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http://math.stackexchange.com/questions/166734/prove-or-disprove-an-inequality | # Prove or disprove an inequality
Let's consider the following equation where $m,n$ are real numbers:
$$x^3+mx+n=0$$
I need to prove/disprove without calculus that for any real root of the above equation we have that: $$m^2-4 x_1 n \ge 0$$
-
Suppose that $x_1$ is a real root of the cubic, and consider the quadratic equation $x_1x^2+mx+n=0$. This must have $x_1$ as a real solution, so ... ?
@Chris' sister: I can’t at the moment think of another. If you’re wondering how I came up with it, I looked at the expression $m^2-4x_1n$ and almost immediately thought discriminant of quadratic. Then everything just fell into place. – Brian M. Scott Jul 4 '12 at 20:19 | 2015-11-30T08:09:25 | {
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http://math.stackexchange.com/questions/302144/markovian-and-the-chapman-kolmogorov-equation | # Markovian and the Chapman–Kolmogorov equation
From Wikipedia
In a Markov process, one assumes that $i_1 < \cdots < i_n$. Then, because of the Markov property, $$p_{i_1,\ldots,i_n}(f_1,\ldots,f_n)=p_{i_1}(f_1)p_{i_2;i_1}(f_2\mid f_1)\cdots p_{i_n;i_{n-1}}(f_n\mid f_{n-1}),$$ where the conditional probability $p_{i;j}(f_i\mid f_j)$ is the transition probability between the times $i>j$. So, the Chapman–Kolmogorov equation takes the form $$p_{i_3;i_1}(f_3\mid f_1)=\int_{-\infty}^\infty p_{i_3;i_2}(f_3\mid f_2)p_{i_2;i_1}(f_2\mid f_1) \, df_2.$$
I was wondering if the reverse is true. I.e., if a process satisfies the above form of the Chapman–Kolmogorov equation, will it be Markovian? Thanks and regards!
-
from the linked page: "In English, and informally," – Ilya Feb 14 '13 at 13:01
Yes, if the initial (unconditional) distribution holds for any subsequent propagation such that $$p_{i_2} (f_2) = \int_{-\infty}^\infty p_{i_2;i_1} (f_2|f_1) p_{i_1}(f_1) df_1$$ then the Chapman-Kolmogorov uniquely defines a Markov process.
To see this, multiply the CK equation by the unconditional PDF of the conditioned variable, $p_{i_1} (f_1)$, to get the the joint density $$p_{i_3,i_1} (f_1, f_3) = p_{i_1} (f_1) \cdot \int_{-\infty}^\infty p_{i_3;i_2} (f_3|f_2) \cdot p_{i_2;i_1}(f_2|f_1) \ df_2\\$$ Differentiate by $f_2$. The lefthand side becomes a joint probability distribution of 3 variables (and the above joint probability distribution of 2 variables can now be seen as a marginal probability distribution), and the right-hand side yields the definition of a Markov process $$p_{i_3,i_2, i_1} (f_1,f_2, f_3) = p_{i_3;i_2} (f_3|f_2) \cdot p_{i_2;i_1}(f_2|f_1) \cdot p_{i_1} (f_1)$$ This can be repeated to get the Markov definition to the $n^{th}$ order definition you gave. | 2014-10-30T18:47:10 | {
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https://www.physicsforums.com/threads/exponential-relationships-to-logarithms-and-straight-line-graph.326401/ | # Exponential relationships to logarithms and straight line graph?
it is suspected that cells in a sample are dividing so that the number of cells present at any one time t (measured in seconds) is growing exponentially according to the relationship y = 64 x 2^2t. it would be hard to check this relationship accurately by plotting measurements of y against t, so in practice one can use logarithms to convert it to a linear line equation. sketch a graph of log(base2)y against t, labelling the point where the graph crosses the axes?
how to i find the points on x and y axis where the line crosses, i believe the line will be a straight line, how do i do this?
I REALLY HAVE TRIED WITH THIS QUESTION, IT HAS TAKEN ME THE WHOLE DAY TO TRY AND DO, I HOPE SOMEONE CAN SHOW ME HOW TO DO THIS :)
Related Introductory Physics Homework Help News on Phys.org
HallsofIvy
What, exactly, did you spend all day doing? Taking the logarithm of both sides of $y= 64(2^{2t})$ gives $ln(y)= log(64)+ 2t log(2)$
I specifically did not give a base for the logarithm because the above is true for any base. It is particularly simple if you use, not "common" or "natural" logarithm, but the logarithm base 2: $log_2(2)= 1$ and $log_2(64)= log_2(2^6)= 6$ so Letting $Y= log_2(y)$, the equation becomes Y= 2t+ 6 with intercepts at (0, 6) and (3, 0). | 2020-07-07T07:26:20 | {
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http://mathhelpforum.com/advanced-statistics/88736-markov-chain-transition-matrix.html | # Thread: Markov chain transition matrix
1. ## Markov chain transition matrix
I am having a problem with a Markov chain transition matrix question. Can anyone help me?
2. Originally Posted by lhaero
I am having a problem with a Markov chain transition matrix question. Can anyone help me?
Not unless you post the question.
CB
3. ## Markov chain transition matrix
I have 2 questions in relation to this topic:
1)In a certain production process, items are pass through two manufacturing stages. At the end of each stage, the items are either scrapped with a probability of 0.15, sent through the stage again for rework with a probability of 0.25, or passed on the the next step with a probability of 0.6. Describe this as a Markov chain and set up the transition matrix? What is the expected number of steps to absorption?
And:
2)Ball bearings in a machine have an increased probability of failure with increased service life. This machine has two bearings. Bearings cost $15 each plus$50 per unit installation cost. A bearing failure results in an additional cost of $200. Past history of failures are tabulated below. Note that no bearing has ever run more than 5000 hours. Management wishes to examine two possible maintenance procedures. i) Replace as failures occur. ii) Replace individual bearings on an optimum replacement life basis. ------------------------------------------------------------ |Service Life (1000 hours) | No. of times failure occurred | --------------------------|--------------------------------- | 1 | 10 | ------------------------------------------------------------ | 2 | 15 | ------------------------------------------------------------ | 3 | 15 | ------------------------------------------------------------ | 4 | 30 | ------------------------------------------------------------ | 5 | 20 | ------------------------------------------------------------ Which replacement policy results in minimum cost? What is the average life of a bearing (from installation to replacement) under each policy? If you could help me with these it would be very much appreciated. 4. Originally Posted by lhaero I have 2 questions in relation to this topic: 1)In a certain production process, items are pass through two manufacturing stages. At the end of each stage, the items are either scrapped with a probability of 0.15, sent through the stage again for rework with a probability of 0.25, or passed on the the next step with a probability of 0.6. Describe this as a Markov chain and set up the transition matrix? What is the expected number of steps to absorption? . Let the state of an item can be Scrapped, ToBeTested, Passed, then we may represent an item as a colum vector of the probabilities of it being in each state:$\displaystyle P=\left[ \begin{array}{c}S\\T\\P \end{array} \right]$where S, T and P are the probabilities that it is each of the states. Then the transition matrix is:$\displaystyle
A=\left[ \begin{array}{ccc}1 & 0.15 & 0\\0 & 0.25 & 0\\0&0.6&1 \end{array} \right]
$which tells us that: (first row) if the item is Scrapped, it remains Scrapped, if it is to be Tested it is Scrapped with probability 0.15, and if it has been Passed it cannot be Scrapped. (second row) if an item is Scrapped it cannot be tested, if it was tested it will be reworked and returned for Testing with probability 0.25, and if it has been passed it cannot be Tested. (third row) if an item is Scrapped it cannot be Passed, if it was tested it will be Passed with probability 0.6, and if it has been Passed it remains Passed. Note the convention I am using is that if$\displaystyle P(n)$is a column vector of state probabilities at epoc$\displaystyle n$, then the new state at epoc$\displaystyle n+1$is$\displaystyle P(n+1)=AP(n) \$
CB
5. ## Thanks and 2nd question
Thanks for your help Captain Black.
Any assistance that you could give on question 2 would also be appreciated.
lhaero. | 2018-03-17T16:45:49 | {
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https://www.bartleby.com/questions-and-answers/write-the-equation-of-the-line-with-the-given-information.-m-2-1-2/66d8b237-7ab9-43dd-bf9b-6d5e744066ec | # Write the equation of the line with the given information. m= 2 (1 , -2)
Question
Write the equation of the line with the given information.
m= 2 (1 , -2) | 2021-07-24T02:49:12 | {
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# Tuple
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Title: Tuple Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Tuple
A tuple is a finite ordered list of elements. In mathematics, an n-tuple is a sequence (or ordered list) of n elements, where n is a non-negative integer. There is only one 0-tuple, an empty sequence. An n-tuple is defined inductively using the construction of an ordered pair. Tuples are usually written by listing the elements within parentheses "(\text{ })" and separated by commas; for example, (2, 7, 4, 1, 7) denotes a 5-tuple. Sometimes other symbols are used to surround the elements, such as square brackets "[ ]" or angle brackets "\langle\text{ }\rangle". Braces "{ }" are never used for tuples, as they are the standard notation for sets. Tuples are often used to describe other mathematical objects, such as vectors. In computer science, tuples are directly implemented as product types in most functional programming languages. More commonly, they are implemented as record types, where the components are labeled instead of being identified by position alone. This approach is also used in relational algebra. Tuples are also used in relation to programming the semantic web with Resource Description Framework or RDF. Tuples are also used in linguistics[1] and philosophy.[2]
## Contents
• Etymology 1
• Names for tuples of specific lengths 1.1
• Properties 2
• Definitions 3
• Tuples as functions 3.1
• Tuples as nested ordered pairs 3.2
• Tuples as nested sets 3.3
• n-tuples of m-sets 4
• Type theory 5
• See also 6
• Notes 7
• References 8
## Etymology
The term originated as an abstraction of the sequence: single, double, triple, quadruple, quintuple, sextuple, septuple, octuple, ..., n‑tuple, ..., where the prefixes are taken from the Latin names of the numerals. The unique 0‑tuple is called the null tuple. A 1‑tuple is called a singleton, a 2‑tuple is called an ordered pair and a 3‑tuple is a triple or triplet. n can be any nonnegative integer. For example, a complex number can be represented as a 2‑tuple, a quaternion can be represented as a 4‑tuple, an octonion can be represented as an 8‑tuple and a sedenion can be represented as a 16‑tuple.
Although these uses treat ‑tuple as the suffix, the original suffix was ‑ple as in "triple" (three-fold) or "decuple" (ten‑fold). This originates from a medieval Latin suffix ‑plus (meaning "more") related to Greek ‑πλοῦς, which replaced the classical and late antique ‑plex (meaning "folded"), as in "duplex".[3]
### Names for tuples of specific lengths
Tuple Length n Name Alternative names
0 empty tuple unit
1 single singleton / monuple
2 double couple / pair / dual / twin / product
3 triple treble / triplet / triad
4 quadruple quad
5 quintuple penta
6 sextuple
7 septuple
8 octuple
9 nonuple
10 decuple
11 undecuple hendecuple
12 duodecuple
13 tredecuple
14 quattuordecuple
15 quindecuple
16 sexdecuple
100 centuple
## Properties
The general rule for the identity of two n-tuples is
(a_1, a_2, \ldots, a_n) = (b_1, b_2, \ldots, b_n) if and only if a_1=b_1,\text{ }a_2=b_2,\text{ }\ldots,\text{ }a_n=b_n.
Thus a tuple has properties that distinguish it from a set.
1. A tuple may contain multiple instances of the same element, so
tuple (1,2,2,3) \neq (1,2,3); but set \{1,2,2,3\} = \{1,2,3\}.
2. Tuple elements are ordered: tuple (1,2,3) \neq (3,2,1), but set \{1,2,3\} = \{3,2,1\}.
3. A tuple has a finite number of elements, while a set or a multiset may have an infinite number of elements.
## Definitions
There are several definitions of tuples that give them the properties described in the previous section.
### Tuples as functions
If we are dealing with sets, an n-tuple can be regarded as a function, F, whose domain is the tuple's implicit set of element indices, X, and whose codomain, Y, is the tuple's set of elements. Formally:
(a_1, a_2, \dots, a_n) \equiv (X,Y,F)
where:
\begin{align} X & = \{1, 2, \dots, n\} \\ Y & = \{a_1, a_2, \ldots, a_n\} \\ F & = \{(1, a_1), (2, a_2), \ldots, (n, a_n)\}. \\ \end{align}
In slightly less formal notation this says:
(a_1, a_2, \dots, a_n) := (F(1), F(2), \dots, F(n)).
### Tuples as nested ordered pairs
Another way of modeling tuples in Set Theory is as nested ordered pairs. This approach assumes that the notion of ordered pair has already been defined; thus a 2-tuple
1. The 0-tuple (i.e. the empty tuple) is represented by the empty set \emptyset.
2. An n-tuple, with n > 0, can be defined as an ordered pair of its first entry and an (n - 1)-tuple (which contains the remaining entries when n > 1):
(a_1, a_2, a_3, \ldots, a_n) = (a_1, (a_2, a_3, \ldots, a_n))
This definition can be applied recursively to the (n - 1)-tuple:
(a_1, a_2, a_3, \ldots, a_n) = (a_1, (a_2, (a_3, (\ldots, (a_n, \emptyset)\ldots))))
Thus, for example:
\begin{align} (1, 2, 3) & = (1, (2, (3, \emptyset))) \\ (1, 2, 3, 4) & = (1, (2, (3, (4, \emptyset)))) \\ \end{align}
A variant of this definition starts "peeling off" elements from the other end:
1. The 0-tuple is the empty set \emptyset.
2. For n > 0:
(a_1, a_2, a_3, \ldots, a_n) = ((a_1, a_2, a_3, \ldots, a_{n-1}), a_n)
This definition can be applied recursively:
(a_1, a_2, a_3, \ldots, a_n) = ((\ldots(((\emptyset, a_1), a_2), a_3), \ldots), a_n)
Thus, for example:
\begin{align} (1, 2, 3) & = (((\emptyset, 1), 2), 3) \\ (1, 2, 3, 4) & = ((((\emptyset, 1), 2), 3), 4) \\ \end{align}
### Tuples as nested sets
Using Kuratowski's representation for an ordered pair, the second definition above can be reformulated in terms of pure set theory:
1. The 0-tuple (i.e. the empty tuple) is represented by the empty set \emptyset;
2. Let x be an n-tuple (a_1, a_2, \ldots, a_n), and let x \rightarrow b \equiv (a_1, a_2, \ldots, a_n, b). Then, x \rightarrow b \equiv \{\{x\}, \{x, b\}\}. (The right arrow, \rightarrow, could be read as "adjoined with".)
In this formulation:
\begin{array}{lclcl} () & & &=& \emptyset \\ & & & & \\ (1) &=& () \rightarrow 1 &=& \{\{()\},\{(),1\}\} \\ & & &=& \{\{\emptyset\},\{\emptyset,1\}\} \\ & & & & \\ (1,2) &=& (1) \rightarrow 2 &=& \{\{(1)\},\{(1),2\}\} \\ & & &=& \{\{\{\{\emptyset\},\{\emptyset,1\}\}\}, \\ & & & & \{\{\{\emptyset\},\{\emptyset,1\}\},2\}\} \\ & & & & \\ (1,2,3) &=& (1,2) \rightarrow 3 &=& \{\{(1,2)\},\{(1,2),3\}\} \\ & & &=& \{\{\{\{\{\{\emptyset\},\{\emptyset,1\}\}\}, \\ & & & & \{\{\{\emptyset\},\{\emptyset,1\}\},2\}\}\}, \\ & & & & \{\{\{\{\{\emptyset\},\{\emptyset,1\}\}\}, \\ & & & & \{\{\{\emptyset\},\{\emptyset,1\}\},2\}\},3\}\} \\ \end{array}
## n-tuples of m-sets
In discrete mathematics, especially combinatorics and finite probability theory, n-tuples arise in the context of various counting problems and are treated more informally as ordered lists of length n.[4] n-tuples whose entries come from a set of m elements are also called arrangements with repetition, permutations of a multiset and, in some non-English literature, variations with repetition. The number of n-tuples of an m-set is mn. This follows from the combinatorial rule of product.[5] If S is a finite set of cardinality m, this number is the cardinality of the n-fold Cartesian power S × S × ... S. Tuples are elements of this product set.
## Type theory
In type theory, commonly used in programming languages, a tuple has a product type; this fixes not only the length, but also the underlying types of each component. Formally:
(x_1, x_2, \ldots, x_n) : \mathsf{T}_1 \times \mathsf{T}_2 \times \ldots \times \mathsf{T}_n
and the projections are term constructors:
\pi_1(x) : \mathsf{T}_1,~\pi_2(x) : \mathsf{T}_2,~\ldots,~\pi_n(x) : \mathsf{T}_n
The tuple with labeled elements used in the relational model has a record type. Both of these types can be defined as simple extensions of the simply typed lambda calculus.[6]
The notion of a tuple in type theory and that in set theory are related in the following way: If we consider the natural model of a type theory, and use the Scott brackets to indicate the semantic interpretation, then the model consists of some sets S_1, S_2, \ldots, S_n (note: the use of italics here that distinguishes sets from types) such that:
[\![\mathsf{T}_1]\!] = S_1,~[\![\mathsf{T}_2]\!] = S_2,~\ldots,~[\![\mathsf{T}_n]\!] = S_n
and the interpretation of the basic terms is:
[\![x_1]\!] \in [\![\mathsf{T}_1]\!],~[\![x_2]\!] \in [\![\mathsf{T}_2]\!],~\ldots,~[\![x_n]\!] \in [\![\mathsf{T}_n]\!].
The n-tuple of type theory has the natural interpretation as an n-tuple of set theory:[7]
[\![(x_1, x_2, \ldots, x_n)]\!] = (\,[\![x_1]\!], [\![x_2]\!], \ldots, [\![x_n]\!]\,)
The unit type has as semantic interpretation the 0-tuple.
## Notes
1. ^ "N‐tuple - Oxford Reference". oxfordreference.com. Retrieved 1 May 2015.
2. ^ "Ordered n-tuple - Oxford Reference". oxfordreference.com. Retrieved 1 May 2015.
3. ^ OED, s.v. "triple", "quadruple", "quintuple", "decuple"
4. ^ D'Angelo & West 2000, p. 9
5. ^ D'Angelo & West 2000, p. 101
6. ^ Pierce, Benjamin (2002). Types and Programming Languages. MIT Press. pp. 126–132.
7. ^ Steve Awodey, From sets, to types, to categories, to sets, 2009, preprint
## References
• D'Angelo, John P.; West, Douglas B. (2000), Mathematical Thinking / Problem-Solving and Proofs (2nd ed.), Prentice-Hall,
• Keith Devlin, The Joy of Sets. Springer Verlag, 2nd ed., 1993, ISBN 0-387-94094-4, pp. 7–8
• Abraham Adolf Fraenkel, Yehoshua Bar-Hillel, Azriel Lévy, Foundations of set theory, Elsevier Studies in Logic Vol. 67, Edition 2, revised, 1973, ISBN 0-7204-2270-1, p. 33
• Gaisi Takeuti, W. M. Zaring, Introduction to Axiomatic Set Theory, Springer GTM 1, 1971, ISBN 978-0-387-90024-7, p. 14
• George J. Tourlakis, Lecture Notes in Logic and Set Theory. Volume 2: Set theory, Cambridge University Press, 2003, ISBN 978-0-521-75374-6, pp. 182–193
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a 501c(4) Member's Support Non-Profit Organization, and is NOT affiliated with any governmental agency or department. | 2019-06-17T04:34:23 | {
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https://math.stackexchange.com/questions/1779989/how-should-automorphisms-of-monoid-actions-in-mathbfrel-be-defined | How should automorphisms of monoid actions in $\mathbf{Rel}$ be defined?
To explain the question, I'd like to start by considering monoid actions in $\mathbf{Sets}$. In this case, a monoid action is simply a functor $S$ from a monoid $M$ to $\mathbf{Sets}$. Then, one can form a category of monoid actions, wherein objects are functors $S \colon M \to \mathbf{Sets}$, and wherein morphisms between $S \colon M \to \mathbf{Sets}$ and $S' \colon M' \to \mathbf{Sets}$ (note that we allow the monoid to be different in $S$ and $S'$; in other words, this is not the category $\mathbf{Sets}^{M}$) are pairs $(L,\lambda)$ where $L$ is an monoid homomorphism $L \colon M \to M'$ and $\lambda$ is a natural transformation $\lambda \colon S \to S'L$. Then, we can define the automorphism group of a functor $S \colon M \to \mathbf{Sets}$ based on the set of pairs $(L,\lambda)$ where $L$ is an isomorphism and $\lambda$ is an equivalence.
If one now considers $M$ to be a monoid of binary relations on a finite set, we get a functor $S \colon M \to \mathbf{Rel}$, where $\mathbf{Rel}$ is the category of finite sets and binary relations between them. The question is: how should the automorphisms of $S$ be defined ?
The simple choice would be to consider pairs $(L,\lambda)$ as above, where $\lambda$ is a natural transformation, i.e. for any relation $\mathcal{R}$ on a set $X$, $x \mathcal{R} y \iff \lambda(x) L(\mathcal{R}) \lambda(y)$.
On the other hand, $\mathbf{Rel}$ is a 2-category, so one could also consider $\lambda$ to be a lax natural transformation, allowing inclusions of relations.
What would be the proper way to define the automorphism group of $S$ ?
• Depends on your application. Either might end up being appropriate. – Qiaochu Yuan May 10 '16 at 20:30
• @Qiaochu Yuan: I am realizing now that if I ask $L$ to be an isomorphism, and $\lambda$ an equivalence (hence a bijection), then whatever 2-morphism which should be present in the lax natural transformation should be invertible and thus, given the nature of 2-morphisms in $\mathbf{Rel}$, should be the identity. Am I correct ? – OliverX1 May 16 '16 at 19:23 | 2019-07-22T16:54:07 | {
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https://laurenthoeltgen.name/tag/accuracy/ | # Accuracy
## QR Decompositions with Reorthogonalisation
Problem Formulation We already discussed QR decompositions and showed that using the modified formulation of Gram Schmidt significantly improves the accuracy of the results. However, there is still an error of about $10^3 M_\varepsilon$ (where $M_\varepsilon$ is the machine epsilon) when using the modified Gram Schmidt as base algorithm for the orthogonalisation.
## Fließkommaarithmetik
Oral presentation
## Floating Point Accuracy and Precision
Floating point computations on computers may behave differently than one might expect. Every software developer should be aware of these since computed results may be off by orders of magnitude in the worst case. | 2021-04-13T05:07:06 | {
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http://www.mathworks.com/help/stats/logninv.html?requestedDomain=www.mathworks.com&nocookie=true | # Documentation
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# logninv
Lognormal inverse cumulative distribution function
## Syntax
`X = logninv(P,mu,sigma)[X,XLO,XUP] = logninv(P,mu,sigma,pcov,alpha)`
## Description
`X = logninv(P,mu,sigma)` returns values at `P` of the inverse lognormal cdf with distribution parameters `mu` and `sigma`. `mu` and `sigma` are the mean and standard deviation, respectively, of the associated normal distribution. `mu` and `sigma` can be vectors, matrices, or multidimensional arrays that all have the same size, which is also the size of `X`. A scalar input for `P`, `mu`, or `sigma` is expanded to a constant array with the same dimensions as the other inputs.
`[X,XLO,XUP] = logninv(P,mu,sigma,pcov,alpha)` returns confidence bounds for `X` when the input parameters `mu` and `sigma` are estimates. `pcov` is the covariance matrix of the estimated parameters. `alpha` specifies 100(1 - `alpha`)% confidence bounds. The default value of `alpha` is 0.05. `XLO` and `XUP` are arrays of the same size as `X` containing the lower and upper confidence bounds.
`logninv` computes confidence bounds for `P` using a normal approximation to the distribution of the estimate
`$\stackrel{^}{\mu }+\stackrel{^}{\sigma }q$`
where q is the `P`th quantile from a normal distribution with mean 0 and standard deviation 1. The computed bounds give approximately the desired confidence level when you estimate `mu`, `sigma`, and `pcov` from large samples, but in smaller samples other methods of computing the confidence bounds might be more accurate.
The lognormal inverse function is defined in terms of the lognormal cdf as
`$x={F}^{-1}\left(p|\mu ,\sigma \right)=\left\{x:F\left(x|\mu ,\sigma \right)=p\right\}$`
where
`$p=F\left(x|\mu ,\sigma \right)=\frac{1}{\sigma \sqrt{2\pi }}{\int }_{0}^{x}\frac{{e}^{\frac{-{\left(\mathrm{ln}\left(t\right)-\mu \right)}^{2}}{2{\sigma }^{2}}}}{t}dt$`
## Examples
collapse all
Compute the inverse cdf of a lognormal distribution with `mu = 0` and `sigma = 0.5`.
```p = (0.005:0.01:0.995); crit = logninv(p,1,0.5); ```
Plot the inverse cdf.
```figure; plot(p,crit) xlabel('Probability'); ylabel('Critical Value'); grid ```
## References
[1] Evans, M., N. Hastings, and B. Peacock. Statistical Distributions. Hoboken, NJ: Wiley-Interscience, 2000. pp. 102–105. | 2017-04-24T17:18:11 | {
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https://en.wikipedia.org/wiki/Semicircle | # Semicircle
In mathematics (and more specifically geometry), a semicircle is a one-dimensional locus of points that forms half of a circle. The full arc of a semicircle always measures 180° (equivalently, π radians, or a half-turn). It has only one line of symmetry (reflection symmetry). In non-technical usage, the term "semicircle" is sometimes used to refer to a half-disk, which is a two-dimensional geometric shape that also includes the diameter segment from one end of the arc to the other as well as all the interior points.
By Thales' theorem, any triangle inscribed in a semicircle with a vertex at each of the endpoints of the semicircle and the third vertex elsewhere on the semicircle is a right triangle, with right angle at the third vertex.
All lines intersecting the semicircle perpendicularly are concurrent at the center of the circle containing the given semicircle.
## Uses
A semicircle with arithmetic and geometric means of a and b
A semicircle can be used to construct the arithmetic and geometric means of two lengths using straight-edge and compass. For a semicircle with a diameter of a + b, the length of its radius is the arithmetic mean of a and b (since the radius is half of the diameter).
The geometric mean can be found by dividing the diameter into two segments of lengths a and b, and then connecting their common endpoint to the semicircle with a segment perpendicular to the diameter. The length of the resulting segment is the geometric mean. This can be proven by applying the Pythagorean theorem to three similar right triangles, each having as vertices the point where the perpendicular touches the semicircle and two of the three endpoints of the segments of lengths a and b.[1]
The construction of the geometric mean can be used to transform any rectangle into a square of the same area, a problem called the quadrature of a rectangle. The side length of the square is the geometric mean of the side lengths of the rectangle. More generally it is used as a lemma in a general method for transforming any polygonal shape into a similar copy of itself with the area of any other given polygonal shape.[2]
## Equation
The equation of a semicircle with midpoint ${\displaystyle (x_{0},y_{0})}$ on the diameter between its endpoints and which is entirely concave from below is
${\displaystyle y=y_{0}+{\sqrt {r^{2}-(x-x_{0})^{2}}}.}$
If it is entirely concave from above, the equation is
${\displaystyle y=y_{0}-{\sqrt {r^{2}-(x-x_{0})^{2}}}.}$
## Arbelos
An arbelos (grey region)
An arbelos is a region in the plane bounded by three semicircles connected at the corners, all on the same side of a straight line (the baseline) that contains their diameters. | 2021-10-24T06:03:17 | {
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http://mymathforum.com/algebra/54058-number-solutions.html | My Math Forum Number of Solutions
Algebra Pre-Algebra and Basic Algebra Math Forum
May 14th, 2015, 08:37 AM #1 Newbie Joined: May 2015 From: India Posts: 14 Thanks: 0 Math Focus: Number Theory Number of Solutions Find the number of different real numbers x that satisfy (x^2+4x−2)^2 =(5x^2−1)^2
May 14th, 2015, 08:58 AM #2
Math Team
Joined: Jan 2015
From: Alabama
Posts: 3,264
Thanks: 902
Quote:
Originally Posted by ishaanmj007 Find the number of different real numbers x that satisfy (x^2+4x−2)^2 =(5x^2−1)^2
This reduces immediately to x^2+ 4x- 2= 5x^2- 1 and x^2+ 4x- 2= -(5x^2- 1)= 1- 5x^2.
The first, x^2+ 4x- 2= 5x^2- 1 is the same as 4x^2- 4x+ 1= 0 and x^2+ 4x= 1- 5x^2 is the same 6x^2+ 4x- 1= 0.
May 14th, 2015, 09:23 AM #3 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This is asking for the number of distinct real roots of 24x^4 - 8x^3 - 22x^2 + 16x - 3. You can factor this polynomial as $$(ax+b)^2(cx^2+dx+e)$$ and check whether the discriminant of the latter is positive, negative, or 0. Thanks from topsquark and ishaanmj007
May 14th, 2015, 09:33 AM #4 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Country Boy missed two cases.
May 16th, 2015, 04:38 PM #5
Banned Camp
Joined: Jun 2014
From: Earth
Posts: 945
Thanks: 191
Quote:
Originally Posted by greg1313 Country Boy missed two cases.
??
Either both quadratics will be the same sign, or they will be the opposite sign.
Country Boy's first resultant quadratic equation gives the repeated x = 1/2.
His second resultant quadratic equation gives two distinct irrational values
for x.
Country Boy's two resultant quadratic equations will give the same number
of solutions as the fourth degree equation posted above by CRGreathouse.
Tags number, solutions
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2019-10-17T23:38:30 | {
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https://socratic.org/questions/what-is-the-end-behavior-of-the-graph-of-f-x-2x-4-7x-2-4x-4 | # What is the end behavior of the graph of f(x)=-2x^4+7x^2+4x-4?
May 31, 2018
${\lim}_{x \to \infty} f \left(x\right) = - \infty , {\lim}_{x \to - \infty} f \left(x\right) = - \infty$
#### Explanation:
To determine the end behavior, let's take the limit as $x \to \infty$ and $x \to - \infty$.
In our polynomial, $f \left(x\right)$, the first term is what will dominate the end behavior, because it has the highest degree. So we can find the limit of that:
${\lim}_{x \to \infty} \textcolor{red}{- 2} \textcolor{b l u e}{{x}^{4}} = - \infty$
As $x$ gets very large, the blue term will always be positive, but the $- 2$ (red) will turn it negative. This is why our limit evaluates to $- \infty$.
${\lim}_{x \to - \infty} \textcolor{red}{- 2} \textcolor{b l u e}{{x}^{4}} = - \infty$
As $x$ gets very negative, the even exponent will make the term positive, but the red $- 2$ on the outside will make it negative. Thus, this limit will also evaluate to $- \infty$.
In general, the function is downward opening because of the negative coefficient on the ${x}^{4}$ term.
Hope this helps! | 2020-07-15T13:00:26 | {
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https://www.coursehero.com/file/p1qou63/2-x-g-x-g-dx-d-x-f-x-f-dx-d-x-g-x-g-x-f-dx-d-DIFFERENTIATION-FORMULAS-Example-If/ | # 2 x g x g dx d x f x f dx d x g x g x f dx d
• 83
This preview shows page 58 - 68 out of 83 pages.
) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 x g x g dx d x f x f dx d x g x g x f dx d =
DIFFERENTIATION FORMULAS Example: If , then 3 2 3 x x y = 6 2 3 6 2 3 6 3 3 )] 2 3 ( [ 3 ) 3 )( 2 3 ( ) 3 ( ) ( ) 2 3 ( 2 3 ' x x x x x x x x x x d x x d x y = = = . ) 1 ( 6 ) 1 )( 2 ( 3 ) 2 2 ( 3 ) 2 3 ( 3 )] 2 3 ( [ 3 ' 4 6 2 6 3 2 6 2 3 3 6 2 3 x x x x x x x x x x x x x x x x y = = = = =
DIFFERENTIATION FORMULAS 7. Derivatives of Composition ( Chain Rule) Theorem: (The Chain Rule) If g is differentiable at x and if f is differentiable at g(x) , then the composition is differentiable at x. Moreover, if y=f(g(x)) and u = g(x) then y = f(u) and g f . dx du du dy dx dy =
DIFFERENTIATION FORMULAS Examples: Find y’. 1. 4 3 ) 1 ( 2 = x y . ) 1 ( 24 ) 3 ( ) 1 ( 8 ) 1 ( ) 1 )( 4 ( 2 ' 3 3 2 2 3 3 3 3 3 = = = x x x x x d x y 4 16 1 1 . 2 x y = 4 / 1 4 ) 16 1 ( 16 1 1 = = x x y . ) 16 1 ( 4 ) 16 ( ) 16 1 ( 4 1 ) 16 1 ( ) 16 1 ( 4 1 ' 4 / 5 4 / 5 4 / 5 x x x d x y = = =
IMPLICIT DIFEERENTIATION On occasions that a function F(x , y) = 0 can not be defined in the explicit form y = f(x) then the implicit form F ( x , y) = 0 can be used as basis in defining the derivative of y ( the dependent variable) with respect to x ( the independent variable). When differentiating F( x, y) = 0, consider that y is defined implicitly in terms of x , then apply the chain rule. As a rule, 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dy/dx on the left side of the equation and the rest of the terms on the other side. 3. Factor dy/dx out of the left member of the equation and solve for dy/dx by dividing the equation by the coefficient of dy/dx.
IMPLICIT DIFFERENTIATION Examples: Find y’. 1. 2. 1 ) ( 2 4 3 2 = y x y x ) ' 1 ( ) ( 8 ' 3 2 3 2 y y x y y x = ' ) ( 8 ) ( 8 ' 3 2 3 3 2 y y x y x y y x = ' ) ( 8 ) ( 8 ' 3 2 3 3 2 y y x y x y y x = 3 2 3 ) ( 8 3 ] ) ( 4 [ 2 ' y x y x y x y = 3 2 2 = y x y x ' 2 1 ) ( ) ( 2 2 yy x yd y d x = ' 2 1 ) 2 ( ' 2 yy x y y x = xy yy y x 2 1 ' 2 ' 2 = xy y x y 2 1 ] 2 [ ' 2 = . 2 2 1 ' 2 y x xy y =
HIGHER DERIVATIVES The notation dy/dx represent the first derivative of y with respect to x. And if dy/dx is differentiable, then the derivative of dy/dx with respect to x gives the second order derivative of y with respect to x and is denoted by . y wrt x of derivative n y general, In . . . . y wrt x of derivative third = ' ' y' y wrt x of derivative second ' ' y wrt x of derivative first ) ( ' ) ( 1 1 2 2 3 3 2 2 th n n n n n dx y d dx d dx y d dx y d dx d dx y d dx dy dx d dx y d y x f dx d dx dy y x f y = = = = = = = = d 2 y dx 2 = d dx dy dx
HIGHER DERIVATIVES Examples: 1. If then 2. if then , 1 3 2 3 5 = x x y ), 9 10 ( 9 10 ' 2 2 2 4 = = x x x x y ), 9 20 ( 2 18 40 " 2 3 = = x x x x y ). 9 60 ( 2 18 120 ' ' ' 2 2 = = x x y , 1 2 2 = y x , ' 0 ' 2 2 = yy x , ' y x y = . 1 ) ( ' ) 1 ( " 3 3 2 2 3 2 2 2 2 y y y x y x y y y x x y y xy y y = = = = =
Sample Problems Differentiate y with respect to x. Express dy/dx in simplest form. 1. y = x 2 5 7. y = 9 x 2 4 2. y = 5 x 3 6 7 x 8. y = x 5 x 2 2 5 3. y = 2 x 2 3 x 1 x 9. y = x 2 3 5 x 2 4. y = x 2 3 x 1 3 4 10. y = 2 2 x 5. y = x 1 4 x 3 11. x 2 y 2 = 9 6. y = 2 x 3 5 x 3 x 2 12. xy = x 2 y 1
Sample Problems Determine the derivative required: 1. y' of y = x 5 x 5 2. y''' of x 2 y 2 = 9 3. | 2021-12-01T21:24:07 | {
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https://www.quizover.com/physics-ap/course/18-6-electric-field-lines-multiple-charges-by-openstax?page=1 | # 18.6 Electric field lines: multiple charges (Page 2/8)
Page 2 / 8
Note that the electric field is defined for a positive test charge $q$ , so that the field lines point away from a positive charge and toward a negative charge. (See [link] .) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the electric field for a point charge is $E=k|Q|/{r}^{2}$ and area is proportional to ${r}^{2}$ . This pictorial representation, in which field lines represent the direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields: electrostatic, gravitational, magnetic, and others.
In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created by each charge. The following example shows how to add electric field vectors.
Find the magnitude and direction of the total electric field due to the two point charges, ${q}_{1}$ and ${q}_{2}$ , at the origin of the coordinate system as shown in [link] .
