidal crystal. While, as expected,
the crystalline ordering initially decreases with an increasing fraction of strong particles, the crystalline order is surprisingly recovered for sufficiently large fractions. We confirm our
experimental results with particle-based simulations, which permit us to elucidate how this non-monotonic behavior results from the competition between the particle-potential and particle–particle
Perfect crystalline structures are not commonly found in Nature, because, even in the absence of impurities, structural defects occur spontaneously and disrupt the periodicity of the crystalline
lattice.1 [For example, when a melt is cooled down, multiple]
crystallites grow with degenerate orientations.2[Since the ]
coar-sening time of these crystallites diverges with size, structural defects appear and prevent the emergence of global order.3,4 While the existence of these defects is a challenge when growing
single crystals, it can also be an opportunity when engineering the properties of materials; indeed, control over defects enables the development of solid-state devices with fine-tuned mechanical
resilience, optical properties, and heat and electrical conductivity.5–9 In atomic crystals, engineering structural defects is an experimental challenge for two reasons:10
first, current visualization techniques at the atomic scale do not provide a high spatial or time resolution;11,12second, no current technique can control the density of defects in a systematic
manner.13The first challenge can be overcome studying colloidal crystals as models for atomic systems,14,15[where colloidal ]
parti-cles can be individually tracked using standard digital video microscopy techniques,16–18[and have in fact also been used to]
study crystallisation and melting of colloidal crystals in the presence of extended laser fields.19,20Here, we demonstrate that the second challenge can be solved combining a binary colloidal mixture
and an optical random potential generated by a speckle light pattern. This permits us to control the density of structural defects in the resulting 2D colloidal crystal and to explore a surprising
non-monotonic behavior of their ordering and stability.
We use a binary colloidal suspension of equally-sized poly-styrene (refractive index nPS E 1.59) and silica (nSi E 1.42)
spherical particles with diameters dPS = 6.24 0.22 mm and
dSi= 6.73 0.22 mm, respectively. The particles interactions are
hard-sphere like but the following results can be reproduced with soft interactions as well (see ESI†).21 To characterize the composition of the mixture, we use the molar fraction of polystyrene
particles defined as w = Nps/Nt where Nps is the
Centro de Fı´sica Teo´rica e Computacional and Departamento de Fı´sica, Faculdade de Cieˆncias, Universidade de Lisboa, P-1749-016 Lisboa, Portugal. E-mail: [email protected]
b[Department of Physics, Bilkent University, Cankaya, 06800 Ankara, Turkey] c[Department of Physics, Bilkent University and UNAM, Cankaya, 06800 Ankara,]
d[Department of Chemistry, University College London, 20 Gordon Street,] London WC1H 0AJ, UK
e[Department of Physics, University of Gothenburg, 41296 Gothenburg, Sweden] †Electronic supplementary information (ESI) available. See DOI: 10.1039/d0sm00208a ‡Contributed equally.
Received 4th February 2020, Accepted 3rd April 2020 DOI: 10.1039/d0sm00208a
number of polystyrene particles and Ntis the total number of
particles. We let these particles sediment at the bottom surface of a homemade sample chamber so that they are effectively confined in a quasi-2D space (see the Section Materials and methods). We
illuminate from above with a speckle pattern, which we generate by mode-mixing a laser beam in a multi-mode optical fibre (see Fig. 5).22–24 Speckle patterns form rough, disordered optical potentials
characterized by wells whose depths are exponentially distributed, with spatial corre-lations that are Gaussian with an average width (grain size) set by diffraction. As proposed in ref. 25, to
characterize the strength and correlation length of the optically generated random field, we first identify the ‘‘bright spots’’ and then fit a Gaussian to each spot, using the code in ref. 26. We
found s = 2.7 0.2 mm, which is less than half the diameter of the particles. Furthermore, the fibre imposes a Gaussian envelope (beam waist sG= 72.5 0.2 mm) to the speckle pattern, which
attracts the particles towards the center of the speckle pattern effectively confining them in space. Since the optical forces acting on the particles increase for larger mismatches between their
refractive index and that of the surrounding medium (here water, nw E 1.33),27 the optical forces acting on the
polystyrene (strong) particles are about 2 higher than those exerted on silica (weak) particles (estimated using the FORMA method28). Importantly, the optical forces at the deepest local minima of
the speckle potential are strong enough to trap the strong particles, but not the weak ones (see ESI† for an estimation of the strength of the optical traps21).
We start with a low concentration of particles (1.4 107[ml]1[)]
and switch on the optical potential. The particles are attracted towards its center by the Gaussian envelope. When only weak particles are present (w = 0), they eventually form a compact structure
with hexagonal order, as shown in Fig. 1a. When we introduce strong particles, these get trapped in the local minima of
the disordered potential and introduce defects that reduce the hexagonal order. Already with only 20% of strong particles (w = 0.2), the presence of structural defects is clearly visible (see Fig.
1b). The impact is even more pronounced when 50% of the particles (w = 0.5) are strongly interacting with the potential (Fig. 1c). Thus, strong particles act as defects in the crystalline structure
of the weak ones, compromising global order. We were able to determine that the deformation of the structure was not caused by particle bidispersity since when subject only to a Gaussian envelope the
particles formed a crystalline structure independently of the number of strong particles present (see ESI†).21The experimental results are confirmed by particle-based simulations, as shown in Fig.
1d–f (see Section Materials and methods). As we will see in more detail below, we can control the density of defects by adjusting w as well as the intensity and grain size of the pattern.
To quantify the order of the crystalline structure, we mea-sure the six-fold bond-order parameter,hf6i, defined as29
hf6i ¼ 1 6Nc XNc l XNb j ei6ylj ; (1)
where the out sum is over the Ncparticles within 7.5 particle
diameters from the center of the potential (the area shown in Fig. 1), which is the area where the aggregate is formed and does not include the boundary particles. The inner sum is over the
Nbneighbors of a particle in the Voronoi tessellation, and
yljis the angle between the x-axis and the line connecting the
centers of particles j and l. hf6i = 1 for perfect hexagonal
crystals (in practice, it is never exactly one, because of thermal fluctuations and other transient perturbations to the periodic order) and it decreases with the number of structural defects. Fig. 2
showshf6i obtained experimentally and numerically as a
function of the molar fraction w. For w = 0,hf6iE 1, consistent
with the formation of an hexagonal periodic structure. As expected, as w increases, the value ofhf6i decreases due to
the formation of structural defects. The snapshots in the top rows of Fig. 2 show the final configurations (first row), the corresponding Voronoi tessellations (second row), and the spatial Fourier
transform (third row), for different values of w. Surprisingly, the data reported in Fig. 2 show that hf6i
reaches a minimum at wminE 0.6, and that the global order
increases for w 4 wmin. In particular, for w = 1, the strong
particles self-assemble into a hexagonal crystal, despite the presence of the underlying random potential. This non-monotonic dependence is also observed at higher densities. In Fig. S5 of the
ESI,†21 we shown that the same behavior is observed numerically in a system with a number of particles that is 25% higher. This result is corroborated by the Voronoi tessellation of the final
configurations and by the respective spatial Fourier transforms. From this analysis, we can see that the number of Voronoi cells with a number of neighbors different from six becomes higher near the
minimum ofhf6i,
even though the Voronoi-cell size in both experiments and simulations does not vary significantly compared with the particle size (see Fig. S6 from the ESI†).21Also, The Fourier Fig. 1 Colloidal
crystals with tunable degree of disorder. Final
configura-tions obtained in (a–c) experiments and (d–f) simulaconfigura-tions, for different molar fractions w of strong particles. The weak (silica) particles are light gray, and the strong
(polystyrene) particles are dark gray. The illumination for the images is delivered by an optical fibre which produces the vignetting effect observed in the experimental images.
transforms display dimmer intensity peaks near the minimum ofhf6i.
In order to shed light on the non-monotonic behavior, we first analyze the trajectories obtained by particle-based simulations, without the Gaussian envelope and a particle density 10 lower than that
of maximal packing, to study the interactions between the two particle species and the local minima in the potential. Fig. 3(a) shows individual trajectories of weak (light gray) and strong (dark
gray) particles at various w. In all cases, the weak particles can hop between minima, while the strong particles are readily trapped in them. This qualitative analysis for a lower density elucidates
the possible underlying
mechanisms at higher densities. In the presence of the Gaussian envelope, particles are dragged to the center and the strong particles quickly populate the minima that are sufficiently deep to prevent
their escape. At low w, the number of strong particles is lower than the number of such minima so they remain there for the entire simulation time, because this configuration is energe-tically
favorable (Fig. 3b and c); therefore, the number of spatial defects increases monotonically with the number of the trapped strong particles, leading to a decrease ofhf6i with increasing w.
At large w, the number of strong particles is greater than the number of potential minima and thus it becomes energetically favorable to have more than one strong particle in one minimum (Fig. 3d).
This allows the spatial rearrangement of the particles since the energy of the interaction with the speckle is no longer strong enough to localize the particles, a large-scale crystalline structure
is favorable, consistent with the increase inhf6i observed
in Fig. 2. When w = 1, all particles are strong and thus the hexagonal crystalline structure is recovered. We also counted the number of strong and weak particles situated in minima of the random
potential as a function of w. As shown in Fig. S8 of the ESI,†21
the minima are mainly populated by strong particles and the average number of particles is larger than one for values of w above the one at which the six-fold bond order parameter is the minimum. In
order to explore how robust the non-monotonic dependence ofhf6i as a function of w is, we studied numerically how it depends
on the properties of the underlying speckle pattern. The speckle is characterized by a strength V corresponding to the average potential depth (in units of kBT, where kBis the Boltzmann constant and
T is
the absolute temperature of the sample) and by a spatial correlation s (in units of the particle diameter), which corresponds to the average grain size. Fig. 4(a) showshf6i for different V. Although
curves in the range 1.51o V r 18.8 feature one minimum, its position and intensity vary with V: the number of minima that can trap particles is expected to increase with V. Thus, the fraction of
particles that can be trapped also increases and the corresponding value of wminshifts to the right while the minimum becomes deeper.
For V 4 18.8, the behavior seems to become independent of the molar fraction (and always disordered), because the weak particles are also strongly trapped. Fig. 4(b) showshf6i for different values
of s. A pronounced minimum is only observed for intermediate values of s, close to unity (particle diameter). If s c 0.5 or s{ 0.5, the optical forces are negligible for different reasons: for s c
0.5, the gradient of the optical potential is very small on the scale of the particle; and for s{ 0.5, the optical potential varies on a length scale smaller than the particle size and thus its
gradient averages to zero over the particle cross-section (see Fig. S9, ESI†).21In the latter case, the optical force on a particle is the sum of the contributions over the particle’s cross-section,
which can be described by an effective random potential that differs from the one originally applied (Fig. S10 and S11, ESI†).21
In conclusion, we have shown that the order in a two-dimensional binary colloidal crystal can be controlled by an Fig. 2 Crystalline order for different molar fractions of strong particles.