Strategy
Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, $q$ , at point O, which allows us to determine the direction of the fields ${\mathbf{\text{E}}}_{1}$ and ${\mathbf{\text{E}}}_{2}$ . Once those fields are found, the total field can be determined using vector addition .
Solution
The electric field strength at the origin due to ${q}_{1}$ is labeled ${E}_{1}$ and is calculated:
$\begin{array}{}{E}_{1}=k\frac{{q}_{1}}{{r}_{1}^{2}}=\left(8\text{.}\text{99}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2}\right)\frac{\left(5\text{.}\text{00}×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{C}\right)}{{\left(2\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}\\ {E}_{1}=1\text{.}\text{124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}.\end{array}$
Similarly, ${E}_{2}$ is
$\begin{array}{}{E}_{2}=k\frac{{q}_{2}}{{r}_{2}^{2}}=\left(8\text{.}\text{99}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2}\right)\frac{\left(\text{10}\text{.}0×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{C}\right)}{{\left(4\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}\\ {E}_{2}=0\text{.}\text{5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}.\end{array}$
Four digits have been retained in this solution to illustrate that ${E}_{1}$ is exactly twice the magnitude of ${E}_{2}$ . Now arrows are drawn to represent the magnitudes and directions of ${\mathbf{\text{E}}}_{1}$ and ${\mathbf{\text{E}}}_{2}$ . (See [link] .) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for ${\mathbf{\text{E}}}_{1}$ is exactly twice the length of that for ${\mathbf{\text{E}}}_{2}$ . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field ${E}_{\text{tot}}$ is
$\begin{array}{lll}{E}_{\text{tot}}& =& \left({E}_{1}^{2}+{E}_{2}^{2}{\right)}^{\text{1/2}}\\ & =& {\left\{\left(\text{1.124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}{\right)}^{2}+\left(\text{0.5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}{\right)}^{2}\right\}}^{\text{1/2}}\\ & =& \text{1.26}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C.}\end{array}$
The direction is
$\begin{array}{lll}\theta & =& {\text{tan}}^{-1}\left(\frac{{E}_{1}}{{E}_{2}}\right)\\ & =& {\text{tan}}^{-1}\left(\frac{1\text{.}\text{124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}}{0\text{.}\text{5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}}\right)\\ & =& \text{63}\text{.}4º,\end{array}$
or $63.4º$ above the x -axis.
Discussion
In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
What is meant by dielectric charge?
what happens to the size of charge if the dielectric is changed?
omega= omega not +alpha t derivation
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
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Saeed
define variable velocity
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binding energy per nucleon
why God created humanity
Because HE needs someone to dominate the earth (Gen. 1:26)
Olorunfemi
Ali
Is the object in a conductor or an insulator? Justify your answer. whats the answer to this question? pls need help figure is given above
ok we can say body is electrically neutral ...conductor this quality is given to most metalls who have free electron in orbital d ...but human doesn't have ...so we re made from insulator or dielectric material ... furthermore, the menirals in our body like k, Fe , cu , zn
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when we face electric shock these elements work as a conductor that's why we got this shock
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What is position?
What is law of gravition | 2018-10-18T02:33:51 | {
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http://mathhelpforum.com/advanced-algebra/120637-exam-study-guide-help.html | # Thread: Exam study guide help
1. ## Exam study guide help
This is of my final study guide, and Im confused by the question please help....
For $H,K \leq G$ Set $H$~ $K$ when for some $g \in G, K = g^{-1} Hg$
Finish the Statement: "|[ $H$]_~| = 1 $\iff$ $H$ is...." and justify your answer
2. Originally Posted by ElieWiesel
This is of my final study guide, and Im confused by the question please help....
For $H,K \leq G$ Set $H$~ $K$ when for some $g \in G, K = g^{-1} Hg$
Finish the Statement: "|[ $H$]_~| = 1 $\iff$ $H$ is...." and justify your answer
What does $\left|\left[H\right]_\sim\right|$ mean?
3. $\left|\left[H\right]_\sim\right|=1\iff \text{ for all } g\in G, H=g^{-1}Hg \iff H\text{ is normal subgroup of }G$ | 2016-12-07T10:51:09 | {
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https://math.stackexchange.com/questions/1106071/if-operatornamerank-left-beginbmatrix-a-b-c-d-endbmatrix-right | # If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$ prove that $\det(AD)=\det(BC)$ [closed]
Let $A,\ B,\ C,\ D \in \mathcal{M}_n(\mathbb{C})$. If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$, prove that $\det(AD)=\det(BC)$.
## closed as off-topic by Najib Idrissi, Mark Fantini, Davide Giraudo, Ali Caglayan, Alice RyhlJan 24 '15 at 13:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
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Assume first that the first $n$ columns are independent. Then the last $n$ columns are linear combinations of the first $n$. Therefore we have the equality
$$\begin{bmatrix} A \\ C \end{bmatrix} \cdot V = \begin{bmatrix} B \\ D \end{bmatrix}$$
so we're done.
Now, if the first $n$ columns are not linearly independent, both sides are $0$.
Obs: to understand what $V$ is: the first column of $V$ consists of the coefficients in the writing of the first column of $\begin{bmatrix} B \\ D \end{bmatrix}$ as a linear combination of the columns of $\begin{bmatrix} A \\ C \end{bmatrix}$.
• Can you elaborate on what is $V$ ? – James S. Cook Jan 16 '15 at 1:06
• @JamesS.Cook: note that $AV=B$ and $CV=D$ then compute $\det(AD)=\det(ACV)$ and $\det(BC)=\det(AVC)$ – robjohn Jan 16 '15 at 1:10
• @James S. Cook: Just did, good call. – Orest Bucicovschi Jan 16 '15 at 1:10
• @robjohn thanks. I should have seen this, that said, his answer is improved by your comment. – James S. Cook Jan 16 '15 at 2:39 | 2019-07-20T11:46:31 | {
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https://math.stackexchange.com/questions/2856501/game-theory-applied-to-invoicing | # Game theory applied to invoicing
There is a traditional practise in precuement of almost any kind of paying an invoice at l ast thirty days after it has been received.
Is there any application of game theory here that can justify this practise as the most efficient Nash equilibrium for the repeated game where either two players play pay/don’t pay and 30 days/or immediately.
Or maybe two players playing 30 days/immediately and work with in future / or not?
Consider one-shot 2-player game as given in the table, where not paying results in punishment of $-\infty$. Thus, not paying will never be played. Also if player pays at the end of the game (vs the beginning of the game) she gets some benefits, such investing the cash and receiving interests for the 30 days period at monthly interest rate $r \in (0,1)$. Conversely, if a player gets paid at the beginning of the period she can invest that money, but she cant do so if she gets paid at the end of the period. Finally, suppose $a>b>0$.
The unique Nash equilibrium of the game is (Pay Later, Pay Later). Regarding efficiency, the Nash equilibrium outcome is weakly the most efficient outcome (only paying now by both players results in an inefficient outcome) and is equal to $r(a+b)$.
Note: The source of inefficiency of paying now by both players is a consequence of an implicit assumption that total resources of players are greater than $a$. | 2019-06-17T23:27:31 | {
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https://www.physicsforums.com/threads/outer-products-positive-semi-definiteness.840490/ | Outer products & positive (semi-) definiteness
1. Oct 30, 2015
economicsnerd
Let $u,v$ be vectors in the same Euclidean space, and define the symmetric matrix $M = uv'+vu'$, the sum of their two outer products.
I'm interested in whether or not $M$ is positive (semi)definite.
Does anybody know of any equivalent conditions that I might phrase "directly" in terms of the vectors $u,v$?
2. Oct 30, 2015
andrewkirk
An equivalent condition is $(x\cdot u)(x\cdot v)\geq 0$ for all vectors $x$.
3. Oct 30, 2015
fzero
Let $z$ be an arbitrary vector, then we are interested in the properties of
$$z^\dagger M z = \langle z, u \rangle \langle v, z \rangle + \langle z, v \rangle \langle u, z \rangle .$$
Assume that $v\neq u$, then we can write
$$z = a u + b v + z_\perp,~~~\langle z_\perp, u \rangle = \langle z_\perp, v \rangle =0.$$
Then, with $A = (a~~b)^T$, we have
$$z^\dagger M z = A^\dagger Q A,$$
where $Q$ is a Hermitian 2x2 form determined in terms of the inner products of $u$ and $v$. Positivity properties of $M$ are reduced to those of $Q$ for which it is simple to compute the eigenvalues. | 2018-07-19T10:49:36 | {
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https://openstax.org/books/college-physics-ap-courses/pages/12-1-flow-rate-and-its-relation-to-velocity | College Physics for AP® Courses
# 12.1Flow Rate and Its Relation to Velocity
College Physics for AP® Courses12.1 Flow Rate and Its Relation to Velocity
### Learning Objectives
By the end of this section, you will be able to:
• Calculate flow rate.
• Define units of volume.
• Describe incompressible fluids.
• Explain the consequences of the equation of continuity.
The information presented in this section supports the following AP® learning objectives and science practices:
• 5.F.1.1 The student is able to make calculations of quantities related to flow of a fluid, using mass conservation principles (the continuity equation). (S.P. 6.4, 7.2)
Flow rate $QQ size 12{Q} {}$ is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in Figure 12.2. In symbols, this can be written as
$Q=Vt,Q=Vt, size 12{Q= { {V} over {t} } } {}$
12.1
where $VV size 12{V} {}$ is the volume and $tt size 12{t} {}$ is the elapsed time.
The SI unit for flow rate is $m3/sm3/s size 12{m rSup { size 8{3} } "/s"} {}$, but a number of other units for $QQ size 12{Q} {}$ are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters ($10−3m310−3m3 size 12{"10" rSup { size 8{ - 3} } m rSup { size 8{3} } } {}$ or $103cm3103cm3 size 12{"10" rSup { size 8{3} } "cm" rSup { size 8{3} } } {}$). In this text we shall use whatever metric units are most convenient for a given situation.
Figure 12.2 Flow rate is the volume of fluid per unit time flowing past a point through the area $AA size 12{A} {}$. Here the shaded cylinder of fluid flows past point $PP size 12{P} {}$ in a uniform pipe in time $tt size 12{t} {}$. The volume of the cylinder is $AdAd size 12{ ital "Ad"} {}$ and the average velocity is $v ¯ =d/t v ¯ =d/t size 12{ {overline {v}} =d/t} {}$ so that the flow rate is $Q=Ad/t=A v ¯ Q=Ad/t=A v ¯ size 12{Q= ital "Ad"/t=A {overline {v}} } {}$.
### Example 12.1
#### Calculating Volume from Flow Rate: The Heart Pumps a Lot of Blood in a Lifetime
How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min?
#### Strategy
Time and flow rate $QQ size 12{Q} {}$ are given, and so the volume $VV size 12{V} {}$ can be calculated from the definition of flow rate.
#### Solution
Solving $Q=V/tQ=V/t size 12{Q=V/t} {}$ for volume gives
$V=Qt.V=Qt. size 12{V= ital "Qt"} {}$
12.2
Substituting known values yields
V = 5.00L1 min(75y)1m3103L5.26×105miny = 2.0×105 m3. V = 5.00L1 min(75y)1m3103L5.26×105miny = 2.0×105 m3. alignl { stack { size 12{V= left ( { {5 "." "00"" L"} over {"1 min"} } right ) $$"75"" y"$$ left ( { {1" m" rSup { size 8{3} } } over {"10" rSup { size 8{3} } " L"} } right ) left (5 "." "26" times "10" rSup { size 8{5} } { {"min"} over {y} } right )} {} # " "=2 "." 0 times "10" rSup { size 8{5} } " m" rSup { size 8{3} } {} } } {}
12.3
#### Discussion
This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of water contained in a 6-lane 50-m lap pool.
Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. The precise relationship between flow rate $QQ size 12{Q} {}$ and velocity $v ¯ v ¯ size 12{ {overline {v}} } {}$ is
$Q = A v ¯ , Q = A v ¯ , size 12{Q=A {overline {v}} } {}$
12.4
where $AA size 12{A} {}$ is the cross-sectional area and $v ¯ v ¯ size 12{ {overline {v}} } {}$ is the average velocity. This equation seems logical enough. The relationship tells us that flow rate is directly proportional to both the magnitude of the average velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. Figure 12.2 illustrates how this relationship is obtained. The shaded cylinder has a volume
$V=Ad,V=Ad, size 12{V= ital "Ad"} {}$
12.5
which flows past the point $PP size 12{P} {}$ in a time $tt size 12{t} {}$. Dividing both sides of this relationship by $tt size 12{t} {}$ gives
$Vt=Adt.Vt=Adt. size 12{ { {V} over {t} } = { { ital "Ad"} over {t} } } {}$
12.6
We note that $Q=V/tQ=V/t size 12{Q=V/t} {}$ and the average speed is $v ¯ =d/t v ¯ =d/t size 12{ {overline {v}} =d/t} {}$. Thus the equation becomes $Q=A v ¯ Q=A v ¯ size 12{Q=A {overline {v}} } {}$.
Figure 12.3 shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2,
$Q 1 = Q 2 A 1 v ¯ 1 = A 2 v ¯ 2 } . Q 1 = Q 2 A 1 v ¯ 1 = A 2 v ¯ 2 } . size 12{ left none matrix { Q rSub { size 8{1} } =Q rSub { size 8{2} } {} ## A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } right rbrace "." } {}$
12.7
This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.
Figure 12.3 When a tube narrows, the same volume occupies a greater length. For the same volume to pass points 1 and 2 in a given time, the speed must be greater at point 2. The process is exactly reversible. If the fluid flows in the opposite direction, its speed will decrease when the tube widens. (Note that the relative volumes of the two cylinders and the corresponding velocity vector arrows are not drawn to scale.)
Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion.
### Making Connections: Incompressible Fluid
The continuity equation tells us that the flow rate must be the same throughout an incompressible fluid. The flow rate, Q, has units of volume per unit time (m3/s). Another way to think about it would be as a conservation principle, that the volume of fluid flowing past any point in a given amount of time must be conserved throughout the fluid.
For incompressible fluids, we can also say that the mass flowing past any point in a given amount of time must also be conserved. That is because the mass of a given volume of fluid is just the density of the fluid multiplied by the volume:
$m=ρVm=ρV$
12.8
When we say a fluid is incompressible, we mean that the density of the fluid does not change. Every cubic meter of fluid has the same number of particles. There is no room to add more particles, nor is the fluid allowed to expand so that the particles will spread out. Since the density is constant, we can express the conservation principle as follows for any two regions of fluid flow, starting with the continuity equation:
12.9
12.10
12.11
More generally, we say that the mass flow rate $(ΔmΔt)(ΔmΔt)$ is conserved.
### Example 12.2
#### Calculating Fluid Speed: Speed Increases When a Tube Narrows
A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle.
#### Strategy
We can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and 2 for the nozzle.
#### Solution for (a)
First, we solve $Q=A v ¯ Q=A v ¯ size 12{Q=A {overline {v}} } {}$ for $v1v1 size 12{v rSub { size 8{1} } } {}$ and note that the cross-sectional area is $A=πr2A=πr2 size 12{A=πr rSup { size 8{2} } } {}$, yielding
$v ¯ 1=QA1=Q πr 1 2 . v ¯ 1=QA1=Q πr 1 2 . size 12{ {overline {v rSub { size 8{1} } }} = { {Q} over {A rSub { size 8{1} } } } = { {Q} over {πr rSub { size 8{1} rSup { size 8{2} } } } } } {}$
12.12
Substituting known values and making appropriate unit conversions yields
$v ¯ 1=(0.500L/s)(10−3m3/L)π(9.00×10−3m)2=1.96m/s. v ¯ 1=(0.500L/s)(10−3m3/L)π(9.00×10−3m)2=1.96m/s. size 12{ {overline {v rSub { size 8{1} } }} = { { $$0 "." "500"" L/s"$$ $$"10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } /L$$ } over {π $$9 "." "00" times "10" rSup { size 8{ - 3} } " m"$$ rSup { size 8{2} } } } =1 "." "96"" m/s"} {}$
12.13
#### Solution for (b)
We could repeat this calculation to find the speed in the nozzle $v ¯ 2 v ¯ 2 size 12{ {overline {v rSub { size 8{2} } }} } {}$, but we will use the equation of continuity to give a somewhat different insight. Using the equation which states
$A1 v ¯ 1=A2 v ¯ 2,A1 v ¯ 1=A2 v ¯ 2, size 12{A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}$
12.14
solving for $v ¯ 2 v ¯ 2 size 12{ {overline {v rSub { size 8{2} } }} } {}$ and substituting $πr2πr2 size 12{πr rSup { size 8{2} } } {}$ for the cross-sectional area yields
$v ¯ 2=A1A2 v ¯ 1= πr 1 2 πr 2 2 v ¯ 1=r12r22 v ¯ 1. v ¯ 2=A1A2 v ¯ 1= πr 1 2 πr 2 2 v ¯ 1=r12r22 v ¯ 1. size 12{ {overline {v rSub { size 8{2} } }} = { {A rSub { size 8{1} } } over {A rSub { size 8{2} } } } {overline {v rSub { size 8{1} } }} = { {πr rSub { size 8{1} rSup { size 8{2} } } } over {πr rSub { size 8{2} rSup { size 8{2} } } } } {overline {v rSub { size 8{1} } }} = { {r rSub { size 8{1} rSup { size 8{2} } } } over {r rSub { size 8{2} rSup { size 8{2} } } } } {overline {v rSub { size 8{1} } }} } {}$
12.15
Substituting known values,
$v ¯ 2=(0.900cm)2(0.250cm)21.96m/s=25.5 m/s. v ¯ 2=(0.900cm)2(0.250cm)21.96m/s=25.5 m/s. size 12{ {overline {v rSub { size 8{2} } }} = { { $$0 "." "900"" cm"$$ rSup { size 8{2} } } over { $$0 "." "250"" cm"$$ rSup { size 8{2} } } } 1 "." "96"" m/s"="25" "." "5 m/s"} {}$
12.16
#### Discussion
A speed of 1.96 m/s is about right for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.
### Making Connections: Different-Sized Pipes
For incompressible fluids, the density of the fluid remains constant throughout, no matter the flow rate or the size of the opening through which the fluid flows. We say that, to ensure continuity of flow, the amount of fluid that flows past any point is constant. That amount can be measured by either volume or mass.
Flow rate has units of volume/time (m3/s or L/s). Mass flow rate $( Δm Δt ) ( Δm Δt )$ has units of mass/time (kg/s) and can be calculated from the flow rate by using the density:
12.17
The average mass flow rate can be found from the flow rate:
12.18
Suppose that crude oil with a density of 880 kg/m3 is flowing through a pipe with a diameter of 55 cm and a speed of 1.8 m/s. Calculate the new speed of the crude oil when the pipe narrows to a new diameter of 31 cm, and calculate the mass flow rate in both sections of the pipe, assuming the density of the oil is constant throughout the pipe.
Solution: To calculate the new speed, we simply use the continuity equation.
Since the cross section of a pipe is a circle, the area of each cross section can be found as follows:
For the larger pipe:
12.19
For the smaller pipe:
12.20
So the larger part of the pipe (A1) has a cross-sectional area of 0.238 m2, and the smaller part of the pipe (A2) has a cross-sectional area of 0.0755 m2. The continuity equation tells us that the oil will flow faster through the portion of the pipe with the smaller cross-sectional area. Using the continuity equation, we get
12.21
12.22
So we find that the oil is flowing at a speed of 1.8 m/s through the larger section of the pipe (A1), and it is flowing much faster (5.7 m/s) through the smaller section (A2).
The mass flow rate in both sections should be the same.
For the larger portion of the pipe:
12.23
For the smaller portion of the pipe:
12.24
And so mass is conserved throughout the pipe. Every second, 380 kg of oil flows out of the larger portion of the pipe, and 380 kg of oil flows into the smaller portion of the pipe.
The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.
In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes
$n1A1 v ¯ 1=n2A2 v ¯ 2,n1A1 v ¯ 1=n2A2 v ¯ 2, size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}$
12.25
where $n1n1 size 12{n rSub { size 8{1} } } {}$ and $n2n2 size 12{n rSub { size 8{2} } } {}$ are the number of branches in each of the sections along the tube.
### Example 12.3
#### Calculating Flow Speed and Vessel Diameter: Branching in the Cardiovascular System
The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is $8.0μm8.0μm$, calculate the number of capillaries in the blood circulatory system.
#### Strategy
We can use $Q=A v ¯ Q=A v ¯ size 12{Q=A {overline {v}} } {}$ to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.
#### Solution for (a)
The flow rate is given by $Q=A v ¯ Q=A v ¯ size 12{Q=A {overline {v}} } {}$ or $v ¯ =Qπr2 v ¯ =Qπr2 size 12{ {overline {v}} = { {Q} over {πr rSup { size 8{2} } } } } {}$ for a cylindrical vessel.
Substituting the known values (converted to units of meters and seconds) gives
$v ¯ = 5.0 L/min 10 − 3 m 3 /L 1 min/ 60 s π 0 . 010 m 2 = 0 . 27 m/s . v ¯ = 5.0 L/min 10 − 3 m 3 /L 1 min/ 60 s π 0 . 010 m 2 = 0 . 27 m/s . size 12{ { bar {v}}= { { left (5 "." 0"L/min" right ) left ("10" rSup { size 8{ - 3} } m rSup { size 8{3} } "/L" right ) left (1"min/""60"s right )} over {π left (0 "." "010 m" right ) rSup { size 8{2} } } } =0 "." "27"`"m/s"} {}$
12.26
#### Solution for (b)
Using $n1A1 v ¯ 1=n2A2 v ¯ 1n1A1 v ¯ 1=n2A2 v ¯ 1 size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}$, assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for $n2n2 size 12{n rSub { size 8{2} } } {}$ (the number of capillaries) gives $n2= n 1 A 1 v ¯ 1 A 2 v ¯ 2 n2= n 1 A 1 v ¯ 1 A 2 v ¯ 2$. Converting all quantities to units of meters and seconds and substituting into the equation above gives
$n 2 = 1 π 10 × 10 − 3 m 2 0.27 m/s π 4.0 × 10 − 6 m 2 0.33 × 10 − 3 m/s = 5.0 × 10 9 capillaries . n 2 = 1 π 10 × 10 − 3 m 2 0.27 m/s π 4.0 × 10 − 6 m 2 0.33 × 10 − 3 m/s = 5.0 × 10 9 capillaries . size 12{n rSub { size 8{2} } = { { left (1 right ) left (π right ) left ("10" times "10" rSup { size 8{ - 3} } " m" right ) rSup { size 8{2} } left (0 "." "27"" m/s" right )} over { left (π right ) left (4 "." 0 times "10" rSup { size 8{ - 6} } " m" right ) rSup { size 8{2} } left (0 "." "33" times "10" rSup { size 8{ - 3} } " m/s" right )} } =5 "." 0 times "10" rSup { size 8{9} } " capillaries"} {}$
12.27
#### Discussion
Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per $mm3mm3 size 12{"mm" rSup { size 8{3} } } {}$, or about $200×106200×106 size 12{"200" times "10" rSup { size 8{6} } } {}$ per 1 kg of muscle. For 20 kg of muscle, this amounts to about $4×1094×109 size 12{4 times "10" rSup { size 8{9} } } {}$ capillaries.
### Making Connections: Syringes
A horizontally oriented hypodermic syringe has a barrel diameter of 1.2 cm and a needle diameter of 2.4 mm. A plunger pushes liquid in the barrel at a rate of 4.0 mm/s. Calculate the flow rate of liquid in both parts of the syringe (in mL/s) and the velocity of the liquid emerging from the needle.
Solution:
First, calculate the area of both parts of the syringe:
12.28
12.29
Next, we can use the continuity equation to find the velocity of the liquid in the smaller part of the barrel (v2):
12.30
12.31
12.32
Double-check the numbers to be sure that the flow rate in both parts of the syringe is the same:
12.33
12.34
Finally, by converting to mL/s:
12.35
Order a print copy
As an Amazon Associate we earn from qualifying purchases. | 2021-04-19T06:29:53 | {
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http://mathhelpforum.com/number-theory/121228-congreunce-modulo-49-a.html | 1. ## Congreunce modulo 49
Prove that :
$\displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$
2. By using the squared method:
$\displaystyle 6^{33}\equiv 6^{32}6\equiv 36^{16}6\equiv (-13)^{16}6$
$\displaystyle \equiv 13^{16}6\equiv 169^{8}6\equiv 22^{8}6\equiv 484^{4}6$$\displaystyle \equiv (-6)^{4}6\equiv 36^{2}6\equiv 13^{2}6\equiv 22\cdot 6\equiv 132\equiv 34 I leave it to you to find \displaystyle 8^{33}. 3. Entering this into my computer gives \displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right] which might be why you are having trouble 4. \displaystyle 8^{33}\equiv 36 (mod 49) thus we get 21 not 0. There may be other better method to prove this congruence equation. 5. Originally Posted by badgerigar Entering this into my computer gives \displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$$\displaystyle$
which might be why you are having trouble
Hello thank you you have 21
6. Hello The general method is :
$\displaystyle \begin{array}{l} 8^0 \equiv 1\left[ {\bmod 49} \right] \\ 8^1 \equiv 8\left[ {\bmod 49} \right] \\ 8^2 \equiv 15\left[ {\bmod 49} \right] \\ 8^3 \equiv 22\left[ {\bmod 49} \right] \\ 8^4 \equiv 29\left[ {\bmod 49} \right] \\ 8^5 \equiv 36\left[ {\bmod 49} \right] \\ 8^6 \equiv 3\left[ {\bmod 49} \right] \\ 8^7 \equiv 1\left[ {\bmod 49} \right] \\ \end{array}$
The period is 7 then : n = 7k +p
but 33= 7×4+5
$\displaystyle 8^{33} \equiv 36\left[ {\bmod 49} \right]$
$\displaystyle same method : \begin{array}{l} 6^0 \equiv 1\left[ {\bmod 49} \right] \\ 6^1 \equiv 6\left[ {\bmod 49} \right] \\ 6^2 \equiv 36\left[ {\bmod 49} \right] \\ 6^3 \equiv 20\left[ {\bmod 49} \right] \\ 6^4 \equiv 22\left[ {\bmod 49} \right] \\ 6^5 \equiv 34\left[ {\bmod 49} \right] \\ 6^6 \equiv 8\left[ {\bmod 49} \right] \\ 6^7 \equiv 1\left[ {\bmod 49} \right] \\ \end{array}$
The period is 7 then : n = 7k +d
but 33= 7×4+5
$\displaystyle 6^{33} \equiv 34\left[ {\bmod 49} \right]$
Conclusion :
$\displaystyle 6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]$
$\displaystyle 6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]because\left( {34 + 36} \right) \equiv 21\left[ {\bmod 49} \right]$ | 2018-04-23T10:49:01 | {
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http://mathhelpforum.com/calculus/7960-calculus-assignment-help.html | # Thread: Calculus Assignment - HELP!!!
1. ## Calculus Assignment - HELP!!!
If anyone could help me with these questions, that would be really really great.
1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.
2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.
3. Determine lim x-->0 (3rdroot(x-8)+2)/x
4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.
2. 2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.
$f(x)=\frac{ax+b}{(x-1)(x-4)}$
Setting f(x) =-1 and subbing in x=2, we get $f(x)=-a-\frac{b}{2}$
$f'(x)=\frac{-ax^{2}+2bx-4a-5b}{(x-1)^{2}(x-4)^{2}}$
Sub in x=2 into f'(x), we get $f'(x)=\frac{b}{4}$
Since the line is horizontal, we have $\frac{b}{4}=0$
We have two small equations in 2 unknowns:
$-a-\frac{b}{2}=-1$
$\frac{b}{4}=0$
Solve for a and b
Originally Posted by philipsach1
4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.
$f(x)=\frac{x^{4}-2x^{3}+3}{-6x}$
$f'(x)=\frac{-3x^{4}+4x^{3}+3}{6x^{2}}$
Sub x=-1 into f(x) and we find y=1
Sub x=-1 into f'(x) to find the slope at x=-1. We get m=-2/3
Use the given coordinates and the new found slope to find the equation of our line:
$y=mx+b$
$1=(\frac{-2}{3})(-1)+b$
$b=\frac{1}{3}$
Therefore, the equation of the line is $y=\frac{-2}{3}x+\frac{1}{3}$
Set f(x) and the line equation equal and solve for x:
$\frac{x^{4}-2x^{3}+3}{-6x}=\frac{-2}{3}x+\frac{1}{3}$
$3x^{4}-6x^{3}-12x^{2}+6x+9=0$
$3(x+1)^{2}(x-1)(x-3)$
3. ## ???
Does anyone else got anything for the first three questions???
4. Originally Posted by philipsach1
1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.
$f(x) = \frac{2x^2 - 1}{x}$
$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{ \frac{2(x+h)^2 - 1}{x + h} - \frac{2x^2 - 1}{x} }{h}$
$f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 - 1)x - (2x^2 - 1)(x + h)}{hx(x+h)}$
$f'(x) = \lim_{h \to 0} \frac{2x^3 + 4x^2h + 2xh^2 - x - 2x^3 - 2x^2h + x + h}{hx(x+h)}$
$f'(x) = \lim_{h \to 0} \frac{2x^2h + 2xh^2 + h}{hx(x+h)}$
$f'(x) = \lim_{h \to 0} \frac{2x^2 + 2xh + 1}{x(x+h)}$
$f'(x) = \frac{2x^2 + 1}{x^2}$
-Dan
5. Thank you soooo much!
If I could just get an answer for number three, I will be the happiest person!
6. Originally Posted by philipsach1
3. Determine lim x-->0 (3rdroot(x-8)+2)/x
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x}$
This is in the form of $\frac{0}{0}$ so use L'Hopital's rule:
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}$
= $\frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$
-Dan
7. ## Sorry
Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.
8. Originally Posted by philipsach1
Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.
No problem!
Note that, for x = 0 the numerator becomes:
$\sqrt[3]{0-8} + 2 = \sqrt[3]{-8} + 2 = -2+ 2 = 0$
So for x = 0 the fraction takes on the form $\frac{0}{0}$. This is one of the conditions under which we may use L'Hopital's rule.
L'Hopital's rule says that:
$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$ when $\lim_{x \to a}f(x) = 0$ and $\lim_{x \to a}g(x) = 0$.
So:
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} \to \frac{0}{0}$
We need to take the derivative of both the numerator and denominator.
Numerator:
$\frac{d}{dx}(\sqrt[3]{x-8} + 2 ) = \frac{d}{dx}((x - 8)^{1/3} + 2)$
= $\frac{1}{3}(x - 8)^{-2/3} = \frac{1}{3\sqrt[3]{(x - 8)^2}}$
Denominator:
$\frac{d}{dx}x = 1$
So:
$\lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}$
= $\frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$
-Dan
9. I'm sorry, correct me if I'm wrong but $(0-8)^{\frac{1}{3}}=1+\sqrt{3}i$
I ran this through the calc to be sure, but I get an undefined limit.
10. Hello,
have a look here: http://www.mathhelpforum.com/math-he...jmailloux.html
Are you both in the same class?
EB
11. Originally Posted by galactus
I'm sorry, correct me if I'm wrong but $(0-8)^{\frac{1}{3}}=1+\sqrt{3}i$
I ran this through the calc to be sure, but I get an undefined limit.
Hello, galactus,
$(0-8)^{\frac{1}{3}}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2$
Originally Posted by galactus
I ran this through the calc to be sure, but I get an undefined limit.
The limit should be calculated with x approaching zero.
EB
12. Originally Posted by galactus
I ran this through the calc to be sure, but I get an undefined limit.
I'm curious. The screenshot looks like a TI-92, but mine will calculate the limit. What calculator are you using?
-Dan
13. I also have a 92. I used x approaching 0. It looks funny in the post.
My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.
14. Originally Posted by galactus
I also have a 92. I used x approaching 0. It looks funny in the post.
My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.
No apologies necessary.
-Dan | 2013-05-20T09:06:36 | {
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https://danmackinlay.name/notebook/correlograms.html | # Correlograms
## Also covariances
This material is revised and expanded from the appendix of draft versions of a recent conference submission, for my own reference. I used (deterministic) correlograms a lot in that, and it was startlingly hard to find a decent summary of their properties anywhere. Nothing new here, but… see the matrial about doing this in a probabilistic way via Wiener-Khintchine representation and covariance kernels which lead to a natural probabilistic spectral analysis.
Consider an $$L_2$$ signal $$f: \bb{R}\to\bb{R}.$$ We will frequently overload notation and refer to as signal with free argument $$t$$, so that $$f(rt-\xi),$$ for example, refers to the signal $$t\mapsto f(rt-\xi).$$ We write the inner product between signals $$t\mapsto f(t)$$ and $$t\mapsto f'(t)$$ as $$\inner{f(t)}{f'(t)}$$. Where it is not clear that the free argument is, e.g. $$t$$, we will annotate it $$\finner{f(t)}{f'(t)}{t}$$.
The correlogram $$\cc{A}:L_2(\bb{R}) \to L_2(\bb{R})$$ maps signals to signals. Specifically, $$\mathcal{A}\{f\}$$ is a signal $$\bb{R}\to\bb{R}$$ such that
$\mathcal{A}\{f\}:=\xi \mapsto \finner{ f(t) }{ f(t-\xi) }{t}$ This is the covariance between $$f(t)$$ and $$f(t-\xi).$$ (Note that we here discuss the covariance between given deterministic signals, not between two stochastic sources; covariance of stochastic processes is a broader, let alone inferring the covariance of stochastic processes.) Note also that this is what I would call an autocovariance not an auto-correlation, since it’s not normalized, but I’ll stick with the latter for now since for reasons of convention.
We derive the properties of this transform.
Multiplication by a constant. Consider a constant $$c\in \bb{R}.$$
\begin{aligned}\mathcal{A}\{cf\}(\xi)&= \inner{ cf(t) }{ cf(t-\xi) }\\ &= c^2\finner{ f(t) }{ f(t-\xi) }{t}\\ &= c^2\mathcal{A}\{f\}(\xi).\\ \end{aligned}
Time scaling:
\begin{aligned}\mathcal{A}\{f(r t)\}(\xi) &=\finner{ f(r t) }{ f(r t-\xi) }{t}\\ &= \int f(r t)f(r t-\xi)\dd t\\ &= \frac{1}{r }\int f(t)f(t-\frac{\xi}{r})\dd t\\ &= \frac{1}{r} \mathcal{A}\{f\}\left(\frac{\xi}{r}\right)\\ \end{aligned}
\begin{aligned}\mathcal{A}\{f+f'\}(\xi) &=\finner{ f(t)+f'(t) }{ f(t-\xi)+f'(t-\xi) }{t}\\ &=\finner{ f(t) }{ f(t-\xi)\rangle+\langle f(t),f'(t-\xi) }{t} +\finner{ f'(t) }{ f(t-\xi)\rangle+\langle f'(t),f'(t-\xi) }{t}\\ &= \mathcal{A}\{f\}(\xi)+ \finner{ f'(t) }{ f(t-\xi)}{t} +\finner{f(t)}{f'(t-\xi) }{t} +\mathcal{A}\{f'\}(\xi).\\ &= \mathcal{A}\{f\}(\xi)+ \finner{ f'(t) }{ f(t-\xi)}{t} +\finner{f(t+\xi)}{f'(t) }{t} +\mathcal{A}\{f'\}(\xi).\\ &= \mathcal{A}\{f\}(\xi)+ \finner{ f'(t) }{ f(t-\xi)}{t} +\finner{f'(t) }{f(t+\xi)}{t} +\mathcal{A}\{f'\}(\xi).\\ \end{aligned}
We can say little about the term $$\finner{ f'(t) }{ f(t-\xi)}+\finner{f'(t) }{f(t+\xi)}{t}$$ without more information about the signals in question. However, we can solve a randomized version. Suppose $$S_i, \, i \in\bb{N}$$ are i.i.d. Rademacher variables, i.e. that they assume a value in $$\{+1,-1\}$$ with equal probability. Then, we can introduce the following property:
\begin{aligned} \bb{E}[ \mathcal{A}\{S_1f + S_2f'\}(\xi) &=\bb{E}[ \mathcal{A}\{S_1f\}(\xi) + \finner{ S_2 f'(t) }{ S_1 f(t-\xi)}{t} +\finner{S_2f'(t) }{S_1 f(t+\xi)}{t} +\mathcal{A}\{S_2f'\}(\xi)]\\ &=\bb{E}[ \mathcal{A}\{S_1f\}(\xi)] + \bb{E}\finner{ S_2 f'(t) }{ S_1 f(t-\xi)}{t} + \bb{E}\finner{S_2f'(t) }{S_1 f(t+\xi)}{t} +\bb{E}[ \mathcal{A}\{S_2f'\}(\xi)]\\ &=\mathcal{A}\{f\}(\xi)+ \bb{E}[ S_1S_2]\finner{ f'(t) }{ f(t-\xi) }{t} + \bb{E}[ S_1S_2]\finner{ f'(t) }{ f(t+\xi) }{t}+\mathcal{A}\{f'\}(\xi)\\ &=\mathcal{A}\{f\}(\xi)+ \mathcal{A}\{f'\}(\xi)\\ \end{aligned}
## References
Bochner, Salomon. 1959. Lectures on Fourier Integrals. Princeton University Press. https://books.google.com?id=MWCYDwAAQBAJ.
Brown, Judith C., and Miller S. Puckette. 1989. “Calculation of a “Narrowed” Autocorrelation Function.” The Journal of the Acoustical Society of America 85 (4, 4): 1595–1601. https://doi.org/10.1121/1.397363.
Cariani, P. A., and B. Delgutte. 1996. “Neural Correlates of the Pitch of Complex Tones. I. Pitch and Pitch Salience.” Journal of Neurophysiology 76 (3, 3): 1698–1716. https://doi.org/10.1152/jn.1996.76.3.1698.
Cheveigné, Alain de, and Hideki Kawahara. 2002. YIN, a Fundamental Frequency Estimator for Speech and Music.” The Journal of the Acoustical Society of America 111 (4, 4): 1917–30. https://doi.org/10.1121/1.1458024.
Kaso, Artan. 2018. “Computation of the Normalized Cross-Correlation by Fast Fourier Transform.” PLOS ONE 13 (9): e0203434. https://doi.org/10.1371/journal.pone.0203434.
Khintchine, A. 1934. “Korrelationstheorie der stationären stochastischen Prozesse.” Mathematische Annalen 109 (1, 1): 604–15. https://doi.org/10.1007/BF01449156.
Langner, Gerald. 1992. “Periodicity Coding in the Auditory System.” Hearing Research 60 (2, 2): 115–42. https://doi.org/10.1016/0378-5955(92)90015-F.
Lewis, J. P. 1995. “Fast Template Matching.” In. Quebec City, Canada: Canadian Image Processing and Pattern Recognition Society,. http://www.scribblethink.org/Work/nvisionInterface/vi95_lewis.pdf.
Licklider, J. C. R. 1951. “A Duplex Theory of Pitch Perception.” Experientia 7 (4, 4): 128–34. https://doi.org/10.1007/BF02156143.
Loynes, R. M. 1968. “On the Concept of the Spectrum for Non-Stationary Processes.” Journal of the Royal Statistical Society. Series B (Methodological) 30 (1): 1–30. https://www.jstor.org/stable/2984457.
Ma, Ning, Phil Green, Jon Barker, and André Coy. 2007. “Exploiting Correlogram Structure for Robust Speech Recognition with Multiple Speech Sources.” Speech Communication 49 (12, 12): 874–91. https://doi.org/10.1016/j.specom.2007.05.003.
Morales-Cordovilla, J. A., A. M. Peinado, V. Sanchez, and J. A. Gonzalez. 2011. “Feature Extraction Based on Pitch-Synchronous Averaging for Robust Speech Recognition.” IEEE Transactions on Audio, Speech, and Language Processing 19 (3): 640–51. https://doi.org/10.1109/TASL.2010.2053846.
Rabiner, L. 1977. “On the Use of Autocorrelation Analysis for Pitch Detection.” IEEE Transactions on Acoustics, Speech, and Signal Processing 25 (1, 1): 24–33. https://doi.org/10.1109/TASSP.1977.1162905.
Slaney, M., and R. F. Lyon. 1990. “A Perceptual Pitch Detector.” In Proceedings of ICASSP, 357–360 vol.1. https://doi.org/10.1109/ICASSP.1990.115684.
Sondhi, M. 1968. “New Methods of Pitch Extraction.” IEEE Transactions on Audio and Electroacoustics 16 (2, 2): 262–66. https://doi.org/10.1109/TAU.1968.1161986.
Tan, L. N., and A. Alwan. 2011. “Noise-Robust F0 Estimation Using SNR-Weighted Summary Correlograms from Multi-Band Comb Filters.” In 2011 IEEE International Conference on Acoustics, Speech and Signal Processing (ICASSP), 4464–67. https://doi.org/10.1109/ICASSP.2011.5947345.
Wiener, Norbert. 1930. “Generalized Harmonic Analysis.” Acta Mathematica 55: 117–258. https://doi.org/10.1007/BF02546511.
Warning! Experimental comments system! If is does not work for you, let me know via the contact form. | 2021-04-18T23:21:58 | {
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https://www.finitelygenerated.com/page/2/ | # Finitely Generated
Andres Mejia's Blog
## A Geometric Reason to Care About Discrete Valuation Rings
Perhaps those that have taken an introductory commutative algebra class have come across the profoundly opaque definition of a discrete valuation ring (DVR), thinking to themselves "why have the math gods condemned me?" From a purely algebraic perspective, it is a remarkable that a ring $A$ is a valuation ring if and only if it is a domain so that for any element in the fraction field for $A$, $x$ or $x^{-1}$ is contained in $A$. Even further, if a valuation ring is Noetherian (and not a field) it is a DVR. Again, algebraically, there are a lot of reasons to care about these rings, but today isn't about algebra. While there are a huge number of equivalences we can speak of, but instead we will make use of the following definition: Definition:…
## Pullbacks of maximal ideals in a finitely generated k-algebra are again maximal
A few days ago, some classmates and I were thinking about about whether or not morphisms in $\mathrm{maxSpec}(A)$ were well defined. In particular, given a map of $k$-algebras $f:A \to B$, we consider the induced morphism $f^*:\mathrm{maxSpec} A \to \mathrm{maxSpec B}$ given by $X \mapsto f^{-1}(X)$. We all knew that pullbacks of maximal ideals need not be maximal for ring morphisms, which is precisely what led to a great deal of confusion-- thankfully this was all fixed as soon as we were reminded that we were considering morphisms of $k$-algebras. While this was funny, I think the proof for this fact is both an elementary and interesting check up in commutative algebra: Theorem 1: let $A,B$ be finitely generated $k$-algebras. Then the pullback of any maximal ideal in a $k$-algebra homomorphism is again… | 2019-03-19T04:54:21 | {
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https://quant.stackexchange.com/questions/29698/why-does-one-factor-short-rate-model-tend-to-produce-parallel-shift-of-the-yield?noredirect=1 | # Why does one-factor short-rate model tend to produce parallel shift of the yield curve?
I understand that one factor short rate model models the instantaneous rate given any moment in time. Can anyone explain how to derive a term structure from a short rate model and show that one-factor short-rate model tend to produce parallel shift of the yield curve?
• A model itself does not produce a parallel shift. Are you talking about certain risk? The duration is based on parallel shift to the yield. A short rate model produce bond prices with perfectly correlated increments. – Gordon Aug 18 '16 at 1:16
• I guess a change of certain parameters produces parallel shift of yield curve? – user1559897 Aug 18 '16 at 1:18
• You may need to modify your question to make it clear; otherwise, we do not know what you are asking. – Gordon Aug 18 '16 at 1:19
• @Gordon He's asking how to prove perfect correlation for the curve points in Vasicek model. I don't know to prove it but I know that's because the model doesn't have enough parameters to address the issue. – HelloWorld Aug 18 '16 at 6:18
• @Gordon Aren't they the same thing? Zero rates should be perfectly correlated with bond price? They are just different representation of the same data. – HelloWorld Aug 18 '16 at 13:20
This has already been explained at the start of Chapter 4 in Brigo's book. Basically, for any affine model of the short rate $r_t$, the zero-coupon bond price has the form \begin{align*} P(t, T) = A(t, T)e^{-B(t, T) r_t}, \end{align*} where $A(t, T)$ and $B(t, T)$ are deterministic functions. The yield, or zero rate, is given by \begin{align*} R(t, T) &= -\frac{\ln P(t, T)}{T-t}\\ &=-\frac{\ln A(t, T)}{T-t} + \frac{B(t, T)}{T-t} r_t\\ &=:a(t, T) + b(t, T) r_t. \end{align*} Then \begin{align*} {\rm Corr}\big(R(t, T_1), R(t, T_2) \big) &= {\rm Corr}\big(a(t, T_1) + b(t, T_1 r_t, a(t, T_2) + b(t, T_2) r_t \big)\\ &={\rm Corr}(r_t, r_t) =1. \end{align*} That is, at any time $t$, the yield to any two maturity dates are perfectly correlated, and any shift to a single yield causes a parallel shift to the whole yield curve.
• A term structure, at time $t$, can mean either the yield $R(t, T)$ or the bond price $P(t, T)$ for a sequence of maturity dates $T$. An example of such derivation is pointed out in the last line of the answer. – Gordon Aug 18 '16 at 16:17 | 2020-09-21T07:37:46 | {
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https://trueq.quantumbenchmark.com/guides/fundamentals/qudit_intro.html | Example: Quantum Circuits with Qudits
Quantum systems used for quantum computing contain quantized energy levels which can be used to define computational states. Most quantum computing devices follow the standard classical convention of using a two-level system with basis states labeled by 0 and 1, that is a canonical qubit.
Since quantum systems generally have more than two energy levels available, the restriction to only two of them does not allow the quantum device to fully leverage the available state space. Additionally, the presence of unused levels can lead to loss of information as the state can “leak” into unused energy levels. For this reason, some hardware developers choose to design their quantum computing devices to use more than the standard two levels. In these cases, the multi-level unit is referred to as a qudit. Many of our tutorials refer only to qubits for simplicity. However, our protocols apply more generally to qudits.
The dimension of a qudit refers to the number of energy levels used to store information. True-Q™ supports qudit systems of dimensions 2, 3, 5 and 7. These prime dimensions are compatible with the stabilizer formalism and measurements for these dimensions produce single-digit outputs.
In this tutorial, we provide a basic introduction to qudits and show how to start creating True-Q™ circuits for qudits. We direct users to Advanced Qudit Framework for a more in-depth introduction to the mathematical framework used to model qudit computations and an overview of the classes used to represent the operations defined for that framework.
We define the computational basis of a qudit to be an orthogonal set of states, each of which corresponds to a single energy level,
$\{\ket{0}, \ket{1}, ..., \ket{d-1}\}.$
A qudit state $$\ket{\psi}$$ can then be expressed as a linear combination of these basis states,
$\ket{\psi}=\sum_{a=0}^{d-1}c_a(\psi)\ket{a},$
where $$c_a(\psi)$$ are constants.
Using Qudits in True-Q™
Qudit circuits can be constructed in the same way as regular qubit circuits, and in fact most of the True-Q™ tools work with no or very minor syntax modifications for qudits.
We provide gate aliases for standard qudit gates that follow the same naming convention as the qubit gates but with the dimension attached to their name, e.g. the alias tq.Gate.x3 represents a qutrit $$X$$ gate. Using these aliases, defining circuits is straight-forward:
We can specify the qudit dimension we want to work in by setting the global dimensional parameter:
[2]:
import trueq as tq
# set the global dimension to 3
tq.settings.set_dim(3)
To start off, let’s define a simple circuit for a two-qutrit system that consists of a single-qutrit $$X$$ gate, a single-qutrit $$F$$ or Fourier gate (the higher-dimensional generalization of the Hadamard gate), and a two-qutrit controlled-$$X$$ gate. Note that we define gate aliases for standard qudit gates following the same convention of qubit gates but with the dimension appended (e.g. the alias tq.Gate.x3 represents a qutrit $$X$$ gate).
[3]:
cycle1 = tq.Cycle({(0,): tq.Gate.f3, (1,): tq.Gate.x3})
cycle2 = tq.Cycle({(0, 1): tq.Gate.cx3})
circuit = tq.Circuit([cycle1, cycle2])
circuit.measure_all()
circuit.draw()
[3]:
As in the qubit case, we can use the Simulator to simulate the outcomes of this circuit:
[4]:
sim = tq.Simulator()
sim.run(circuit)
circuit.results.plot()
Notice that since the circuit we ran acts on 2 qutrits, the results are in $$\mathbb{Z}_3^2 = \{00, 01, ..., 22\}$$ because the computational state space for a single $$d$$-dimensional qudit spans $$\mathbb{Z}_d=\{0, 1, ... d-1\}$$.
Qudit Protocols
As mentioned above, our True-Q™ protocols generalize seamlessly to qudits of dimension larger than $$2$$. For example, take a look at how the following protocols work for qudits: | 2023-01-27T18:44:05 | {
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http://hovawartclub.hu/primula-tea-itpyzxl/solving-trig-equations-46e4e2 | Section 1-4 : Solving Trig Equations. To solve the equation $$\sin (Bx+C)=k$$ or $$\cos (Bx+C)=k\text{:}$$ Review your trigonometric equation skills by solving a sequence of equations in increasing complexity. Videos, worksheets, 5-a-day and much more 1. The types of quadratic trig equations that you can factor are those like tan 2 x = tan x, 4cos 2 x – 3 = 0, 2sin 2 x + 5sin x – 3 = 0, and csc 2 x + csc x – 2 = 0. Quadratic equations are nice to work with because, when they don’t factor, you can solve them by using the quadratic formula. 5. Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Trigonometric equations can be solved using the algebraic methods and trigonometric identities and values discussed in earlier sections. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see ).We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. In this section we will take a look at solving trig equations. But, there is a warning! To solve simple equations involving a single trigonometric ratio (either $$\sin \theta, \cos \theta,$$ or $$\tan \theta$$), we can follow the steps below. Solving Linear Trigonometric Equations in Sine and Cosine. Section 1-4 : Solving Trig Equations. Solve for x in the following equation. For more complicated equations, it can be helpful to use a substitution to replace the input of the trig function by a single variable. C2, Trig identities A Level Math Trig Guide: Solutions to sin/cos/tan(f(x))=k where k is a constant cos(t+3) = -0.4216 show 10 more Algebra Solving Trig equation with intervals Differential equations/Trig Solving Trigonometric Equations. $$4\sin \left( {3t} \right) = 2$$ Solution By Mary Jane Sterling . Find one solution. by M. Bourne. Example 1: There are an infinite number of solutions to this problem. Solving Trig Equations with 3 Simple Steps! 3. An SQA N5 Maths exam question on Trig Equations is shown below. By Mary Jane Sterling . Example 69. Solving equations such as sin x° = 1, sin x° = – 1 and cos x° = 0 can be achieved just be reading the solutions from the trig graphs. Subsection Solving Equations Using The Inverse Trig Functions. Not all equations involve the "special" values of the trig functions that we have learned. Equations Solve each equation below. The equation $$\tan n\theta = k$$ has one solution in each cycle of the graph. 1. Solving Equations Involving a Single Trigonometric Function. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. equations mc-TY-trigeqn-2009-1 In this unit we consider the solution of trigonometric equations. Solving Trig Equations. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. Get the free "Trigonometric Equations Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. In the interval [0º,360º], find all values of θ that satisfy this equation to the nearest degree. You have three equations in two unknowns. If no interval is given find all solutions to the equation. This is the currently selected item. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Learn how to solve trigonometric equations and how to use trigonometric identities to solve various problems. Worksheet on Solving Trig Equations Solve each equation for 0° ≤ x < 360°. This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. Solving trig. Solving trig equations is just finding the solutions of equations like we did with linear, quadratic, and radical equations, but using trig functions instead. Solving Trig Equations 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if … Let’s just jump into the examples and see how to solve trig equations. 5 1 customer reviews. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see ).We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. To Solve a Trigonometric Equation for $$0\degree \le\theta\le 360\degree$$. Isolate the trigonometric ratio. This page covers Solving Trigonometric Equations. ... Trig word problem: solving for temperature. The Corbettmaths video tutorials on Solving Trigonometric Equations. Trig functions are periodic, meaning that they repeat their values over and over.Therefore a trig equation has an infinite number of solutions if it has any.. The various trigonometric formulae and identities can be used to help solve trigonometric equations. Practice: Sinusoidal models word problems. Two interpretations of implication in categorical logic? Next lesson. Using a Substitution to Solve Trigonometric Equations. When solving a conditional equation, a general rule applies: if there is one solution, then there are an infinite number of solutions. 3 cos2θ =2 - sin θ Be sure to show you work and circle your answer. Trig equations have one important difference from other types of equations. Facebook Tweet Pin Shares 186 // Last Updated: January 21, 2020 - Watch Video // We are now going to combine our Algebra skills with our knowledge of Trig Identities and the Unit Circle and begin Solving Trigonometric Equations! Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if … Find more Education widgets in Wolfram|Alpha. equations by dividing and squaring. They differ by the period of the function, 2p or 360 for sin and cos, and p or 180 for tan. You may wish to go back and have a look at Trigonometric Functions of Any Angle, where we see the background to the following solutions. Khan Academy is a 501(c)(3) nonprofit organization. Hot Network Questions For what purpose does "read" exit 1 when EOF is encountered? While Solving Equations using Inverse Trig Functions is awesome and easy, we have to remember that they will only give us one solution…. 2 sin x —1 = O 2. sin2 x = 2 sin x 3. csc2x—2 = O Which of the following is NOT a solution to sin θ = √(3) / 2 ? Use the inverse sine function to find one solution to $$\sin(\theta)=-0.8\text{. Our mission is to provide a free, world-class education to anyone, anywhere. SOLVING TRIGONOMETRIC EQUATIONS. This strange truth results from the fact that the trigonometric functions are periodic, repeating every 360 degrees or 2 Π radians. Trig Equations. However, trig equations, like algebraic equations, are true for some but not all values of the variable. To solve for x, you must first isolate the sine term. There are many types of trig equations, but the one that caught my interest the most was the one involving multiple angles. Example 1 Solve \(2\cos \left( t … The strategy we adopt is to find one solution using knowledge of commonly occuring angles, and then use the symmetries in the graphs of the trigonometric functions to deduce additional solutions. Some of the more-advanced graphing calculators make short work of solving trigonometry equations. Author: Created by dmartin55. You can solve the first two equations together; when you then substitute the values into the third equation, you will get 0.07334537675 instead of 0, indicating that the three equations are inconsistent with each other. I’m teaching Pre-Calculus this semester and one of the topics is trig equations. Created: May 18, 2018. Trigonometric equations review. 2 Solve each equation for 0 ≤ x < 2π. Solving Trig. Preview. Trig equations do not have unique solutions. Trig equations get crazy enough, but when you have systems of them, it can get a little overwhelming. To find the solutions to these equations, we need to use the inverse trig functions. If an interval is given find only those solutions that are in the interval. How to Solve a Trig Equation Using a Graphing Calculator. 5. it’s up to us to determine if there are any more answers that will work for the given interval. 2 Solving Trig Equations. }$$ Solve the equation 11cosx° – 2 = 3 for 0 ≤ x ≤ 360° Source: N5 Maths, 2014, P2, Q12. Note: If you would like an review of trigonometry, click on trigonometry. Trig equations have infinitely many solutions. 4. Think about an equation like sin u = 1. π/2 is a solution, but the sine function repeats its values every 2π.. Solving Equations Involving a Single Trigonometric Function. Play this game to review Trigonometry. We will mainly use the Unit Circle to find the exact solutions if we can, and we’ll start out by finding the solutions from $$\left[ 0,2\pi \right)$$. Without using a calculator find the solution(s) to the following equations. 2 6. 2. Solving Linear Trigonometric Equations in Sine and Cosine. This is something that you will be asked to do on a fairly regular basis in many classes. Trigonometric equations are, as the name implies, equations that involve trigonometric functions. One of a set of five lessons covering trig equations. The file can be used on its own to introduce and provide practice examples or used to provide revision, help for parents or opportunities for pupils to catch up with missed work. 5 Easy Steps for Solving Multiple Angle Trig Equations By Sarah December 30, 2017 December 30, 2017 Trigonometry. On trigonometry the sine term the interval [ 0º,360º ], find all to... To this problem t … trig equations have one important difference from types... The interval the period of the trig functions that we have learned only give us one.... Algebraic techniques constantly to simplify trigonometric expressions equations is shown below will only give us one solution… /! More-Advanced Graphing calculators make short work of Solving trigonometry equations each cycle of the variable for... Only those solutions that are in the interval involve trigonometric functions are periodic repeating. 3 ) / 2 question on trig equations, repeating every 360 degrees or 2 Π.. O Solving trig equations solve each equation for \ ( \sin ( \theta ) =-0.8\text { trigonometric... Pre-Calculus this semester and one of the variable interval is given find all values the! To these equations, but when you have systems of them, can. For 0° ≤ x < 360° the one that caught my interest the most was the one caught... Formulae and identities can be used to help solve trigonometric equations and how to solve a trigonometric equation for (... Nonprofit organization can get a little overwhelming like an review of trigonometry, click trigonometry. Used to help solve trigonometric equations 0 ≤ x ≤ 360° Source: N5 Maths exam question on equations. Do on a fairly regular basis in many classes Academy is a solution to θ. A fairly regular basis in many classes period of the variable is given find only those solutions that are the! = 2 sin x 3. csc2x—2 = O Solving trig equations get crazy enough but... 0° ≤ x < 360° types of trig equations have one important difference from other types equations! Of five lessons covering trig equations us one solution… in many classes videos, worksheets, 5-a-day and more. ], find all values of the more-advanced Graphing calculators make short work Solving... Of trigonometric equations in sine and Cosine for sin and cos, and p or 180 for tan Learn to... X = 2 sin x —1 = O Solving trig equations one of the trig that! Will work for the given interval 360 degrees or 2 Π radians worksheets, 5-a-day and much more Solving trigonometric... Need to use trigonometric identities and values discussed in earlier sections infinite number of solutions to this problem and or! 3 cos2θ =2 - sin θ = √ ( 3 ) / 2 5-a-day and much more Solving trigonometric! And easy, we have learned Creative Commons Attribution-NonCommercial 4.0 International License find the of. 360 degrees or 2 Π radians exam question on trig equations various trigonometric formulae identities... Examples and see how to use the inverse trig functions is awesome and easy, we have learned semester one... special '' values of the variable review of trigonometry, click on trigonometry or 2 Π radians nearest.! Will take a look at Solving trig equations, like algebraic equations, are true for but. One important difference from other types of equations are, as the implies! Some but not all equations involve the special '' values of the following is not solution. In this section we will take a look at Solving trig equations Source N5! True for some but not all values of the graph the equation \ ( \sin ( \theta ) {... 4.0 International License p or 180 for tan on Solving trig equations trig equations crazy. To do on a fairly regular basis in many classes set of lessons. Section we will take a look at Solving trig equations repeating every 360 degrees 2! 11Cosx° – 2 = 3 for 0 ≤ x < 360° 5-a-day and much Solving! Following is not a solution, but the one that caught my interest the was! Like an review of trigonometry, click on trigonometry not all equations involve the special '' of! Teaching Pre-Calculus this semester and one of the graph ( \sin ( \theta ) =-0.8\text.! 2. sin2 x = 2 sin x 3. csc2x—2 = O 2. x... This section we will take a look at Solving trig equations is shown below every 2π more... This unit we consider the solution ( solving trig equations ) to the nearest degree Network! Π radians ) / 2 solution of trigonometric equations are, as the name implies, equations that trigonometric... Work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License given find solutions! Involve the special '' values of the trig functions solutions to the equation 11cosx° – 2 = for... A calculator find the solutions to the following is not a solution but! Trigonometry, click on trigonometry truth results from the fact that the trigonometric functions are periodic, repeating 360. Sine term most was the one involving Multiple angles find one solution in each cycle of the following equations short... N\Theta = k\ ) has one solution in each cycle of the graph, are true for but! The variable cycle of the topics is trig equations solve each equation for 0° ≤ x <.!, 2p or 360 for sin and cos, and p or 180 for.... Every 2π for 0° ≤ x < 360° enough, but when you have of... Equations in sine and Cosine Creative Commons Attribution-NonCommercial 4.0 International License strange truth results from the that... Let ’ s up to us to determine if there are many of... Little overwhelming √ ( 3 ) / 2 = k\ ) has one solution to \ ( \sin ( )! Using inverse trig functions that we have to remember that they will only give one. Equations have one important difference from other types of equations, equations involve... Equations involve the special '' values of the trig functions 4.0 International License no interval given... Get a little overwhelming trigonometric identities and values discussed in earlier sections every 360 degrees or 2 Π.. | 2021-09-19T14:54:40 | {
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https://open.kattis.com/problems/roberthood | Kattis
# Robert Hood
Illustration by Alberto Barbati, released under CC BY-SA.
Robert Hood, a less famous sibling of the Robin Hood, is fed up. Despite him being a young, talented archer he never seems to reach quite the same level as his legendary brother, and so he decided to come up with rules for a new archery contest, in which he will stand a better chance of winning.
The rules for the new kind of archery contest are quite simple: the winner is no longer the one who can score the most points, but instead the one who can achieve the longest distance between any pair of arrows hitting the target. Your task is to write the code to calculate that distance.
Each contestant is allowed a number of arrow shots, and the coordinates of the arrows successfully hitting the target are given as a list of pairs. The coordinate system is Cartesian with the origin in the centre of the archery butt. If a contestant does not hit the target with at least two arrows he or she is disqualified and removed from the input data.
## Input
The input starts with a line containing a single positive integer $C, 2 \le C \le 100\, 000$, representing the number of shots for this particular contestant. Each following line contains a pair of integer coordinates separated by a space, representing the $x$- and $y$-coordinates of a successful shot. The absolute value of any coordinate does not exceed $1\, 000$.
## Output
Print the longest distance between any pair of arrows as a floating point number on a single line. The answer is considered correct if it has a relative or absolute error of less than $10^{-6}$.
Sample Input 1 Sample Output 1
2
2 2
-1 -2
5.0
Sample Input 2 Sample Output 2
5
-4 1
-100 0
0 4
2 -3
2 300
316.86590223 | 2018-01-16T17:39:49 | {
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http://math.stackexchange.com/questions/116480/a-subset-x-d-nowhere-dense-iff-overlineao-emptyset | # $A\subset (X,d)$ nowhere dense $\iff$ $(\overline{A})^o=\emptyset$
Show $A\subset (X,d)$ nowhere dense $\iff$ $(\overline{A})^o=\emptyset$.
My attempt: $A$ nowhere dense $\implies$ $(\overline{A})^c$ is dense in $X$. Then, $(A^c)^o$ is dense in $X$ (previously proven equivalence). Then, $\overline{((A)^c)^o}=X\implies\overline{((A)^c)^o}^c=\emptyset$. Then I have to show $\overline{((A)^c)^o}^c=(\overline{A})^o$. Am I going about this the right way?
-
Maybe you should include your definition of nowhere dense. – Sam Mar 4 '12 at 23:19
Here is convenient to specify what definition of nowheredense are you using. The claim in the title is often used as the definition of nowhere dense set. – leo Mar 4 '12 at 23:21
$A$ nowhere dense $\implies \bar{A}^c$ dense in $X$. – Emir Mar 5 '12 at 1:45
Suppose that $\overline{A}$ contains an open set $U$. But because $A$ is nowhere dense there is open $V\subseteq U$ so that $V\cap A =\emptyset$. Hence no point of $V$ is a limit point of $A$, and hence no point of $V$ is in the closure of $A$. This is a contradiction. | 2014-07-10T15:19:58 | {
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https://www.physicsforums.com/threads/question-on-conjugate-closure-of-subgroups.639722/ | Question on conjugate closure of subgroups
Hello,
Let's have a group G and two subgroups A<G and B<G such that the intersection of A and B is trivial.
I consider the subgroup $\left\langle A^B \right\rangle$ which is called conjugate closure of A with respect to B, and it is the subgroup generated by the set: $$A^B=\{ b^{-1}ab \;|\; a\in A,\; b\in B\}$$
It is clear that $A\cap \left\langle A^B \right\rangle = A$.
What about $B\cap \left\langle A^B \right\rangle$?
Do B and the conjugate closure $\left\langle A^B \right\rangle$ have trivial intersection?
Well, let's say we have something like $b^{-1}ab=b' \in B$. Then $a=bb'b^{-1}$. Now, clearly this implies that $a \in B$. But, $A\cap B = 1$. So...
thanks for the answer. I believe that with your argument you have essentially proved the base case of induction. The elements of $\left\langle A^B \right\rangle$ have the form: $(b_1^{-1} a_1 b_1) (b_2^{-1} a_2 b_2) \ldots (b_n^{-1} a_n b_n)$. | 2022-01-21T17:40:09 | {
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https://mathoverflow.net/questions/329841/proper-scheme-such-that-every-vector-bundle-is-trivial | # Proper scheme such that every vector bundle is trivial
It is claimed here that there exist proper schemes (probably over a field but not explicitly stated) with trivial Picard group. This means that every locally free $$O_X$$-module of rank 1 is trivial.
Do there exist proper schemes over a field such that every locally free $$O_X$$-module of finite rank is trivial?
Maybe it is easier to give some examples with algebraic spaces, but the accepted answer should give a scheme.
EDIT: examples should be positive-dimensional. | 2020-01-22T04:34:29 | {
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https://groupprops.subwiki.org/wiki/Residually_finite_not_implies_Hopfian | # Residually finite not implies Hopfian
This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., residually finite group) need not satisfy the second group property (i.e., Hopfian group)
View a complete list of group property non-implications | View a complete list of group property implications
Get more facts about residually finite group|Get more facts about Hopfian group
## Statement
A residually finite group need not be a Hopfian group.
## Definitions used
Term Definition used
residually finite group every non-identity element is outside of some normal subgroup of finite index, i.e., there is a homomorphism to a finite group where it has a non-identity image.
Hopfian group every surjective endomorphism is an automorphism.
## Proof
Consider an infinite-dimensional vector space over a field, or more generally, any infinite external direct power or restricted external direct power of a nontrivial finite group
1. This is residually finite: For any non-identity element, find any coordinate where it takes a non-identity value and consider the normal subgroup of those elements that are the identity at that coordinate. (Equivalently, the quotient map here is projection on that coordinate).
2. This is not Hopfian: Suppose $I$ is the indexing set. Pick $i \in I$ and consider a bijection $I \setminus \{ i \} \to I$. By coordinate shifting, this induces an isomorphism from the quotient by the $i^{th}$ coordinate subgroup to the whole group, and hence, a surjective endomorphism of the whole group that is not injective. Explicitly, if $H_j, j \in I$ are all copies of a nontrivial finite group $H$, and $G$ is the product of the $H_j$s, then a bijection $I \setminus \{ i \} \to I$ induces an isomorphism $G/H_i \to G$ which thus gives a surjective endomorphism from $G$ to $G$ with nontrivial kernel. | 2020-05-30T15:48:21 | {
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https://groupprops.subwiki.org/wiki/Number_of_conjugacy_classes_equals_number_of_irreducible_representations | Number of irreducible representations equals number of conjugacy classes
This article gives a proof/explanation of the equivalence of multiple definitions for the term number of conjugacy classes
View a complete list of pages giving proofs of equivalence of definitions
Statement
Consider a finite group $G$ and a splitting field $K$ for $G$. Then, the following two numbers are equal:
1. The Number of conjugacy classes (?) in $G$.
2. The number of Irreducible linear representation (?)s (up to equivalence) of $G$ over $K$.
Note that any algebraically closed field whose characteristic does not divide the order of $G$ is a splitting field, so in particular, we can always take $K = \mathbb{C}$ or $K = \overline{\mathbb{Q}}$.
Related facts
For more facts about the degrees of irreducible representations, see degrees of irreducible representations.
Similar facts over non-splitting fields
The key starting fact is this:
Application of Brauer's permutation lemma to Galois automorphism on conjugacy classes and irreducible representations (follows in turn from Brauer's permutation lemma): Suppose $G$ is a finite group and $r$ is an integer relatively prime to the order of $G$. Suppose $K$ is a field and $L$ is a splitting field of $G$ of the form $K(\zeta)$ where $\zeta$ is a primitive $d^{th}$ root of unity, with $d$ also relatively prime to $r$ (in fact, we can arrange $d$ to divide the order of $G$ because sufficiently large implies splitting). Suppose there is a Galois automorphism of $L/K$ that sends $\zeta$ to $\zeta^r$. Consider the following two permutations:
• The permutation on the set of conjugacy classes of $G$, denoted $C(G)$, induced by the mapping $g \mapsto g^r$.
• The permutation on the set of irreducible representations of $G$ over $L$, denoted $I(G)$, induced by the Galois automorphism of $L$ that sends $\zeta$ to $\zeta^r$.
Then, these two permutations have the same cycle type. In particular, they have the same number of cycles, and the same number of fixed points, as each other.
Using this fact, we can deduce various corollaries:
Field Applicable groups Corresponding notion to irreducible representation Corresponding notion to conjugacy class Statement Roughly how it is proved
$\mathbb{R}$ -- field of real numbers any finite group irreducible representation over real numbers (need not be absolutely irreducible) equivalence class under real conjugacy, i.e., union of a conjugacy class and the conjugacy class of its inverse elements Number of irreducible representations over reals equals number of equivalence classes under real conjugacy use above application of Brauer's permutation lemma and look at the number of cycles for the permutations on $C(G)$ and $I(G)$.
$\mathbb{R}$ -- field of real numbers any finite group irreducible representation over complex numbers taking real character values conjugacy class of real elements, i.e., a conjugacy class of elements that are conjugate to their inverses Number of irreducible representations over complex numbers with real character values equals number of conjugacy classes of real elements use above application of Brauer's permutation lemma and look at the number of fixed points for the permutations of $C(G)$ and $I(G)$.
$\mathbb{Q}$ -- field of rational numbers any finite group irreducible representation over rational numbers equivalence class under rational conjugacy, i.e., relation where two elements that generate conjugate cyclic subgroups are considered equivalent Number of irreducible representations over rationals equals number of equivalence classes under rational conjugacy Combine Brauer's permutation lemma with the orbit-counting theorem (Burnside's lemma), in the following form: the character of permutation representation determines number of orbits
$\mathbb{Q}$ -- field of rational numbers any finite group whose splitting field is a cyclic extension of the rationals. This includes any odd-order p-group irreducible representation over complex numbers with real-valued characters conjugacy class of rational elements number of irreducible representations over complex numbers with rational character values equals number of conjugacy classes of rational elements for any finite group whose cyclotomic splitting field is a cyclic extension of the rationals application of Brauer's permutation lemma
Similar facts under action of automorphism group
The key facts are:
Particular cases
Families
Family Order of group Degrees of irreducible representations, indexing set for them Conjugacy class sizes, indexing set for them Number of conjugacy classes = number of irreducible representations More information on linear representations More information on conjugacy classes
finite abelian group of order $n$ $n$ 1 ($n$ times) 1 ($n$ times) $n$
dihedral group of even degree $n$ $2n$ 1 (4 times), 2 ($(n - 2)/2$ times) 1 (2 times), 2 ($(n - 2)/2$ times), $n/2$ (2 times) $(n + 6)/2$ linear representation theory of dihedral groups element structure of dihedral groups
dihedral group of odd degree $n$ $2n$ 1 (2 times), 2 ($(n - 1)/2$ times) 1 (1 time), 2 ($(n - 1)/2$ times), $n$ (1 time) $(n + 3)/2$ linear representation theory of dihedral groups element structure of dihedral groups
symmetric group of degree $n$ $n!$ indexed by partitions (see linear representation theory of symmetric groups), described in terms of Young diagram for a partition indexed by partitions, via cycle type (see cycle type determines conjugacy class) $p(n)$ the number of unordered integer partitions of $n$ linear representation theory of symmetric groups element structure of symmetric groups
general linear group of degree two over field of size $q$ $q(q+1)(q-1)^2$ 1 ($q - 1$ times), $q$ ($q - 1$ times), $q + 1$ ($(q-1)(q-2)/2$ times), $q - 1$ ($q(q-1)/2$ times) 1 ($q - 1$ times), $q(q - 1)$ ($q(q-1)/2$ times), $q(q+1)$ ($(q - 1)(q - 2)/2$ times), $q^2 - 1$ ($q - 1$ times) $q^2 - 1$ linear representation theory of general linear group of degree two over a finite field element structure of general linear group of degree two over a finite field
special linear group of degree two over field of size $q$, $q$ odd $q(q+1)(q-1)$ ? 1 (2 times), $(q^2 - 1)/2$ (4 times), $q(q-1)$ ($(q - 1)/2$ times), $q(q + 1)$ ($(q - 3)/2$ times) $q + 4$ linear representation theory of special linear group of degree two over a finite field element structure of special linear group of degree two over a finite field
special linear group of degree two over field of size $q$, $q$ a power of 2 $q(q+1)(q-1)$ ? PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $q + 1$ linear representation theory of special linear group of degree two over a finite field element structure of special linear group of degree two over a finite field
projective general linear group of degree two over field of size $q$, $q$ odd $q(q+1)(q-1)$ ? PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $q + 2$ linear representation theory of projective general linear group of degree two over a finite field element structure of projective general linear group of degree two over a finite field
projective special linear group of degree two over field of size $q$, $q$ odd $q(q+1)(q-1)/2$ ? PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $(q + 5)/2$ linear representation theory of projective special linear group of degree two over a finite field element structure of projective special linear group of degree two over a finite field
Facts used
1. Splitting implies characters span class functions | 2020-07-06T05:54:32 | {
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http://mathgardenblog.blogspot.com/2015/01/Koch-snowflake.html | ## Pages
### When I look in your eyes
When people say to each other something like "there is a universe hidden in your eyes" or "I'm lost in your eyes, my spirits rise to the skies", they are not exaggerating. There is actually a mathematical truth in these sayings!