Six-fold bond order parameterhf6i as a function of the molar fraction w
obtained experimentally (circles) and numerically (squares; the blue line connects the symbols for visual guidance). The error bars show the standard deviation ofhf6i over 500 frames in the
stationary state of the
experiments (i.e., after 30 minutes from the start of the experiments). The numerical results are averages over 100 samples. The top snapshots show the final configurations in the experiments (first
row), the Voronoi tessellation (second row), and the spatial Fourier transform (third row) for w = 0, 0.23, 0.6, and 1. The filled (empty) circles at the center of the Voronoi cells indicate strong
(weak) particles. The cells are colored by the number of nearest neighbors, namely, equal (green), lower (red), greater (blue) than six. See also Supplementary Video 1 (ESI†).
underlying random optical potential. While previous studies19,20
have shown how freezing and melting are influenced by the intensity of the laser field and the particle density. We employ a disordered potential and a binary mixture where some particles interact
strongly with the substrate and others are weakly interacting. This permits us to study a system where disorder and impurities are present, which is highly relevant for applications. Since the
intensity of the optical forces depends on the mismatch of
the indices of refraction of the particles and the surrounding medium, the particles with the larger index mismatch are more responsive (strong particles) than those with the lower mismatch (weak
particles). For the parameters of the optical potential that were considered, only the strong particles respond significantly to the potential. Thus, strong particles tend to occupy the minima of the
potential and nucleate structural defects in the, otherwise, periodic hexagonal structure of the weak particles. Fig. 3 Local dynamics of the interaction between particles and minima in the random
potential. (a) Examples of trajectories of weak (light gray) and strong (dark gray) particles in the presence of a speckle obtained numerically for different values of the molar fraction w. The
particle density is 10 lower than that of maximal packing and the Gaussian envelope is absent. The four simulations were preformed under exactly the same conditions, including the same sequence of
random numbers for the thermostat (see ESI†).21[The black circles on the top left corner indicate the particle size. The random]
potential intensities are in units of kBT and s is one particle diameter. (b) When a weak particle (light gray) is located at a potential minimum and a strong
particle (dark gray) is in its vicinity, it is energetically favorable to exchange the two, but the opposite process (c) is not. (d) The free energy may be significantly reduced when two particles of
the same species share the same potential minimum. See also Supplementary Video 2 (ESI†).
Fig. 4 Dependence of the order parameter on the speckle properties. Six-fold bond order parameter as a function of the molar fraction (w) obtained numerically, for different values of the speckle (a)
strength and (b) spatial correlation s. Results in (a) were obtained for s = 0.5 and in (b) for V = 15.1, and are averages over 100 samples.
The density of defects is controlled by the fraction of strong particles and the statistical properties of the underlying potential. When the number of strong particles increases beyond the number of
local minima that can trap them, the trapping mechanism becomes less effective and the hexagonal order is recovered as the fraction of strong particles increases.
Here, we have considered a random optical potential with Gaussian spatial correlations and a characteristic length that is of the order of the particle size. However, it is technically possible to
generate other optical potentials, e.g. periodic27or with different spatial correlations.30,31 Thus, one can control not only the density of defects but also their spatial distribu-tion. Time-varying
optical potentials or driving forces could also be employed to change the position of strong particles and defects in time, affecting the overall dynamics, what raises several relevant fundamental and
applied questions.18,23,32,33 Understanding how the spatial distribution of defects influ-ences the physical properties of materials is a question of both scientific curiosity and technological
interest that can now be addressed in a systematic way.
A non-monotonic dependence of the density of defects on the particle ratio was also found for a binary mixture of Yukawa particles coupled to a random (quenched) field in ref. 34, where the particles
differ in charge, which impacts the particle–particle interaction, but the response to the external field is identical. By contrast, here the particle–particle interactions are identical for both
species, while their response to the external field is distinct. This difference is key to enable the external control of the density of defects, as proposed here.
Materials and methods
Sample preparation
Diluted aqueous stock solutions of polystyrene and silica colloidal spheres (microparticles GmbH, diameter dPS= 6.24
0.22 mm and dSi= 6.73 0.22 mm, respectively) were used to
prepare binary solutions with different molar fractions of poly-styrene particles from w = 0 to w = 1. The total density of particles was kept constant at 1.4 107[ml]1
. These colloidal solutions were confined in a homemade sample chamber (internal thickness 200 mm), built between a bottom glass slide (made hydrophilic by treatment in a 0.25 M NaOH solution) and a
top flat-terminated fibre coupler (Thorlabs, SM1SMA) held apart by two layers of a thermoplastic spacer, which at the same time was also used for sealing the chamber. The fibre coupler was used to
connect the output end of a multimode optical fiber (core diameter 105 mm, NA = 0.22, length 51 m). See also Fig. 5. Experimental setup
A homemade inverted optical microscope setup was used for carrying out the experimental investigations of structural defects in colloidal crystals formed under random optical potentials, as
schematically shown in Fig. 5.24 An image of the sample with colloidal particles was projected by a micro-scope objective (Nikon Plan Fluorite Imaging Objective, 20,
NA = 0.5, WD = 2.1 mm) onto a monochrome charge-coupled device (CCD) camera with an acquisition rate between 1 and 8 frames per second (fps). The incoherent illumination was provided by a LED lamp at
l = 625 nm coupled into the optical fiber using a dichroic mirror (Thorlabs, DMLP650). The parti-cles were tracked by digital video microscopy.35
The static speckle light pattern with a Gaussian envelope was generated by focusing a laser beam (wavelength l = 976 nm, output power P = 90 mW) into a multimode optical fiber using a plano-convex
lens (focal distance f = 25.4 mm). The output speckle pattern is the result of the multipath interference of the optical waves carrying random phases within the multimode optical fiber.23,24,36The
length of the optical path between the fiber tip and the imaging plane where the colloidal particles lay (i.e., the bottom of the sample chamber) determines the final speckle grain size. The typical
duration of an experiment is about 90 minutes.
The smooth optical potential was obtained by the speckle suppression using a high frequency mechanical oscillator connected to the stretched interval of the optical fiber. The vibrational frequency
was adjusted with DC voltage up to 12 000 rpm.
We performed Brownian dynamics (BD) simulations of a binary mixture of N = 800 particles with several compositions, on a two-dimensional square box with linear size L. The particle species differ in
the strength of their response to the optical potential. The interaction potential between a pair of particles i and j with diameter dp is independent of the species and is
given by the repulsive part of a Lennard-Jones potential:
VijðrÞ ¼ e dp r 12 dp r 6 " # ; (2)
where e sets the energy scale. This is a very steep and short-ranged potential that only affects neighbouring particles within a cut-off distance of rcut= 21/6dp.
The external potential has two contributions. The first contribution is a Gaussian potential that attracts the particles Fig. 5 Schematic representation of the experimental setup and sample chamber.
towards the centre of the simulation box, given by VGaussianðrÞ ¼ VGke ðr7:5Þ 2 sG2 if r 4 7:5 0 if r 7:5 8 > < > : (3)
where sGis the width of the Gaussian and VGksets the scale
of this interaction, which depends on the particle type k; the interaction VGk with the most responsive (strong) particles is
2 that with the least responsive (weak) ones; r is the distance to the centre of the simulation box. By definition, we ensure that the Gaussian potential is zero in the central region of the box, of
radius 7.5, where we carried out the statistical analysis of the system. This potential is used to confine the particles in the centre of the box at sufficiently high densities. The second contribution
to the external potential reproduces the potential generated by a speckle pattern.22We used the Fourier filtering method (FFM) to generate numerically random potentials with Gaussian spatial
correlations.37,38The FFM takes advantage of the fact that the correlation function of a field E(-r), is the inverse Fourier transform of the absolute value of its Fourier coefficients, |E-k|2, as
stated by the Wiener–Khinchin theorem.39This relation
allows us to sample random Fourier coefficients that when trans-formed back into real space describe a random potential with the desired spatial correlations. The depths of those potentials have a
Gaussian distribution. To convert it into an exponential distribu-tion, as measured for the speckle, we used the following procedure: the random surface is discretized in 1024 1024 cells, which we
sort by the intensity of the potential. Then, we produce a sorted list of intensities drawn from an exponential distribution and sub-stitute each cell intensity by the corresponding entry on the
ranked list of intensities. We tested this procedure with Gaussian and power-law correlation functions and confirmed that it does lead to the desired distribution of intensities, without affecting
the nature of the correlation function. The forces due to this potential are then calculated using finite differences. In all simulations, we considered Gaussian correlations with a dispersion s.
When s o dp (where dp is the diameter of the
particles) the speckle features vary on distances shorter than the particle size and we need to consider an effective speckle pattern that is the result of the integration of the speckle intensities
over the particle volume (see below section ‘‘Effective speckle properties’’). The results present in Fig. 1 and 2 were achieved with s = 0.4. The potentials strength ratio is VG/V = 1 in the
simulations presented in Fig. 1, 2, 4(a) and (b).
The motion of a particle i in the surrounding medium is described by the overdamped Langevin equation
gd~ri dt ¼ ~ri X j VijðrÞ þ Vextð~riÞ " # þ ~xi; jai; (4)
where g is the Stokes–Einstein friction coefficient and ~x[i] is a random stochastic term that mimics the thermal noise that results from the interaction with the medium. This term is given by a
normal distribution with zero mean and auto-correlation that is independent of space and time and proportional to the thermostat temperature T, i.e. hxn
i(t)xli(t0)i = 2kBTgidnld(t t0),
where n and l are indices that run over the space dimensions and kB is the Boltzmann constant. The characteristic time is
defined as t = dp2g/kBT. Eqn (4) is integrated following the
algorithm developed by Branka and Heyes,40[i.e. a second-order]
stochastic Runge–Kutta scheme, with a time step of Dt = 104t. We set the diameter of the particle, dp, as the unit length, the
simulation box has linear size L = 50 and the width of the external Gaussian potential is sG = L/2. The energy is given in units of
kBT with e = 10 and VG = 200. The simulations were run for
2 104t and the data used in the calculations was taken in the last 1.5 103[t, when the evolution was found to be in the stationary]
state in the centre of the box. For all data points, we used 100 samples to average the relevant quantities. While we do not expect a strong dependence on the geometry of the experimental setup, in
order to make a direct comparison with the experimental results, rather than using periodic boundary conditions, we considered the same circular confinement with an external potential. This also
allows us to study the initial dynamics that result from increasing the local concentration in the center due to the confining potential.
Conflicts of interest
There are no conflicts to declare.