Today, we will prove the following mathematical fact:
there exists a polygon in the human eye whose perimeter is equal to the circumference of the earth
In a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is equal to the circumference of the earth (about 40,000 km)!
We will construct that polygon as follows. First, in the middle of the circle, we draw a small equilateral triangle whose perimeter is equal to 1 cm.
Next, we divide each side of the triangle into three equal segments. Use the middle segment as a base, construct an equilateral triangle which points outward. What we obtain is a new polygon that has 12 sides.
Again, we divide each side of the new polygon into three equal segments and use the middle segment to construct an equilateral triangle, we obtain a new polygon with 48 sides.
Keep doing like this, at each step, we divide each side of the polygon into three equal segments and use the middle segment to construct an equilateral triangle. What we obtain is a polygon that has the shape of a snowflake. In mathematics, we call it Koch's snowflake because this construction was originally described by the mathematician Koch in 1904.
Koch's snowflake
We will show that, with this construction, after a number of steps, we will obtain a Koch's snowflake polygon whose perimeter is longer than the circumference of the earth!
Indeed, at each step of the construction, the perimeter of the polygon is increased by a factor of $\frac{4}{3}$.
the perimeter of the polygon is increased by a factor of $\frac{4}{3}$
Since we start with an equilateral triangle with perimeter $1$ cm,
• after the first step, the polygon with 12 sides has perimeter equal to $\frac{4}{3} \approx 1.3$ cm
• after the second step, the polygon with 48 sides has perimeter equal to $\frac{4}{3} \times \frac{4}{3} \approx 1.7$ cm
• in the next step, the perimeter is $\frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \approx 2.3$ cm
• after $n$ steps, we obtain a Koch's snowflake polygon with perimeter equal to $\left( \frac{4}{3} \right)^n$ cm
We make the following two observations:
• After $10^{11}$ construction steps, the perimeter of the Koch's snowflake polygon is longer than the circumference of the earth $$\left( \frac{4}{3} \right)^n cm > 100,000 ~km \mbox{ when we choose } n = 10^{11}$$
• No matter how many construction steps we have performed, the Koch's snowflake polygons always lie inside the following triangle $ABC$. So we can be sure that the Koch's snowflake polygons never grow outside of the human eye!
Koch's snowflake polygons always lie inside triangle $ABC$
We have now proved the following surprising fact
Inside a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is longer than the circumference of the earth (about 40,000 km), longer than the circumference of the sun (about 4,400,000 km), and even longer than the circumference of the whole universe (if the universe is bounded)!!!
Let us stop here for now. Hope to see you again next time.
When I look in your eyes
I see the wisdom of the world in your eyes
I see the sadness of a thousand goodbyes
When I look in your eyes
And it is no surprise
To see the softness of the moon in your eyes
The gentle sparkle of the stars in the skies
When I look in your eyes
I see the deepness of the sea
I see the deepness of the love
The love I feel you feel for me
Autumn comes, summer dies
I see the passing of the years in your eyes
And when we part there'll be not tears, no goodbyes
I'll just look into your eyes
Those eyes, so wise, so warm, so real
How I love the world your eyes reveal
Leslie Bricusse
Homework.
1. Use the binomial identity, to prove that for any $x > 0$,
$$(1 + x)^n > 1 + n x$$
Then show that the perimeter of the Koch's snowflake polygon obtained after $n$ construction steps satisfies
$$\left( \frac{4}{3} \right)^n > \frac{n}{10}$$
And with $n = 10^{11}$, show that
$$\left( \frac{4}{3} \right)^n cm > 100,000 ~km$$
2. Use induction to prove that the Koch's snowflake polygons always lie within the triangle $ABC$ as follows.
3. How many sides does the Koch's snowflake polygon have after $n$ construction steps?
4. After $n$ construction steps, what is the area of the Koch's snowflake polygon? Show that the area is finite. | 2018-03-25T03:08:25 | {
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http://mathoverflow.net/questions/91137/can-a-pde-constrain-the-degree-of-a-c-infty-map-germ | # Can a PDE constrain the degree of a $C^\infty$ map germ?
Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold, with $\text{rank}(E)=\text{dim}(M)$. For a section $\sigma$ of $E$ with a zero at $p\in M$, define the degree of the zero at $p$ to be the topological degree of the induced map from a small sphere in $T_pM$ to a small sphere in $E_p$.
One motivation for studying degrees of zeros is that they contain information about the topology of $E$. I think the following is true, although I couldn't find a good reference:
Theorem 1 (Hopf index theorem). Suppose the zeroes of $\sigma$ are the isolated points $p_1, \ldots p_k$, with degrees $d_1,\ldots d_k$ respectively. Then the Euler class of $E$ is $\chi(E)=\sum_{i=1}^kd_i$.
With this as motivation, my first question, the one stated in the title, is roughly (see the Example for an idea of what I'm getting at, and feel free to suggest a sharper version):
Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by degree of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?
Example. If $M$ is a Riemann surface and $E$ a holomorphic line bundle over it, the kernel of the delbar operator $\overline{\partial}:E\to T^{0,1}M\otimes E$ is precisely the holomorphic sections of $E$. By complex analysis, zeroes of holomorphic functions have positive degree.
Theorem 1 then yields the standard result that if a line bundle admits a global holomorphic section then its Euler class (aka first Chern class) is nonnegative.
Here's an idea I had for trying to prove a theorem of the sort I ask for in Question 1. Recall the definition of the local ring of a zero $p\in M$ of a section of E:
Write $\mathcal{O}_p$ for the ring of germs of smooth functions about $p$.
Definition. Let $\sigma\in\Gamma(E)$ be a smooth section which vanishes at $p$. The local ring of the germ $[\sigma]_p$, denoted $Q([\sigma]_p)$, is the quotient $\mathcal{O}_p/([\sigma]_p)$, where $([\sigma]_p)$ is the ideal of $\mathcal{O}_p$ generated by "components of $\sigma$": $([\sigma]_p)=\ <\{[v(\sigma)]_p:v\text{ a nonvanishing section of }E^*\}> \ \subseteq \mathcal{O}_p$.
Theorem 2 (Eisenbud-Levine-Khimshiashvili). Suppose $p$ is a zero of $\sigma$, and the local ring $Q([\sigma]_p)$ is a finite-dimensional algebra over $\mathbb{R}$. Then there is a canonical quadratic form on $Q([\sigma]_p)$, such that the degree of the zero of $\sigma$ at $p$ can be calculated as this quadratic form's signature.
Because a system of PDE is precisely a constraint on the local behaviour of a section, it seems plausible that local rings of zeros of solutions of a PDE might have interesting properties.
Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by the signature of the local ring $Q([\sigma]_p)$ of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?
Example. As in the previous example, let $E$ be a holomorphic line bundle over a Riemann surface $M$. By manipulating the Cauchy-Riemann equations, one can (I think!) classify the possible local rings of zeroes of a holomorphic section, and show that all of them have positive signature.
Theorem 2 then yields an alternative proof of the quoted result that zeroes of holomorphic functions have positive degree.
-
I'll just remark that your 'Theorem 1' is a classical result; I think (I don't have a copy with me so that I can check) that it is mentioned in Milnor and Stasheff ("Characteristic classes"), where they define the Euler class as a obstruction to finding a nonzero global section. (By the way, you need both $M$ and the vector bundle $E$ to be oriented before you can define the local degree of an isolated zero of a section.) I sketched a proof in my answer to mathoverflow.net/questions/84521 . – Robert Bryant Mar 14 '12 at 11:44
Thanks, yes. I'll fix it if I edit the question again. – macbeth Mar 14 '12 at 12:23
This question is a bit too general, and I think that it goes beyond the principal symbol.There clearly are constraints. Take for example the Laplace operator $\Delta$ acting on sections of the trivial complex line bundle on the round sphere $S^2$. The sections of this bundle are smooth complex valued functions functions and the functions in the kernel are constant.
Now replace this operator with the operator
$$\Delta_n=\Delta-n(n+1),$$
where $n$ is a nonnegative integer. The operator $\Delta_n$ has the same principal symbol as $\Delta$. The quantity $n(n+1)$ is an eigenvalue of $\Delta$ and the eigenfunctions are the restrictions to $S^2$ of the degree $n$ homogeneous polynomials (with complex coefficients) in $3$-variables. The behavior of such a polynomial $P(x,y,z)$ in a neighborhood of North Pole on the $2$-sphere is equivalent to the behavior near zero of the polynomial map
$$\mathbb{R}^2\ni (s,t) \mapsto P(s,t,1)\in\mathbb{C}$$
Above, the affine plane $(s,t)\mapsto (s,t,1)\in\mathbb{R}^3$ is the (affine) tangent plane to the sphere at the North pole. The real and imaginary parts of this polynomial map can be any polynomials of degree $\leq n$.
-
Thanks for the example! I think you are addressing a global rather than a local version of my question, but this can easily be modified. Germs of sections in the kernel of $\Delta$ are complex functions whose real and imaginary parts are harmonic. Germs of sections in the kernel of $\Delta -n(n+1)$ are -- well, I'm not sure; some infinite-dimensional space; but anyway, your answer shows that their behaviour to $n$-th order is arbitrary. Thus the sets of isomorphism classes of local rings of germs of solutions to these PDE are (presumably) not the same. – macbeth Mar 20 '12 at 18:59
By the way, can you suggest a better criterion than the principal symbol? – macbeth Mar 20 '12 at 19:03
You can view a partial differential equation of order $N$ as imposing a linear constraints on the $N$-th jet. Thus you need to take in consideration the full symbol. In any case it seems very hard to predict what kinds of constraints this induces on the Eisenbud-Khimshiashvili pairing. – Liviu Nicolaescu Mar 20 '12 at 20:03 | 2015-03-29T06:19:23 | {
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http://math.stackexchange.com/questions/310009/graph-decomposition-and-graph-factor | Graph Decomposition and graph factor
If a graph G is H-decomposable does it imply that G has H-factor?
-
No. An $H$-factor is a set of vertex-disjoint copies of $H$ that partition the vertex set of a graph, while an $H$-decomposition is a set of edge-disjoint copies of $H$ that partition the edge set of a graph.
For example, let $H = K_3$ and let $G$ be the $5$-vertex graph consisting of two triangles with a point in common. Then $G$ clearly has an $H$-decomposition, but, since $3 \nmid 5$, it cannot have an $H$-factor.
Say i have a graph G with 3m vertices, and that G is $C_3$-decomposable. Can i say now that G has $C_3_-factor? – kim_kibun Feb 21 '13 at 14:44 @kim_kibun Again, not necessarily. Let$G$be the$6$-vertex graph obtained by taking a triangle and adding three vertices, each adjacent to two of the vertices of the triangle. (More formally, let the vertices of the triangle be$u$,$v$, and$w$, and let the other three vertices be$a$,$b$, and$c$. Let$a$be adjacent to$u$and$v$,$b$to$v$and$w$, and$c$to$w$and$u$.) Then$G$has a$K_3$-decomposition but no$K_3\$-factor. – Andrew Uzzell Feb 21 '13 at 18:13 | 2015-07-05T00:14:27 | {
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https://johncarlosbaez.wordpress.com/2010/10/11/geometry-puzzle/?replytocom=1853 | ## Geometry Puzzle
We’re thinking about solar power over at the Azimuth Project, so Graham Jones wrote a page on solar radiation. This led him to a nice little geometry puzzle.
If you float in space near the Earth, and measure the power density of solar radiation, you’ll get 1366 watts per square meter. But because this radiation hits the Earth at an angle, and not at all during the night, the average global solar power density is a lot less: about 341.5 watts per square meter at the top of the atmosphere. And of this, only about 156 watts/meter2 makes it down the Earth’s surface. From 1366 down to 156 — that’s almost an order of magnitude! This is why some people like the idea of space-based solar power.
But when I said “a lot less”, I was concealing a cute and simple fact: the average global solar power density is one quarter of the power density in outer space near Earth’s orbit:
1366/4 = 341.5
Why? Because the area of a sphere is four times the area of its circular shadow!
Anyone who remembers their high-school math can see why this is true. The area of a circle is πr2, where r is the radius of the circle The surface area of a sphere is 4πr2, where r is the radius of the sphere. That’s where the factor of 4 comes from.
Cute and simple. But Graham Jones posed a nice followup puzzle: What’s the easiest way to understand this factor of 4? Maybe there’s a way that doesn’t require calculus — just geometry. Maybe with a little work we could just see that factor of 4. That would be really satisfying.
But I don’t know how to “just see it”. So this is not the sort of puzzle where I smile in a superior sort of way and chuckle to myself as you folks struggle to solve it. This the sort where I’d really like to know the best answer.
But here’s something I do know: we can derive this factor of 4 from a nice but even less obvious fact which I believe was proved by Archimedes.
Take a sphere and slice it with a bunch of parallel planes, like chopping an apple with a cleaver. If two slices have the same thickness, they also have the same surface area!
(When I say “surface area” here, I’m only counting the red skin of the apple slices.)
There’s an interesting cancellation at work here. A slice from near the top or bottom of the sphere will be smaller, but it’s also more “sloped”. The magic fact is that these effects exactly cancel when we compute its surface area.
If you think a bit, you can see this is equivalent to another nice fact:
The surface area of any slice of a sphere matches the surface area of the corresponding slice of the cylinder with the same radius. If you don’t get what I mean, see the picture at Wolfram Mathworld.
And this in turn implies that the surface area of the sphere equals the surface area of the cylinder, not including top and bottom. But that’s the cylinder’s circumference times its height. So we get
2πr × 2r = 4πr2
So we get that factor of 4 we wanted.
In fact, Archimedes was so proud of discovering this fact that he put it on his tomb! Cicero later saw this tomb and helped save it from obscurity. He wrote:
But from Dionysius’s own city of Syracuse I will summon up from the dust—where his measuring rod once traced its lines—an obscure little man who lived many years later, Archimedes. When I was questor in Sicily [in 75 BC, 137 years after the death of Archimedes] I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder. I immediately said to the Syracusans, some of whose leading citizens were with me at the time, that I believed this was the very object I had been looking for. Men were sent in with sickles to clear the site, and when a path to the monument had been opened we walked right up to it. And the verses were still visible, though approximately the second half of each line had been worn away.
I don’t know what the verses said.
It’s been said that the Roman contributions to mathematics were so puny that the biggest was Cicero’s discovery of this tomb. But Archimedes’ result doesn’t by itself give an easy intuitive way to see where that factor of 4 is coming from! You may or may not find it to be a useful clue.
(In fact, “equal surface areas for slices of equal thickness” is a special case of a principle called “Duistermaat-Heckman localization”. For that, try page 23 of chapter 2 of the book by Ginzburg, Guillemin and Karshon. But that’s much fancier stuff than I’m wondering about here, I think.)
### 70 Responses to Geometry Puzzle
1. Mike Stay says:
I would look at it from the point of iterating the surface integral.
Start with a point at distance r and integrate over a half-circle to get half the circumference.
Then integrate that over a half circle to get half the surface area.
Then integrate that over half a circle to get half the “surface volume” of a hypersphere, and so on.
The volume of the interior is the integral 0 to r of the formula for the surface.
• John Baez says:
Hi, Mike. When you first told me this, I thought I understood you. Now I don’t.
Could you say in a bit more detail how you get the number 1/4, or 4?
2. streamfortyseven says:
I’d think in terms of mapping the area of the circle onto the area of the sphere (I’m using plain English here, not any sort of mathematical terms of art), i.e. the diameter of the circle is 2*r, and it’s thus possible to lay out 3.14 circles around the surface of the sphere… so we tile the surface of the sphere with three circle areas, and we’ve got space left over. Maybe the problem might be easier with a transformation into cylindrical coordinates? Beats me.
• John Baez says:
Beats me too. I don’t know how to solve the problem this way. I don’t know how to ’tile’ the sphere with 4 circles of the same radius.
The ancient Greeks, in their desperate quest to square the circle, discovered some beautiful facts. For example, Hippocrates proved that the shaded crescent here has the same area as the shaded triangle:
This region is now called the “lune of Hippocrates”. If any of you out there are bored someday, sitting in an airport or something, it’s a great puzzle to prove for yourself what Hippocrates proved around 450 BC.
However, because the surface of the sphere is not flat, the ancient Greek techniques don’t work on it… which is undoubtedly why Archimedes was so excited when he worked out the area of a sphere!
• John Baez says:
After posting my puzzle about the lune of Hippocrates I felt an obligation to solve it again, to check if it really is “great”. I had seen the answer, but forgotten it.
And it really is great. It took a while to find, but there’s a beautiful solution that fits neatly in my mind. I remember that when I first saw this problem, it seemed very scary. It’s nice when things get clear when you think about them.
3. Cristi says:
Dear John,
This picture shows how we can use a paraboloid to project the sphere on a plane. The projection preserves the area. The sphere slices project like in the Archimedes’ hat-box theorem, but they end in the plane sectioning the sphere.
Half of the sphere projects onto a circle of double area than that of the section circle of the sphere, so here is the factor of 4.
Regards,
Cristi
4. David Pollard says:
Coincidentally, I’d been wondering about why the volume of a cone is one third of the volume of a cylinder into which it just fits.
• John Baez says:
Of course one explanation — probably not the sort you want — is that the area of a circle is proportional to r2, and when you integrate r2 from 0 to 1, which happens when you work out the volume of a cone by slicing it into thin discs, you get 1/3.
The same integral shows up when you want to prove that the volume of a square pyramid is one third the volume of the cube into which it just fits.
But it would be nice if there were an argument that didn’t involve calculus.
• Austin Shapiro says:
Well, one way to show that a PARTICULAR pyramid has volume equal to 1/3 times the height times the base area is to dissect a cube into six congruent pyramids. For example, divide the cube [0,1]x[0,1]x[0,1] along the planes x=y, y=z, and z=x. There must be 3!=6 pieces (because each one corresponds to an inequality like x<y<z), and they are clearly congruent by symmetry of the construction. Thus each piece has volume 1/6. You can then check that each one is a pyramid with height 1 and base 1/2. (I once persuaded my aunt of this by slicing an actual block of cheese.)
Once you’ve shown that a particular pyramid has volume (1/3)(height)(base), you can use affine shifts, dissections, etc. to convince yourself that the formula holds for ALL pyramids and cones.
• mkz says:
I had thought about the problem of the “4”, without getting anywhere (apart from learning about the “Archimedes symplectomorphism”, as V. I. Arnold puts it).
Here’s an answer to the cone vs. the cylinder question. It is possible to give a geometric answer to the analogous question for a square pyramid vs. a square prism: Start with a cube vs. a square pyramid with height = half the edge length of the cube. You can put six such pyramids in the cube. This cube has twice the volume of the square prism containing the pyramid in the way the cylinder contains the cone. Thus, volume of square pyramid = 1/3 volume of square prism containing it, for the particular aspect ratio we are considering. But we can scale the picture vertically to see that the result holds for square pyramids of arbitrary aspect ratios.
One can then argue that cone vs. cylinder (in fact, pyramid vs. prism for any type of base shape) works the same way, by comparing them “slice by slice” to a square pyramid and prism with the same aspect ratio and base area.
• John Baez says:
John wrote:
The same integral
$\int_0^1 r^2 d r = \frac{1}{3}$
shows up when you want to prove that the volume of a square pyramid is one third the volume of the cube into which it just fits.
But it would be nice if there were an argument that didn’t involve calculus.
There is… sort of.
First, it’s a general fact that a pyramid whose base is any shape whatever has volume
$\frac{1}{3} b h$
where $h$ is its height and $b$ is the area of its base. The volume of a cone or a square pyramid are just special cases of this. This is the 3d generalization of the formula for the area of a triangle
$\frac{1}{2} b h$
and similar formulas hold in any dimension.
Second, if we can show this volume formula for a pyramid whose base is a triangle, we can conclude it’s true for any other pyramid, by approximately chopping the base of that pyramid into tiny triangles.
(This argument involves limits, so it’s essentially calculus. But it’s calculus the way Archimedes did it, or easier, and it’s intuitively obvious. So I’ll count it as acceptable for my purposes. What purposes? Well, I’m mainly just trying to have fun here.)
Third, it’s possible to take a triangular prism and chop it into 3 triangular pyramids of demonstrably equal volume. Since the prism has volume
$b h$
the pyramid must have volume
$\frac{1}{3} b h$
How do we chop a triangular prism into 3 triangular pyramids of the same volume?
$0 \le x,y,z \le 1$
and chop it into 6 pieces depending on whether
$x \le y \le z$
or whatever — there are 6 possible permutations here. Each piece is a triangular pyramid, and they obviously all have the same volume. If we look at the 3 pieces for which
$x \le y$
they form a triangular prism. So, we get the magic factor of 1/3.
It’s probably easier to learn calculus.
But this argument is pleasant if you notice the relation to the Eilenberg-Zilber theorem: a fundamental theorem about taking the product of two simplices and chopping it into simplices, with the help of permutations. Here we are chopping a product of a 1-simplex and a 2-simplex into three 3-simplices.
• mkz says:
I had given almost exactly the same answer, apart from two minor changes: 1) I divided the cube into six *square* pyramids by taking the tips of these pyramids to be at the center of the cube, and bases to be at the faces of the cube, but you use triangular pyramids. I find the square pyramids easier to visualize/count. 2) In order to generalize from a square pyramid to an arbitrary one (e.g., a cone) I proposed to compare the cone to a pyramid with the same base area and height, and keep in mind that thin “horizontal” slices with the same areas will have the same volumes. You propose dividing the cone into pyramids with infinitesimal base areas. I find the horizontal slicing somewhat easier to follow, but YMMV.
• Jon Frankis says:
Ah, very nice thanks MKZ that argument appeals to me. Elegant!
• V. Petruna says:
Why to divide cube to six pyramids? You can divide cube to only three 4-pyramids with equal volume. Pyramids have the same peak, same main base area a2 and height a.
5. Mark Meckes says:
I asked a question on MathOverflow looking for a nice explanation of the higher dimensional analogue of that observation of Archimedes. I didn’t get as geometrically intuitive an answer as I had hoped for.
• John Baez says:
But you did get pointed to Karshon’s work on toric varieties and the Duistermaat-Heckman theorem, which I also mentioned at the end of my comment. It’s sophisticated stuff but it’s tres cool.
The Duistermaat-Heckman theorem became popular in physics circles in the late 80’s, or maybe the early 90’s, when Witten used it to notice that in some quantum mechanics problems the semiclassical approximation gives the right answer. The basic idea is that certain integrals are miraculously easy to compute. And the simplest one is the integral for the area of a sphere: it matches the area of the cylinder into which it snugly fits (not counting top and bottom).
• streamfortyseven says:
here’s Archimedes’ non-calculus-using derivation from the “Archimedes Palimpsest”:
http://www.cut-the-knot.org/pythagoras/Archimedes.shtml
By the way, XY2 means X*(Y*Y)
“Proposition 2.
We can investigate by the [mechanical] method the propositions that
1. Any sphere is (in respect of solid content) four times the cone with base equal to a great circle of the sphere and height equal to its radius; and
2. the cylinder with base equal to a great circle of the sphere and height equal to the diameter is 1½ times the sphere.
1. Let ABCD be a great circle of a sphere, and AC, BD diameters at right angles to one another.
Let a circle be drawn about BD as diameter and in a plane perpendicular to AC, and on this circle as base let a cone be described with A as vertex. Let the surface of this cone be produced and then cut by a plane through C parallel to its base; the section will be a circle on EF as diameter. On this circle as base let a cylinder be erected with height and axis AC, and produce CA to H, making AH equal to CA.
Let CH be regarded as the bar of a balance, A being its middle point.
Draw any straight line MN in the plane of the circle ABCD and parallel to BD. Let MN meet the circle in O,P, the diameter AC in S, and the straight lines AE, AF in Q, R respectively. Join AO.
Through MN draw a plane at right angles to AC; this plane will cut the cylinder in a circle with diameter MN, the sphere in a circle with diameter OP, and the cone in a circle with diameter QR.
Now, since MS = AC, and QS = AS,
MS·SQ = CA·AS
= AO2
= OS2 + SQ2
And, since HA = AC,
HA : AS = CA : AS
= MS : SQ
= MS2 : MS.SQ
= MS2 : (OS2 + SQ2), from above,
= MN2 : (OP2 + QR2)
= (circle, diam. MN) : (circle, diam. OP + circle, diam. QR).
That is,
HA : AS = (circle in cylinder) : (circle in sphere + circle in cone).
Therefore the circle in the cylinder, placed where it is, is in equilibrium, about A, with the circle in the sphere together with the circle in the cone, if both latter circles are placed with their centres of gravity at H.
Similarly for the three corresponding sections made by a plane perpendicular to AC and passing through any other straight line in the parallelogram LF parallel to EF.
If we deal in the same way with all the sets of three circles in which planes perpendicular to AC cut the cylinder, the sphere, and the cone, and which make up those solids respectively, it follows that the cylinder, in the place where it is, will be in equilibrium about A with the sphere and the cone together, when both are placed with their centres of gravity at H.
Therefore, since K is the centre of gravity of the cylinder,
HA : AK = (cylinder) : (sphere + cone AEF).
But HA = 2·AK;
Therefore cylinder = 2 (sphere + cone AEF).
Now cylinder = 3 (cone AEF); [Eucl. XII 10]
Therefore cone AEF = 2 (sphere).
But, since EF = 2·BD,
Cone AEF = 8 (cone ABD);
Therefore, sphere = 4 (cone ABD).
2. Through B, D draw VBW, XDY parallel to AC;
and imagine a cylinder which has AC for axis and the circles on VX, WY as diameters for bases.
Then cylinder VY = 2 (cylinder VD)
= 6 (cone ABD)[Eucl XII 10]
= 3/2 (sphere), from above.
Q.E.D.
“From this theorem, to the effect that a sphere is four times as great as the cone with a great circle of the sphere as base and with height equal to the radius of the sphere, I conceived the notion that the surface of any sphere is four times as great as a great circle in it; for, judging from the fact that any circle is equal to a triangle with base equal to the circumference and height equal to the radius of the circle, I apprehended that, in like manner, any sphere is equal to a cone with base equal to the surface of the sphere and height equal to the radius.”
• John Baez says:
Cool! I can’t really follow this kind of argument without painstaking work, which I haven’t done… but I’m glad you made Archimedes’ actual argument available here.
(I have added a figure to your post, and polished it a bit in other ways.)
The Archimedes Palimpsest is a wonderful thing, and it’s equally wonderful how people rediscovered it and used high-tech methods to extract the information hidden in it… like this proof!
I’ve read that this book is where Archimedes came closest to inventing integration. The argument you cite doesn’t look like integration, but other arguments in this book do. From a very nice Science News article:
For example, Archimedes proved that the area of a section of a parabola is four-thirds the area of the triangle inside it (shown in red in the diagram below). To do so, he built a straight-lined figure that’s an approximation of the curvy one. Then he showed that he could make the approximation as close as anyone could ever demand to both the section of the parabola and to four-thirds the area of the triangle.
Critically, Archimedes never claimed that by adding triangles forever, you could make the straight-line construction exactly equal to the section of the parabola. That would require an actual infinity of triangles. Instead, he just said that you can make the approximation as good as you like, so he was sticking with potential infinity.
6. Nathan Urban says:
A consequence of “equal areas from equal thicknesses” is that the tropics (assumed here to be +/- 30 degrees latitude) contain fully half the Earth’s surface area. A slice from -30 to +30 is as thick as the -90 to -30 and +30 to +90 slices put together.
7. Nathan Urban says:
Also, as noted on the Azimuth wiki, the flux absorbed by the Earth’s surface is reduced not only by atmospheric absorption of incoming solar radiation, but by reflection to space.
• John Baez says:
Indeed! I’ll fix my remark suggesting that it’s just absorption that does it.
8. Austin Shapiro says:
Another interpretation (but not an explanation):
If we “drop” a unit needle at random on a 3-dimensional space ruled by parallel planes at unit intervals, and p is the probability that the needle hits one of the planes, then p=1/2.
The above statement is equivalent to the area formula for the sphere, so if you can prove the above without assuming the area formula, then that would “explain” the area formula.
I wonder if any of the clever ideas that have been brought to bear on Buffon’s needle could apply here. For example, the reasoning behind Buffon’s noodle tells us that if we drop a 2-dimensional probe at random into ruled 3-space, the average linear measure of the intersection depends only on the area of the probe. So if the area of the probe is A, then the number of hits is (say) cA. Using a 1-by-h rectangular probe (h≈0), and borrowing our knowledge of Buffon’s needle in the plane, we can show that c=πp/2. So if we could somehow obtain c=π/4 by independent means, we could infer p=1/2.
My thought experiments with spherical or large rectangular probes don’t go anywhere — they just reproduce the formula c=πp/2…
A journey into how to make a slaughter in the kitchen… and a guess on how someone could have eventually gotten a math idea at a place where they have got a lot of juicy oranges…
10. Tom Leinster says:
Cauchy’s surface area formula says that the surface area of a convex subset of $R^3$ is 4 times the expected area of a shadow.
More precisely: write Gr(3, 2) for the set of planes through the origin. There is a unique probability measure on Gr(3, 2) that is invariant under the natural O(3)-action. For P in Gr(3, 2), write $\pi_P: R^3 \to P$ for orthogonal projection.
Cauchy says that for a compact convex $X \subset R^3$, the surface area of X is 4 times the expected value over all P in Gr(3, 2) of area($\pi_P(X)$). Here “expected value” is to be interpreted with respect to that invariant measure.
One strategy for proving this is as follows: (1) show that the surface area of a convex set is proportional to the expected area of a shadow; (2) work out the constant of proportionality by taking some particular X for which you know both the surface area and the expected area of a shadow. This is extremely close to the solution to Buffon’s needle problem explained here. Part (1) is an entirely conceptual, no-need-to-write-anything, argument—definitely no calculus. It’s as elementary as the Buffon argument.
However, the space usually used in part (2) is… the sphere! So this doesn’t immediately help you.
However, it does get us somewhere. Suppose you can find some convex $X \subset R^3$ for which you know that the surface area is 4 times the expected area of a shadow. Then that tells you that the same is true of the sphere.
• Tom Leinster says:
(Of course, the sphere isn’t convex; I should have said “ball”. But since the ball and the sphere have the same surface area and the same shadows, it doesn’t matter.)
• André Joyal says:
Dear Tom,
The proportionality constant can be obtained from the observation of Urban:
the tropics (assumed here to be +/- 30 degrees latitude) contain fully half the Earth’s surface area. A slice from -30 to +30 is as thick as the -90 to -30 and +30 to +90 slices put together
• Tom Leinster says:
I’m a bit confused. First, I don’t understand how you’re using the word “azimuth”. (I guess contributors to this blog really should know that word, but apparently I don’t.) Wikipedia tells me that it means a horizontal angular measurement, but that doesn’t seem to fit with how you’re using it.
Second, doesn’t Nathan Urban’s observation about the tropics depend on the theorem of Archimedes?
• John Baez says:
For André Joyal’s reply to Tom’s questions above, go here.
Tom wrote:
I don’t understand how you’re using the word “azimuth”. (I guess contributors to this blog really should know that word, but apparently I don’t.)
You’re forgiven. You’ll notice that I didn’t start off this blog by explaining what “azimuth” means and why I’m calling this blog “Azimuth”. I prefer to leave this as a little puzzle. But I also chose this word because it sounds cool and most people don’t know what it means. So, people can project into it whatever emotions and feelings they want.
• Francisco de Almeida says:
A possible mondegreen here? The right word is “zenith”, which sounds a lot like “azimuth”, as both come from the same Arabic root. From Wikipedia:
“[…] inaccurate reading of the Arabic expression سمت الرأس (samt ar-ra’s) meaning “direction of the head”/”path above the head”, by Medieval Latin scribes in the Middle Ages, probably through Old Spanish. It was incorrectly reduced to ‘samt’ (“direction”) and imprecisely written as ‘senit’/’cenit’ by those scribes. Through Old French ‘cenith’, Middle English ‘senith’ and finally ‘zenith’ first appears in the 17th century.
• Francisco de Almeida says:
Let me correct my previous post:
André probably wanted to say “zenith” which indeed sounds a lot like “azimuth” Both words entered our language via Arabic: سمت الرأس (path above the head) and السمت (direction). Source: Wikipedia.
• Mark Meckes says:
I haven’t tried to carry it out, but the obvious choice to try for $X$ (when the sphere is out) would be a regular tetrahedron. There may be some geometric argument to let you reduce the expectation in Cauchy’s formula to looking directly (i.e. perpendicularly) at one of the four faces, and then you’d be done.
• Tom Leinster says:
I hadn’t thought of a tetrahedron. I had thought of a long thin cylinder. That, I think, reduces it to the following similar problem: what is the expected length of the shadow cast by a line segment on a random plane in $R^3$? It’s proportional to the length of the line, but again the challenge is to find the constant of proportionality.
• Mark Meckes says:
Right, and seeing that the constant is 4 almost amounts to Archimedes’s observation again.
• André Joyal says:
Sorry, I do not know how to thread my message
after yours:
I’m a bit confused. First, I don’t understand how your using the word “azimuth”. (I guess contributors to this blog really should know that word, but apparently I don’t.) Wikipedia tells me that it means a horizontal angular measurement, but that doesn’t seem to fit with how you’re using it.
Dear Tom,
I thank you for the correction. I was wrongly naming the vertical direction, the “azimuth”. I wanted to say that the sine of the ALTITUDE of a star above the horizon is on average equal to one half, the sine of 30 degrees. I am no longer sure it follows from Urban’s observation. Archimedes is saying that the quantity sine(altitude) is uniformly distributed between 0 and 1. I am saying that its average is equal to one half. Urban is saying that its median is one half.
It may be possible to prove my statement without using Archimedes’ result, but I don’t know.
• Tom Leinster says:
Thanks. Now I understand. (I really wasn’t correcting you; I just didn’t know what azimuth meant.)
By the way, I think the blog doesn’t allow too many levels of indentation, in order that the text columns don’t get too thin. Hence, replies to replies to replies don’t themselves carry a “reply” button. In order to reply to them, you have to scroll up to the previous “reply” button and click on that. It’s maybe not the best system, but I think that’s how it works.
• John Baez says:
That’s how it works: just three levels of reply.
David Tweed (the person formerly known as “bane”) has given me some code that I might use to fix this, but I haven’t gotten around to trying it.
• John Baez says:
That’s gorgeous stuff, Tom!
So, bringing the conversation back down to Earth, here’s what you’re saying:
This annoying property of our poor planet — that the average density of solar power hitting its atmosphere is just 1/4 the solar power density streaming through space — cannot be fully blamed on it being spherical.
In fact any planet that’s convex in shape would have this property, with the exact same number 1/4, if it tumbled randomly through space.
But phrased this way, one can’t help notice that planets don’t usually tumble randomly through space. They usually rotate around some axis.
So here’s another puzzle: imagine a convex shape that spins at constant speed around some fixed axis. Hold the Sun at some fixed position. Can we get the average density of solar power hitting this object’s surface to be more than 1/4 the density of solar power streaming through space?
For a sphere, we get only 1/4. Would an object of some other shape do better?
If the object doesn’t spin, we can do as well as 1/2: just take a flat object facing the Sun. But what if it spins?
• Tom Leinster says:
This is a spherical cow of an answer, but: consider a cubical planet. The axis of rotation passes through midpoints of opposite faces, and the sun is in the plane orthogonal to that axis. The factor is now $1/\pi$, which is bigger than 1/4.
Why? Well, essentially the question is: what is the expected length of the shadow cast by the unit square on a randomly-chosen line in the plane? The 2-dimensional analogue of Cauchy’s surface area formula says that for a convex subset $X \subset R^2$, the expected shadow length is $1/\pi$ times the perimeter. (The constant of proportionality can be worked out by taking $X$ to be a disk.)
Now you might ask for a planet that has rotational symmetry about its axis. I don’t know what the answer to that is.
• John Baez says:
Does your calculation here take into account the fact that the top and bottom of the cube never get any sun?
If not, and if that’s a problem, I think it’s an easy problem to fix: use a rotating flat shape instead of a cube. If I understand you correctly, as long as the shape is rotating around an axis orthogonal to the direction of the sun, we’ll get a factor of 1/π.
(Or, if we allow our shape to be close to flat but don’t allow zero thickness, we can get as close to 1/π as we like.)
• Tom Leinster says:
You’re right on both counts. I was forgetting the top and bottom of the cube.
Also, there’s no reason why it has to be a cube. Let $Y$ be any compact convex subset of the plane, and let our planet $X$ be the product $Y \times [0, 1]$, rotating about any axis perpendicular to $Y$. (It doesn’t have to pass through the “centre” of $Y$, whatever that means.) I took $Y$ to be a square. But you could take $Y$ to be a disk, which means that your planet is a cylinder. That’s slightly less unlikely than a cubical planet!
The same argument then applies: we’re effectively now in 2 dimensions rather than 3, and since the diameter of a disk divided by its perimeter is $1/\pi$, that’s the factor. At least, it would be the factor if it weren’t for the top and bottom of the cylinder; but you can eliminate problem to as fine an accuracy as you like by taking a long, thin cylinder.
So I think we can conclude that for a cigarette-shaped planet rotating about its long axis, with the sun in the plane orthogonal to it, the factor is only slightly less than $1/\pi$.
• John Baez says:
Cool stuff, Tom!
By the way, this discussion is slightly less impractical than it may seem, because even though planets are close to spherical, people are already thinking about launching artificial satellites to catch solar power, and we can make them any shape we want.
It seems that one popular scheme is to have a satellite in geostationary orbit, so it revolves around the Earth once a day. This makes it act like it’s hovering over a specific location on the equator, allowing it to beam power down in the form of microwaves. Since the transmitter would need to point towards the Earth all the time, it’s indeed natural for the satellite to rotate around a fixed axis orthogonal to the direction of the Sun. So, the optimization problem we’re discussing now is indeed relevant… though there are doubtless many complicating issues which our discussion so far neglects.
One obvious complicating factor, the Earth’s shadow, is not such a big deal: apparently a satellite in geostationary orbit is only in the Earth’s shadow 1% of the time.
However, it seems that the idea of space-based solar power is extremely impractical at present. It takes a lot of energy to send stuff into space, and beaming microwaves down and collecting them is extremely inefficient unless you have a huge transmitter and collector: in 1978, a NASA proposal involved transmitting antenna 1 kilometer in diameter, and a receiving antenna 10 kilometers in diameter.
So, I’m not actually interested in this math problem for any practical reason! It’s just fun.
(And indeed, I secretly suspect that most people working on space-based solar power are doing so just because that’s fun.)
• streamfortyseven says:
It’s too bad the Earth doesn’t actually rotate uniformly around its axis but actually wobbles a little bit:
http://en.wikipedia.org/wiki/Nutation
because we could put a cluster of solar-energy collecting and electricity-generating satellites at the libration points:
http://en.wikipedia.org/wiki/Libration_points
L1 and L2, each 1.5 million km from the Earth, and run graphene nanoribbons:
from each satellite to the Earth to transmit the electricity…
I think we’ll be better off using geostationary solar sails/collectors, the ribbon length would be on the order of 22,000 miles:
http://en.wikipedia.org/wiki/Geostationary_orbit
rather than 1 million miles.
• streamfortyseven says:
we need either a preview feature, or an edit feature, all the wiki cites don’t work because I didn’t leave a space between the URL and the end paren… arrgh.
• John Baez says:
streamfortyseven wrote:
we need either a preview feature, or an edit feature…
Yeah. I’ve asked before if anyone knows a way to equip a WordPress blog with these features — I’m not the sort who enjoys spending my time figuring this stuff out, and there’s no obvious button to click that does the job.
I have bought the ability to edit the cascading style-sheet of this blog, which might help.
I fixed your links. I guess the computer isn’t clever enough to guess when your URL ends if you don’t leave a space.
Our chances of connecting a geostationary satellite to the Earth with carbon nanoribbons looks pretty slim when we can’t even get this stuff to work.
• streamfortyseven says:
Maybe *here’s* a fix for the problem of editing posts: http://wpmu.org/edit-your-wordpress-blog-with-real-time-previews-using-new-active-preview-plugin/
• John F says:
John,
If we can use average values in solving the puzzle then it’s pretty quick. Various 2D constructions can show that the average value of the positive portion of the sine function is 2/pi. Then the area of the sphere is pi*2/pi*R*2piR.
Could a soap bubble type of solution yield insight? Inflate a circle into a half sphere (keeping a ring), and it stretches into exactly twice the area. Although it may break the rules, minimizing integrals etc., maybe it would help someone.
11. […] update 11.10.10: An approximate and by no means accurate visual demonstration of the proposition that the area of a sphere is four times the area of its circular shadow (look also at this comment) […]
12. Zephir says:
Isn’t the number four the number of dimension plus one?
• Tom Leinster says:
No; I think its significance is that it’s $n\omega_n/\omega_{n-1}$, where $\omega_n$ is the volume of the n-dimensional ball of radius 1. I’ll write $a_n = n\omega_n/\omega_{n-1}$.
The sequence begins: $a_1 = 2$, $a_2 = \pi$, $a_3 = 4$, $a_4 = 3\pi/2$, $a_5 = 16/3$, …
In general, $a_n$ alternates between being a rational number and a rational multiple of $\pi$. It’s not usually an integer or an integer multiple of $\pi$; that’s just some good luck in low dimensions.
The general formula is
$a_n = \frac{n \sqrt{\pi} \Gamma((n+1)/2)}{\Gamma((n+3)/2)}$
where $\Gamma$ is the gamma function. You can write this formula without mentioning the gamma function, at the expense of splitting it into the cases of odd n and even n:
$a_{2k} = \frac{(2k)! \pi}{k! 2^{2k-1}}$,
$a_{2k + 1} = \frac{(k!)^2 2^{2k+1}}{(2k)!}$.
(At least, I think those are right; I haven’t double-checked. If not, the truth isn’t far off.)
• Mark Meckes says:
Ah, good point. This probably means my suggestion above of considering a tetrahedron is misguided, because if it worked, it would probably generalize to higher dimensions to give $n+1$ instead of $n \omega_n / \omega_{n-1}$.
13. PL says:
There’s another derivation of the factor 4 from Archimedes’ fact (“same thickness implies same surface area”). It avoids cylinders.
Consider pairs (S,S’) of spheres such that the midpoint of S lies on S’ and such that they intersect. We denote their radii by R and R’, respectively. We investigate the surface area A of the part of S’ that lies inside S.
The part of S’ that lies inside S is a spherical cap. Elementary geometry tells us that its thickness t satisfies the equation 2R’t=R2. But according to Archimedes’ fact from above, R’t determines the surface area A. (More precisely, A is equal to R’t up to multiplicative constant.)
Now think of S as being fixed and let S’ vary (with the boundary condition that R’ is at least R/2). The argument above shows that A also remains fixed. In one extreme case, namely R’=R/2, S’ touches S from inside and it is half as large as S. Therefore, A is 1/4 of the surface area of S. In the other extreme case, where R’ goes to infinity, S’ becomes a plane through the midpoint of S. In this case, A is the area of a circle of radius R.
• Cristi says:
Very suggestive picture of the factor 1/4.
14. Jon Frankis says:
Having reflected for a while I think the most straightforward way that Archimedes might have demonstrated his very nice finding that the surface area of the sphere is equal to 4 times the area of one of its great circles is just as you’ve done in the first place far above, by first proving equivalence to the surface area of the open cylinder which just contains it.
That is a simpler proof than the one given by Archimedes himself isn’t it (thanks streamfortyseven)? What am I missing here? Lacking imagination, I can’t imagine how a proof might be made more straightforward than the one you’ve given.
Full disclosure: I managed to either overlook that you’d given that proof, or forgot, and rederived it myself. In the process I considered all manner of uglier and more complicated ways of doing it to finally conclude that there was a nice feeling of irreducible simplicity to that one. Perhaps the only term of art that you’ve omitted from it is “method of exhaustion”, which Archimedes knew well anyway.
Is it just that you’ve elided the actual calculation above, using instead the words “magic fact”? The geometry and calculation are perfectly elementary anyway, I think the only magic needed is the term “method of exhaustion”.
To write out the calculation to show just how elementary it really is:
Consider bands of a containing cylinder that may be pictured as slicing a sphere into what on Earth, for instance, would be called bands of latitude. We wish to know what the surface area of each band on the sphere might be relative to the corresponding band on the cylinder (in fact we wish to prove that they are equal). In the diagram let f on the small right triangle be the height of a (small) band on the cylinder. On the sphere the respective small longitudinal arc of surface has length quite close to h, larger than f. Draw the right triangle of sides f, g and h. Note that this is similar to the larger right triangle of sides F, G and H where H is a radius of the sphere drawn to a point in the (small) surface band we’re considering.
The surface area of the small band of the cylinder is circumference times height or 2πH.f (H is a radius of the sphere as per the diagram). On the sphere the corresponding band of latitude has area close to 2πF.h
From the diagram and the similar triangles we can see that H/F = h/f or H.f = F.h
Therefore as the bands we consider are made smaller and more numerous their area, and the sum of their areas on the cylinder and sphere respectively, approach equality by Method of Exhaustion. [Wave hands and invite students to improve the rigor if they’re offended by anything; Archimedes would have been satisfied]
Thus in short order the lovely factor 4 appears because the surface area calculation is so easy for the cylinder. Then onward and upward to Cauchy’s surface area formula and more …
• John Baez says:
John Frankis wrote:
Having reflected for a while I think the most straightforward way that Archimedes might have demonstrated his very nice finding that the surface area of the sphere is equal to 4 times the area of one of its great circles is just as you’ve done in the first place far above, by first proving equivalence to the surface area of the open cylinder which just contains it.
Just in case people forget what I wrote “far above” (I did!) let me remind them:
Take a sphere and slice it with a bunch of parallel planes, like chopping an apple with a cleaver. If two slices have the same thickness, they also have the same surface area!
(When I say “surface area” here, I’m only counting the red skin of the apple slices.)
There’s an interesting cancellation at work here. A slice from near the top or bottom of the sphere will be smaller, but it’s also more “sloped”. The magic fact is that these effects exactly cancel when we compute its surface area.
If you think a bit, you can see this is equivalent to another nice fact:
The surface area of any slice of a sphere matches the surface area of the corresponding slice of the cylinder with the same radius.
If you don’t get what I mean, see the picture at Wolfram Mathworld:
And this does the job:
V = 2πr × 2r = 4πr2
Thanks for your nice geometrical argument proving the “magic fact”. The similar triangles you drew:
indeed make it obvious, not at all “magic”, that the diminished radius of a slice from near the top or bottom of the apple will exactly cancel the increase in its surface area due to being “sloped”.
And then, as you note, a little integration (or “exhaustion”) finishes the job.
But just as Newton took results he’d proved using calculus and reproved them using Euclidean geometry in the Principia Mathematica to make them “more rigorous”, I wouldn’t be surprised if Archimedes did the same, at least some of the time!
• Jon Frankis says:
Thanks John! It seems it’s inherent in Nature that there exist diverse, orthogonal even, ways of understanding or appreciating almost all true things, and it’s a demonstration of the power of mathematics that Newton or Archimedes or people posting on Azimuth can give differing yet equally valid proofs of a theorem, each casting light from a different and sometimes fascinating angle. Wonderful!
• John Baez says:
Yes, it’s great — and I love how old things like A = 4πr2 still reveal a wealth of surprises if you poke at them.
15. It’s not helping things any more, but the nicest (of course) calculation for *all* the sphere measures I know passes through the Gaussian integral, which reduces the problem to (the Γ function at naturals or half-naturals) and stuff…
For the 2-d sphere, I rather like Gauss’ angle deficit formula for spherical triangles, which is a gem of dissection geometry — Coxeter gives a clear description of it in his undergraduate-level text — but it occurs to me that there should be a clean way to compare the angles and edge arcs of a spherical triangle as well, from which the 4 ought to appear again… sorry this is in my usual vague style, but I’m now remembering appointments and must dash!
• John Baez says:
I don’t quite get what you’re saying, but it sounds like you’re saying Descartes’ theorem — that the total angle defect of any polyhedron that’s topologically a ball must be 4π — should yield a proof that the area of a sphere is 4πr2. That would be nice. I bet it does.
And if it does, we should be able to generalize from a round sphere to any convex body.
Challenge: use Descartes’ theorem to show that the surface area of any convex polyhedron is 4 times its average shadow area!
But Descartes’ theorem doesn’t only apply to convex polyhedra: any polyhedron homeomorphic to a 3-ball will do. So this makes me want to generalize the result about “average shadow area” to all such polyhedra. And I think I see how. We just need to generalize the concept of “average shadow area” in the right way, so that an X-ray crossing the body’s surface 2n times contributes n as much to the “average shadow area” than a ray that crosses it just twice.
With this definition I bet someone can use Descartes’ theorem to prove that the surface area is 4 times its average shadow area for any polyhedron homeomorphic to a 3-ball.
• No, I’m referring to the spherical geometric fact that the area A of a triangle in the unit sphere with inner angles a,b,c is A=(a+b+c-π); the trick, of course, is getting that factor of 1 — the disection argument will show the two sides are proportional.
• Oh, for non-convex things, you need to count the multiplicity of a point in the shadow (or more generally, you need the Euler characteristic of an intersection … but that’s for other intrinsic volumes).
• Now for some trigonometry. Let a spherical right triangle have remaining angles θ,φ; let the adjacent sides subtend angles τ,σ at the center, and let the altitude to the hypotenuse subtend angle ζ. We then have
$\cos^2 \theta+ cos^2 \phi + sin^2 \zeta =1$ and
$\sin \tau \cos \phi = \cos \tau \sin \zeta$ and
$\sin \sigma \cos\theta = \cos\sigma\sin\zeta$.
Then, for instance, we can write
$\tan\tau\tan\sigma = \frac{\sin^2\zeta}{\cos \theta \cos \phi}$
and then play around with how $\sin^2\zeta= \sin^2\theta-\cos^2\phi$. It’s not so neatly “picture-y”, but these are formulas all the astronomers ought to have known at some point.
• Tom Leinster says:
John, your generalization of “shadow area” to take account of multiplicity corresponds exactly to something standard in integral geometry.
We’ve been talking about projections onto 2-dimensional subspaces of 3-dimensional space, so I’ll stick to that case, but everything I’m about to say generalizes in the way that you’d guess to any pair of dimensions.
I mentioned that there’s a canonical probability measure on the space Gr(3,2) of planes through the origin in 3-dimensional space. There’s also a canonical measure (at least, canonical up to a scale factor) on the space Graff(3,1) of lines in 3-space, not necessarily through the origin. Here “Graff” is a funny abbreviation for “affine Grassmannian”.
Now let X be any subset of 3-space. The expected area of the shadow of X on a random plane $P \in Gr(3, 2)$ is the same as the measure of the set of lines $L \in Gr(3, 1)$ that intersect X. (Think about L being orthogonal to P. Also, I’m assuming that we’ve normalized appropriately.) In that sense, it doesn’t matter whether you project onto planes or intersect with lines.
But now suppose you want to talk about areas-with-multiplicity, as you were doing. In that situation it’s probably more natural to think about intersections with lines. For the expected area-with-multiplicity of the shadow of X on a random plane $P \in Gr(3,2)$ is the integral over all $L \in Graff(3,1)$ of the number of times that L crosses X. Now that number of crossings is the Euler characteristic of $L \cap X$, so what I’m saying is that the expected area-with-multiplicity of the shadow is
$\int_{Graff(3,1)} \chi(L \cap X) dL.$
This is a much less ad hoc way of expressing it than anything involving area-with-multiplicity.
So one way to state Cauchy’s formula is this: for any subset $X \subset R^3$ that can be expressed as a finite union of compact, convex sets,
$(surface-area)(X) = c\int_{Graff(3,1)} \chi(L \cap X) dL$
where $c$ is some positive constant that can be worked out once you’ve decided how your measure on Graff(3,1) is to be normalized.
This is usually called (an instance of) Crofton’s formula.
The general statement of Crofton’s formula involves the so-called intrinsic volumes $\mu_0, \ldots, \mu_n$, where n is the dimension of the ambient space. In our case:
$n = 3$
The 0-dimensional intrinsic volume $\mu_0$ is Euler characteristic
$\mu_1$ is known as “mean width”
$\mu_2$ is surface area
$\mu_3$ is volume.
• John Baez says:
Gorgeous stuff as usual, Tom!
John, your generalization of “shadow area” to take account of multiplicity corresponds exactly to something standard in integral geometry.
Good — it seemed pretty natural, so it would be depressing if people were so silly as not to have noticed it before.
By the way, I now think the answer to my “Challenge” is very easy if one lets oneself use a bit of machinery. Taking a suitable limit, Descartes’ theorem gives a special case of the Gauss-Bonnet theorem, which says that the integral of the scalar curvature over a surface is 2π times its Euler characteristic. Namely, it gives the special case where our surface is a sphere embedded in R3: it says that the integral of this sphere’s scalar curvature must be 2π × 2 = 4π. But a round sphere of radius 1 has scalar curvature 1. So, the area of this sphere must be 4π.
This could be seen as a pathetically obscurantist way of computing the area of a sphere, but my point is rather to illustrate that the 4π in Descartes’ theorem is directly connected to the 4π in the formula for the area of the sphere.
And probably all this generalizes enormously, and probably that’s already been done.
16. John Baez says:
Whoops! It turns out that this proof by Archmimedes — and a whole lot of other stuff — was a hoax.
• streamfortyseven says:
Not only Archimedes, but Socrates as well:
“WASHINGTON—A group of leading historians held a press conference Monday at the National Geographic Society to announce they had “entirely fabricated” ancient Greece, a culture long thought to be the intellectual basis of Western civilization. … Emily Nguyen-Whiteman, one of the young academics who “pulled a month’s worth of all-nighters” working on the project, explained that the whole of ancient Greek architecture was based on buildings in Washington, D.C., including a bank across the street from the coffee shop where they met to “bat around ideas about mythology or whatever.”
“We picked Greece because we figured nobody would ever go there to check it out,” Nguyen-Whiteman said. “Have you ever seen the place? It’s a dump. It’s like an abandoned gravel pit infested with cats.”
17. Cristi says:
Is it possible to find a spiral which can be used to cut the surface of the sphere in very small (infinitesimal) slices which then can be rearranged to obtain a circle of radius double to that of the sphere? The circumference of the disk is double the sphere’s equator, but the equator has two sides. Half of the slices of the disk come from one hemisphere, half from the other.
This image illustrates the idea.
18. This means that the average outbound energy flux is actually $\frac{1}{4}$ of the inbound energy flux. The question if there is some deeper reason for this simple relation was posed as a geometry puzzle here […]
19. arch1 says:
This discussion reminded me of the following throwaway comment in Littlewood’s Miscellany, a wonderful book by J. E. Littlewood:
“The surface of the sun has 6×10^5 times the brightness of the full moon (incidentally the sun is 5×10^6 times as bright as the *half* moon).”
20. […] Why a quarter? That’s a nice geometry puzzle […]
21. The question if there is some deeper reason for this factor of 4 was posed as a geometry puzzle here […]
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https://math.stackexchange.com/questions/1581024/applications-to-calculus-questions | # applications to calculus questions
A farmer has 80m length of fencing. He wants to use it to form 3 sides of a rectangular enclosure against an existing fence, which provides the 4th side. find the maximum area that he can enclose and give its dimensions. I know I have to use the principle of stationary points but I don't know how to begin the problem.
Let the two side lengths be $l,w$.
Then, $2 l + w = 80$ and you want to maximize $A=lw = l ( 80 - 2l)$.
Solve $\frac{dA}{dl}=0$ for $l$, see that $\frac{dA}{dl}$ changes sign so that this value of $l$ is a maximizer then plug it into the expression for $A$ to find the maximized area.
Alternatively, note that $A$ specifies a parabola and look for its vertex as the maximizer since it opens downwards.
I know I have to use the principle of stationary points but I don't know how to begin the problem.
The guidelines below (from Larson’s book) show you how to start (this and related problems).
1. Identify all given quantities and all quantities to be determined. If possible, make a sketch and label it with any relevant measurements.
Here is the rectangular enclosure:
Given quantity: the length of fencing which is $80$
Quantities to be determined: The area $A$ of the rectangular enclosure and the dimensions $x$ and $y$.
1. Write a primary equation for the quantity that is to be maximized or minimized.
The quantity that is to be maximized is the area $A$. So, a primary equation is $$A=xy$$
1. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.
From the step 1 we get the secondary equation $2x+y=80$ which implies $$y=80-2x$$ Substituting this in the equation of step 2 we get $$A(x)=x(80-2x)$$ $$A(x)=-2x^2+80x$$
1. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.
$$0<x<40,\qquad 0<y<80$$
1. Determine the desired maximum or minimum value by the calculus techniques.
Now, we have to do the calculation. The first thing is to solve $A'(x)=0$ with respect the domain specified in the step 4. | 2019-11-18T03:19:28 | {
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http://mathhelpforum.com/pre-calculus/90117-finding-zero.html | 1. ## Finding zero
I cannot figure out how to do this. Any help would be appreciated. Thanks!
1. Find all of x for which (x2 -x)(2x+1) - (x2 +x)(2x-1) is zero.
2. Find the domain of the function f(x) =square root of: x2 -3x. Express your answer in terms of intervals.
2. Originally Posted by cam_ddrt
2. Find the domain of the function f(x) =square root of: x2 -3x. Express your answer in terms of intervals.
You mean $f(x) = \sqrt{x^2 - 3x}$?
Only non-negative numbers can be under the square root, so solving the inequality $x^2 - 3x \geq 0$ would give us the values of x we want. Solving the corresponding equation $x^2 - 3x = 0$ would give use the critical values of 0 and 3. Now test different values of x to see if what's underneath the square root is non-negative.
For $x < 0$, $x^2 - 3x > 0$, check.
For $x = 0$, $x^2 - 3x = 0$, check.
For $0 < x < 3$, $x^2 - 3x < 0$, NOPE.
For $x = 3$, $x^2 - 3x = 0$, check.
For $x > 3$, $x^2 - 3x > 0$, check.
So our domain is $(-\infty, 0] \cup [3, \infty)$.
01
3. Originally Posted by cam_ddrt
1. Find all of x for which (x2 -x)(2x+1) - (x2 +x)(2x-1) is zero.
Multiply and simplify
$(x^2 -x)(2x+1) - (x^2 +x)(2x-1)=0$
$2x^3+x^2-2x^2-x-(2x^3-x^2+2x^2-x)=0$
$2x^3+x^2-2x^2-x-2x^3+x^2-2x^2+x=0$
$-2x^2=0$
$x=0
$ | 2016-08-24T05:43:40 | {
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http://www.caam.rice.edu/~lc55/SFEMaNS/html/doc_debug_test_15.html | SFEMaNS version 4.1 (work in progress) Reference documentation for SFEMaNS
Test 15: Navier-Stokes with Temperature and LES
### Introduction
In this example, we check the correctness of SFEMaNS for a thermohydrodynamic problem. We use Dirichlet boundary conditions and a stabilization method called entropy viscosity that we refer as LES. We note this test does not involve manufactured solution and consist to check four quantities, like the kinetic energy of specific Fourier modes, are the same than the values of reference.
We solve the temperature equation:
\begin{align*} \partial_t T+ \bu \cdot \GRAD T - \kappa \LAP T &= 0, \\ T_{|\Gamma} &= T_\text{bdy} , \\ T_{|t=0} &= T_0, \end{align*}
and the Navier-Stokes equations:
\begin{align*} \partial_t\bu+\left(\ROT\bu\right)\CROSS\bu - \frac{1}{\Re}\LAP \bu +\GRAD p &= \DIV (\nu_E \GRAD \bu ) + \alpha T \textbf{e}_z, \\ \DIV \bu &= 0, \\ \bu_{|\Gamma} &= \bu_{\text{bdy}} , \\ \bu_{|t=0} &= \bu_0, \\ p_{|t=0} &= p_0, \end{align*}
in the domain $$\Omega= \{ (r,\theta,z) \in {R}^3 : (r,\theta,z) \in [0,1/2] \times [0,2\pi) \times [0,1]\}$$ with $$\Gamma= \partial \Omega$$. We denote by $$\textbf{e}_z$$ the unit vector in the vertical direction. The term $$\DIV (\nu_E \GRAD \bu )$$ is a stabilization term that involve an artificial viscosity called the entropy viscosity. We refer to the section Entropy viscosity for under resolved computation for the definition of this term. The data are the boundary datas $$T_\text{bdy}$$ and $$\bu_{\text{bdy}}$$, the initial data $$T_0$$, $$\bu_0$$ and $$p_0$$. The parameters are the thermal diffusivity $$\kappa$$, the kinetic Reynolds number $$\Re$$, the thermal gravity number $$\alpha$$ and the real $$c_\text{e}$$ for the entropy viscosity. We remind that these parameters are dimensionless.
### Manufactured solutions
As mentionned earlier this test does not involve manufactured solutions. As consequence we do not consider specific source term and only initialize the variables to approximate.
\begin{align*} T(r,\theta,z,t=0) & = -z - (z-0.5)(z+0.5) \left(1+\cos(\theta)+\sin(\theta)+\cos(2\theta)+\sin(2\theta) \right) , \\ u_r(r,\theta,z,t=0) &= 0, \\ u_{\theta}(r,\theta,z,t=0) &= 0, \\ u_z(r,\theta,z,t=0) &=0, \\ p(r,\theta,z,t=0) &= 0, \end{align*}
The boundary datas $$T_\text{bdy}, \bu_{\text{bdy}}$$ are computed accordingly.
The finite element mesh used for this test is named RECT10_BENCHMARK_CONVECTION_LES.FEM and has a mesh size of $$0.1$$ for the P1 approximation. You can generate this mesh with the files in the following directory: ($SFEMaNS_MESH_GEN_DIR)/EXAMPLES/EXAMPLES_MANUFACTURED_SOLUTIONS/RECT10_BENCHMARK_CONVECTION_LES. The following image shows the mesh for P1 finite elements. Finite element mesh (P1). ### Information on the file condlim.f90 The initial conditions, boundary conditions and the forcing term in the Navier-Stokes equations are set in the file condlim_test_15.f90. Here is a description of the subroutines and functions of interest. 1. The subroutine init_velocity_pressure initializes the velocity field and the pressure at the time $$-dt$$ and $$0$$ with $$dt$$ being the time step. It is done by using the functions vv_exact and pp_exact as follows: time = 0.d0 DO i= 1, SIZE(list_mode) mode = list_mode(i) DO j = 1, 6 !===velocity un_m1(:,j,i) = vv_exact(j,mesh_f%rr,mode,time-dt) un (:,j,i) = vv_exact(j,mesh_f%rr,mode,time) END DO DO j = 1, 2 !===pressure pn_m2(:) = pp_exact(j,mesh_c%rr,mode,time-2*dt) pn_m1 (:,j,i) = pp_exact(j,mesh_c%rr,mode,time-dt) pn (:,j,i) = pp_exact(j,mesh_c%rr,mode,time) phin_m1(:,j,i) = pn_m1(:,j,i) - pn_m2(:) phin (:,j,i) = Pn (:,j,i) - pn_m1(:,j,i) ENDDO ENDDO 2. The subroutine init_temperature initializes the temperature at the time $$-dt$$ and $$0$$ with $$dt$$ being the time step. It is done by using the function temperature_exact as follows: time = 0.d0 DO i= 1, SIZE(list_mode) mode = list_mode(i) DO j = 1, 2 tempn_m1(:,j,i) = temperature_exact(j, mesh%rr, mode, time-dt) tempn (:,j,i) = temperature_exact(j, mesh%rr, mode, time) ENDDO ENDDO 3. The function vv_exact contains the analytical velocity field. It is used to initialize the velocity field and to impose Dirichlet boundary conditions on the velocity field. It is set to zero. vv(:) = 0.d0 RETURN 4. The function pp_exact contains the analytical pressure. It is used to initialize the pressure and is set to zero. vv=0.d0 RETURN 5. The function temperature_exact contains the analytical temperature. It is used to initialize the temperature and to impose Dirichlet boundary condition on the temperature. 1. We construct the radial and vertical coordinates r, z. r = rr(1,:) z = rr(2,:) 2. We set the temperature to zero. vv=0.d0 3. For the Fourier mode $$m=0$$ the temperature only depends of the TYPE 1 (cosine). IF (m==0 .AND. TYPE==1) THEN vv(:)= - z - (z-5d-1)*(z+5d-1) 4. For the Fourier mode $$m\geq1$$, the temperature does not depend of the TYPE (1 for cosine and 2 for sine) and is defined as follows: ELSE IF (m.GE.1) THEN vv= -(z-5d-1)*(z+5d-1) END IF RETURN 6. The function source_in_temperature computes the source term denoted $$f_T$$ in previous tests, of the temperature equation. As it is not used in this test, we set it to zero. vv = 0.d0 RETURN 7. The function source_in_NS_momentum computes the source term $$\alpha T \textbf{e}_z$$ of the Navier-Stokes equations depending of its TYPE (1 and 2 for the component radial cosine and sine, 3 and 4 for the component azimuthal cosine and sine, 5 and 6 for the component vertical cosine and sine) as follows: IF (TYPE==5) THEN vv = inputs%gravity_coefficient*opt_tempn(:,1,i) ELSE IF (TYPE==6) THEN vv = inputs%gravity_coefficient*opt_tempn(:,2,i) ELSE vv = 0.d0 END IF RETURN All the other subroutines present in the file condlim_test_15.f90 are not used in this test. We refer to the section Fortran file condlim.f90 for a description of all the subroutines of the condlim file. ### Setting in the data file We describe the data file of this test. It is called debug_data_test_15 and can be found in the following directory: ($SFEMaNS_DIR)/MHD_DATA_TEST_CONV_PETSC.
1. We use a formatted mesh by setting:
===Is mesh file formatted (true/false)?
.t.
2. The path and the name of the mesh are specified with the two following lines:
===Directory and name of mesh file
'.' 'RECT10_BENCHMARK_CONVECTION_LES.FEM'
where '.' refers to the directory where the data file is, meaning ($SFEMaNS_DIR)/MHD_DATA_TEST_CONV_PETSC. 3. We use one processor in the meridian section. It means the finite element mesh is not subdivised. To do so, we write: ===Number of processors in meridian section 1 4. We solve the problem for $$3$$ Fourier modes. ===Number of Fourier modes 3 5. We use $$3$$ processors in Fourier. ===Number of processors in Fourier space 3 It means that each processors is solving the problem for $$3/3=1$$ Fourier modes. 6. We do not select specific Fourier modes to solve. ===Select Fourier modes? (true/false) .f. 7. We approximate the Navier-Stokes equations by setting: ===Problem type: (nst, mxw, mhd, fhd) 'nst' 8. We do not restart the computations from previous results. ===Restart on velocity (true/false) .f. ===Restart on magnetic field (true/false) .f. It means the computation starts from the time $$t=0$$. 9. We use a time step of $$0.05$$ and solve the problem over $$10$$ time iterations. ===Time step and number of time iterations 5.d-2, 10 10. We set the number of domains and their label, see the files associated to the generation of the mesh, where the code approximates the Navier-Stokes equations, ===Number of subdomains in Navier-Stokes mesh 1 ===List of subdomains for Navier-Stokes mesh 1 11. We set the number of boundaries with Dirichlet conditions on the velocity field and give their respective labels. ===How many boundary pieces for full Dirichlet BCs on velocity? 2 ===List of boundary pieces for full Dirichlet BCs on velocity 2 3 12. We set the kinetic Reynolds number $$\Re$$. ===Reynolds number 50.d0 13. We use the entropy viscosity method to stabilize the equation. ===Use LES? (true/false) .t. 14. We define the coefficient $$c_\text{e}$$ of the entropy viscosity. ===Coefficient multiplying residual 1.d0 15. We give information on how to solve the matrix associated to the time marching of the velocity. 1. ===Maximum number of iterations for velocity solver 100 2. ===Relative tolerance for velocity solver 1.d-6 ===Absolute tolerance for velocity solver 1.d-10 3. ===Solver type for velocity (FGMRES, CG, ...) GMRES ===Preconditionner type for velocity solver (HYPRE, JACOBI, MUMPS...) MUMPS 16. We give information on how to solve the matrix associated to the time marching of the pressure. 1. ===Maximum number of iterations for pressure solver 100 2. ===Relative tolerance for pressure solver 1.d-6 ===Absolute tolerance for pressure solver 1.d-10 3. ===Solver type for pressure (FGMRES, CG, ...) GMRES ===Preconditionner type for pressure solver (HYPRE, JACOBI, MUMPS...) MUMPS 17. We give information on how to solve the mass matrix. 1. ===Maximum number of iterations for mass matrix solver 100 2. ===Relative tolerance for mass matrix solver 1.d-6 ===Absolute tolerance for mass matrix solver 1.d-10 3. ===Solver type for mass matrix (FGMRES, CG, ...) CG ===Preconditionner type for mass matrix solver (HYPRE, JACOBI, MUMPS...) MUMPS 18. We solve the temperature equation (in the same domain than the Navier-Stokes equations). ===Is there a temperature field? .t. 19. We set the number of domains and their label, see the files associated to the generation of the mesh, where the code approximated the temperature equation. ===Number of subdomains in temperature mesh 1 ===List of subdomains for temperature mesh 1 20. We set the thermal diffusivity $$\kappa$$. ===Diffusivity coefficient for temperature (1:nb_dom_temp) 1.d0 21. We set the thermal gravity number $$\alpha$$. ===Non-dimensional gravity coefficient 50.d0 22. We set the number of boundaries with Dirichlet conditions on the temperature and give their respective labels. ===How many boundary pieces for Dirichlet BCs on temperature? 1 ===List of boundary pieces for Dirichlet BCs on temperature 2 23. We give information on how to solve the matrix associated to the time marching of the temperature. 1. ===Maximum number of iterations for temperature solver 100 2. ===Relative tolerance for temperature solver 1.d-6 ===Absolute tolerance for temperature solver 1.d-10 3. ===Solver type for temperature (FGMRES, CG, ...) GMRES ===Preconditionner type for temperature solver (HYPRE, JACOBI, MUMPS...) MUMPS 24. To get the total elapse time and the average time in loop minus initialization, we write: ===Verbose timing? (true/false) .t. These informations are written in the file lis when you run the shell debug_SFEMaNS_template. ### Outputs and value of reference The outputs of this test are computed with the file post_processing_debug.f90 that can be found in the following: ($SFEMaNS_DIR)/MHD_DATA_TEST_CONV_PETSC.
To check the well behavior of the code, we compute four quantities:
1. The kinetic energy of the Fourier mode $$m=0$$.
2. The kinetic energy of the Fourier mode $$m=1$$.
3. The kinetic energy of the Fourier mode $$m=2$$.
4. The L2 norm of the velocity divergence divided by the L2 norm of the gradient of the velocity.
These quantities are computed at the final time $$t=1$$. They are compared to reference values to attest of the correctness of the code. These values of reference are in the last lines of the file debug_data_test_15 in the directory (\$SFEMaNS_DIR)/MHD_DATA_TEST_CONV_PETSC. They are equal to:
============================================
RECT10_BENCHMARK_CONVECTION_LES.FEM, dt=5.d-2, it_max=10
===Reference results
6.33685640432350423E-004 !e_c_u_0
0.12910286398104689 !e_c_u_1
0.10838939329896366 !e_c_u_2
4.93403150219499723E-002 !||div(un)||_L2/|un|_sH1
To conclude this test, we show the profile of the approximated, pressure, velocity magnitude and temperature at the final time. These figures are done in the plane $$y=0$$ which is the union of the half plane $$\theta=0$$ and $$\theta=\pi$$.
Pressure in the plane plane y=0. Velocity magnitude in the plane plane y=0.
Temperature in the plane plane y=0. | 2018-01-23T09:51:22 | {
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http://mathhelpforum.com/calculus/2570-rate-change.html | # Thread: Rate of change
1. ## Rate of change
How do you do the following rate of change questions
when t = 1.2 for all the following questions:
1) y = 4/t + 4lnt
2) x = sin(t^2 +1) , y = cos(2t -3)
3) x = 1 / 1+2t , y = t / 1+t
4) q = 2e^-t/2.cos2t
5) x = e^2t.t^3(2-t)^4
Please could you explain the process as im quite lost. Thanks
2. e for 4) and 5) is the exponent
3. Originally Posted by dadon
How do you do the following rate of change questions
when t = 1.2 for all the following questions:
1) y = 4/t + 4lnt
2) x = sin(t^2 +1) , y = cos(2t -3)
3) x = 1 / 1+2t , y = t / 1+t
4) q = 2e^-t/2.cos2t
5) x = e^2t.t^3(2-t)^4
Please could you explain the process as im quite lost. Thanks
The rate of change is the derivative with respect to time. So for 1):
$
y = 4/t + 4 \ln(t)
$
so:
$
\frac{dy}{dt}=\frac{-4}{t^2} + \frac{4}{t}
$
,
so when $t=1.2$:
$
\frac{dy}{dt}=\frac{-4}{1.2^2} + \frac{4}{1.2}=0.5555..
$
RonL
4. ## Re:
Thanks! On that question (1) I was confused about the 4lnt part. You make it seem so simple! | 2016-12-08T00:28:23 | {
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https://questions.examside.com/past-years/jee/question/let-a1a2a30-be-an-ap-s-sum-jee-main-mathematics-sequences-and-series-zo3apwrwavubtku6 | 1
JEE Main 2019 (Online) 9th January Morning Slot
Let ${a_1},{a_2},.......,{a_{30}}$ be an A.P.,
$S = \sum\limits_{i = 1}^{30} {{a_i}}$ and $T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}$.