Andre´ S. Nunes, Margarida M. Telo da Gama and Nuno A. M. Arau´jo acknowledge financial support from the Portuguese Foundation for Science and Technology (FCT) under Contracts no. EXCL/FIS-NAN/0083/
2012, UIDB/00618/2020, UIDP/00618/ 2020, SFRH/BD/119240/2016 and PTDC/FIS-MAC/28146/2017 (LISBOA-01-0145-FEDER-028146). Margarida M. Telo da Gama and Nuno Arau´jo would like to thank the Isaac Newton
Institute for Mathematical Sciences for support and hospitality during the program ‘‘The mathematical design of new materials’’ where the final version of this manuscript was completed. This program
was supported by EPSRC Grant Number: EP/R014604/1. Giorgio Volpe acknowledges support from the Royal Society under grant RG150514. Iryna Kasianiuk acknowledges partial support of Tu¨bitak grant
115F401. Denis Kasyanyuk acknowledges partial support of Tu¨bitak grant 116F111. Agnese Callegari acknowledges partial support of Tu¨bitak grant 115F401 and 116F111. We also acknowledge Parviz Elahi
for his help with the experimental setup.
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dyna-mics computer simulations: Multivariable case, Phys. Rev. E: Stat. Phys., Plasmas, Fluids, Relat. Interdiscip. Top., 1999, 60, 2381. | {"url":"https://9lib.net/document/4yr0g0oy-ordering-of-binary-colloidal-crystals-by-random-potentials.html","timestamp":"2024-11-14T09:53:40Z","content_type":"text/html","content_length":"184812","record_id":"<urn:uuid:a04ec36d-91bb-4ad5-bf8b-40fc0f0486d0>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00680.warc.gz"} |
Construction Cost Estimation Software Quiz | Test Your Knowledge
🏗️ Construction Cost Estimation Software Quiz
Take our Construction Cost Estimation Software Quiz and test your knowledge on the commonly used tools in the industry. Find out which software offers an all-in-one solution, provides markup and
collaboration tools, helps contractors estimate project costs, and allows for easy automation and streamlining of the estimating process.
Construction Cost Estimation Software Quiz
Test your knowledge on the commonly used construction cost estimation software.
Understanding the cost of construction projects can be a complex process, requiring precise calculations and accurate data. To simplify this task, a variety of construction cost estimation software
are available on the market, each with their unique features and benefits. The software you choose can greatly impact the efficiency and accuracy of your cost estimation process.
ProEst, for instance, is a comprehensive solution that integrates estimating and bidding into one platform. This all-in-one software simplifies the process, making it an ideal choice for businesses
looking for a streamlined approach. On the other hand, Bluebeam Revu stands out for its suite of markup and collaboration tools, facilitating team communication and enhancing project coordination.
For contractors seeking a quick and accurate way to estimate project costs, STACK is a viable option. It's designed to speed up the estimation process without compromising on accuracy. Meanwhile,
PlanSwift shines in its ability to automate and streamline the estimating process, reducing manual tasks and increasing productivity.
Choosing the right software for your business depends on your specific needs and objectives. If you're new to the field, our comprehensive guide for beginners on cost estimation can provide valuable
insights. For a more detailed look at the different types of cost estimation, our article on which type of cost estimation is right for your business can be a helpful resource.
Remember, the key to successful cost estimation lies not only in the software you use but also in your understanding of the cost estimation techniques in the construction industry. By combining the
right tools with a solid knowledge base, you can make informed financial decisions and ensure the success of your construction projects.
At Cost Of, we're committed to helping you navigate the complexities of cost estimation. Whether you're a first-time homeowner or a seasoned contractor, our resources are designed to provide accurate
and reliable information to guide your financial decisions. Explore our content today to learn more about cost estimation and its crucial role in construction projects. | {"url":"https://costof.com/quiz/construction-cost-estimation-software-quiz","timestamp":"2024-11-04T23:12:37Z","content_type":"text/html","content_length":"73883","record_id":"<urn:uuid:45adbc0d-03cf-482a-b8dd-2348c31689a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00371.warc.gz"} |
Determining the surroundings of a flight trajectory - Datascience.aero
What does knowing a “trajectory” mean? Does this entail knowing the position of a plane at every minute? Or, more precisely, should we know its position every 30 seconds? 10? The answer to this
problem really depends on the use case, since we have to assume that infinite precision is impossible and that, as a result, we have to identify the minimum sampling ratio for each application.
However, for some particular cases, we can use “tricks”, like not working with the trajectory itself but with its surroundings.
Therefore, given a set of discrete points (remember, we don’t have infinite precision) that represent a trajectory, we may want to know the passed areas near to said trajectory. This information can
be useful for many reasons, like for studying the weather in these surroundings.
A common option is to use grid-based geographic indexing systems, such as Uber’s H3 hexagonal grid system (if you want to know more about it, we have already discussed it here).
But if we work directly with the original discrete points, we may encounter a problem: if two points are too far apart, we end up skipping some cells and gain inconsistent information.
This can be solved by interpolating the trajectory between the real points, which gives us a higher resolution. However, this operation is computationally expensive and results in the total time
being drastically incremented, especially if many flights are processed.
The above trajectory is a clear example of what we have already mentioned. The red markers correspond to the real trajectory and the blue markets to the interpolated points. The gray cells are the
ones obtained with the real trajectory (and therefore can also be obtained with the interpolated trajectory), while the purple cells only appear if we use the interpolated trajectory; we miss them if
we only used the real trajectory.
By zooming into a specific area, we can see what happens: the original red points are located quite close to the missing cells, but since they were not originally connected, they skipped these cells
because of the sampling frequency.
A different approach to solve this problem that doesn’t involve interpolation would involve incrementing cell resolution (using smaller cells) and considering a higher degree of neighbors. Instead of
just having large cells where the point are located (with the gaps corresponding to the purple cells), we would have smaller cells with their neighbors up to a number of degrees (for example, two
degrees of neighbors). This would smooth our result since the gaps would be filled. And most importantly, the computations required for these actions are not intensive, so the processing time
wouldn’t increase as much as with the interpolation. | {"url":"https://datascience.aero/determining-surroundings-flight-trajectory/","timestamp":"2024-11-02T20:34:44Z","content_type":"text/html","content_length":"50248","record_id":"<urn:uuid:2de9ce36-42d7-4835-b506-9fe637ad4bdc>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00682.warc.gz"} |
Fractions on a Number Line
Rational Numbers on the Number Line
Integers (Number Lines, Adding & Subtracting)
Topic 4 Pythagorean Theorem Vocabulary Quiz
Mod 12 - Pythagorean Theorem Review
Compare and Ordering Real Numbers
Real numbers on the number line/irrational numbers
Basic Middle School Math Skills Diagnostic Test
Adding and Subtracting Integers
Irrational & Rational Numbers
Irrational Numbers on the Number Line
Plot, Compare, and Order Fractions
Adding Negative & Positive Integers
Estimating Square Roots on a Number line
Explore Fractions on a Number Line Worksheets by Grades
Explore Other Subject Worksheets for class 8
Explore printable Fractions on a Number Line worksheets for 8th Class
Fractions on a Number Line worksheets for Class 8 are an excellent resource for teachers looking to enhance their students' understanding of fractions and their applications in mathematics. These
worksheets provide a variety of exercises and problems that help students grasp the concept of fractions, as well as their representation on number lines. With the use of Fraction Models, students
can visualize the relationship between different fractions and gain a deeper understanding of their properties. Teachers can incorporate these worksheets into their lesson plans to provide a
comprehensive learning experience for their Class 8 students. By using these Math worksheets, teachers can ensure that their students develop a strong foundation in fractions and are well-prepared
for more advanced mathematical concepts. Fractions on a Number Line worksheets for Class 8 are an indispensable tool for teachers who want to make fractions an engaging and accessible topic for their
Quizizz is an innovative platform that offers a wide range of educational resources, including Fractions on a Number Line worksheets for Class 8, to help teachers create interactive and engaging
learning experiences for their students. In addition to worksheets, Quizizz also provides teachers with access to a vast library of quizzes, games, and other learning materials that cover various
topics in Math, including Fractions and Fraction Models. Teachers can easily customize these resources to suit the needs of their Class 8 students and track their progress through detailed reports
and analytics. With Quizizz, teachers can seamlessly integrate technology into their lesson plans and create a dynamic learning environment that caters to the diverse needs of their students. By
incorporating Quizizz into their teaching strategies, teachers can ensure that their Class 8 students have a solid understanding of fractions and are well-equipped to tackle more complex mathematical
concepts in the future. | {"url":"https://quizizz.com/en/fractions-on-a-number-line-worksheets-class-8?page=1","timestamp":"2024-11-08T14:43:51Z","content_type":"text/html","content_length":"159410","record_id":"<urn:uuid:e6c1e280-5a55-481d-b084-b81daa142bd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00646.warc.gz"} |
Logic and Set Theory
The last section of this unit deals with the Venn Diagram. Instead of using capital letters and brackets to denote sets we shall now use a pictorial method.
In a Venn diagram there is a rectangle which shows the universal set. Circles within the rectangle show subsets of the universal set.
(figure available in print form)
If you have the universal set { 1,2,3,4,5} and Set A= {1,2,3} and set B = {3,4} then the two circles are shown to intersect since 3 is a member of both set A and set B. The diagram above shows the 3
where the circles intersect. Also shown are the 1 and 2 in A only and the 4 in B only. The number 5 is part of the Universal set but is neither an element of A or B.
Draw a Venn diagram to show the following:
1. U= {1,2,3,4,5}
A= {2,3}
B= {1,2,4}
(figure available in print form)
2. U= { 2,4,6,8,12}
A= {2,6,4,12}
B= {2,12}
C= {6}
(figure available in print form)
Sample lesson Plan 2 describes how the Venn diagram can be used to simplify problems dealing with intersection.
The use of diagrams can indeed make what appears to be a difficult problem rather easy. Students should be encouraged to use the Venn diagram as well as other pictures and charts to help make their
word problems and problem solving easier and more meaningful. | {"url":"https://teachersinstitute.yale.edu/curriculum/units/1980/7/80.07.04/15","timestamp":"2024-11-10T00:08:03Z","content_type":"text/html","content_length":"39415","record_id":"<urn:uuid:1d8d8a79-97d7-485e-b7bf-d26356c68c62>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00039.warc.gz"} |
The Stacks project
Lemma 15.31.1. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ be an Koszul-regular sequence. Then the extended alternating Čech complex $R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}} \to \bigoplus
\nolimits _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r}$ from Section 15.29 only has cohomology in degree $r$.
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The number type. This type must fulfill the requirements on FieldNumberType
The point type.
The segment type.
The iso-rectangle type.
Function object. Must provide the operator Point_2 operator()(Segment_2 seg, int i), which returns the source or target of seg. If i modulo 2 is 0, the source is returned, otherwise the target is
Function object. Must provide the operator Segment_2 operator()(Point_2 p, Point_2 q), which introduces a segment with source p and target q. The segment is directed from the source towards the
Function object. Must provide the operator Iso_rectangle_2 operator()(Point_2 left, Point_2 right, Point_2 bottom, Point_2 top), which introduces an iso-oriented rectangle fo whose minimal $x$
coordinate is the one of left, the maximal $x$ coordinate is the one of right, the minimal $y$ coordinate is the one of bottom, the maximal $y$ coordinate is the one of top.
Function object. Must provide the operator double operator()(FT), which computes an approximation of a given number of type FT. The precision of this operation is of not high significance, as it is
only used in the implementation of the heuristic technique to exploit a cluster of kd-trees rather than just one.