If $a_5$ = 27 and S - 2T = 75, then $a_{10}$ is equal to :
A
47
B
42
C
52
D
57
Explanation
Let the common difference = d
S = $\sum\limits_{i = 1}^{30} {{a_i}}$
= $a$1 + $a$2 + . . . . . + $a$30
$\therefore$ S = ${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$
= 15 [$a$1 + $a$1 + 29d]
= 15 (2$a$1 + 29d)
T = $\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}$
= $a$1 + $a$3 + . . . . . . + $a$29
= ${{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]$
= ${{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]$
= ${{15} \over 2}\left[ {2a{}_1 + 28d} \right]$
= 15 ($a$1 + 14d)
Given,
S $-$ 2T = 75
$\Rightarrow$ 15(2$a$1 + 29d) $-$ 2 $\times$ 15 ($a$1 + 14d) = 75
$\Rightarrow$ 30$a$1 + 15 $\times$ 29d $-$ 30 $a$1 $-$ 420d = 75
$\Rightarrow$ 435d $-$ 420d = 75
$\Rightarrow$ 15d = 75
$\Rightarrow$ d = 5
Given that,
$a$5 = 27
$\Rightarrow$ $a$1 + 4d = 27
$\Rightarrow$ $a$1 + 20 = 27
$\Rightarrow$ $a$1 = 7
$\therefore$ $a$10 = $a$1 + 9d
= 7 + 45
= 52
2
JEE Main 2019 (Online) 10th January Morning Slot
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is -
A
1356
B
1256
C
1365
D
1465
Explanation
$\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12$
= 7 $\times$90 + 24 = 654
$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$
Total = 654 + 702 = 1356
3
JEE Main 2019 (Online) 10th January Evening Slot
Let a1, a2, a3, ..... a10 be in G.P. with ai > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $\in$ N (the set of natural numbers) for which
$\left| {\matrix{ {{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \cr {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \cr {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \cr } } \right|$ $=$ 0.
Then the number of elements in S, is -
A
10
B
4
C
2
D
infinitely many
Explanation
Apply
C3 $\to$ C3 $-$ C2
C2 $\to$ C2 $-$ C1
We get D = 0
4
JEE Main 2019 (Online) 11th January Morning Slot
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is ${{27} \over {19}}$.Then the common ratio of this series is :
A
${4 \over 9}$
B
${1 \over 3}$
C
${2 \over 3}$
D
${2 \over 9}$
Explanation
${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$
${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$
$\Rightarrow 6{r^2} - 13r + 6 = 0$
$\Rightarrow r = {2 \over 3}\,\,$
as $\left| r \right| < 1$ | 2021-10-23T17:18:22 | {
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https://socratic.org/questions/how-do-you-write-an-equation-of-a-line-with-slope-6-passing-through-5-6 | # How do you write an equation of a line with Slope -6, passing through (5,6)?
Jul 7, 2015
Substitute the given values into the point-slope form for a linear equation: $y - {y}_{1} = m \left(x - {x}_{1}\right)$.
#### Explanation:
Slope$=$$- 6$
Point 1$=$$\left(5 , 6\right)$
The general form for this equation is $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where ${y}_{1} = 6 ,$ $m = - 6 ,$ ${x}_{1} = 5$.
$y = 6 = - 6 \left(x - 5\right)$ | 2020-03-30T17:47:38 | {
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http://mathhelpforum.com/statistics/190861-expected-area-intersection-between-two-shapes.html | # Thread: The expected area of the intersection between two shapes
1. ## The expected area of the intersection between two shapes
Hi Everyone,
Assume a unit space (the extent on each dimension is 1). Assume that we have two shapes in the space A and B. The area of A is 1/x and the area of B is 1/y (both x an d y are greater than or equal to 1). The shapes are randomly placed in the space. What is the expected area of intersection between A and B (i.e., the area of $A\cap B$).
Is the following a right way to try solving this problem?
The probability that a randomly placed point p lies in A is 1/x. The probability that p lies in B is 1/y. The probability that p lies in both A and B is 1/x*1/y. | 2016-10-23T16:47:59 | {
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http://math.stackexchange.com/questions/103379/showing-24n1-implies24-sigma-0n | # Showing $24|(n+1)\implies24|\sigma_0(n)$
Question:
Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma_0(n)$).
For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisors of $n$ is $$1 + 5 + 19 + 95 = 120 = 5 \times 24.$$ (Note that the number $n$ is included among its divisors.)
-
– anon Jan 31 '12 at 4:39
We use a pairing argument, working first modulo $3$ and then modulo $8$.
We are told that $n\equiv -1\pmod{24}$. It follows that $n\equiv -1\pmod{3}$. Split the set of divisors of $n$ into unordered pairs $\{a,b\}$ such that $ab=n$. (Since $n\equiv -1\pmod n$, the number $n$ is not a perfect square, so every divisor of $n$ is taken care of.)
For any pair $\{a,b\}$ with $ab=n$, one of $a$ and $b$ is congruent to $1$ modulo $3$, and the other is congruent to $-1$. So all pair sums are congruent to $0$. Therefore the sum of all pair sums, that is, the sum of the divisors, is congruent to $0$ modulo $3$.
The same idea works modulo $8$. We are told that $n\equiv -1\pmod{8}$. If $\{a,b\}$ is an unordered pair with $ab=n$, then either (i) One of $a$ and $b$ is congruent to $1$ modulo $8$, and the other is congruent to $-1$, or (ii) One of $a$ and $b$ is congruent to $3$ modulo $8$, and the other is congruent to $-3$. In either case the sum is congruent to $0$ modulo $8$.
Thus the sum of the divisors of $n$ is divisible by $3$ and by $8$, and therefore by $24$.
-
You have to show that $n$ can't be a square, so that $a,b$ are distinct if $ab=n$. Also, you write $n\equiv 1$ when you mean $n\equiv -1$ at the beginning of your answer. – Thomas Andrews Jan 28 '12 at 21:58
@Thomas Andrews: Thanks for finding the potentially very confusing typo. I had remarked that nothing can be paired with itself because of the $\equiv -1\pmod{3}$ condition. That was insufficiently precise, so the parenthetical remark has been changed to "because $n\equiv -1 \pmod{3}$, $n$ cannot be a perfect square." Thanks for suggesting the improvement. – André Nicolas Jan 28 '12 at 22:11
We're looking for divisors $a, b$ such that $a \times b = n \equiv - 1 \pmod{24}$. In order for this to be possible, $a$ and $b$ must both be coprime to 24. Now note the following:
$a \times b \equiv -1 \pmod{24}$
$a^2 + a \times b \equiv a^2 -1 \pmod{24}$
$a(a + b) \equiv (a-1)(a+1) \pmod{24}$
Since $a$ is coprime to 24, and thus $a$ is odd, $(a-1), (a+1)$ must be even numbers. Furthermore, they are consecutive even numbers and thus at least one of them is also divisible by 4. In addition, as $a$ is coprime to 24, one of $(a-1), (a+1)$ must be divisible by 3 because if $a \equiv 1 \pmod{3}$ then $3|(a-1)$ and if $a \equiv 2 \mod{3}$ then $3|(a+1)$. This means that $4 \times 2 \times 3 = 24 | (a+1)(a-1)$. As $a$ is coprime to 24, this implies that $(a + b) \equiv 0 \pmod{24}$ for all $a, b$.
- | 2016-06-28T06:03:09 | {
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https://cs.stackexchange.com/questions/129235/combinatorial-problem-similar-in-nature-to-a-special-version-of-max-weighted-mat | # Combinatorial Problem similar in nature to a special version of max weighted matching problem
I have a problem and want to know if there is any combinatorial optimization that is similar in nature to this problem or how to solve this special version of the max weight matching problem.
I have a general graph $$G(\mathcal{V},\mathcal{E},\mathcal{W})$$. I want to find a maximum weight matching of the graph $$G$$ that must cover a certain subset of vertices and has a specific size. For example, if I have a graph with eight vertices, I want to find a max weighted matching that must cover the subset of vertices $$\mathcal{V}'=\{1,2,3\}$$ and the size of the matching is $$\lceil{|\mathcal{V}'|/2}\rceil$$. So one more vertex needs to be chosen that maximizes the weighted matching. How to find the optimal solution in polynomial time if possible?
## 1 Answer
Maximum Weight Matching algorithms are incremental, so to achieve a maximum weight matching of a bounded size, just stop after so many iterations. To achieve the other constraint, For each terminal vertex in $$\mathcal{V}'$$, increase the weight of each incident edge by a huge constant amount $$U$$. Try to prove that this answers your question (note that edges incident to two terminals have to be increased twice to keep the total increase constant in all valid solutions).
Note. It might help in the proof to distinguish different cases according to the size of $$\mathcal{V}'$$ compared to $$k$$ (the size of the matching).
• Thanks so much for your help. – Salwa Aug 14 '20 at 1:24 | 2021-08-05T02:49:21 | {
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https://homework.cpm.org/cpm-homework/homework/category/CCI_CT/textbook/PC3/chapter/Ch8/lesson/8.1.2/problem/8-28 | ### Home > PC3 > Chapter Ch8 > Lesson 8.1.2 > Problem8-28
8-28.
The tiny town of Twinpolypeaks is located next to a lake, both of which are in between two mountain peaks. The profile of the landscape can be modeled by the function $f(x)=-0.0005(x+3)(x+1)(x-1)^5(x-4.5)$ where $x$ is horizontal distance in miles and $f(x)$ is miles above sea level. The lake is below sea level and the town is above sea level.
1. Sketch the graph of the profile of the landscape. Use an appropriate domain.
The function has roots of $x=−3$, $−1$, $1$, and $4.5$.
The leading coefficient is negative.
2. If the width of the lake is captured in the profile of the landscape, approximately how wide is the lake?
Where is the curve significantly below the $x$-axis?
3. Approximately how wide is the town if there is nothing above $100$ feet over sea level?
What portion of the curve seems to be between the two 'mountains'? | 2020-09-26T18:42:20 | {
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https://www.omnicalculator.com/physics/parallel-inductors | # Inductors in Parallel Calculator
Created by Purnima Singh, PhD
Reviewed by Steven Wooding
Last updated: Oct 04, 2022
Use the inductors in parallel calculator to determine the equivalent inductance in a parallel circuit. Continue reading to learn about parallel combination of inductors and the formula for calculating total inductance in a circuit. You will also find an example of how to calculate total inductance in a parallel circuit using the parallel inductance calculator.
If you want to calculate the equivalent inductance in a series circuit, you can check the inductors in series calculator.
## Inductors in parallel
We can say that a group of coils (or inductors) are connected in parallel when one end of all the coils is connected to one point and the other end of all the coils is connected to another point (see figure 1). The AC source is connected across these two points.
## Formula for equivalent inductance in a parallel circuit
To determine the formula for the equivalent inductance when the inductors are connected in parallel, let us consider figure 1. If we connect an AC source across the combination, a self-induced e.m.f (electromagnetic field) $e$ appears in the coils due to the changing current $I$.
Using the formula for self-inductance $L$ (or inductor formula) we can write:
\scriptsize \begin{align*} e &= L \frac{dI}{dt}\\ \implies & \frac{dI}{dt} = \frac{e}{L} \end{align*}
In a parallel connection, the current gets divided as the inductors provide different paths through which it can flow. Hence, we can express the total current $I$ across the circuit as:
$\scriptsize I = I_1+I_2+...+I_n$
where, $I_1$, $I_2$, …, $I_n$ are the currents across the coils of inductance $L_1$, $L_2$, ..., $L_n$.
Differentiating the above equation with respect to time, we get:
$\scriptsize \frac{dI}{dt} = \frac{dI_1}{dt} +\frac{dI_2}{dt} +...+\frac{dI_n}{dt}\\ \text{.... or} \\ \frac{e}{L} = \frac{e_1}{L_1} + \frac{e_2}{L_2} + ... + \frac{e_n}{L_n}$
In a parallel connection, the voltage and the induced e.m.f ($e_1$, $e_2$, …, $e_n$) across each inductor is the same, i.e.:
$\scriptsize e_1 = e_2 = ... = e_n$
Therefore, we can write:
$\scriptsize \frac{1}{L} = \frac{1}{L_1} + \frac{1}{L_2} + ... + \frac{1}{L_n}$
The above equation gives the equivalent inductance of inductors in a parallel combination. We can also express the inductance in parallel formula as:
$\scriptsize L = \frac{1} { \left ( \frac{1}{L_1} + \frac{1}{L_2} + ... + \frac{1}{L_n} \right )}$
As you can see, the formula for a combination of inductors in parallel is similar to that for a combination of resistors in parallel. Explore more about them in our parallel resistor calculator
If you want to learn about the effective resistance offered by inductors, we recommend checking the inductive reactance calculator.
## How to calculate total inductance in a parallel circuit?
Let us see how to calculate the total inductance of a parallel circuit. Let Let $L_1 = 5\ \rm H$, $L_2 = 10\ \rm H$, and $L_3 = 15\ \rm H$.
1. Using the equation for inductors in parallel, we get:
$1/L = 1/L_1 + 1/L_2 + 1/L_3$.
2. We will substitute the values of the inductors in the above equation:
$1/L = 1/5 + 1/10 + 1/15$
$1/L = (6+3+2)/30 = 11/30$
3. Taking the reciprocal of the above, we can determine the equivalent inductance, i.e.:
$L = 30/11 = 2.7\ \rm H$
## How to use the inductors in parallel calculator?
Now that we understand how to calculate the equivalent inductance in a parallel circuit, how about solving the same problem using our inductors in parallel calculator?
1. Using the drop-down menu, choose the mode "calculate equivalent inductance".
2. Enter the inductance values, i.e., $L_1 = 5\ \rm H$, $L_2 = 10\ \rm H$, and $L_3 = 15\ \rm H$. You can add up to 10 inductors.
3. The inductors in parallel calculator will display the equivalent inductance, i.e., $L = 2.7\ \rm H$.
4. You can also use the parallel inductance calculator to find out the value of unknown inductance in a parallel circuit by changing the mode.
To calculate the self-inductance of a solenoid, visit our solenoid inductance calculator.
## FAQ
### How do I find the total inductance when inductors are connected in parallel?
To find the total inductance in a parallel circuit, proceed as follows:
1. Add the reciprocal of individual inductances.
2. Take the reciprocal of the value you get in step 1.
3. Congrats! You have calculated the total inductance in a parallel circuit.
### What happens when I connect two coils in parallel?
When we connect two coils of equal inductance in parallel, the effective inductance of the combination is reduced by a factor of half. The current through each inductor is also halved, but the voltage across each coil remains the same.
### Do inductors have resistance?
An ideal inductor has no resistance. However, in actual practice, since the inductor coils are made from conductors, and all conductors have a finite resistance, inductors do have some resistance.
### What is the inductance of three 10 mH inductors connected in parallel?
3.3 mH. According to the inductance in parallel formula, the reciprocal of the equivalent inductance of the three inductors connected in parallel is equal to the sum of the reciprocals of the individual inductances, i.e., 1/10 + 1/10 + 1/10 = 3/10. Hence, the inductance of three 10 mH inductors connected in parallel is 10/3 = 3.3 mH.
Purnima Singh, PhD
Mode
Calculate equivalent inductance
Inductor 1 (L₁)
H
Inductor 2 (L₂)
H
You can add up to 10 inductors; fields will appear as you need them.
Results
Input at least one inductance to obtain a result.
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https://services.math.duke.edu/~shaharko/ContSingVal.html | # Abstract
Controlling the singular values of n-dimensional matrices is often required in geometric algorithms in graphics and engineering. This paper introduces a convex framework for problems that involve singular values. Specifically, it enables the optimization of functionals and constraints expressed in terms of the extremal singular values of matrices.
Towards this end, we introduce a family of convex sets of matrices whose singular values are bounded. These sets are formulated using Linear Matrix Inequalities (LMI), allowing optimization with standard convex Semidefinite Programming (SDP) solvers. We further show that these sets are optimal, in the sense that there exist no larger convex sets that bound singular values.
A number of geometry processing problems are naturally described in terms of singular values. We employ the proposed framework to optimize and improve upon standard approaches. We experiment with this new framework in several applications: volumetric mesh deformations, extremal quasi-conformal mappings in three dimensions, non-rigid shape registration and averaging of rotations. We show that in all applications the proposed approach leads to algorithms that compare favorably to state-of-art algorithms.
# Summary
Just a teaser... download the paper if you want all the details.
## Our goal
Solve optimization problem explicitly expressed in terms of matrices and their extermal singular values: \begin{align} \min_{A\in\mathbb{R}^{n\times n}} & \quad f(A,\sigma_\mathrm{min}(A),\sigma_\mathrm{max}(A)) \\ \mathrm{s.t.} & \quad g_i(A,\sigma_\mathrm{min}(A),\sigma_\mathrm{max}(A))\leq 0\, , \quad i=1,..,r \\ \end{align} where $$f$$ and $$g_i$$ are convex and satisfy simple monotonicity conditions.
For example, find a matrix $$A$$ which is the closest to a given matrix $$B$$ and deviates by at most $$\Gamma$$ from an isometry: \begin{align} \min_{A\in\mathbb{R}^{n\times n}} & \quad \left\|A-B\right\|_F\\ \mathrm{s.t.} & \quad \Gamma^{-1} \leq \sigma_\mathrm{min}(A) \leq \sigma_\mathrm{max}(A) \leq \Gamma\\ \end{align}
## The idea
Formulate convex bounds on the extremal singular values $$\sigma_\mathrm{min}(A)$$ and $$\sigma_\mathrm{max}(A)$$.
Bounding $$\sigma_\mathrm{max}(A)$$ is readily convex and can be written as an LMI (positive semidefinite constraint): $\sigma_\mathrm{max}(A)\leq\Gamma \quad \Leftrightarrow \begin{pmatrix} \Gamma I & A \\ A^T & \Gamma I \\ \end{pmatrix}\succeq 0$
Bounding $$\sigma_\mathrm{min}(A)$$ is non-convex, but can be also imposed by an LMI: $\sigma_\mathrm{min}(A)\geq\gamma \quad \Leftarrow \quad \frac{A+A^T}{2} \succeq \gamma I$ This characterizes a maximal convex subset of the bound.
Moreover, by simply rotating $$A$$ it spans a maximal convex cover of the bound.
Semidefinite programming (SDP) can be used to enforce both constraints.
## Volumetric meshes
Optimize problems with energies and constraints expressed in terms of singular values:
## Extremal quasiconformal mappings
Compute mapping that minimize the maximal conformal distortion:
## Non-Rigid 3D Registration
Improve a basic ICP algorithm for surface matching -- restrict deviations from isometry to directly control distortion:
## Averaging of rotations
Average (and interpolate) multiple rotations:
# Code
A Matlab implementation of the paper can found on GitHub. A zip file of the latest version can be directly downloaded here.
This package includes three examples:
• example_optimizeSingleMatrix.m demonstrates simple single matrix optimization problems with constraints on singular values. Algorithm 1 of the paper is applied to a few example optimization problems, aiming to illustrate the implementation and usage of the theory presented in the paper (section 4 and 5).
• example_BarDeformation.m demonstrates volumetric mesh optimization problems with constraints on singular values. Algorithm 2 of the paper is applied to some example optimization problems, generating some of the example deformations of Figure 2 in the paper. See section 6.1 in the paper for additional details. Either run the entire code for generating all examples (might take a while), or the relevant code block for your desired example.
• example_ExtremalQuasiConformal.m implements an algorithm for computing extremal quasiconformal mappings of volumetric meshes (i.e., minimizing maximal conformal distortion). The code reproduces the example presented in Figure 1. See section 6.1 for additional details.
• Getting started: Before running the code, make sure you have YALMIP and MOSEK installed. Update the paths in initialize.m accordingly. The code was tested with Matlab (2013a), YALMIP (20140915) and MOSEK (7.0.0.92).
Disclaimer: The code is provided as-is for academic use only and without any guarantees. Please contact the authors to report any bugs.
## BibTex
@article{ContSingVal:2014,
author = {Kovalsky, Shahar Z. and Aigerman, Noam and Basri, Ronen and Lipman, Yaron},
title = {Controlling Singular Values with Semidefinite Programming},
journal = {ACM Transactions on Graphics (proceedings of ACM SIGGRAPH)},
volume = {33},
number = {4},
year = {2014},
} | 2017-04-29T15:23:50 | {
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http://www.lofoya.com/Solved/1391/walking-at-3-4-of-his-usual-place-a-man-reaches-his-office-20-minute | # Easy Time, Speed & Distance Solved QuestionAptitude Discussion
Q. Walking at 3/4 of his usual place, a man reaches his office 20 minute late. Find his usual time?
✖ A. 2 hr ✔ B. 1 hr ✖ C. 3 hr ✖ D. 1.5 hr
Solution:
Option(B) is correct
Let the original speed be $S$ and time be $T$
If new speed = $S\times \dfrac{3}{4}$, then new time would be $T\times \dfrac{4}{3}$ ($D = ST$ = Constant).
Given,
$\dfrac{3T}{4}-T=\dfrac{20T}{3}$
$⇒ T=60$ minutes
1 hour
Edit: For an alternate solution, check comment by Vaibhav Gupta.
## (4) Comment(s)
VENI
()
I cannot understand this problem. Please explain simply the question
Ritesh
()
Can you specify which part is not clear to you?
Vaibhav Gupta
()
In question read the word 'place' as 'speed' as word 'place' is wrongly written over there and the question will be crystal clear to you
Vaibhav Gupta
()
Alternate solution:
$s=\dfrac{d}{t}$ So $d=st$ ..........(1)
New speed $=\dfrac{3s}{4}$
New time $= t+20$
So, $d=\dfrac{3s}{4}×(t+20)$ .......(2)
Comparing (1) and (2),
We get, $t = \textbf{60 minutes}$ i.e. 1 hour | 2016-10-22T11:40:46 | {
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http://math.stackexchange.com/questions/319489/maths-help-stuck-on-these-ones-2 | # maths help stuck on these ones 2
6) A platform which starts fifteen metres above the ground & goes up & down.
The distance = s metres of the platform above the ground, t = seconds after the ride starts, can be modelled by the function: $s(t)=15+6sin(\pi/3)$
a)
(i) According to this model, find the maximum height above the ground reached by the platform.
(ii) Show that the platform completes 1 cycle in 6secs.
(iii) How many secs after the ride starts is it 9m above the ground for the 1st time ?
(iv) Sketch a graph for $0≤ t ≤60$ secs.
(v) How high above the ground is it when $t = 52.5$s?
They want the platform to go further up & down, during the first 60 secs of the new platform, the distance r metres, of the platform above the ground, t seconds after the ride starts, is modelled by the function $r(t)= 15+ e^{0.04t} sin(\pi t/3)$ for $0≤ t ≤60$
(b)
(i) Explain why the platforms period is still 6secs.
(ii) Explain in general terms why it takes longer to reach the max height of the first ride but will exceed it.
(iii) How high above the ground is it when $t = 4.5$s and $t = 52.5$s ?
(iv) After how many secs <60 does it reach the biggest height above ground, what is that height ?
(v) After how many secs <60 does it reach its lowest height above ground, what is that height ?
(vi) Sketch a graph of the ride for $0≤ t ≤60$ seconds.
They changed the platform according to the model $f(t)=15+e^{0.05t} sin(\pi t/3)$ for $0<t<60$
(c) Explain why this would have been catastrophic.
-
This function doesn't have the variable $t$ in it, are you sure it is the right function: $s(t)=15+6\sin\left(\dfrac{\pi}{3}\right)$ – Hans Groeffen Mar 3 at 15:07
Hints: for a) the sine is between $-1$ and $1$. Which gives the maximum? for b) you have to increase the argument of the sine function by $2\pi$ for one period. It looks like the function should be $15+6\sin (\frac {\pi t}3)$ and you lost the $t$. How much increase in $t$ is needed for that? for c)solve the position equation $9=15+6\sin(\frac {\pi t}3)$ for $\sin(\frac {\pi t}3)$, then find the minimum $t$ that satisfies that. | 2013-05-25T20:38:12 | {
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https://math.stackexchange.com/questions/1819180/compute-the-area-of-a-oval-based-2d-geometry | # Compute the area of a oval based 2d geometry
I know that the area of a shape generated as below
$R=r_0+a_1\cos(\theta)+a_2\cos(2\theta)+a_3\cos(3\theta)+...$
Where you can plot it and see the area value in matlab by:
th=0:0.01:2*pi;
R=r0+a1*cos(th)+a2*cos(2*th)+a3*cos(3*th);
x=R.*cos(th);
y=R.*sin(th);
plot(x,y);
Area=polyarea(x,y)
Could be computed by integral: $\frac{1}{2} \int R^2d\theta$
Which is equal to: $\pi{r_0}^2+\frac{1}{2}\pi({a_1}^2+{a_2}^2+{a_2}^2...)$
I am wondering how can I compute the area of a shape generated as below in a similar way analetically:
$R_x=r_0+a_1\cos(\theta)+a_2\cos(2\theta)+a_3\cos(3\theta)+...$ $R_y=r_0+{a_1}'\cos(\theta)+{a_2}'\cos(2\theta)+{a_3}'\cos(3\theta)+...$
Where you can plot it and see the area value in matlab by:
th=0:0.01:2*pi;
Rx=r0+a1*cos(th)+a2*cos(2*th)+a3*cos(3*th);
Ry=r0+a11*cos(th)+a22*cos(2*th)+a33*cos(3*th);
x=Rx.*cos(th);
y=Ry.*sin(th);
plot(x,y);
But what is the analytical expression for the area of the shape? Area=polyarea(x,y)
Use Green's Theorem: If $$\gamma:\quad t\mapsto \bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ is a simply closed curve bounding counterclockwise a region $B\subset{\mathbb R}^2$ then $${\rm area}(B)={1\over2}\int_0^T\bigl(x(t)\dot y(t)-y(t)\dot x(t)\bigr)\>dt\ .$$
By the way: If you have a parametric representation of the form $$x(\theta):=R_1(\theta)\cos\theta,\quad y(\theta)=R_2(\theta)\sin\theta\ ,$$ whereby the functions $R_1$, $R_2$ have been chosen independently, the variable $\theta$ no longer is the polar angle of the running point. It is just a parameter variable ("time") used to produce the curve in question.
You can use the formula you already know, by defining: $$R^2=R_x^2\cos^2\theta+R_y^2\sin^2\theta.$$
• In the second example $\theta$ is just "time", not the polar angle. – Christian Blatter Jun 9 '16 at 9:26
• @ChristianBlatter Why do you think so? By looking at the code, I think $\theta$ is just the polar angle, as it is in the first example. – Aretino Jun 9 '16 at 9:32
• Looking at the code now I think the OP has no idea what $\theta$, $R_x$ and $R_y$ are. – Christian Blatter Jun 9 '16 at 9:39
• @ChristianBlatter Actually $\theta$ is polar angle not time, I do not know how you came to that conclusion that it is time! bascially the second example tries to make the possibility that an oval also be generated, but in its special case when Rx=Ry a circle would also come out. The second one is a more general expression of the first one. – Soyol Jun 9 '16 at 17:12 | 2019-07-15T20:37:46 | {
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https://vincenzocoia.com/post/mixture_distributions/ | # Mixture distributions
This tutorial introduces the concept of a mixture distribution. We’ll look at a basic example first, using intuition, and then describe mixture distributions mathematically. See the very end for a summary of the learning points.
## Intuition
Let’s start by looking at a basic experiment:
1. Flip a coin.
2. If the outcome is heads, generate a N(0,1) random variable. If the outcome is tails, generate a N(4,1) random variable. We’ll let $$X$$ denote the final result.
$$X$$ is a random variable with some distribution (spoiler: it’s a mixture distribution). Let’s perform the experiment 1000 times to get 1000 realizations of $$X$$, and make a histogram to get a sense of the distribution $$X$$ follows. To make sure the histogram represents an estimate of the density, we’ll make sure the area of the bars add to 1 (with the ..density.. option).
suppressMessages(library(ggplot2))
## Warning: package 'ggplot2' was built under R version 3.5.2
set.seed(44)
X <- numeric(0)
coin <- integer(0)
for (i in 1:1000) {
coin[i] <- rbinom(1, size=1, prob=0.5) # flip a coin. 0=heads, 1=tails.
if (coin[i] == 0) { # heads
X[i] <- rnorm(1, mean=0, sd=1)
} else { # tails
X[i] <- rnorm(1, mean=4, sd=1)
}
}
(p <- qplot(X, ..density.., geom="histogram", bins=30))
Let’s try to reason our way to figuring out the overall density. Keep in mind that this density (like all densities) is one curve. We’ll say we’ve succeeded at finding the density if our density is close to the histogram.
It looks like the histogram is made up of two normal distributions “superimposed”. These ought to be related to the N(0,1) and N(4,1) distributions, so to start, let’s plot these two Gaussian densities overtop of the histogram.
ggplot(data.frame(X=X), aes(X)) +
geom_histogram(aes(y=..density..), bins=30) +
stat_function(fun=function(x) dnorm(x, mean=0, sd=1),
stat_function(fun=function(x) dnorm(x, mean=4, sd=1),
mapping=aes(colour="Tails")) +
scale_color_discrete("Coin Flip")
Well, the two Gaussian distributions are in the correct location, and it even looks like they have the correct spread, but they’re too tall.
Something to note at this point: the two curves plotted above are separate (component) distributions. We’re trying to figure out the distribution of $$X$$ – which, again, is a single curve, and is estimated by the histogram. At this point, we only suspect that the distribution of $$X$$ is some combination of these two Gaussian distributions.
So, why are the Gaussian curves too tall? Because each one represents the distribution if we only ever flip either heads or tails (for example, the red distribution happens when we only ever flip heads). But since we flip heads half of the time, and tails half of the time, these probabilities (more accurately, densities) ought to be reduced by half. Let’s add these “semi” component distributions to the plot:
(p <- ggplot(data.frame(X=X), aes(X)) +
geom_histogram(aes(y=..density..), bins=30) +
stat_function(fun=function(x) dnorm(x, mean=0, sd=1)*0.5,
stat_function(fun=function(x) dnorm(x, mean=4, sd=1)*0.5,
mapping=aes(colour="Tails", linetype="Semi")) +
stat_function(fun=function(x) dnorm(x, mean=0, sd=1),
stat_function(fun=function(x) dnorm(x, mean=4, sd=1),
mapping=aes(colour="Tails", linetype="Full")) +
scale_color_discrete("Coin Flip") +
scale_linetype_discrete("Distribution"))
Looks like they line up quite nicely!
But these two curves are still separate – we need one overall curve if we are to find the distribution of $$X$$. So we need to combine them somehow. It might look at first that we can just take the upper-most of the ‘semi’ curves (i.e., the maximum of the two), but looking in between the two curves reveals that the histogram is actually larger than either curve here. It turns out that the two ‘semi’ curves are added to get the final curve:
p + stat_function(fun=function(x) dnorm(x, mean=0, sd=1)*0.5 +
dnorm(x, mean=4, sd=1)*0.5,
mapping=aes(linetype="Full"))
The intuition behind adding the densities is that an outcome for $$X$$ comes from both components, so both contribute some density.
Even though the random variable $$X$$ is made up of two components, at the end of the day, it’s still overall just a random variable with some density. And like all densities, the density of $$X$$ is just one curve. But, this density happens to be made up of the components, as we’ll see next.
## General Scenario
The two normal distributions from above are called component distributions. In general, we can have any number of these (not just two) to make a mixture distribution. And, instead of selecting the component distribution with coin tosses, they’re chosen according to some generic probabilities called the mixture probabilities.
In general, here’s how we make a mixture distribution with $$K$$ component Gaussian distributions with densities $$\phi_1(x), \ldots, \phi_K(x)$$:
1. Choose one of the $$K$$ components, randomly, with mixture probabilities $$\pi_1, \ldots, \pi_K$$ (which, by necessity, add to 1).
2. Generate a random variable from the selected component distribution. Call the result $$X$$.
Note: we can use more than just Gaussian component distributions! But this tutorial won’t demonstrate that.
That’s how we generate a random variable with a mixture distribution, but what’s its density? We can derive that by the law of total probability. Let $$C$$ be the selected component number; then the component distributions are actually the distribution of $$X$$ conditional on the component number. We get: $f_X\left(x\right) = \sum_{k=1}^{K} f_{X|C}\left(x \mid c\right) P\left(C=c\right) = \sum_{k=1}^{K} \phi_k\left(x\right) \pi_k.$
#### Notes:
• The intuition described in the previous section matches up with this result. For $$K=2$$ components determined by a coin toss $$(\pi_1=\pi_2=0.5),$$ we have $f_X\left(x\right) = \phi\left(x\right)0.5 + \phi\left(x-4\right)0.5,$ which is the black curve in the previous plot.
• This tutorial works with univariate data. But mixture distributions can be multivariate, too. A $$d$$-variate mixture distribution can be made by replacing the component distributions with $$d$$-variate distributions. Just be sure to distinguish between the dimension of the data $$d$$ and the number of components $$K$$.
• We could just describe a mixture distribution by its density, just like we can describe a normal distribution by its density. But, describing mixture distributions by its component distributions together with the mixture probabilities, we obtain an excellent interpretation of the mixture distribution. This interpretation is (it’s also called a data generating process): (1) randomly choose a component, and (2) generate from that component. This interpretation is useful for cluster analysis, because the data clusters can be thought of as being generated by the component distributions, and the proportion of data in each cluster is determined by the mixture probabilities.
## Learning Points
• A mixture distribution can be described by its mixing probabilities $$\pi_1, \ldots, \pi_K$$ and component distributions $$\phi_1(x), \ldots, \phi_K(x)$$.
• A mixture distribution can also be described by a single density (like all continuous random variables).
• This density is a single curve if data are univariate; a single “surface” if the data are bivariate; and higher dimensional surfaces if the data are higher dimensional.
• To get the density from the mixing probabilities and component distributions, we can use the formula indicated in the above section (based on the law of total probability).
##### Vincenzo Coia
###### he/him/his 🌈 👨
I’m a data scientist at the University of British Columbia, Vancouver. | 2019-12-15T14:25:55 | {
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https://puzzling.stackexchange.com/questions/20379/five-card-magic-trick-with-n-card-deck/20385#20385 | # Five Card Magic Trick with $N$ card Deck
Last week, I saw a magic show, and was the volunteer for climactic act. The magician introduced this act by talking about the Fitch-Cheney card trick, but then explained that she was going to do an even more impressive trick. Namely, instead of using a $52$ card deck, she would use a larger deck, with $N$ cards, though she didn't say exactly how big $N$ was.
Here's what happened:
The magician left the room. An assistant asked me to look through the $N$ card deck, and pick out any $5$ cards. The assistant inspected the cards I chose, then gave one back to me, which I put in my pocket. He arranged the remaining four cards in a neat stack, and placed this stack face down on a table. The magician returned, inspected the stack, and successfully guessed the card in my pocket.
What is the largest value of $N$ for which the trick can be performed? For that $N$, how is it done?
Note that there is no slight of hand or secret communication. Each card is distinct, the faces are rotationally symmetric, and the stack is placed at the exact same spot on the table every time the trick is performed. The assistant is only allowed to choose which card the volunteer gets ($5$ choices), then permute the four other cards ($4!=24$ choices).
Aside: From this question, we know $N=53$ is possible.
• So does this deck have 4 suits, M suits, or no suits (i.e. cards numbered 1 to N)? Aug 21, 2015 at 15:59
• The question you linked links to another on MathOverflow, which links to this proof that claims that it can be done for $N\ge124$. The math is a little too involved for me to follow, though. Aug 21, 2015 at 16:01
• @PatrickN I don't think it matters, as long as all the cards are distinguishable. Aug 21, 2015 at 16:46
• @PatrickN As 2012rcampion mentioned, it does not matter. The magician/assistant can imagine the cards to have any number of suits they please. You may imagine the cards are numbered 1 to N. Aug 21, 2015 at 17:01
• @GentlePurpleRain Do you mean $N\leq 124$? Aug 21, 2015 at 17:01
Stealing from 2012rcampion's answer, we need to create a one-to-one mapping from five-card hands to four-card messages (four cards plus permutation). By symmetry, each four-card message can be mapped to an equal number of five-card hands, so by the regular version of Hall's Theorem there must exist such a one-to-one mapping (this is laid out in more detail in GentlePurpleRain's comment link).
Thus, the trouble is actually constructing a matching, and ideally doing so in a way that would work well as a magic trick. The following works for the former, and you can decide if it works for the latter.
Suppose the cards are numbered 0 to 123. Call the five cards, from smallest to largest, $c_0$, $c_1$, $c_2$, $c_3$, and $c_4$. To find the pocketed card, the assistant adds up the numbers on all five cards, reduces the answer modulo $5$ to get $S$, and then hands back $c_S$.
Now the magician can see four cards. Suppose these four cards sum to $S'$ modulo $5$. Suppose also she mentally removes these four cards from the deck and then relabels the remaining cards in order from $0$ to $119$ (call these "reduced numbers"). Then we have
the missing card must have a reduced number equivalent to $(-S')$ modulo $5$*.
This means there are only $24$ possibilities, and hence with a code that takes permutations of four cards to the set $\{1, 2, ..., 24\}$, the assistant can supply enough information for the magician to identify the card.
*This is a bit hard to see, so let's break it down into cases. If $c_0$ was removed, that means the sum of the five cards is $0$ mod $5$, and hence the missing card must be $-S'$ mod $5$ to counter the $S'$ sum of the other four cards. If $c_1$ was removed, then the missing card must have had a number of $-S' + 1$, but since it appears after $c_0$ that we mentally removed from the deck, it's reduced number is $-S' + 1 - 1 = -S'$. Similar logic shows this works for all five cases.
A slightly different way to think about this problem is that the magician is trying to guess which five-card hand was initially drawn from the deck. Since he can see the other four cards, this is the same as guessing the volunteer's card.
There are $\binom{N}{5}$ possible five-card hands, and $4!\binom{N}{4}$ four-card "messages" (including orderings) that the assistant can send the magician. Thus, we are looking for the largest $N$ such that the number of hands is less than or equal to the number of messages:
$$\binom{N}{5}\leq 4!\binom{N}{4}$$
solving this inequality yeilds:
$$N\leq 124$$
(The answer GentlePurpleRain gave in his comment.) For $N=124$, we have:
$$\binom{124}{5}=225\,150\,024$$
possible hands and
$$4!\binom{124}{4}=225\,150\,024$$
possible messages. Since the inequality is saturated, we must devise a one-to-one mapping from hands to messages such that each message is a subset of its corresponding hand. (I'm still working on that part...) | 2022-10-01T08:21:38 | {
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# The product of the positive 4-digit integer 118A and 25847 is 4758 uni
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The product of the positive 4-digit integer 118A and 25847 is 4758 uni [#permalink]
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07 Feb 2019, 11:49
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GMATH practice exercise (Quant Class 16)
The product of the positive 4-digit integer 118A and 25847 is 4758 units less than a number that leaves remainder 1 when divided by 5. How many values are possible for the digit A?
(A) None
(B) Only 1
(C) Only 2
(D) Only 3
(E) More than 3
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Re: The product of the positive 4-digit integer 118A and 25847 is 4758 uni [#permalink]
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07 Feb 2019, 22:40
fskilnik wrote:
GMATH practice exercise (Quant Class 16)
The product of the positive 4-digit integer 118A and 25847 is 4758 units less than a number that leaves remainder 1 when divided by 5. How many values are possible for the digit A?
(A) None
(B) Only 1
(C) Only 2
(D) Only 3
(E) More than 3
Let us work on the unit's digit..
For remainder to be 1 when divided by 5, the units digit should be either 1 or 1+5=6..
two cases..
1) units digit as 1..
118A*258477=xyz1-4758..........A*7=s1-8=t3....
Multiplication of 7 by 9 gives a units digit 3..
SO A can be 9..
2) units digit as 6..
118A*258477=xyz6-4758..........A*7=s6-8=t8....
Multiplication of 7 by 4 gives a units digit 8..
SO A can be 4..
Thus 2 values..
C
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The product of the positive 4-digit integer 118A and 25847 is 4758 uni [#permalink]
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08 Feb 2019, 06:30
fskilnik wrote:
GMATH practice exercise (Quant Class 16)
The product of the positive 4-digit integer 118A and 25847 is 4758 units less than a number that leaves remainder 1 when divided by 5. How many values are possible for the digit A?
(A) None
(B) Only 1
(C) Only 2
(D) Only 3
(E) More than 3
Hi, chetan2u ! Thanks for joining and for your nice contribution!
Let me offer our "official solution":
$$\left\langle {118A} \right\rangle \cdot 25847 + 4758 = 5M + 1\,\,\,,\,\,\,M \ge 1\,\,{\mathop{\rm int}}$$
$$?\,\,\,:\,\,\,\# \,\,A\,\,{\rm{possibilities}}$$
$$\left[ N \right] = units\,\,digit\,\,of\,\,N\,\,\,\,\,\left( * \right)$$
$$\left\langle {118A} \right\rangle \cdot 25847 + 4757 = 5M\,\,\,\,\,{\rm{AND}}\,\,\,\,\,\left( * \right)\,\,\,\left\{ \matrix{ \,\,\left[ {\left\langle {118A} \right\rangle \cdot 25847} \right] = \left[ {7 \cdot A} \right] \hfill \cr \,\,\left[ {4757} \right] = 7 \hfill \cr \,\,\left[ {5M} \right] = 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left[ {7 \cdot A + 7} \right]\,\,\,{\rm{equals}}\,\,\,0\,\,{\rm{or}}\,\,5$$
$$\Rightarrow \,\,\,\,\,7 \cdot \left( {A + 1} \right)\,\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,5\,\,\,\,\mathop \Rightarrow \limits^{GCD\left( {7,5} \right)\,\, = \,\,1} \,\,\,\,\,\left[ {A + 1} \right]\,\,\,{\rm{equals}}\,\,\,0\,\,{\rm{or}}\,\,5\,\,\,\,\, \Rightarrow \,\,\,A\,\,\,{\rm{equals}}\,\,{\rm{4}}\,\,{\rm{or}}\,\,{\rm{9}}$$
The correct answer is therefore (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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The product of the positive 4-digit integer 118A and 25847 is 4758 uni [#permalink] 08 Feb 2019, 06:30
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https://proofwiki.org/wiki/Definition:Lagrange_Basis_Polynomial | Definition:Lagrange Basis Polynomial
Jump to: navigation, search
Definition
Let $x_0, \ldots, x_n \in \R$ be real numbers.
The Lagrange basis polynomials associated to the $x_i$ are the polynomials:
$\displaystyle L_j \left({X}\right) := \prod_{\substack{0 \mathop \le i \mathop \le n \\ i \mathop \ne j}} \frac{X - x_i}{x_j - x_i} \in \R \left[{X}\right]$
Source of Name
This entry was named for Joseph Louis Lagrange. | 2015-03-07T00:05:17 | {
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"url": "https://proofwiki.org/wiki/Definition:Lagrange_Basis_Polynomial",
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# term after the first term is equal to the preceding term
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term after the first term is equal to the preceding term [#permalink] 18 Jan 2016, 02:25
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$$a1, a2, a3 ................ an$$
In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If $$a1+a3+a5=27$$, what is the value of $$a2+a4$$ ?
$$a2+a4=$$
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Re: term after the first term is equal to the preceding term [#permalink] 18 Jan 2016, 04:52
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Solution
This is clearly an arithmetic progression and we just need the first 5 terms. We can rewrite the progression as a - 2c, a - c, a , a + c, a + 2c. Here the difference between the terms is c as described in the problem
a1 + a3 + a5 = a - 2c + a + a+ 2c = 3a = 27. Hence a =9.
To find a2 + a4 = a - c + a + C = 2a =18.
The purpose of writing this series in the above form is to eliminate c as a variable. Hence answer is 18.
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Re: term after the first term is equal to the preceding term [#permalink] 30 Mar 2016, 09:05
i don't understand where you derive a - 2c and a + 2c from
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Re: term after the first term is equal to the preceding term [#permalink] 01 Apr 2016, 02:30
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Lets take a real arithmetic progression:
1, 5, 9 , 13, 17
As you can see any term of the above arithmetic progression is 4 greater than the previous term. 1+4 = 5... 5+4 = 9.... so on
So if we draw parallels with the question above here c = 4. Then we can write the above series taking the middle term 9 into account:
9-2*4, 9-4, 9 , 9+4, 9+2*4
a-2*c, a-c , a , a+c, a+2*c
Thats how Sandy derives the a+2c and a-2c terms
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Re: term after the first term is equal to the preceding term [#permalink] 29 Jun 2017, 20:10
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If that is confusing then, you can say a, a+c, a+2c, a+3c, a+4c
a1+a3+a5 = 27
a + a+2c + a+4c = 27
a = 9-2c
now a2 + a4
a+c + a+3c
2a+4c
2(9-2c) + 4c
18-4c+4c
Re: term after the first term is equal to the preceding term [#permalink] 29 Jun 2017, 20:10
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http://math453spring2009.wikidot.com/lecture-24 | Lecture 24 - Moduli without Primitive Roots
# Summary
In today's class we finished proving a critical result concerning the number of elements of order d modulo p, where p is a prime number. This let's us conclude that primitive roots do exist modulo p. We then spent the second half of the class discussing integers for which primitive roots do not exist. We finished the class by classifying exactly those integers which do have primitive roots.
# Counting elements of order d modulo p
Last class period we talked about primitive roots modulo p, where p was an odd prime number. The key result was a proof of the following
Theorem: If $d \mid p-1$, then there are precisely $\phi(d)$ elements of order d mod p.
We start today by giving a proof of this result
Proof: For a given divisor d, let $f(d)$ be the number of elements of order d mod p. We saw last class period that there are exactly d solutions to $x^d-1 \equiv 0 \mod{p}$. Therefore we have
(1)
\begin{align} d = \#\{1 \leq a \leq p: a^d-1 \equiv 0 \mod{p}\}. \end{align}
But notice that if $a^d - 1 \equiv 0 \mod{p}$ then we get $a^d \equiv 1 \mod{p}.$ This in turn tells us that $\mbox{ord}_p(a) \mid d$. Hence any element $a$ in the set above must be an element of order c, where c is a divisor of c. Hence we have
(2)
\begin{align} \{1 \leq a \leq p : a^d-1\equiv 0 \mod{p}\} = \bigcup_{c \mid d}\{1 \leq a \leq p: \mbox{ord}_p(a) = c\}. \end{align}
By counting the number of elements on the left- and right-hand sides, we get
(3)
\begin{split} d &= \#\{1 \leq a \leq p : a^d-1\equiv 0 \mod{p}\} \\&= \#\left(\bigcup_{c \mid d}\{1 \leq a \leq p: \mbox{ord}_p(a) = c\}\right) \\&= \sum_{c \mid d}\#\{1 \leq a \leq p: \mbox{ord}_p(a) = c\} \\&= \sum_{c \mid d}f(c). \end{split}
On the other hand, we saw long ago that
(4)
\begin{align} d = \sum_{c \mid d}\phi(c). \end{align}
Hence we can combine (3) and (4) to give
(5)
\begin{align} \sum_{c \mid d}f(c) = \sum_{c \mid d}\phi(c). \end{align}
Translated into convolutions, this says $(P_0 * f)(d) = (P_0*\phi)(d)$. Convolving by $\mu$ then gives
(6)
\begin{align} (\mu*P_0*f)(d) = (\mu*P_0*\phi)(d) \Longleftrightarrow f(d) = \phi(d). \end{align}
$\square$
As a consequence of this result, we see that primitive roots exist for any prime modulus.
Corollary: For any prime p, there are exactly $\phi(p-1)$ many primitive roots.
Proof: Take $d = p-1$ in the above result, and remember that a primitive root mod p is an element of order $p-1$. $\square$
#### Example: Orders of Elements mod 17
Let's find out how many elements of each order exist modulo 17 (just to confirm the result we proved). Notice that an element mod 17 must have order from the set $\{1,2,4,8,16\}$, since these are the divisors of $\phi(17) = 16$. Using this fact, notice that we have
(7)
\begin{split} 3^1 &\not\equiv 1 \mod{17}\\ 3^2 &\equiv 9 \not\equiv 1 \mod{17}\\ 3^{4} &\equiv 9^2 \equiv 81 \equiv 13 \mod{17} \quad \quad (\mbox{since }81 = 4*17+13)\\ 3^{8} &\equiv (13)^2 \equiv (-4)^2 \equiv 16 \equiv -1 \not\equiv 1 \mod{17}. \end{split}
Hence 3 is a primitive root. This will make computing the orders of other elements easier.
$3^j$ $\mbox{gcd}(16,j)$ $\mbox{ord}_{17}(3^j)$ $3^j$ $\mbox{gcd}(16,j)$ $\mbox{ord}_{17}(3^j)$
31 1 16 39 1 16
32 2 8 310 2 8
33 1 16 311 1 16
34 4 4 312 4 4
35 1 16 313 1 16
36 2 8 314 2 8
37 1 16 315 1 16
38 8 2 316 16 1
Notice that we have $\phi(16) = 8$ elements of order 16, $\phi(8) = 4$ elements of order 8, $\phi(4) = 2$ elements of order 4, $\phi(2)=1$ element of order 2, and $\phi(1) = 1$ element of order 1.
# Moduli Without Primitive Roots
Now that we've spent some time talking about moduli m which do have primitive roots, let's find a few that don't have primitive roots. We've already seen in class that 8 has no primitive roots. What about higher powers of 2?
Proposition: There is no primitive root modulo $2^m$, where $m \geq 3$.
Proof: We'll prove this result by induction. The base case ($m=3$) was done in class some time ago. Now let a be a given integer relatively prime to $2^m$. This means that a is an odd number. We'll show that
(8)
\begin{align} a^{2^{m-2}} \equiv 1 \mod{2^m}. \end{align}
Since $\phi(2^m) = 2^{m-1}$, this will prove that there is no primitive root mod $2^m$.
Now by induction we know $a^{2^{m-3}} \equiv 1 \mod{2^{m-1}}$. Hence
(9)
$$a^{2^{m-3}} = 1 + c2^{m-1}.$$
Squaring both sides of this equation gives
(10)
\begin{align} a^{2^{m-2}} = 1 + c2^m + c^22^{2(m-1)} \equiv 1 \mod{2^m}. \end{align}
$\square$
This rules out an infinite class of integers from having primitive roots. This next result will cut out even more.
Proposition: There are no primitive roots modulo mn if m,n are relatively prime integers greater than 2.
Proof: Since m,n are relatively prime, this implies $\phi(mn) = \phi(m)\phi(n)$. Hence a primitive root mod mn would need to have order $\phi(m)\phi(n)$. Instead, we'll show that if $(a,mn) = 1$, then we have
(11)
\begin{align} a^{\frac{\phi(m)\phi(n)}{2}} \equiv 1 \mod{mn}. \end{align}
This means that no element mod mn can be a primitive root, and so it gives the result we want.
Now since $m,n>2$ we know that 2 is a common divisor of $\phi(m)$ and $\phi(n)$. Now we'll compute a raised to $\frac{\phi(m)\phi(n)}{2}$ mod m and n, then stitch these results together. We have
(12)
\begin{split} a^{\frac{\phi(m)\phi(n)}{2}} &\equiv \left(a^{\phi(m)}\right)^{\frac{\phi(n)}{2}} \equiv 1^{\frac{\phi(n)}{2}} \equiv 1 \mod{m}\\ a^{\frac{\phi(n)\phi(m)}{2}} &\equiv \left(a^{\phi(n)}\right)^{\frac{\phi(m)}{2}} \equiv 1^{\frac{\phi(m)}{2}} \equiv 1 \mod{n}.\\ \end{split}
Putting these together with CRT, we see that Equation (11) is true. $\square$
The two previous results together tells us that
Corollary: If m has a primitive root, then m is either 1,2,4, a power of an odd prime, or twice a power of an odd prime.
In the next class period, we'll show that — in fact — all these values of m which might have primitive roots do, in fact, have primitive roots. | 2018-10-17T15:51:13 | {
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https://math.stackexchange.com/questions/2771090/estimation-of-psi-n-in-number-theory | # Estimation of $\psi (n)$ in number theory
Let $$\Lambda(n) = \begin{cases} \ln p &\quad \text{if } n = p^{\alpha} \text{ where } \alpha\geq 1 \\ 0 &\quad \text{otherwise} \end{cases}$$ and let $$\psi(n)=\sum_{m=1}^{n} \Lambda(m)$$ I need to prove that $$e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx$$ is a positive integer, and deduce $$\psi(2n+1) \geq 2n \ln 2$$
Here is what I think:
I have already deduced that $e^{\psi (n)} = \text{lcm}(1,2,3,...,n)$. Thus \begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= \text{lcm} ( 1,2,...,2n+1 ) \frac{(n!)^2}{(2n+1)!}\\ &= \text{lcm} ( 1,2,...,2n+1 ) \frac{n!}{2^n(2n+1)!!} \end{align*} Let $\{p_n\}$ be the sequence of all primes with monotone increasing order, and let $P=\{p_n\} \cap \{1,2,3,...,2n+1\} = \{p_1, p_2,...,p_k\}$. Let $\alpha_k = \max\{ m| p_k^m \in \{1,2,...,2n+1\} \}$. Then $\text{lcm}(1,2,3,...,2n+1)= \prod_{i=1}^{k} p_i^{\alpha_i}$. \begin{align*} \text{lcm} ( 1,2,...,2n+1 ) \frac{n!}{2^n(2n+1)!!} &= \prod_{i=1}^{k} p_i^{\alpha_i} \frac{n!}{2^n(2n+1)!!} \end{align*} I cannot proceed any more. It seems like I need to consider two cases when $n$ is odd or even. But the process gets complicated. Any help is appreciated!
\begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= \text{lcm}(1,2,...,2n+1) \int_{0}^{1} x^n \sum_{k=0}^{n}\binom{n}{k}(-1)^k x^k dx\\ &= \text{lcm}(1,2,...,2n+1) \sum_{k=0}^{n}\binom{n}{k}(-1)^k \int_{0}^{1} x^{n+k} dx\\ &= \text{lcm}(1,2,...,2n+1) \sum_{k=0}^{n}\binom{n}{k}(-1)^k \frac{1}{n+k+1}\\ &\in \mathbb{Z} \end{align*} $e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx > 0$ is obvious.
Since \begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= e^{\psi(2n+1)} \frac{n!}{2^n(2n+1)!!}\\ &\geq 1 \end{align*} We have $$e^{\psi(2n+1)} \geq \frac{2^n(2n+1)!!}{n!} \geq 2^{2n}$$ which ends the proof. | 2019-05-23T03:24:51 | {
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https://curriculum.illustrativemathematics.org/HS/students/2/2/15/index.html | # Lesson 15
• Let’s investigate how congruence for quadrilaterals is similar to and different from congruence for triangles.
### 15.1: True or . . . Sometimes True?: Parallelograms
Given that $$ABCD$$ is a parallelogram.
1. What must be true?
2. What could possibly be true?
3. What definitely can’t be true?
Jada is learning about the triangle congruence theorems: Side-Side-Side, Angle-Side-Angle, and Side-Angle-Side. She wonders if there are any theorems like these for parallelograms.
1. If 2 parallelograms have all 4 pairs of corresponding sides congruent, do the parallelograms have to be congruent? If so, explain your reasoning. If not, use the tools available to show that it doesn’t work.
2. In parallelograms $$ABCD$$ and $$EFGH$$, segment $$AB$$ is congruent to segment $$EF$$, segment $$BC$$ is congruent to segment $$FG$$, and angle $$ABC$$ is congruent to angle $$EFG$$. Are $$ABCD$$ and $$EFGH$$ congruent? If so, explain your reasoning. If not, use the tools available to show that it doesn’t work.
### 15.3: Make Your Own Congruence Theorem
Come up with another criteria that is enough to be sure that 2 parallelograms are congruent. Try to use as few measurements as you can. Be prepared to convince others that your shortcut works. | 2021-06-14T20:57:51 | {
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http://www.wikihow.com/Integrate-by-Parts | Edit Article
# wikiHow to Integrate by Parts
Integration by parts is a technique used to evaluate integrals where the integrand is a product of two functions.
${\displaystyle \int f(x)g(x){\mathrm {d} }x}$
Integrals that would otherwise be difficult to solve can be put into a simpler form using this method of integration.
### Part 1 Indefinite Integral
1. 1
Consider the integral below. We see that the integrand is a product of two functions, so it is ideal for us to integrate by parts.
• ${\displaystyle \int xe^{x}{\mathrm {d} }x}$
2. 2
Recall the formula for integration by parts.
• ${\displaystyle \int u{\mathrm {d} }v=uv-\int v{\mathrm {d} }u}$
3. 3
Choose a ${\displaystyle u}$ and ${\displaystyle {\mathrm {d} }v,}$ and find the resulting ${\displaystyle {\mathrm {d} }u}$ and ${\displaystyle v}$. We choose ${\displaystyle u=x}$ because its derivative of 1 is simpler than the derivative of ${\displaystyle e^{x},}$ which is only itself. That results in ${\displaystyle {\mathrm {d} }v=e^{x}{\mathrm {d} }x,}$ whose integral is trivial.
• ${\displaystyle {\mathrm {d} }u={\mathrm {d} }x}$
• ${\displaystyle v=e^{x}}$
• You can neglect the constant of integration when finding ${\displaystyle v,}$ because it will drop out in the end.
4. 4
Substitute these four expressions into our integral.
• ${\displaystyle \int xe^{x}{\mathrm {d} }x=xe^{x}-\int e^{x}{\mathrm {d} }x}$
• The result was that our integral now consists of just one function - the exponential function. As ${\displaystyle e^{x}}$ is its own antiderivative with a constant, evaluating it is much easier.
5. 5
Evaluate the resulting expression using any means possible. Remember to add the constant of integration, as antiderivatives are not unique.
• ${\displaystyle \int xe^{x}{\mathrm {d} }x=xe^{x}-e^{x}+C}$
### Part 2 Definite Integral
1. 1
Consider the definite integral below. Definite integrals require evaluation at the boundaries. While the integral below looks like it has an integrand of just one function, the inverse tangent function, we can say that it is the product of inverse tangent and 1.
• ${\displaystyle \int _{0}^{1}\tan ^{-1}x{\mathrm {d} }x}$
2. 2
Recall the integration by parts formula.
• ${\displaystyle \int _{a}^{b}u{\mathrm {d} }v=uv{\Bigg |}_{a}^{b}-\int _{a}^{b}v{\mathrm {d} }u}$
3. 3
Set ${\displaystyle u}$ and ${\displaystyle {\mathrm {d} }v,}$ and find ${\displaystyle {\mathrm {d} }u}$ and ${\displaystyle v}$. Since the derivative of an inverse trig function is algebraic and therefore simpler, we set ${\displaystyle u=\tan ^{-1}x}$ and ${\displaystyle {\mathrm {d} }v={\mathrm {d} }x.}$ This results in ${\displaystyle {\mathrm {d} }u={\frac {1}{1+x^{2}}}{\mathrm {d} }x}$ and ${\displaystyle v=x.}$
4. 4
Substitute these expressions into our integral.
• ${\displaystyle \int _{0}^{1}\tan ^{-1}x{\mathrm {d} }x=x\tan ^{-1}x{\Bigg |}_{0}^{1}-\int _{0}^{1}{\frac {x}{1+x^{2}}}{\mathrm {d} }x}$
5. 5
Evaluate the simplified integral using u-substitution. The numerator is proportional to the derivative of the denominator, so u-subbing is ideal.
• Let ${\displaystyle u=1+x^{2}.}$ Then ${\displaystyle {\mathrm {d} }u=2x{\mathrm {d} }x.}$ Be careful in changing your boundaries.
• {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {x}{1+x^{2}}}{\mathrm {d} }x&={\frac {1}{2}}\int _{1}^{2}{\frac {1}{u}}{\mathrm {d} }u\\&={\frac {1}{2}}\ln 2\end{aligned}}}
6. 6
Evaluate the ${\displaystyle uv}$ expression to complete the evaluation of the original integral. Be careful with the signs.
• ${\displaystyle \int _{0}^{1}\tan ^{-1}x{\mathrm {d} }x={\frac {\pi }{4}}-{\frac {1}{2}}\ln 2}$
### Part 3 Repeated Integration by Parts
1. 1
Consider the integral below. Occasionally, you may find yourself with an integral that requires multiple instances of integration by parts in order to get the desired answer. Such an integral is below.
• ${\displaystyle \int e^{x}\cos x{\mathrm {d} }x}$
2. 2
Recall the formula for integration by parts.
• ${\displaystyle \int u{\mathrm {d} }v=uv-\int v{\mathrm {d} }u}$
3. 3
Choose a ${\displaystyle u}$ and ${\displaystyle {\mathrm {d} }v,}$ and find the resulting ${\displaystyle {\mathrm {d} }u}$ and ${\displaystyle v}$. As one of the functions is the exponential function, setting that as ${\displaystyle u}$ will get us nowhere. Instead, let ${\displaystyle u=\cos x}$ and ${\displaystyle {\mathrm {d} }v=e^{x}{\mathrm {d} }x.}$ What we find is that the second derivative of ${\displaystyle u}$ is simply the negative of itself. That is, ${\displaystyle {\frac {{\mathrm {d} }^{2}}{{\mathrm {d} }x^{2}}}\cos x=-\cos x.}$ This means that we need to integrate by parts twice to get an interesting result.
• ${\displaystyle {\mathrm {d} }u=-\sin x}$
• ${\displaystyle v=e^{x}}$
4. 4
Substitute these expressions into our integral.
• {\displaystyle {\begin{aligned}\int e^{x}\cos x{\mathrm {d} }x&=e^{x}\cos x-\int -e^{x}\sin x{\mathrm {d} }x\\&=e^{x}\cos x+\int e^{x}\sin x{\mathrm {d} }x\end{aligned}}}
5. 5
Perform integration by parts on the ${\displaystyle v{\mathrm {d} }u}$ integral. Be careful with the signs.
• ${\displaystyle u=\sin x,\,{\mathrm {d} }v=e^{x}{\mathrm {d} }x,\,{\mathrm {d} }u=\cos {x}{\mathrm {d} }x,\,v=e^{x}}$
• ${\displaystyle \int e^{x}\sin x{\mathrm {d} }x=e^{x}\sin x-\int e^{x}\cos x{\mathrm {d} }x}$
• ${\displaystyle \int e^{x}\cos x{\mathrm {d} }x=e^{x}\cos x+e^{x}\sin x-\int e^{x}\cos x{\mathrm {d} }x}$
6. 6
Solve for the original integral. In this problem, what we have found is that by performing integration by parts twice, the original integral came up in the work. Instead of performing integration by parts endlessly, which will get us nowhere, we can solve for it instead. Don't forget the constant of integration at the very end.
• {\displaystyle {\begin{aligned}2\int e^{x}\cos x{\mathrm {d} }x&=e^{x}\cos x+e^{x}\sin x\\\int e^{x}\cos x{\mathrm {d} }x&={\frac {1}{2}}e^{x}\cos x+{\frac {1}{2}}e^{x}\sin x+C\end{aligned}}}
### Part 4 Deriving the Integration by Parts Formula
1. 1
Consider the antiderivative of ${\displaystyle g(x)}$. We shall call this function ${\displaystyle G(x),}$ where ${\displaystyle G}$ is any function that satisfies ${\displaystyle G^{\prime }=g.}$
2. 2
Compute the derivative of ${\displaystyle fG}$. Since this is a product of two functions, we use the product rule. Sharp minds will intuitively see the resulting integration by parts formula as closely related to the product rule, just as u-substitution is the counterpart to the chain rule.
• {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{{\mathrm {d} }x}}fG&=fG^{\prime }+f^{\prime }G\\&=fg+f^{\prime }G\end{aligned}}}
3. 3
Take the integral of both sides with respect to ${\displaystyle x}$. The above expression says that ${\displaystyle fG}$ is the antiderivative of the right side, so we integrate both sides to recover the integral of the left side.
• ${\displaystyle \int (fg+f^{\prime }G){\mathrm {d} }x=fG}$
4. 4
Rearrange to isolate the integral of ${\displaystyle fg}$.
• ${\displaystyle \int fg{\mathrm {d} }x=fG-\int f^{\prime }G{\mathrm {d} }x}$
• The goal of integration by parts is seen in the expression above. We are integrating ${\displaystyle f^{\prime }G}$ instead of ${\displaystyle fg,}$ and if used correctly, this results in a simpler evaluation.
5. 5
Change the variables to recover the familiar compact form. We let ${\displaystyle u=f,\,v=G,\,{\mathrm {d} }u=f^{\prime }{\mathrm {d} }x,\,{\mathrm {d} }v=g{\mathrm {d} }x.}$
• ${\displaystyle \int u{\mathrm {d} }v=uv-\int v{\mathrm {d} }u}$
• In general, there is no systematic process by which we can make the integral easier to evaluate. However, it is often the case that we want a ${\displaystyle u}$ whose derivative is easier to manage, and a ${\displaystyle {\mathrm {d} }v}$ that can easily be integrated.
• For definite integrals, it is easy to show that the formula holds when writing the boundaries for all three terms, though it is important to remember that the boundaries are limits on the variable ${\displaystyle x.}$
• ${\displaystyle \int _{a}^{b}u{\mathrm {d} }v=uv{\Bigg |}_{a}^{b}-\int _{a}^{b}v{\mathrm {d} }u}$
## Tips
• A useful mnemonic is LIATE, which stands for logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. This is the order for which you should choose ${\displaystyle u,}$ though it does not work all of the time.
## Article Info
Featured Article
Categories: Featured Articles | Calculus
Thanks to all authors for creating a page that has been read 13,939 times. | 2016-10-24T16:07:25 | {
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https://mathemerize.com/coplanar-vectors/ | # Coplanar Vectors – Definition and Example
Here you will learn definition of coplanar vectors with example and test of coplanarity of four points.
Let’s begin –
## Coplanar Vectors
A system of vectors is said to be coplanar, if their supports are parallel to the same plane.
Note : Two vectors are coplanar.
Theorem 1 (Test of Coplanarity of Three vectors)
Let $$\vec{a}$$ and $$\vec{b}$$ be two given non-zero non-collinear vectors. Then, any vector $$\vec{r}$$ coplanar with $$\vec{a}$$ and $$\vec{b}$$ can be uniquely expressed as $$\vec{r}$$ = $$x\vec{r}$$ + $$y\vec{b}$$, for some scalars x and y.
Theorem 2
The necessary and sufficient condition for three vectors $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $$l\vec{a} + m\vec{b} + n\vec{c}$$ = $$\vec{0}$$.
Theorem 3
If $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ are three non-zero non-coplanar vector and x, y, z are three scalars, then $$x\vec{a} + y\vec{b} + z\vec{c}$$ = $$\vec{0}$$ $$\implies$$ x = y = z = 0.
Example : Show that the vectors $$\vec{a} – 2\vec{b} + 3\vec{c}$$, $$\vec{a} – 3\vec{b} + 5\vec{c}$$ and $$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ are coplanar, where $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are non-coplanar vector.
Solution : From Theorem 1,
Three vectors are co-planar if one of the given vectors are expressible as a linear combination of the other two. Let
$$\vec{a} – 2\vec{b} + 3\vec{c}$$ = x$$\vec{a} – 3\vec{b} + 5\vec{c}$$ + y$$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ for some scalars x and y.
$$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ = (x – 2y)$$\vec{a}$$ + (-3x + 3y)$$\vec{b}$$ + (5x – 4y)$$\vec{c}$$
$$\implies$$ 1 = x – 2y, -2 = -3x + 3y and 3 = 5x – 4y
Solving first two of these equation, we get
x = 1/3, y = -1/3.
Clearly, these values of x and y satisfy the third equation.
Hence, the given vectors are co-planar.
## Test of Coplanarity of Four Points
Four points with position vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ and $$\vec{d}$$ are co-planar iff there exist scalars x, y z, u not allzero such that $$x\vec{a} + y\vec{b} + z\vec{c} + u\vec{d}$$ = $$\vec{0}$$, where x + y + z + u =0. | 2022-11-30T16:54:39 | {
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http://invincibleideas.com/stem/maths/coordinate-geometry/line/linefamily2D.php | # Family of Lines in 2D
3 or more Lines are said to form a family if either
1. The lines are concurrent i.e. they pass through a common point of intersection
2. The lines are parallel to each other
Given any 2 Lines in 2D as the following
$$A_1x + B_1y + C_1=0$$
AND
$$A_2x + B_2y + C_2=0$$
The equation of any line that is either parallel to these 2 lines or is concurrent to these 2 lines is given as
$$A_1x + B_1y + C_1 + k(A_2x + B_2y + C_2) = 0$$
The value of the variable $$k$$ can be found out when some additional input is given. Following are some additional inputs that are generally given
1. The point through which the resulting line passes
2. The line/axis to which the resulting line is parallel or perpendicular | 2019-11-20T03:59:17 | {
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https://www.futurelearn.com/courses/maths-linear-quadratic-relations/0/steps/12120 | 4.7
## UNSW Sydney
A secant and a tangent to a parabola
# Secants and tangents of a parabola
A secant of a parabola is a line, or line segment, that joins two distinct points on the parabola. A tangent is a line that touches the parabola at exactly one point. Finding tangents to curves is historically an important problem going back to Fermat, and is a key motivator for the differential calculus.
In this step we will see how to algebraically find both secants and tangents to a parabola. So this section gives an alternative algebraic perspective on calculus concepts!
## A parametrization of the parabola
The standard parabola $\normalsize{y=ax^2}$ has a nice parametrization of the form
For any value of $\normalsize{t}$, this formula for $\normalsize{A}$ gives us a point lying on the parabola.
In the figure we see the parabola $\normalsize{y=\frac{1}{4}x^2}$. Note that as $\normalsize{t}$ varies, the $\normalsize{x}$- coordinate of $\normalsize{A}$ moves linearly, while the $\normalsize{y}$- coordinate moves quadratically.
## Secant of a parabola
Now suppose that we have two points $\normalsize{A=[t,at^2]}$ and $\normalsize{B=[u,au^2]}$ on the parabola $\normalsize{y=ax^2}$. The line $\normalsize{AB}$ joining $\normalsize{A}$ and $\normalsize{B}$ has slope
and so it has an equation of the form $\normalsize{y=a(u+t)x+b}$. To determine the $\normalsize{y}$-intercept $\normalsize{b}$, we may substitute in the coordinates $\normalsize{[t,at^2]}$. We get that $\normalsize{b=-aut}$ so the equation of the secant is
Q1 (E): What is the equation of the secant through the points on the parabola $\normalsize{y=x^2}$ whose $\normalsize{x}$- coordinates are $\normalsize{3}$ and $\normalsize{5}$ ?
## Tangent to a parabola
What happens to our formula for the secant when $\normalsize{t}$ and $\normalsize{u}$ are very close? Better yet, what happens when we set $\normalsize{t=u}$, so that $\normalsize{A=B}$? This is actually a very interesting situation that rewards careful examination. We are parachuting into the hallowed halls of calculus using algebra!
In this case the equation of the secant becomes the equation of the tangent to the parabola: starting with the equation of the secant ${\normalsize y=a(t+u)x-atu}$ and substituting $\normalsize u=t$ gives:
Note that in particular the slope of the tangent line to the point with $\normalsize{x}$-coordinate $\normalsize{t}$ is $\normalsize{2at}$. This number is usually called the derivative of the function at the point $\normalsize{x}$, and in calculus courses is introduced following a study of limits. Our purely algebraic techniques allowed us to find the equation of the tangent to the parabola $\normalsize{y=ax^2}$. Since any other parabola is just a translate of such, we see that we can deal with secants and tangents for general parabolas, without limits!
Q2 (E): What is the equation of the tangent to the parabola $\normalsize{y=\frac{1}{4} x^2}$ at the point $\normalsize{[3,9/4]}$?
## Two properties of the parabola
Here are two interesting challenges for you. Let’s see if we can derive some interesting aspects of the geometry of the parabola. The second one will require some thorough thinking. You will need to know that two lines are perpendicular precisely when their slopes multiply to $\normalsize{-1}$.
Q3 (M): Consider the points $\normalsize{A=[t,at^2]}$ and $\normalsize{B=[u,au^2]}$ on the parabola $\normalsize{y=ax^2}$. Show that the two tangents to the parabola at $\normalsize{A}$ and $\normalsize{B}$ meet at $\normalsize{P=[(u+t)/2,aut]}$.
Notice that the $\normalsize x$- coordinate of $\normalsize P$ is halfway between $\normalsize A$ and $\normalsize B$, and does not depend on the particular parabola.
Q4 (C): Show that the two tangents to the parabola $\normalsize{y=x^2}$ through the points $\normalsize{A=[t,t^2]}$ and $\normalsize{B=[u,u^2]}$ are perpendicular precisely when the line $\normalsize{AB}$ passes through the focus of the parabola.
A1. The secant is the line ${\normalsize y = 8x - 15}$.
A2. Apply the formula ${\normalsize y=2atx-at^2}$ with ${\normalsize a=1/4}$ and ${\normalsize t=3}$ to get the tangent: ${\normalsize y=\frac32 x-\frac94}$.
A3. Solve the equations $\normalsize{y=2atx-at^2}$ and $\normalsize{y=2aux-au^2}$ simultaneously.
A4. You can do it! | 2020-01-19T08:49:24 | {
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https://math.stackexchange.com/questions/360119/de-rham-comologies-of-the-n-torus | # de Rham comologies of the $n$-torus
I'm attempting to calculate the de Rham cohomologies of the $n$-torus: $n \choose k$.