Function object. Must provide the operator Comparison_result operator()(Point_2 p, Point_2 q) which returns SMALLER, EQUAL or LARGER according to the $x$-ordering of points p and q.
Function object. Must provide the operator Comparison_result operator()(Point_2 p, Point_2 q) which returns SMALLER, EQUAL or LARGER according to the $y$-ordering of points p and q.
Rounds a point to a center of a pixel (unit square) in the grid used by the Snap Rounding algorithm. Note that no conversion to an integer grid is done yet. Must have the syntax void operator()(
Point_2 p,FT pixel_size,FT &x,FT &y) where $p$ is the input point, pixel_size is the size of the pixel of the grid, and $x$ and $y$ are the $x$ and $y$-coordinates of the rounded point respectively.
Convert coordinates into an integer representation where one unit is equal to pixel size. For instance, if a point has the coordinates $\left(3.7,5.3\right)$ and the pixel size is $0.5$, then the
new point will have the coordinates of $\left(7,10\right)$. Note, however, that the number type remains the same here, although integers are represented. Must have the syntax Point_2 operator()(
Point_2 p,NT pixel_size) where $p$ is the converted point and pixel_size is the size of the pixel of the grid.
Returns the vertices of a polygon, which is the Minkowski sum of a segment and a square centered at the origin with edge size pixel edge. Must have the syntax void operator()(std::list<Point_2>&
vertices_list, Segment_2 s, NT unit_square) where vertices_list is the list of the vertices of the Minkowski sum polygon, $s$ is the input segment and unit_square is the edge size of the pixel.
The following functions construct the required function objects occasionally referred as functors listed above.
Construct_vertex_2 traits.construct_vertex_2_object ()
traits.construct_segment_2_object ()
traits.construct_iso_rectangle_2_object ()
Compare_x_2 traits.compare_x_2_object ()
Compare_y_2 traits.compare_y_2_object ()
Snap_2 traits.snap_2_object ()
traits.integer_grid_point_2_object ()
traits.minkowski_sum_with_pixel_2_object () | {"url":"https://doc.cgal.org/Manual/3.2/doc_html/cgal_manual/Snap_rounding_2_ref/Concept_SnapRoundingTraits_2.html","timestamp":"2024-11-08T18:50:58Z","content_type":"text/html","content_length":"18314","record_id":"<urn:uuid:6884237d-271e-4517-a9f9-be1a51f63735>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00352.warc.gz"} |
Re: write loop over multiple variables and flag if conditions are met
hello! I am trying to write a loop to flag two criteria. First if the difference between lag(timeperiod) - timeperiod >=4 and then to compare if diag codes between time periods is the same of
different. I want to run the loop over each patientID until one of these criteria's are met.
E.g. data
patientID, timeperiod, combined
wh7,1, diag1_diag2_diag3
wh7, 4, diag1_diag2_diag3_diag4
wh7, 10, diag1_diag2
wh7 15, diag4_diag10
wh4, 2, diag5_diag11_diag16
wh4, 4, diag5_diag11
07-11-2022 03:53 PM | {"url":"https://communities.sas.com/t5/SAS-Programming/write-loop-over-multiple-variables-and-flag-if-conditions-are/m-p/822894/highlight/true","timestamp":"2024-11-08T09:49:51Z","content_type":"text/html","content_length":"318583","record_id":"<urn:uuid:ce10b950-c525-4dc5-81c3-35e82946da4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00354.warc.gz"} |
How Many Seconds in a Day？[Answer + Calculation + Converter]
How many seconds in a day? This common question arises in various contexts, from scientific calculations to everyday life. In this article, we will explore the total number of seconds in a day,
explain the calculation process, and provide online tools for quick conversions. Whether you’re a student, a professional, or just curious, this guide will help you understand the concept of time
measurement in seconds.
Part 1: How Many Seconds in a Day?
There are 86,400 seconds in a day.
To understand how many seconds are in a day, let’s break it down.
The Calculation Process
The calculation of seconds in one day has roots in various timekeeping systems used by ancient civilizations. While our modern concept of time is based on the atomic clock, which is perfectly
consistent, historical methods relied on the solar day and the Earth’s rotation. A solar day is approximately 24 hours, but due to leap years and the sidereal day, which is slightly shorter, the
number of seconds can vary.
Therefore, the calculation is straightforward:
Step 1. Understanding the Units:
One hour consists of 60 minutes, and one minute contains 60 seconds.
Step 2. Calculate Seconds in One Hour:
Since there are 60 seconds in one minute, we can calculate the number of seconds in one hour as follows:
Step 3. Calculate Seconds in One Day:
There are 24 hours in one day. Therefore, to find the total number of seconds in a day, we multiply the number of seconds in one hour by the number of hours in a day:
This means there are 86,400 seconds in a day.
Why Are there 86,400 Seconds in a Day？
The reason there are 86,400 seconds in a day can be understood through several key concepts related to time measurement and the Earth’s rotation. A day is traditionally defined as the time it takes
for the Earth to complete one full rotation on its axis, resulting in the cycle of day and night.
In the 19th century, the development of accurate pendulum clocks and later atomic clocks allowed scientists to measure the Earth’s rotation more precisely. Through these advancements, it was
determined that the length of a day is not exactly 86,400 seconds but varies slightly due to factors such as gravitational interactions and the Earth’s axial tilt.
To break this down further:
1. Seconds in one Minute: There are 60 seconds in one minute.
2. Minutes in one Hour: There are also 60 minutes in one hour. Therefore, the calculation for the total number of seconds in one day is:
1. Mean Solar Day: The mean solar day is the average length of one day, accounting for variations due to the Earth’s elliptical orbit and axial tilt. The mean solar day averages out to 24 hours,
which is the basis for our timekeeping.
2. Current Definition: The current definition of a second is based on atomic clocks, which measure time with incredible precision by counting the vibrations of atoms in their ground state. This
method allows for accurate timekeeping that aligns closely with the Earth’s rotation.
3. Leap Year and Leap Seconds: To maintain synchronization with the Earth’s position in its orbit, we occasionally add an extra day (February 29) in a leap year. Additionally, leap seconds may be
introduced to account for variations in the Earth’s rotation speed, ensuring that our clocks remain aligned with the mean solar time.
Part 2: Online Calculator
Finding out how many seconds in a day can be simplified with online calculators. If you need quick conversions, here are three recommended tools. Using these tools, you can easily determine how many
seconds are in any given time frame, enhancing your understanding of time measurement.
UnitConverters.net – Convert seconds to days
URL: http://UnitConverters.net
This website offers conversions between various units, including time units. Users can quickly convert seconds to days by inputting a value. It features a user-friendly interface and supports a wide
range of unit conversions, such as length, weight, and temperature.
Gallery – How many seconds in a day?
URL: https://coda.io/@hales/simple-online-calculator-for-dates-and-times/how-many-seconds-in-a-day-33
This online calculator focuses on calculating the number of seconds in a day, specifically showing that there are 86,400 seconds in a day. It explains how to derive this result through mathematical
INCH CALCULATOR – Seconds to Days Converter
URL: https://www.inchcalculator.com/convert/second-to-day/
This site provides a dedicated converter for seconds to days, allowing users to input a number of seconds to get the corresponding days. It also offers detailed explanations of the conversion process
and examples.
Part 3: Solved Examples on How Many Seconds in a Day
To clarify the concept further, here are some examples that illustrate how many seconds correspond to various time units.
Example 1: How Many Seconds in 2 Days?
Question How many seconds are in 2 days?
Answer 2days×86,400seconds/day=172,800seconds
Explanation Since there are 86,400 seconds in one day, multiplying by 2 gives the total number of seconds in two days.
Example 2: How Many Seconds in 3 Hours?
Question How many seconds are in 3 hours?
Answer 3hours×3,600seconds/hour=10,800seconds
Explanation Each hour has 3,600 seconds. Therefore, multiplying 3 hours by 3,600 seconds gives the total seconds in three hours.
Example 3: How Many Seconds in 15 Minutes?
Question How many seconds are in 15 minutes?
Answer 15minutes×60seconds/minute=900seconds
Explanation Since there are 60 seconds in each minute, multiplying 15 minutes by 60 seconds gives the total number of seconds in fifteen minutes.
Example 4: How Many Seconds in a Week?
Question How many seconds are in one week?
Answer 1week×604,800seconds/week=604,800seconds
Explanation There are 604,800 seconds in a week (7 days × 86,400 seconds). Thus, one week equals 604,800 seconds.
Example 5: How Many Seconds in a Month (30 Days)?
Question How many seconds are in 30 days?
Answer 30days×86,400seconds/day=2,592,000seconds
Explanation Multiplying the number of seconds in a day (86,400) by 30 gives the total seconds in a 30-day month.
If your child has problems with the above calculation process, you can take the WuKong Mathematics online 1-on-1 course for free to learn these mathematical calculation processes.
Part 4: How Many Seconds are in Various Time Units? (Conversion Chart)
Understanding time measurement requires knowing how to convert between different units and other units that measure time. Here’s a brief introduction to some common conversions.
• 1 Minute: 60 seconds
• 1 Hour: 3,600 seconds
• 1 Day: 86,400 seconds
• 1 Week: 604,800 seconds
• 1 Year: 31,536,000 seconds
Conversion Chart
Unit of Time Seconds Minutes Hours Days Weeks Months* Years* Decades* Centuries*
1 Second 1 second 1/60 minute 1/3600 hour 1/86400 day 1/604800 week 1/2,592,000 month 1/31,536,000 year 1/315,360,000 decade 1/3,153,600,000 century
1 Minute 60 seconds 1 minute 1/60 hour 1/1440 day 1/10080 week 1/43,200 month 1/525,600 year 1/5,256,000 decade 1/52,560,000 century
1 Hour 3600 seconds 60 minutes 1 hour 1/24 day 1/168 week 1/720 month 1/8,760 year 1/87,600 decade 1/876,000 century
1 Day 86400 seconds 1440 minutes 24 hours 1 day 1/7 week 1/30 month 1/365 year 1/3,650 decade 1/36,500 century
1 Week 604800 seconds 10080 minutes 168 hours 7 days 1 week 1/4.34812 month 1/52.1775 year 1/521.775 decade 1/5217.75 century
1 Month 2,592,000 seconds 43,200 minutes 720 hours 30 days 4.34812 weeks 1 month 1/12 year 1/1.2 decade 1/120 century
1 Year 31,536,000 seconds 525,600 minutes 8,760 hours 365 days 52.1775 weeks 12 months 1 year 1/10 decade 1/100 century
1 Decade 315,360,000 seconds 5,256,000 minutes 87,600 hours 3,650 days 521.775 weeks 120 months 10 years 1 decade 1/10 century
1 Century 3,153,600,000 seconds 52,560,000 minutes 876,000 hours 36,500 days 5,217.75 weeks 1200 months 100 years 10 decades 1 century
Key Tips for Time Conversions
1. Always multiply or divide based on the base unit (seconds).
2. Use online calculators for quick and accurate conversions.
3. Familiarize yourself with common conversions for efficiency.
FAQ about How Many Seconds in a Day
Q1: How Many Days in a Million Seconds?
A: There are approximately 11.57 days in one million seconds (1,000,000 seconds ÷ 86,400 seconds/day).
Q2: How Many Seconds Are There in a Week?
A: There are 604,800 seconds in a week (7 days × 86,400 seconds).
Q3: How Many Seconds Are There in a Year?
A: There are about 31,536,000 seconds in a year (365 days × 86,400 seconds).
Q4: How Many Seconds in a Month?
A: On average, there are about 2,592,000 seconds in a month (30 days × 86,400 seconds).
Q5: How Many Seconds in One Hour?
A: There are 3,600 seconds in one hour. This is calculated based on the following relationship between time units:
Step 1. Define the Units: Let S be the number of seconds in one hour.
Step 2. Set Up the Equation:
Step 3. Substitute the Values:
Step 4: Perform the Multiplication
Q6: What’s the Definition of a Second and a Day?
1. What is a Second?
A second is the base unit of time in the International System of Units (SI) and is defined based on the vibrations of cesium atoms in an atomic clock.
2. What is a Day?
A day is traditionally defined as the time it takes for the Earth to complete one full rotation on its axis, resulting in the cycle of day and night.
How many seconds in a day? There are 86,400 seconds in a day, a fundamental concept in time measurement. This article has explored the calculation process, provided online tools, solved examples, and
offered a conversion chart. Understanding how to convert between time units enhances our grasp of time and its measurement in everyday life. Learn more about How Many Hours in a Month.
Discovering the maths whiz in every child,
that’s what we do.
Suitable for students worldwide, from grades 1 to 12.
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A Bite at the Apple: The Arithmetic of Minority Teachers
Both Democrats and Republicans like to talk about increasing the number of minority teachers. The stated reasons vary, but let’s get right to the specifics: in Minnesota, an education coalition is
asking the state for $80m to help increase the number of teachers of color. Their goal is to bump representation from 4% to 5% statewide, or by about 630 of the state’s approximately 63,000 teachers.
To be sure, it looks like only a fraction of this would get funded, at least given the budget proposals from the governor ($16m) and the House ($37m), and also given the fact that the Republican-led
Senate has other plans. But still: $16m, $37m—this is a lot of real money that our politicians are offering up, at least to start negotiations. And for what?
Let’s start with some simple arithmetic. $80,000,000 divided by 630 new teachers is over an eighth of a million dollars—about $127,000—per supposed new teacher. That’s enough to pay for all of these
630 teachers’ salaries for two or three years. (Or, every teacher’s salary and benefits for almost two years.) Regardless, that’s a lot of money to spend on recruitment and grants, even if it were
funded at about fifty cents on the dollar, as the Democrat/DFL-controlled House proposed.
But that’s not the point, really. The whole discussion of the number of minority teachers seems to ignore a couple of key facts, and uncontroversial facts, at that. First, different racial groups
have different birth rates, which impacts the ratio of adults to children over generations. Second, teachers often stay in the profession for a long time, meaning that, even in a perfect world,
racial composition of teachers will not match a society with changing demographics.
These two factors, hidden in plain sight, are sufficient to explain why there are so few people of color in front of Minnesota’s classrooms. Most interesting, these reasons don’t involve factors like
systemic racism or lack of educational opportunities for minorities. Friends, it’s simpler than that. (And I do mean simple—I’m doing some quick, back-of-the-envelope calculations here, but you’ll
see the bigger point, regardless.)
Let’s start with the bigger of our two issues: demographic trends. Census data show that Minnesota has been getting significantly less white for decades, to the tune of about 4% less every Census.
(White people made up 94% of the population in 1990, 89% in 2000, and 85% in 2010.) Minnesota Compass says that about 20% of Minnesotans are people of color today, which fits the trend.
Of course, patterns like this point to birth rates. Non-white Minnesotans are reproducing faster than white ones, perhaps for a variety of reasons, but perhaps only for the well-documented reason
that lower-income people have more children. Indeed, Minnesota Compass says that people of color make up 6% of retirees (65+), but fully 31% of preschoolers (0-4). That means that, despite being only
20% of the population, people of color account for at least 30% of the births in Minnesota today.
This is the landscape for our discussion of teacher shortages. The population is growing, and groups of color are growing the fastest. It’s unsurprising, then, that people of color have the direst
shortages of teachers: compared to white people, they’ve got fewer adults (the teachers) per kid (the students).
The effects of such demographic change appear quickly, too. Imagine a young person of color who wants to be a teacher. Say she’s a sophomore in college, and has just decided to major in education.
That means she was born in about the year 2000. (Feel old? Join the club.)
Given the trends described above, let’s figure that in the year 2000, birth rates for white people were in the mid-80s, percentage-wise, meaning that birth rates of people of color were about 15%.
That means that, even if “millennial babies” of all colors were equally likely to become teachers, the fresh crop of teachers nowadays would be 15% of color.
Compare this to the demographic of kids today. Imagine that our newbie teacher wants to teach kindergarten in a couple of years. As discussed, that kindergarten class will be about 30% of color. That
means that the percentage of new kindergarten teachers of color will differ from students of color by a factor of two (15% people of color then vs. 30% today).
In other words, accounting only for birth rates, we already have a shortage of teachers of color compared to the population as a whole, and a significant shortage, at that. Again, this calculation
does not rely on any form of racism or oppression, whether in our education system or during the recruitment and training of teachers. Maybe those things are real, and important. But these numbers
show significant divergence by relying only on obvious demographic facts backed by hard data.
But now let’s add another factor: the average age of teachers. By and large, teaching is still a stable profession, akin to the old model of “forty years and a gold watch.” Yes, teachers may be
leaving the profession at a higher rate as of late, but overall, teachers are often career teachers. As evidence, consider the glut of baby boomers still in the profession, whose retirements are
exacerbating teaching shortages state- and nationwide.
Teacher age compounds the disparity between the racial compositions of the teachers and students. That is, if the new crop of 22-year-old teachers looks different than today’s students, consider how
different the older crops of teachers look.
Let’s simplify and say that teachers often teach for forty years, from the time they’re 22 to 62. The median age of teachers in this case would be somewhere around 42. That means that the median
teacher in Minnesota was born, say, sometime in the late 1970s, when people of color represented only about 3-3.5% of Minnesotans.
(Of course, there are other factors at play here. Two examples: the population has grown over time, meaning the number of teachers has increased the most in recent years, meaning that the
demographics probably shifted to more diverse crops of teachers. However, teacher shortages may have been more prevalent in later years, too, meaning that older teachers [and their demographics]
might be a bit overrepresented. For our simple calculations here, we’ll call it a wash—we’re just trying to make a broader point, anyway.)
So, the typical teacher statewide was born at a time when the state was… let’s be generous… 4% people of color. And now we want to spend millions on increasing the percentage of minority teachers
from… 4% people of color?
Seems like a stretch.
And that doesn’t consider other factors that also depress the number of minority teachers, like the fact that minorities as a whole are less likely to go to college at all, let alone graduate from
college, let alone graduate with a degree in education. In this light, the fact that we have 4% teachers of color today is pretty impressive, considering some educational hurdles they face.
We want to spend more money to fix this… problem? Thinking demographically, is this even a “problem” at all?
Should the state spend money to recruit members of a profession, even a state-mandated profession like teaching, in order to match current demographics? That seems to be an uphill battle. Regardless,
it would probably be a losing battle. As mentioned, even if our youngest teachers represented Minnesota’s demographics when they were born, they wouldn’t represent Minnesota now. Also as mentioned,
even those youngest teachers would misrepresent people of color by a factor of two.
As a side note, demographic trends like this are literally unrelated to whether the underrepresented group is even in the minority. Imagine that white people, all of a sudden, increased their birth
rates to exceed the rates of people of color. In such a case, white people—despite being in the majority—would be underrepresented in Minnesota’s classrooms. As a simple case, imagine that white
people are 80 percent of Minnesotans today, but started having 90% of children. In five years, the kindergartners would be 90% white, but the new teachers would be about 80% white. It’s the same
thing: whoever reproduces the fastest will flood the pool of new students, skewing the ratios compared to all teachers born generations earlier.
The government wouldn’t step in then, obviously. Nor should it.
So, people of color aren’t underrepresented in classrooms because they’re minorities, necessarily—they’re underrepresented in no small part because they are reproducing fastest. And that’s not the
state’s business. It can’t be. I mean, where would that stop?
As long as different racial groups have different birth rates, and as long as teachers teach for a long time, and as long as teachers don’t start teaching until their mid-twenties, we’ll
always—always—have disparities between the demographics of teachers and students. Even if this is a problem, it’s an inevitable “problem” that has no “solution.”
Proposals that aim to “solve” mere demographic ratios—that aim to “solve” the inevitable—shouldn’t be in anyone’s budget. Regardless how well-meaning they are, they just don’t add up.
To any number of millions.
P. A. Jensen is editor of RuralityCheck.com.
He lives in northern Minnesota with his wife and son. | {"url":"https://www.ruralitycheck.com/p/a-bite-at-the-apple-the-arithmetic-of-minority-teachers","timestamp":"2024-11-09T05:50:41Z","content_type":"text/html","content_length":"145222","record_id":"<urn:uuid:0e2b34fa-23a1-440a-a44c-6197584d0e57>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00224.warc.gz"} |
Expressions List | Rainbird
You can use a variety of functions that can be used when writing an expression to compare or transform data during the processing of a rule.
The following functions are currently supported within an expression:
Comparison functions
Comparing data
These will compare one instance to another and will evaluate to either True or False.
These operators are not available for all data types. The table below shows what operators can be used with what data. Operators support either a symbol representation (e.g. =) or a natural language
equivalent. For comparing dates, see the date functions.
Operator (and aliases) String (text) Number Date Truth
Comparing lists
These will compare one list of instances against another list and evaluate to either True or False.
The isSubset function enables you to confirm whether items in one list are present in another.
For example, you could check that the skills required for a job role is a subset of the skills listed for a candidate.
To do this, the function needs to be told where to gather this data from. This is done by specifying which two relationships contain this information and which side of the relationship (the subject
or object) will be compared. These relationships should be plural. Using the example above, we might have a rule that looks like this:
Rule: Candidate (%S) > is suitable for > Job role (%O)
Expression: isSubset(%O, has essential skills, *, %S meets, *)
At runtime, if we had a candidate of Simone and the rule was processing a job role of Retail Financial Advisor, the expression would do the following to determine that the 1st list is a subset of the
2nd list and return True from the expression.
Follow our Academy course on comparative expressions for detailed examples and how you can combine multiple comparative expressions.
Mathematical functions
The following mathematical functions are available to use with number concepts only.
Data type of output: number
Mathematical functions can be combined using brackets.
Mathematical functions can be combined using brackets. Where mathematical calculations are being performed you should observe correct usage of brackets, otherwise the calculation will be read left to
Date functions
Use within an expression to compare or transform dates.
These functions allow you to:
Retrieve the current date/time
Get an index for a given date
Add or subtract from a date
Calculate the difference between dates
Compare two dates to determine if one is in the past, present or future from the other
Inputs to these functions must come from a concept set to a date type. An error will be displayed if you try to use these on data that is from a string, number or truth concept.
Date/time format: YYYY-MM-DD HH:MM:SS
These functions may output timestamps, dates, numbers or true/false. The output data type will be mentioned for each function.
Retrieve the current date/time
Functions to get the current date or date/time to use within other functions. For example to work out a person's age you can write an expression to calculate the number of years between a persons
date-of-birth and today's date.
Data type of output: date/time
These functions will output a unix timestamp, which is a machine readable datetime format. Mostly you will not see these timestamps. However, if you do come across them during building and want to
see it in a human-friendly format there’s many websites you can use to convert them such as epoch converter.
Get an index for a given date
Functions that take a date and return an index (number) based on the chosen function.
Data type of output: number
Add or subtract from a date
Add or remove a number of days, weeks, months or years from a date to create a new date.
This function requires you to pass in a date followed by a number you want to add. e.g. addDays(2023-12-22, 3) = 2023-12-25.
To subtract from a date you can use a negative number. e.g. addDays(2023-12-22, -3) = 2023-12-19
These examples use static data, but both arguments of this function can be dynamic by using variables, so long as the data used is of the correct type. E.g. addDays(%date, %number).
Data type of output: date/time
These functions output a timestamp. If the output of this function is assigned to the object of the rule (%O) then the object should be a date type. If the object is a number, the timestamp will be
displayed in the result and/or the evidence.
Calculate the difference between dates
Returns the difference between two dates as a number.
This function requires you to pass in two dates. Regardless of the order, it will always output a positive number.
Data type of output: number
Comparing dates
Pass two dates into the function to check if one is in the past, present or future from the other.
These functions will check the dates and evaluate to true or false for the expression to either pass or fail.
Data type of output: True/false | {"url":"https://docs.rainbird.ai/rainbird/knowledge-modelling/modelling-features/relationships/rules/expressions-list","timestamp":"2024-11-13T06:26:40Z","content_type":"text/html","content_length":"919563","record_id":"<urn:uuid:0693d7c2-510f-40e7-90dc-e92e0dd717b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00404.warc.gz"} |
how can I know the private and public key of the node ?
An address is just a hash of a public key, so it's more compact for sending over email, etc...
You can see the public and private key for an address in the wallet using the validateaddress and dumpprivkey commands.
I used validateaddress but it returns only public key
"isvalid" : true,
"address" : "1FyS12GHVX7HS4wXgRfEy7BZwGDTtTbZydBKCM",
"ismine" : true,
"iswatchonly" : false,
"isscript" : false,
"pubkey" : "020d3f8502cf86111bd3ddf6f8d8e80e1ae89ccfb936533dec5723f8a7dea92bd3",
"iscompressed" : true,
"account" : "",
"synchronized" : true | {"url":"https://www.multichain.com/qa/48100/how-can-i-know-the-private-and-public-key-of-the-node","timestamp":"2024-11-03T13:44:02Z","content_type":"text/html","content_length":"29638","record_id":"<urn:uuid:46d0a6ca-a18e-4551-9955-435489af6c86>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00402.warc.gz"} |
How To Calculate Volume analysis using Clojure?
Volume analysis typically refers to analyzing the trading volume of a security over a specific period to identify trends and make trading decisions. In Clojure, you can calculate the volume of a
security by summing the trading volume for each period.
Here is a step-by-step guide on how to calculate volume analysis using Clojure:
1. Define a list of trading volume data for the security over a specific period. For example, you can create a vector of trading volume data for each day or time period:
1 (def trading-volume [10000 15000 12000 20000 18000])
1. Calculate the total trading volume by summing all the trading volume data in the list. You can use the reduce function to sum the elements in the list:
1 (def total-volume (reduce + trading-volume))
1. Calculate the average trading volume by dividing the total trading volume by the number of data points in the list:
1 (def avg-volume (/ total-volume (count trading-volume)))
1. Calculate the maximum and minimum trading volume in the list using the apply function with max and min:
1 (def max-volume (apply max trading-volume))
2 (def min-volume (apply min trading-volume))
1. Print out the calculated values:
1 (println "Total Volume:" total-volume)
2 (println "Average Volume:" avg-volume)
3 (println "Maximum Volume:" max-volume)
4 (println "Minimum Volume:" min-volume)
By following these steps, you can calculate volume analysis using Clojure. You can further analyze the trading volume data by comparing it with price movements or other technical indicators to make
informed trading decisions. | {"url":"https://forum.finquota.com/thread/how-to-calculate-volume-analysis-using-clojure","timestamp":"2024-11-09T23:03:52Z","content_type":"text/html","content_length":"122043","record_id":"<urn:uuid:d006388a-6bff-4564-aa93-c627071e2061>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00184.warc.gz"} |
Analysis Projects
Analysis Projects for New Members
[A0] Project #A0 - Download MATLAB
[A1] Project #A1 - MATLAB and Coding Basics
• Open MATLAB (orange and blue conical icon on bottom). Also, go to the EMC Google Drive page and "$ Student Reports Summer 2016". This will the be the folder containing all the materials for the
remaining projects. Go to "Project #A1".
• If you are new to MATLAB or would like a refresher before you start, go to: http://www.mathworks.com/help/matlab/getting-started-with-matlab.html?s_cid=learn_doc and look through the tutorials to
the desired level of detail.
• Open the Excel file called “Sample_NEMO_Output.” This is a typical output from our main analysis program, containing the (x,y) coordinates of 21 evenly spaced points along the worm (worm is split
into 20 even segments and coordinates of nodes are stored).
• Find the distance between two arbitrary points (x1,y1) and (x2,y2). (For all the assignments in A1 you may choose to write a MATLAB function to perform the task, or do it through your Command
• It may be helpful to look through these spreadsheet-specific tutorials/documentation before you proceed with the assignment: http://www.mathworks.com/help/matlab/spreadsheets.html
• Estimate the length of a worm for a single frame (any frame is fine).
• Relevant documentation for next portion: http://www.mathworks.com/help/matlab/2-and-3d-plots.html
• Now find the average length of a worm over a time period (multiple frames) and plot length as a function of time in that time period. Include error bars, and please label the plots and display
the units of measurement.
[A2] Project #A2 - Image Processing with Matrices
• Read this helpful article on images in MATLAB: http://www.mathworks.com/help/matlab/creating_plots/working-with-images-in-matlab-graphics.html, and then look around in the more general image
documation here: http://www.mathworks.com/help/matlab/images_btfntr_-1.html
• Make an 800X800 pixels 2D binary image of a white circular disk of radius 200 pixels on black background.
• Go to “A2”. It should contain a binary image and a binary video.
• Find the centroid of the binary worm image, then convert this to spatial coordinates in mm (1mm=320 pixels) with the lower left corner of image as origin (as opposed to top left for row column
image coordinates).
• Using the 2D binary video of a worm, plot the positions (X vs Y) of the worm’s center of mass in mm, this display is referred to as the track of the worm for the video. Account for the moving
stage with the given stage coordinates of the respective video. The NEMO function which incorporates the stage data (you may want to look to this for guidance, is called segment_data and the
stage data is stored in the array poffset).As always, please label the plots and display the units of measurement.
[A3] Project #A3 - Data Fitting
• Helpful reading: http://www.mathworks.com/discovery/data-fitting.html
• Go to "A3". It should contain a single sample laser data file, and a folder.
• With the sample laser intensity data (outside the folder), show the data and the fitted Gaussian on the same plot. Remember to always put units of measurement!
• Extract all the possible parameters from your Gaussian fit and output then in an excel file.
• Using the multiple laser intensity data (varied over wavelength), produce a plot of each of the extracted parameters as a function of wavelength.
[A4] Project #A4 - Choosing your Program
• At this point, you have completed the general data analysis training and have developed skills that will come in handy in your scientific career.
• Speak with a senior analysis team member to learn about the specific software we use in our lab, and you may begin getting familiar with anything that interests you.
Congratulations! You have survived after extensive training to become an official member of the analysis team of the Elegant Mind Club. Now you are fully prepared to start your own project. Please go
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An ellipsoid has radii with lengths of 2 , 8 , and 5 . A portion the size of a hemisphere with a radius of 3 is removed form the ellipsoid. What is the volume of the remaining ellipsoid? | HIX Tutor
An ellipsoid has radii with lengths of #2 #, #8 #, and #5 #. A portion the size of a hemisphere with a radius of #3 # is removed form the ellipsoid. What is the volume of the remaining ellipsoid?
Answer 1
The volume is $= 278.6 {u}^{3}$
Volume of ellipsoid is #=4/3piabc#
Volume of henisphere is #=2/3pir^3#
Remaining volume #=4/3piabc-2/3pir^3#
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Answer 2
To find the volume of the remaining ellipsoid after removing a hemisphere, use the formula for the volume of an ellipsoid:
Volume = (4/3) * π * a * b * c
Where 'a', 'b', and 'c' are the semi-axes lengths of the ellipsoid. In this case, 'a' = 2, 'b' = 8, and 'c' = 5.
First, calculate the volume of the entire ellipsoid using the given semi-axes lengths. Then, calculate the volume of the removed hemisphere with a radius of 3.
Finally, subtract the volume of the hemisphere from the volume of the entire ellipsoid to find the volume of the remaining ellipsoid.
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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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NPUs and TPUs
Neural processing units (NPUs) and tensor processing units (TPUs) are specialized hardware accelerators that are designed to accelerate machine learning and artificial intelligence (AI) workloads.
NPUs and TPUs are optimized for the mathematical operations that are commonly used in machine learning, such as matrix multiplications and convolutions, and they can be used to accelerate a wide
range of machine learning tasks, including image classification, object detection, natural language processing, and speech recognition.
Both NPUs and TPUs are highly efficient and powerful resources for machine learning, but they do have some limitations, such as availability, compatibility, cost, and flexibility. In general, NPUs
and TPUs are important tools that can be used to improve the performance and efficiency of machine learning applications.
• Neural processing units (NPUs) and tensor processing units (TPUs) are both types of hardware accelerators that are designed to accelerate machine learning and artificial intelligence (AI)
workloads. Both NPUs and TPUs are optimized for the mathematical operations that are commonly used in machine learning, such as matrix multiplications and convolutions, and they can be used to
accelerate a wide range of machine learning tasks.
• There are some differences between NPUs and TPUs. One key difference is that TPUs are specifically designed to accelerate deep learning tasks, while NPUs can accelerate a broader range of machine
learning algorithms. TPUs are also developed by Google and are only available on the Google Cloud Platform, while NPUs can be developed and used by any company or organization.
• In terms of performance, both NPUs and TPUs are highly efficient and powerful resources for machine learning. However, TPUs may have a slight performance advantage due to their specific
optimization for deep learning tasks. It is also worth noting that the specific performance of an NPU or TPU will depend on its design and implementation.
• Overall, both NPUs and TPUs are valuable resources for machine learning and AI, and they can be used to improve the performance and efficiency of machine learning applications. The choice between
an NPU and a TPU will depend on the specific needs and goals of the application and the available resources.
What is Neural Processing Unit (NPU)?
An NPU, or Neural Processing Unit, is a type of specialized hardware accelerator that is designed to perform the mathematical operations required for machine learning tasks, particularly those
involving neural networks. NPUs speed up the training and inference phases of deep learning models, allowing them to run more efficiently on many devices.
NPUs are similar to other hardware accelerators, such as GPU (Graphics Processing Unit) and TPU (Tensor Processing Unit), but they are specifically optimized for tasks related to artificial neural
networks. They are typically used with a central processing unit (CPU) to provide additional processing power for machine learning tasks.
NPUs can be found in a variety of devices, including smartphones, tablets, laptops, and other types of computing devices. They are often used to improve the performance of machine learning
applications such as image and speech recognition, natural language processing, and other types of artificial intelligence workloads.
NPUs are optimized for the mathematical operations commonly used in machine learning, such as matrix multiplications and convolutions. These operations are used in many machine learning algorithms,
including deep learning algorithms, which are a type of neural network that is made up of multiple layers of interconnected nodes.
Matrix multiplications and convolutions are used to process and analyze large datasets, and they are computationally intensive operations that require a lot of processing power. NPUs are designed to
efficiently execute these operations, making them well-suited for machine learning tasks that involve large amounts of data.
In addition to matrix multiplications and convolutions, NPUs can also support other types of mathematical operations, such as element-wise operations and activation functions. Element-wise operations
involve applying a mathematical operation to each element in an array or matrix, and activation functions are used to introduce nonlinearity in neural networks.
NPUs can also support other types of machine learning algorithms, such as support vector machines (SVMs) and decision trees, which involve different types of mathematical operations. Intel Nervana
processor is below.Â
Source: Intel Newsroom
• High performance: NPUs are designed to be highly efficient and performant, allowing them to speed up the training and inference phases of deep learning models.
• Specialized design: NPUs are specifically optimized for tasks related to artificial neural networks, such as image and speech recognition, natural language processing, and other machine learning
• Power efficiency: NPUs are designed to be power efficient, allowing them to run for long periods without consuming much power.
• Hardware acceleration: NPUs can accelerate the performance of machine learning tasks, providing a significant boost in performance compared to using a CPU alone.
• Flexibility: NPUs can be used in a variety of devices, including smartphones, tablets, laptops, and other types of computing devices, making them versatile hardware accelerators.
While NPUs have many benefits and can significantly improve the performance of machine learning tasks, they do have some limitations:
• Limited availability: NPUs are not as widely available as other hardware accelerators, such as GPUs (Graphics Processing Units). This means that not all devices may have an NPU available for use.
• Compatibility: NPUs may not be compatible with all machine learning software and frameworks. For example, some NPUs may only be compatible with certain types of neural network architectures or
may require the use of specific software libraries.
• Cost: NPUs can be expensive to produce, which may make them cost-prohibitive for some users.
• Complexity: NPUs can be complex to design and implement, requiring specialized knowledge and expertise.
• Limited scalability: NPUs may not be able to scale as easily as other types of hardware accelerators, such as GPUs, which can be used in distributed computing environments. This may limit their
ability to handle very large and complex machine-learning tasks.
What is Tensor Processing Unit (TPU)?
A tensor processing unit (TPU) is a specialized hardware designed to accelerate machine learning and artificial intelligence (AI) workloads. It is a type of accelerator that is specifically optimized
to perform the mathematical operations that are used in deep learning algorithms, which are a type of neural network that is made up of multiple layers of interconnected nodes.
TPUs are designed to be highly efficient at executing matrix multiplications and convolutions, two of the most computationally intensive operations in deep learning. They are also able to support
other types of mathematical operations, such as element-wise operations and activation functions, and they can be used to accelerate a wide range of machine learning tasks, including image
classification, object detection, natural language processing, and speech recognition.
TPUs are developed by Google to accelerate the training and inference of deep learning models on the Google Cloud Platform. They are an important part of Google’s infrastructure for machine learning
and AI, and they have been used to train some of the largest and most accurate deep learning models. Here is a Google TPU.
Source: Google Cloud Blog
Tensor processing units (TPUs) are specialized pieces of hardware that are designed to accelerate machine learning and artificial intelligence (AI) workloads. Some of the key features of TPUs
• High performance: TPUs are optimized for the mathematical operations commonly used in deep learning algorithms, such as matrix multiplications and convolutions. They can perform these operations
quickly, making them an efficient resource for training and running machine learning models.
• Energy efficiency: TPUs are designed to be energy efficient, which makes them well-suited for large-scale machine learning tasks that require a lot of computing power.
• Scalability: TPUs can be used individually, or they can be connected to form a TPU pod, which is a group of TPUs that can be used to scale up machine learning tasks. This allows users to scale
their machine learning workloads up or down as needed.
• Programmability: TPUs can be programmed using TensorFlow, an open-source machine learning framework developed by Google. This makes it easy for developers to build and deploy machine learning
models on TPUs.
• Custom architecture: TPUs have a custom architecture optimized for machine learning workloads. They have a high memory bandwidth and are designed to support a large number of concurrent
operations, which makes them well-suited for running machine learning models.
While TPUs are very powerful and efficient at executing the mathematical operations that are commonly used in deep learning algorithms, they do have some limitations. Some of the main limitations of
TPUs include:
• Availability: TPUs are developed by Google and are currently only available on the Google Cloud Platform. This means that they are not widely available to users who do not have access to the
Google Cloud Platform.
• Compatibility: TPUs are designed to work with TensorFlow, an open-source machine learning framework developed by Google. This means they may not be compatible with other machine-learning
frameworks or libraries.
• Cost: TPUs are specialized and powerful pieces of hardware, which can be more expensive than other types of hardware.
• Flexibility: TPUs are optimized for deep learning tasks and are not as flexible as other types of hardware that are more general-purpose. This means they may not be well-suited for tasks that do
not involve deep learning.
• Limitations on the types of models: TPUs are optimized for deep learning models, and they may not be as effective at running other types of machine learning models.
What is Matrix Multiplication?
In mathematics, matrix multiplication is a binary operation that takes two matrices as inputs and produces another matrix as output. It is defined as follows: given two matrices A and B, the matrix
product C is a matrix such that its entry in row i and column j is the dot product of row i of matrix A and column j of matrix B.
For example, suppose that A is a 3×2 matrix and B is a 2×3 matrix. The matrix product C would be a 3×3 matrix, and its entries would be computed as follows:
C[i][j] = sum(A[i][k] * B[k][j]) for all k in the range 0 to 1
In this example, the dot product of row i of matrix A and column j of matrix B would be computed for each value of k from 0 to 1, and the resulting products would be summed to produce the entry
in row i and column j of matrix C.
Matrix multiplications are a fundamental operation in linear algebra and are widely used in many fields, including machine learning, computer graphics, and scientific computing. They are
computationally intensive operations that require a lot of processing power, and they are often accelerated using specialized hardware such as NPUs.
What is Convolution?
In mathematics, convolution is a mathematical operation that combines two functions to produce a third function. It is defined as the integral of the product of the two functions after one is
reversed and shifted.
In machine learning, convolution is a type of operation that is used to extract features from data. It involves applying a small matrix called a “kernel” or “filter” to the data and computing the dot
product of the kernel with a small region of the data. This process is repeated for every possible data region, and the resulting dot products are used to create a new set of features.
Convolution is often used in image processing and computer vision tasks, where it is used to detect patterns and features in images. For example, a convolutional neural network (CNN) is designed to
process data using convolutions. CNN’s are commonly used for tasks such as image classification, object detection, and segmentation. They are made up of multiple layers of interconnected nodes, and
each layer applies a set of convolutions to the data to extract features.
Convolution is a computationally intensive operation that requires a lot of processing power, and it is often accelerated using specialized hardware such as NPUs.
Vendor Ecosystem
Several vendors provide neural processing units (NPUs) and tensor processing units (TPUs). Some vendors that provide NPUs include:
• Intel
• Qualcomm
• Huawei
• Samsung
• MediaTek
Tensor processing units (TPUs) are developed by Google and are only available on the Google Cloud Platform.
It is worth noting that the specific NPU or TPU offerings from these vendors may vary in terms of features, performance, and compatibility. It is advisable to research the specific offerings from
each vendor to determine which one is the best fit for your needs and goals.
In conclusion, neural processing units (NPUs) and tensor processing units (TPUs) are specialized hardware accelerators that are designed to accelerate machine learning and artificial intelligence
(AI) workloads. They are optimized for the mathematical operations that are commonly used in machine learning, such as matrix multiplications and convolutions, and they can be used to accelerate a
wide range of machine learning tasks. Both NPUs and TPUs are highly efficient and powerful resources for machine learning, but they do have some limitations.
As machine learning and AI continue to evolve, likely, NPUs and TPUs will also evolve to become even more powerful and efficient. There is ongoing research and development in hardware acceleration
for machine learning, and new technologies and approaches will likely be developed in the future. It is also possible that NPUs and TPUs will become more widely available and affordable, making them
more accessible to a wider range of users. Overall, NPUs and TPUs are important tools that will continue to play a significant role in the advancement of machine learning and AI. | {"url":"https://www.bizety.com/2023/01/03/ai-chips-npu-vs-tpu/","timestamp":"2024-11-08T21:51:01Z","content_type":"text/html","content_length":"282404","record_id":"<urn:uuid:1345c68a-1c7d-4104-b9b0-a2b6af431182>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00049.warc.gz"} |
Lorentz velocity addition - iSoulrag
This post follows on the Gamma factor post here. The form of velocity addition based on the Lorentz transformation is related to a combination of additive and harmonic addition. Galilei velocity
addition is Lorentz velocity addition is which equals In this way the Lorentz transformation attempts to combine arithmetic and harmonic addition.
Michelson Morley Experiment Re-examined
The Michelson-Morley experiment, compared the longitudinal and transverse cases of reflected light, expecting to detect an ether wind (Figure 1). Figure 1. Michelson-Morley apparatus They explain:
“Let sa … be a ray of light which is partly reflected in ab, and partly transmitted in ac, being returned by the mirrors b and c, along ba
Ancient Greek means
The ancient Greeks defined ten means in terms of the following proportions (see here): Let a > m > b > 0. Then m represents (1) the arithmetic mean of a and b if (2) the geometric mean of a and b if
(3) the harmonic mean of a and b if (4) the contraharmonic
Introduction to logic
This post follows others about logic, such as here. Purpose The purpose of logic is to ensure that one’s discourse makes sense. The most important part of making sense is avoiding contradictions,
which would both affirm and deny a proposition. Some propositions may be affirmed in part and denied in part; that is different. The
Rates of motion in time and distance domains
The rates of motion for translational (linear) and rotational (angular) motion in the time domain and the distance domain are as follows: Time Domain Variable Translational Motion Rotational Motion
Location x θ Velocity v = dx/dt ω = dθ/dt Acceleration a = dv/dt α = dω/dt Distance Domain Variable Translational Motion Rotational Motion Chronation z
N-ary distinctions
The ground of each distinction is an indistinct mass or state or condition, a kind of whole without parts or at least without parts that have been discerned. Every instance of the whole is at first,
an instance of one mass or state or condition. A unary distinction is a discernment of something out of
Moral and civil law
Everyone should understand the distinction between what is moral and what is not moral. I have written briefly about that here. What is legal is not necessarily moral. What is moral is not
necessarily legal. What is the relation between the moral law and the civil law? That is something every society must decide for
Algebra and calculus of ratios
Ratio Algebra Let us define an algebra of ratios. A ratio consists of two numeric expressions separated by a colon, and for clarity enclosed in parentheses, i.e., (a : b) with a, b ∈ ℝ. The
expression on the left is the antecedent, and the expression on the right is the consequent. (0 : 0) is
Complete Galilei Group
The following is based on Lévy-LeBlond’s Galilei Group and Galilean Invariance, §2 (Nuovo Cimento, Jan. 1973). Let Ω be the complete Newtonian space, the points (events) of which we label by their
coordinates in some complete Galilean frame, using the notation y = (x(t), z(s)). (1) The complete proper Galilei group G (or Galilei
Gamma factor between means
Consider the mean between two quantities: The arithmetic mean is The harmonic mean is where which equals the gamma factor of the Lorentz transformation. The geometric mean is Then the factor γ2
transforms a harmonic mean into an arithmetic mean: The inverse γ factor transforms an arithmetic mean into a geometric mean: so that The | {"url":"https://www.isoul.org/author/rg/","timestamp":"2024-11-15T03:21:19Z","content_type":"text/html","content_length":"131531","record_id":"<urn:uuid:1647bcf4-427b-4f19-9e0c-3fa042e79798>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00646.warc.gz"} |
John Nash American Mathematician » Famous Mathematicians Vedic Math School
John Nash American Mathematician
Famous Mathematicians / By Prince Jha / 5 minutes of reading
John Nash Jr. was an American Mathematician who was born on June 13, 1928, in Bluefield, West Virginia, the U.S. His father worked in Appalachian Electric Power Company as an electrical engineer and
His mother was a schoolteacher.
He made contributions to partial differential equations, differential geometry game theory. He worked on those things which help in decision-making inside Collective behavior, Networks, Evolution and
adaptation, Pattern formation, Systems theory, Nonlinear dynamics,
The game theory found in everyday life.
John Nash’s theories are mainly used in economics. At Princeton University, he worked as a senior research mathematician.
He was been awarded the Nobel Memorial Prize in Economic Sciences in 1994. In 2015, he was also awarded Abel Prize for his amazing work on nonlinear partial differential equations.
He is the only mathematician who was been rewarded with both the Abel Prize and the Nobel Memorial Prize in Economic Sciences.
John , During April and May of 1959, was diagnosed with paranoid schizophrenia.
He started facing mild clinical depression, a lack of motivation for life, and auditory and perceptional disturbances. He was undergoing mental illness and to cure that he got admitted to the State
Hospital at Trenton in New York.
On May 23, 2015, Nash and his wife Alicia died in a car accident in New Jersey Turnpike. The report says that neither of them was wearing their seatbelt at the time of the accident.
John Nash
John Education
Attended kindergarten and public school when he was a child
Went to a local community college for his higher mathematical learning.
He attended Carnegie Mellon University after getting George Westinghouse Scholarship to pursue Chemical Engineering.
By the age of 19, He has completed both B.S and Ms and Went to Princeton University.
He has even attended the following universities
1. In, 1944-1945, He attended Bluefield College
2. In 1945, He attended Bluefield High School
3. In 1945- 1948, He attended Carnegie Institute of Technology
Unraveling Sophie Germain’s ingenious legacy—discover the untold brilliance that reshaped mathematics!” Explore more pioneering minds in our related articles
John Nash Books
Some of the famous Books and Research Papers of John Nash are:
1. Open Problems in Mathematics
2. The Sacramental Church: The Story of Anglo-Catholicism John Nash
3. Christianity: the One, the Many: What Christianity Might Have Been and Could Still Become Volume 1
4. The Design, Selection, and Implementation of Accounting Information Systems
5. Cases in Corporate Financial Planning and Control John
6. Essays on Game Theory
7. Accounting Information Systems
8. Equilibrium Points in N-person Games
9. The Bargaining Problem
10. Non-cooperative Games
11. Two-person Cooperative Games
Contributions of John Nash in Mathematics
John Nash made ground-breaking contributions in mathematical areas as diverse as partial differential equations, topology & geometry. He has done immense contributions to game theory. Few of the
theorems and the functions which are famous are:
• Nash equilibrium
• Nash functions
• Nash–Moser theorem
• Nash embedding theorem
• Hilbert’s nineteenth problem
During 1945 and 1996, John published 23 scientific studies on various topics on Advance Game theory and Mathematics.
Discover Blaise Pascal’s groundbreaking contributions and explore more visionary thinkers reshaping our understanding of science and philosophy! Uncover related captivating insights now
Interesting facts about John
Inspired by the life of John Nash, A English Movie Director Named Ron Howard Made a Movie in the year 2001, Named as Beautiful Minds.
In this movie, it shows, In 1956, Nash was in a severe disappointment while he was trying to prove Hilbert’s nineteenth problem, a theorem linking elliptic partial differential equations.
In 1951, Nash was hired by the Massachusetts Institute of Technology (MIT) as a C. L. E. Moore instructor in the mathematics faculty.
In a sting operation targeting homosexual men, Nash was arrested for indecent exposure.
Nash was an atheist yet he got married in Church.
Though he earned a tenured position at MIT, he used to reside in a house as a boarder.
After his death, The New York Times published an article about John Nash and his works.
Unveil John Wallis’ mathematical genius and explore more groundbreaking minds reshaping the landscape of numbers and equations! Discover related captivating insights now.”
Awards and Rewards under the name of John
Few of the Notable Honors by John
1. Abel Prize
2. Class of Fellows of the Institute for Operations Research and the Management Sciences
3. Double Helix Medal
4. INFORMS John von Neumann Theory Prize
5. Leroy P Steele Prize
6. Nobel Memorial Prize in Economic Sciences
Quotes By John Nash
• ~ I’ve made the most important discovery of my life.
• “Classes will dull your mind, destroy the potential for authentic creativity.”
• “What truly is logic?”
• “The only thing greater than the power of the mind is the courage of the heart”
• “I cannot waste time in these classes and these books, memorizing the weak assumptions of lesser mortals.”
• “Perhaps it is good to have a beautiful mind, but an even greater gift is to discover a beautiful heart!”
Other Mathematicians Talking about John Nash
Mikhail Leonidovich Gromov writes about Nash’s work:
Nash was solving classical mathematical problems, difficult problems, something that nobody else was able to do, not even to imagine how to do it. …
But what Nash discovered in the course of his constructions of isometric embeddings is far from ‘classical’ — it is something that brings about a dramatic alteration of our understanding of the
basic logic of analysis and differential geometry.
Judging from the classical perspective, what Nash has achieved in his papers is as impossible as the story of his life … [H]is work on isometric immersions … opened a new world of mathematics
that stretches in front of our eyes in yet unknown directions and still waits to be explored.
Other Mathematicians like John Nash
There are Several Mathematicians of 19th Century and 20th Century who would be known for ever in the history like Shakuntala Devi, Srinivasa Ramanujan, David Hilbert.
FAQ About Nash
How did John and Alicia Nash die?
John Nash and his wife, died in a car accident, while they were returning from the award function, where John was awarded Abel Prize.
Did John and Alicia Nash die on the same day?
Yes, Unfortunately, both John and Her wife died on the Same Day i.e on May 23, 2015.
Did John Nash does work for the government?
Nash was an expert in noncooperative game theory. He worked for the National Security Agency of the US government. He used to crack the code and used to develop codes that can’t be decoded.
Did John Nash solve the Riemann hypothesis?
In 1959, John proclaimed he had the proof for the Riemann Hypothesis, So, he assembled hundreds of mathematicians at the University of Columbia to demonstrate his proof to these mathematicians.
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Elastic Potential Energy and Kinetic Energy in SHM in context of Simple Harmonic Motion (SHM)
27 Aug 2024
Understanding Elastic Potential Energy and Kinetic Energy in Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a fundamental concept in physics that describes the oscillatory motion of an object around its equilibrium position. In this article, we will delve into the concepts
of elastic potential energy and kinetic energy in SHM, exploring their relationships and formulas.
What is Simple Harmonic Motion?
Simple Harmonic Motion (SHM) occurs when an object moves back and forth around a fixed point, with the motion being periodic and oscillatory. The motion is characterized by a restoring force that
pulls the object back towards its equilibrium position. SHM is commonly observed in springs, pendulums, and mass-spring systems.
Elastic Potential Energy (U)
In SHM, elastic potential energy (U) is the energy stored in the spring or system as it compresses or stretches. The elastic potential energy is directly proportional to the square of the
displacement from the equilibrium position. Mathematically, this can be represented by:
U = 1/2 kx^2
• U is the elastic potential energy
• k is the spring constant (a measure of the spring’s stiffness)
• x is the displacement from the equilibrium position
Kinetic Energy (K)
As the object moves through SHM, it also possesses kinetic energy (K), which is the energy of motion. The kinetic energy is directly proportional to the square of the velocity. Mathematically, this
can be represented by:
K = 1/2 mv^2
• K is the kinetic energy
• m is the mass of the object
• v is the velocity of the object
Relationship between Elastic Potential Energy and Kinetic Energy
In SHM, the total energy (E) remains constant, as energy is converted back and forth between elastic potential energy and kinetic energy. At any given point in the motion, the sum of the elastic
potential energy and kinetic energy equals the total energy:
E = U + K
As the object moves from its maximum compression to its maximum extension, the elastic potential energy increases while the kinetic energy decreases. Conversely, as the object moves from its maximum
extension to its maximum compression, the elastic potential energy decreases while the kinetic energy increases.
Key Takeaways
1. Elastic potential energy (U) is directly proportional to the square of the displacement from the equilibrium position.
2. Kinetic energy (K) is directly proportional to the square of the velocity.
3. The total energy (E) remains constant in SHM, as energy is converted between elastic potential energy and kinetic energy.
4. The relationship between elastic potential energy and kinetic energy can be represented by E = U + K.
In this article, we have explored the concepts of elastic potential energy and kinetic energy in Simple Harmonic Motion. Understanding these fundamental principles is crucial for grasping the
behavior of oscillatory systems and predicting their motion. By applying the formulas and relationships discussed above, you can better comprehend the intricate dance between energy and motion in
Related articles for ‘Simple Harmonic Motion (SHM)’ :
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Myeconlab from chapter7 to chapter 18, economics homework help
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