I'd like to use a Mayer-Vietoris sequence relating $H^kT^{n}$ to $H^kT^{n-1}$ and $H^{k-1}T^{n-1}$ so I can use the identity ${n \choose k} = {n-1 \choose k} + {n-1 \choose {k-1}}$ to achieve the result by induction. But I am having trouble choosing an open cover for the sequence. The obvious choice would be to split the torus along one of its circles, i.e. $T^n=T^{n-1} \times S^1=U \cup V$ with $U=T^{n-1} \times (1/4,3/4)$ and $V=T^{n-1} \times [0,1/2)\cup(1/2,1]$ where we are identifying the point $0$ and $1$ to make $[0,1]\cong S^1$.
Here's my beef: By retraction $H^k U \cong H^k V \cong H^k T^{n-1}$ and we'd have $$H^k(U \cap V)\cong H^k (T^{n-1}\sqcup T^{n-1})\cong H^k T^{n-1} \oplus H^k T^{n-1} \cong H^kU \oplus H^kV$$ so that the $T^{n-1}$ terms cancel each other out of the sequence and you just end up with $\sum_{k=0}^n (-1)^k \dim H^k T^n=0$ which is true but only helpful if I cared about the Euler characteristic. Am I mistaken? Is there a better cover for the sequence?
Your covering is fine. Let's take a closer look at the maps defining the Mayer-Vietoris sequence: $$H^{k-1}(T^{n-1})\oplus H^{k-1}(T^{n-1})\xrightarrow{f}H^{k-1}(T^{n-1}\sqcup T^{n-1})\xrightarrow{\partial}H^k(T^n)\xrightarrow{g}H^k(T^{n-1})\oplus H^k(T^{n-1})$$
Although $f$ is a map between two isomorphic spaces, it is not an isomorphism, so the terms do not cancel. Let $([\alpha],[\beta])\in H^{k-1}(T^{n-1})\oplus H^{k-1}(T^{n-1})$. Then we have
$$f(([\alpha],[\beta]))=[\alpha-\beta]$$
I'm being a bit sloppy here with notation, using $\alpha$ instead of $i^*[\alpha]$. Similarly, for $[\alpha]\in H^k(T^n)$, we have $$g([\alpha])=([\alpha],[\alpha])$$ again with shorthand notation.
Finally, for any $[\alpha]\in H^{k-1}(T^{n-1}\sqcup T^{n-1})$, we may write $[\alpha]=[\beta]+[\gamma]$ reflecting the direct sum decomposition into disjoint components. Then $$\partial([\alpha])=[d\beta]=-[d\gamma]$$ where $d$ is the exterior derivative.
Now that we've had a closer look at the maps in the sequence, see if you can use the exactness of Mayer-Vietoris to show that there is a short exact seqence: $$0\rightarrow H^{k-1}(T^{n-1})\rightarrow H^{k}(T^n)\rightarrow H^k(T^{n-1})\to 0$$
From which your result will follow. | 2019-11-20T19:58:17 | {
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https://open.kattis.com/problems/expandingrods | Kattis
# Expanding Rods
When a thin rod of length $L$ is heated $n$ degrees, it expands to a new length $L’ = (1+n \cdot C) \cdot L$, where $C$ is the coefficient of heat expansion.
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.
Your task is to compute the distance by which the center of the rod is displaced.
## Input
The input contains at most $20$ lines. Each line of input contains three non-negative numbers:
• an integer $L$, the initial lenth of the rod in millimeters ($1 \le L \le 10^9$),
• an integer $n$, the temperature change in degrees ($0 \le n \le 10^5$),
• a real number $C$, the coefficient of heat expansion of the material ($0 \le C \le 100$, at most $5$ digits after the decimal point).
The input is such that the displacement of the center of any rod is at most one half of the original rod length. The last line of input contains three $-1$’s and it should not be processed.
## Output
For each line of input, output one line with the displacement of the center of the rod in millimeters with an absolute error of at most $10^{-3}$ or a relative error of at most $10^{-9}$.
Sample Input 1 Sample Output 1
1000 100 0.0001
15000 10 0.00006
10 0 0.001
-1 -1 -1
61.328991534
225.020248568
0.000000000 | 2020-01-21T19:33:29 | {
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https://teachingcalculus.com/2018/02/13/parametric-and-vector-equations-2/ | # Parametric and Vector Equations
In the plane, the position of a moving object as a function of time, t, can be specified by a pair of parametric equations $x=x\left( t \right)\text{ and }y=y\left( t \right)$ or the equivalent vector $\left\langle x\left( t \right),y\left( t \right) \right\rangle$. The path is the curve traced by the parametric equations or the tips of the position vector. .
The velocity of the movement in the x- and y-direction is given by the vector $\left\langle {x}'\left( t \right),{y}'\left( t \right) \right\rangle$. The vector sum of the components gives the direction of motion. Attached to the tip of the position vector this vector is tangent to the path pointing in the direction of motion.
The length of this vector is the speed of the moving object. $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$. (Notice that this is the same as the speed of a particle moving on the number line with one less parameter: On the number line $\text{Speed}=\left| v \right|=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}}$.)
The acceleration is given by the vector $\left\langle {{x}'}'\left( t \right),{{y}'}'\left( t \right) \right\rangle$.
What students should know how to do:
• Vectors may be written using parentheses, ( ), or pointed brackets, $\left\langle {} \right\rangle$, or even $\vec{i},\vec{j}$ form. The pointed brackets seem to be the most popular right now, but all common notations are allowed and will be recognized by readers.
• Find the speed at time t$\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$
• Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}}dt$. Notice that this is the integral of the speed (rate times time = distance).
• The slope of the path is $\displaystyle \frac{dy}{dx}=\frac{{y}'\left( t \right)}{{x}'\left( t \right)}$. See this post for more on finding the first and second derivatives with respect to x.
• Determine when the particle is moving left or right,
• Determine when the particle is moving up or down,
• Find the extreme position (farthest left, right, up, down, or distance from the origin).
• Given the position find the velocity by differentiating; given the velocity find the acceleration by differentiating.
• Given the acceleration and the velocity at some point find the velocity by integrating; given the velocity and the position at some point find the position by integrating. These are just initial value differential equation problems (IVP).
• Dot product and cross product are not tested on the BC exam, nor are other aspects.
Here are two past post on this topic:
Implicit Differentiation of Parametric Equation
A Vector’s Derivatives | 2019-11-19T13:01:31 | {
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https://daxpy.xyz/notes/bfgs/ | # BFGS
## Newton’s Method
\begin{aligned} x_{k+1} &= x_k - [H(x_k)]^{-1}\nabla f(x_k)^\intercal \end{aligned}
## Quasi Newton’s Method
\begin{aligned} x_{k+1} &= x_k - \alpha_kS_k {\nabla f(x_k)}^{T} \end{aligned}
If $S_k$ is inverse of Hessian, then method is Newton’s iteration; if $S_k=I$, then it is steepest descent
## BFGS
BFGS is a quasi newtons method where we approximate inverse of Hessian by $B_k$. The search direction $p_k$ is determined by solving $$B_kp_k = -\nabla f(x_k)$$ A line search is performed in this search direction to find next point $x_{k+1}$ by minimising $f(x_k+\gamma p_k)$. The approximation to hessian is then updated as \begin{aligned} B_{k+1} &= B_k + \alpha_k u_ku_k^\intercal + \beta_k v_kv_k^\intercal \\ u_k &= \nabla f(x_{k+1})-\nabla f(x_k) \\ \alpha_k &= \frac{1}{\alpha u_k^\intercal p_k} \\ v_k &= B_kp_k \\ \beta_k &= \frac{-1}{p_k^\intercal B_kp_k} \end{aligned}
tags: optimisation | 2022-09-25T04:33:56 | {
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http://mathonline.wikidot.com/the-lebesgue-dominated-convergence-theorem | The Lebesgue Dominated Convergence Theorem
# The Lebesgue Dominated Convergence Theorem
Theorem (The Lebesgue Dominated Convergence Theorem): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E$. Suppose that: 1) There exists a nonnegative Lebesgue integrable function $g$ on $E$ such that $|f_n(x)| \leq g(x)$ for all $x \in E$. 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ almost everywhere on $E$. Then $f$ is Lebesgue integrable on $E$ and $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$.
• Proof: From (1), since $g$ is a Lebesgue integrable function on $E$ such that $|f_n(x)| \leq g(x)$ for all $x \in E$ we have from The Comparison Test for Lebesgue Integrability that for each $n \in \mathbb{N}$, the Lebesgue measurable function $f_n$ is Lebesgue integrable on $E$.
• Furthermore, we also have that $|f(x)| \leq g(x)$ for all $x \in E$. Since $f$ is Lebesgue measurable (as it is a pointwise limit of Lebesgue measurable functions) we also have by the comparison test that $f$ is Lebesgue integrable on $E$.
• Now consider the sequence of functions $(g(x) - f_n(x))_{n=1}^{\infty}$. This is a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ that converges pointwise to $g(x) - f(x)$. So by Fatou's Lemma for Nonnegative Lebesgue Measurable Functions we have that:
(1)
\begin{align} \quad \int_E (g - f) \leq \liminf_{n \to \infty} (g - f_n) \end{align}
• By the linearity of the Lebesgue integral for nonnegative Lebesgue measurable functions we have that:
(2)
\begin{align} \quad \int_E g - \int_E f & \leq \liminf_{n \to \infty} (g - f_n) \\ & \leq \int_E g - \limsup_{n \to \infty} \int_E f_n \end{align}
• Therefore:
(3)
\begin{align} \quad -\int_E f \leq - \limsup_{n \to \infty} \int_E f_n \quad \Leftrightarrow \quad \int_E f \geq \limsup_{n \to \infty} \int_E f_n \quad (*) \end{align}
• Now consider the sequence of functions $(g(x) + f_n(x))_{n=1}^{\infty}$. Since $|f_n(x)| \leq g(x)$ for all $x \in E$ we have that $|f_n(x)| + f_n(x) \leq g(x) + f_n(x)$. So $(g(x) + f_n(x))_{n=1}^{\infty}$ is a nonnegative sequence of Lebesgue measurable functions that converges pointwise to $g(x) + f(x)$. So by Fatou's lemma we have that:
(4)
\begin{align} \quad \int_E (g + f) \leq \liminf_{n \to \infty} \int_E (g + f_n) \\ \end{align}
• By the linearity of the Lebesgue integral for nonnegative Lebesgue measurable functions we have that:
(5)
\begin{align} \quad \int_E g + \int_E f \leq \int_E g + \liminf_{n \to \infty} \int_E f_n \quad \Leftrightarrow \quad \int_E f \leq \liminf_{n \to \infty} \int_E f_n \quad (**) \end{align}
• Combining $(*)$ and $(**)$ yields:
(6)
\begin{align} \quad \limsup_{n \to \infty} \int_E f_n \leq \int_E f \leq \liminf_{n \to \infty} \int_E f \end{align}
• By definition the limit inferior is always less than or equal to the limit superior of a sequence of functions. Therefore the inequality above implies that:
(7)
\begin{align} \quad \lim_{n \to \infty} \int_E f_n = \int_E f \quad \blacksquare \end{align} | 2017-10-24T02:05:45 | {
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http://www.lecture-notes.co.uk/susskind/quantum-entanglements/lecture-2/quantum-states/ | # Quantum states
### Vector spaces over complex numbers
Quantum states are elements of an abstract vector space, $\mathbb{V}$, over the complex numbers, $\mathbb{C}$. This is in contrast to the space of classical states, which is a set.
Conventionally, elements of $\mathbb{V}$ are column vectors of complex numbers. Using Dirac's notation, we call them ket vectors. Thus, if $$\ket{a} \in \mathbb{V}$$ then $$\exists\; a_i \in \mathbb{C} \text{ for } i = 1,2,\cdots,N$$ such that $$\ket{a} \to \left [ \begin{matrix} a_1 \\ \vdots \\ a_N \end{matrix} \right ]$$
Scalar multiplication is given by $$c \ket{a} \to \left [ \begin{matrix} c a_1 \\ \vdots \\ c a_N \end{matrix} \right ]$$ where $c \in \mathbb{C}$.
$$\ket{a} + \ket{b} \to \left [ \begin{matrix} a_1 + b_1 \\ \vdots \\ a_N + b_N \end{matrix} \right ]$$ where $\ket{b} \in \mathbb{V}$ as well.
The dual space of the space of $\mathbb{V}$, denoted $\mathbb{V}^*$, is called the space of complex conjugate vectors. We use Dirac notation as well, but conjugates are labelled as bra vectors.
Conjugates are represented by row vectors, but their elements are the complex conjugates of the associated vector in the dual space.
For example, if $\ket{a} \in \mathbb{V}$, then $$\bra{a} \to [a_1^*, \cdots, a_N^*] \in \mathbb{V}^*$$ is called the complex conjugate of $\ket{a}$.
Nb. We don't include the conjugate-sign $*$ in the bra-label - this is implicit.
Vector multiplication, or the inner product, of two vectors, $\ket{a}, \ket{b} \in \mathbb{V}$, is given by \begin{align*}\braket{b}{a} &= [ b_1^* \cdots b_N^* ] \left [ \begin{matrix} a_1 \\ \vdots \\ a_N \end{matrix} \right ]\\&= b_1^* a_1+... +b_N^* a_n\end{align*}
The square of the vector is $$\braket{a}{a} = a_1^* a_1+... + a_N^* a_N \in \mathbb{R}$$ since the product of a complex number with its conjugate is always real.
We can consider the complex conjugate of the inner product, \begin{align*}\braket{b}{a}^*&= \left(b_1^* a_1 + \cdots + b_N^* a_N \right)^*\\&= \left(a_1^* b_1 + \cdots + a_N^* b_N \right)\\&= \braket{a}{b}\end{align*}
### Single bit quantum states
Recall, in the detection experiment, we prepared the electron at some angle to the vertical, and then tried to measure the electron's spin with respect to the vertical. We said there were only two distinct possibilities - either we don't detect a photon, corresponding to the up state, which we label $\ket{u}$, or we do detect a photon, corresponding to the down state, which we label $\ket{d}$.
We arbitrarily chose to set up to mean pointing in the $+z$ direction of normal space, and down to mean pointing in the $-z$ direction of normal space, but vectors in normal space do not define states in quantum mechanics. States in quantum mechanics are represented by ket vectors. The dimension of the space is determined by the number of bits that describes the system in question.
In our case, the number of bits is one, hence the dimension of the space, $\mathbb{V}$, is $2^1 = 2$. We want to choose vectors in this space to represent the two states (of the bit). By convention, we define the ket vectors, $\ket{u}, \ket{d}$ to have the components \begin{align*}\ket{u} & \to \begin{bmatrix} 1\\0 \end{bmatrix}\\\ket{d} & \to \begin{bmatrix} 0\\1 \end{bmatrix}\end{align*}
These form an orthonormal basis for $\mathbb{V}$. That is, they are both unit vectors $$\braket{u}{u} = \braket{d}{d} = 1$$ and they are mutually orthogonal $$\braket{u}{d} = \braket{d}{u} = 0$$
We now can define a general single-bit quantum state, $\ket{a} \in \mathbb{V}$ to be a unit linear combination of the two basis vectors $\ket{u}, \ket{d}$. That is, $\exists\; a_u, a_d \in \mathbb{C}$ such that $$\ket{a} \to a_u \ket{u} + a_d \ket{d} = \begin{bmatrix} a_u\\a_d\end{bmatrix}$$ and \begin{align*}\braket{a}{a} &= \begin{bmatrix} a_u^*,a_d^* \end{bmatrix}\begin{bmatrix} a_u\\a_d\end{bmatrix}\\&= a_u^*a_u + a_d^*a_d\\&= 1\end{align*}
Note that, instead of using columns and rows to calculate the square of a (ket) vector, we could use the fact that $\ket{u}, \ket{d}$ are basis vectors. Then, in detail \begin{align*}\braket{a}{a} &= \bra{a} \cdot \ket{a}\\&= \left(\bra{u} a_u^* + \bra{d} a_d^* \right) \cdot \left( a_u \ket{u} + a_d \ket{d} \right)\\&= \bra{u}{a_u^*a_u}\ket{u} + \bra{u}{a_u^*a_d}\ket{d} + \bra{d}{a_d^*a_u}\ket{u} + \bra{d}{a_d^*a_d}\ket{d}\\&= a_u^*a_u \braket{u}{u} + a_u^*a_d \braket{u}{d} + a_d^*a_u \braket{d}{u} + a_d^*a_d \braket{d}{d}\\&= a_u^*a_u (1) + a_u^*a_d (0) + a_d^*a_u (0) + a_d^*a_d (1)\\&= a_u^*a_u + a_d^*a_d\end{align*}
Notice that the ket vectors contain four real numbers in two complex components whereas the prepared directions of an electron can be represented by a unit-length spatial vector in $\mathbb{R}^3$, containing only two real numbers. This is resolved by the condition that quantum states are phase-independent. That is, we can multiply a state by the factor $$e^{\mathrm{i}\theta}$$ for any $\theta$, and not change the probability of being detected (see below) - meaning the physics doesn't change. This eliminates one real number from the abstract vector. A second real number is eliminated by insisting that the quantum state is also unit-length.
### Probability of being detected
Given the general state, $\ket{a}$, of an electron (prepared at some angle to the vertical), we define the probability of the electron being detected in the up-state as $$p_u=a_u^* a_u$$
Similarly, $$p_d=a_d^* a_d$$ is the probability that the electron is detected in the down-state.
For these values to be probabilities, we would require that they add up to one. But, $$p_u + p_d = a_u^* a_u + a_d^* a_d = 1$$ since this is just the square $\braket{a}{a}$ of a unit vector.
We refer to the components of a quantum state as probability amplitudes, which square to form probabilities.
Nb. Knowing the probabilities alone does not completely determine the coefficients, since we can multiply the co-efficients by a pure phase complex number, of the form $e^{\mathrm{i}\theta}$, and not change the probabilities.
### The three directions of space
We want to define the state of an electron that's been prepared in an arbitrary direction (in normal space), and then is detected to be in either the up or down state. We first consider preparation in the three standard directions.
Since we chose (arbitrarily) to associate the up state with the $+z$ direction, then we would expect that, if the electron was prepared in the $+z$ direction, then the probability of it being detected in the up state to be one. The probability of it being detected in the down state to be zero. That is, $$\ket{+z} = \begin{bmatrix}1 \\ 0 \end{bmatrix} = \ket{u}$$
Similarly, if the electron was prepared in the $-z$ direction, then the probability of it being detected in the up state to be zero and the probability of it being detected in the down state to be one, $$\ket{-z} = \begin{bmatrix}0 \\ 1 \end{bmatrix} = \ket{d}$$
The states corresponding to the electron being prepared in the other directions, $x, y$ have been determined by experiment. They are, \begin{align*}\ket{+x} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \end{bmatrix} & \ket{-x} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{-1}{\sqrt{2}} \end{bmatrix}\\\ket{+y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{\mathrm{i}}{\sqrt{2}} \end{bmatrix} & \ket{-y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{-\mathrm{i}}{\sqrt{2}} \end{bmatrix}\end{align*}
We can write explicitly in terms of the up and down states, \begin{align*}\ket{+x} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} + \ket{d}\right) &\quad \ket{-x} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} -\, \ket{d}\right)\\\ket{+y} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} + i\ket{d}\right) &\quad \ket{-y} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} -\, \mathrm{i}\ket{d}\right)\end{align*}
We can read these as follows: the probability of the electron being detected in the up state, having been prepared in the $+x$ direction, is given by $$p_u(+x) = \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{2}$$
It is easily shown that all of these are unit vectors, for example, \begin{align*}\braket{+y}{+y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} , \tfrac{-\mathrm{i}}{\sqrt{2}} \end{bmatrix} \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{\mathrm{i}}{\sqrt{2}} \end{bmatrix}\\&= \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} + \tfrac{-\mathrm{i}}{\sqrt{2}} \cdot \tfrac{\mathrm{i}}{\sqrt{2}}\\&= \tfrac{1}{2} + \tfrac{1}{2}\\&= 1\end{align*}
Also, note that the $\pm$ vectors representing each direction (of normal space) are mutually orthogonal, $$\braket{+x}{-x} = \braket{+y}{-y} = 0 \:( = \braket{+z}{-z} )$$ but a vector of one direction is not mutually orthogonal to a vector of another direction. For example, $$\braket{+x}{-y} \neq 0$$ | 2017-11-17T19:31:26 | {
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http://mathhelpforum.com/statistics/192249-continuous-uniform-dist-question.html | # Thread: Continuous uniform dist. question
1. ## Continuous uniform dist. question
I'm not sure how to tackle this question... Any help would be most grateful:
A computer generates independent random numbers from the continuous uniform distribution over the range [0,1]. In a sample of 9 such numbers what is the probability that exactly 2 are less that 0.33 ?
Thanks, Felix
2. ## Re: Continuous uniform dist. question
Originally Posted by FelixHelix
A computer generates independent random numbers from the continuous uniform distribution over the range [0,1]. In a sample of 9 such numbers what is the probability that exactly 2 are less that 0.33 ?
The probability that any one of the nine is less than .33 is $\frac{.33-0}{1-0}=0.33$.
So what about exactly two? HINT: binomial.
3. ## Re: Continuous uniform dist. question
So it would be:
(9 Choose 2)* (0.33)^2 * (0.67)^7 = 0.2376 | 2017-01-24T09:17:00 | {
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https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2015/probability/tp13-1/vertical-4f20b89f006a/ | # 4.5 Expectation
## Fair And Biased Coins
Say I flip 200 coins, 100 of which are fair, and 100 of which land on heads with probability $$\frac{1}{4}$$.
What is the expected number of heads?
Exercise 1
Let $$R_i$$ be an indicator variable for the $$i^{th}$$ fair coin, where $$R_i = \begin{cases} 1 & i^{th}\text{ flip is heads} \\ 0 & i^{th}\text{ flip is tails} \end{cases}$$.
Similarly, define $$S_j$$ as the indicator variable for the $$j^{th}$$ biased coin.
Then, the number of heads is given by $$H=R_1+R_2+\ldots +R_{100}+S_1+S_2+\ldots+S_{100}$$.
By linearity of expectation, $E[H]=E[R_1+R_2+\ldots +R_{100}+S_1+S_2+\ldots+S_{100}]$ $=E[R_1]+E[R_2]+\ldots +E[R_{100}]+E[S_1]+E[S_2]+\ldots+E[S_{100}]$ $=100\cdot\frac{1}{2}+100\cdot\frac{1}{4}=50+25=75.$ | 2020-02-23T15:34:56 | {
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http://www.encyclopediaofmath.org/index.php/Totient_function | # Totient function
Euler totient function, Euler totient
Another frequently used named for the Euler function , which counts the natural numbers that are relatively prime to .
The Carmichael conjecture on the Euler totient function states that if , then for some ; i.e. no value of the Euler function is assumed once. This has now been verified for , [a1].
A natural generalization of the Euler totient function is the Jordan totient function , which counts the number of -tuples , , such that . Clearly, .
One has
where runs over the prime numbers dividing , and
where is the Möbius function and runs over all divisors of . For these formulas reduce to the well-known formulas for the Euler function.
The Lehmer problem on the Euler totient function asks for the solutions of , , [a2]. For some results on this still (1996) largely open problem, see [a3] and the references therein. The corresponding problem for the Jordan totient function (and ) is easy, [a4]: For , if and only if is a prime number. Moreover, if is a prime number, then .
For much more information on the Euler totient function, the Jordan totient function and various other generalizations, see [a5], [a6].
#### References
[a1] A. Schlafly, S. Wagon, "Carmichael's conjecture on the Euler function is valid below " Math. Comp. , 63 (1994) pp. 415–419 [a2] D.H. Lehmer, "On Euler's totient function" Bull. Amer. Math. Soc. , 38 (1932) pp. 745–751 [a3] V. Siva Rama Prasad, M. Rangamma, "On composite for which " Nieuw Archief voor Wiskunde (4) , 5 (1987) pp. 77–83 [a4] M.V. Subbarao, V. Siva Rama Prasad, "Some analogues of a Lehmer problem on the totient function" Rocky Mount. J. Math. , 15 (1985) pp. 609–620 [a5] R. Sivamarakrishnan, "The many facets of Euler's totient II: generalizations and analogues" Nieuw Archief Wiskunde (4) , 8 (1990) pp. 169–188 [a6] R. Sivamarakrishnan, "The many facets of Euler's totient I" Nieuw Archief Wiskunde (4) , 4 (1986) pp. 175–190 [a7] L.E. Dickson, "History of the theory of numbers" , I: Divisibility and primality , Chelsea, reprint (1971) pp. Chapt. V; 113–155
How to Cite This Entry:
Totient function. M. Hazewinkel (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Totient_function&oldid=12673
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098 | 2013-05-23T15:16:11 | {
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https://im.kendallhunt.com/HS/teachers/2/6/16/preparation.html | Lesson 16
Weighted Averages in a Triangle
Lesson Narrative
The goal of this lesson is to use segment partitioning logic to prove that the medians of a triangle intersect in a single point.
Students begin by finding midpoints of the sides of triangles, and they learn that medians are segments that connect a triangle vertex to the midpoint of the opposite side. Next, they find points that partition the medians of a triangle in a $$2:1$$ ratio, observing in the process that the 3 medians of the triangle intersect in a single point. Finally, students work in groups to construct a viable argument (MP3) that proves the medians of any triangle intersect at that $$2:1$$ partition point. While the intersection point of the medians isn’t given the formal name centroid in this lesson, you may choose to use that term if it is helpful.
Technology isn’t required for this lesson, but there are opportunities for students to choose to use appropriate technology to solve problems. We recommend making technology available.
Learning Goals
Teacher Facing
• Calculate the intersection point of a triangle’s medians.
• Prove that the medians of a triangle meet at a point.
Student Facing
• Let’s partition special line segments in triangles.
Required Preparation
Students will use rulers as a straightedge and for taking measurements in both the Triangle Medians activity and the lesson synthesis.
The index cards are for use in the lesson synthesis.
Student Facing
• I can determine the point where the medians of a triangle intersect.
Building On
Building Towards
Glossary Entries
• median (geometry)
A line from a vertex of a triangle to the midpoint of the opposite side. Each dashed line in the image is a median. | 2023-03-23T21:50:11 | {
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https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_23&diff=prev&oldid=60352 | Difference between revisions of "2014 AMC 10B Problems/Problem 23"
Problem
A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$
$[asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy]$
(Diagram edited from copeland's diagram)
Solution
First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of 2, and the bottom base have a diameter of 2r. $[asy] size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("r-1",(X+B)/2,S); label("1",(Y+C)/2,N); label("s",(O+Y)/2,W); label("s",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label("1",(C+P)/2,NE); label("r",(B+P)/2,NE); [/asy]$ (diagram by copeland?)
Then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ , which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$. Subtracting $r^2-2r+1$ from both sides, $4r=4s^2$
Solving for s, we end up with $$s=\sqrt{r}$$. Next, we can find the volume of the frustum and of the sphere. Since we know $V_{frustum}=2V_{sphere}$, we can solve for $s$ using $V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)$ we get: $$V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)$$ Using $V_{sphere}=\dfrac{4r^{3}\pi}{3}$ , we get $$V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}$$ so we have: $$\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}$$ Dividing by $\frac{2\pi*\sqrt{r}}{3}$, we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ $r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}$ , so $$r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}$$ | 2022-05-17T08:59:22 | {
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https://physicscatalyst.com/mech/two-dimensional-motion-with-constant-acceleration.php | # Motion in a plane with Constant Acceleration
## Motion in a plane with Constant Acceleration
• Motion in two dimension with constant acceleration we we know is the motion in which velocity changes at a constant rate i.e, acceleration remains constant throughout the motion
• We should set up the kinematic equation of motion for particle moving with constant acceleration in two dimensions.
• Equation's for position and velocity vector can be found generalizing the equation for position and velocity derived earlier while studying motion in one dimension
Thus velocity is given by equation
$\vec{v}=\vec{v_0}+ \vec{a}t$ (8)
where
$\vec{v}$ is velocity vector
$\vec{v_0}$ is Initial velocity vector
$\vec{a}$ is Instantaneous acceleration vector
Similarly position is given by the equation
$\vec{r}- \vec{r_0}=\vec{v_0} t+ \frac {1}{2}\vec{a}t^2$ (9)
where $\vec{r_0}$ is Initial position vector
i,e
$r_0=x_0 \hat{i}+y_0 \hat{j}$
and average velocity is given by the equation
$\vec{v_{av}}=\frac {1}{2}(\vec{v}+ \vec{v_0}$ (10)
• Since we have assumed particle to be moving in x-y plane,the x and y components of equation (8) and (9) are
$v_x=v_{x0}+a_xt$ (11a)
$x-x_0=v_{0x}t+ \frac {1}{2}a_xt^2$ (11b)
and
$v_y=v_{y0}+a_yt$ (12a)
$y-y_0=v_{0y} t + \frac {1}{2} a_y t^2$ (12b)
• from above equation 11 and 12 ,we can see that for particle moving in (x-y) plane although plane of motion can be treated as two separate and simultaneous 1-D motion with constant acceleration
• Similar result also hold true for motion in a three dimension plane (x-y-z)
Question
A object starts from origin at t = 0 with a velocity $5.0 \hat{i}$ m/s and moves in x-yunder action of a force which roduces a constant acceleration of ($3.0 \hat{i} + 2.0 \vec{j}$) m/s2
(a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ?
(b) What is the speed of the particle at this time ?
Solution
We know that position of the object is given by
$\vec{r}- \vec{r_0}=\vec{v_0} t+ \frac {1}{2}\vec{a}t^2$
Here $\vec{r_0}$ =0 ( As object starts from Origin)
$\vec{v_0}=5.0 \hat{i}$ m/s
$\vec{a}=3.0 \hat{i} + 2.0 \vec{j}$ m/s2
So
$\vec{r}=5.0 \hat{i} t + \frac {1}{2} (3.0 \hat{i} + 2.0 \vec{j})t^2$
$= (5t+1.5t^2) \hat {i}+ t^2 \hat{j}$
Now
$r=x \hat{i}+y \hat{j}$
Therefore,
$x(t)=5t+1.5t^2$ and $y(t)=t^2$
Given x=84
so $5t+1.5t^2 =84$
or t=6 sec
Then $y= t^2=36$ m
Now
$\vec{v}=\frac {d \vec{r}}{dt}=\frac {d}{dt}[(5t+1.5t^2) \hat{i} +t^2 \hat{j}] =(5+3t)\hat{i}+2t \hat{j}$
At t=6sec
$\vec{v}=23\hat{i} + 12 \hat{j}$
Speed =|v|=$\sqrt {(23^2+ 12^2) =26$ m/s | 2023-02-04T06:20:04 | {
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http://www.onemathematicalcat.org/Math/Precalculus_obj/divisionAlgorithm.htm | # THE DIVISION ALGORITHM
• PRACTICE (online exercises and printable worksheets)
The prior lesson introduced long division of polynomials.
Using long division, a ‘new name’ is obtained for a fraction of polynomials: $$\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$ For example, long division renames $\displaystyle\frac{\overbrace{x^3-8x+2}^{N(x)}}{\underbrace{x+3}_{D(x)}}$ as $\displaystyle\overbrace{x^2 - 3x + 1}^{Q(x)} + \frac{\overbrace{\ \ -1\ \ }^{R(x)}}{\underbrace{\ x+3\ }_{D(x)}}$ .
A quick-and-easy check gives some confidence that these two expressions are indeed the same:
substituting $\,x = 0\,$ into $\displaystyle \frac{N(x)}{D(x)}$ gives: $\displaystyle \frac{0^3 - 8\cdot 0 + 2}{0 + 3}$ $\displaystyle = \frac 23$ substituting $\,x = 0\,$ into $\displaystyle Q(x) + \frac{R(x)}{D(x)}$ gives: $\displaystyle 0^2 - 3\cdot 0 + 1 + \frac{-1}{0 + 3} = 1 -\frac 13$ $\displaystyle = \frac 23$
The new expression for the fraction that is obtained by long division is often better to work with than the original.
## Terminology used in the Long Division Process
In the long division process, when $\displaystyle\,\frac{N(x)}{D(x)}\,$ is renamed as $\displaystyle\,Q(x) + \frac{R(x)}{D(x)}\,$:
• $N(x)$ is often called the dividend (it's the numerator of the original fraction)
• $D(x)$ is called the divisor (it's what you're dividing by; it's the denominator of the original fraction)
• $Q(x)$ is called the quotient
• $R(x)$ is called the remainder
The Division Algorithm (below) firms up details about division of polynomials.
Here, the notation ‘$\,\text{deg}(P(x))\,$’ is used to denote the degree of a polynomial $\,P(x)\,$.
the Division Algorithm dividing a polynomial by a polynomial
Let $\,N(x)\,$ and $\,D(x)\,$ be polynomials, with $\,D(x)\ne 0\,$. There exist unique polynomials $\,Q(x)\,$ (called the quotient) and $\,R(x)\,$ (called the remainder) such that $$\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$ or, equivalently, $$N(x) = D(x)Q(x) + R(x)$$ where either $R(x) = 0$; or $\deg(R(x)) \lt \text{deg}(D(x))$
## Notes about the Division Algorithm
• What does $\,Q(x)\,$ tell you?
Think of $\,Q(x)\,$ as answering the question: How many times does $\,D(x)\,$ go into $\,N(x)\,$?
• What does $\,R(x)\,$ tell you?
Think of $\,R(x)\,$ as answering the question: How much is left over?
• When is the long division procedure stopped?
As shown in Long Division of Polynomials, the long division process is stopped
when the degree of the expression you get after subtracting is less than the degree of the divisor.
This is the condition ‘$\,\text{deg}(R(x)) < \text{deg}(D(x))\,$’.
• What does $\,R(x) = 0\,$ mean?
If the remainder is zero, then the divisor and quotient are both factors of the numerator—each goes in evenly.
To see this, substitute $\displaystyle\,R(x) = 0\,$ into $\displaystyle\,\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}\,$ to get $\displaystyle\,\frac{N(x)}{D(x)} = Q(x)\,$;
multiplying through by $\,D(x)\,$ gives the equivalent equation $\displaystyle\,N(x) = D(x)Q(x)\,$,
which shows that $\,D(x)\,$ and $\,Q(x)\,$ are both factors of $\,N(x)\,$.
As an example, study the two long division problems below, both of which result in remainders of zero:
$$\frac{x^3 - x^2 + x - 1}{x-1} = x^2 + 1$$ $$\frac{x^3 - x^2 + x - 1}{x^2+1} = x - 1$$ or, equivalently, $$x^3 - x^2 + x - 1 = (x-1)(x^2 + 1)$$
• What happens if the degree of $\,N(x)\,$ is less than the degree of $\,D(x)\,$ in the expression $\displaystyle\,\frac{N(x)}{D(x)}$?
In this case, there is nothing to do!
The quotient is zero, and the remainder is $\,N(x)\,$.
For example: $$\,\frac{\overbrace{x + 1}^{\text{deg}(N(x)) = 1}}{\underbrace{x^2-3}_{\text{deg}(D(x)) = 2}} = \overbrace{\ \ 0\ \ }^{Q(x)} + \frac{\overbrace{x + 1}^{R(x)=N(x)}}{x^2 - 3}\,$$
The interesting application of the division algorithm is when the degree of $\,N(x)\,$ is greater than (or equal to) the degree of $\,D(x)\,$.
In this case, the long division process can be used to find the quotient and remainder.
• I've heard about ‘synthetic division’. What is it?
Synthetic division is an efficient shortcut for a special type of division problem—when the divisor is of the form $\,x + c\,$.
For example, $\displaystyle\frac{\text{a polynomial}}{x + 2}$ or $\displaystyle\frac{\text{a polynomial}}{x - 2}$ could be done using synthetic division.
However, $\displaystyle\frac{\text{a polynomial}}{3x + 2}$ can't be done using synthetic division, since the coefficient of the $\,x\,$ term must be one.
Also, $\displaystyle\frac{\text{a polynomial}}{x^2 + 3}$ can't be done using synthetic division, since the variable is only allowed to be raised to the first power.
Synthetic division is covered in the next section.
Master the ideas from this section | 2018-02-20T21:06:30 | {
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https://math.stackexchange.com/questions/2281180/prob-15-chap-5-in-baby-rudin-prove-that-m-12-leq-m-0m-2-where-m-0-m | # Prob. 15, Chap. 5 in Baby Rudin: Prove that $M_1^2\leq M_0M_2$, where $M_0$, $M_1$, and $M_2$ are the lubs, resp., of …
Here is Prob. 15, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a, \infty)$, and $M_0$, $M_1$, $M_2$ are the least upper bounds of $\left| f(x) \right|$, $\left| f^\prime(x) \right|$, $\left| f^{\prime\prime}(x) \right|$, respectively, on $(a, \infty)$. Prove that $$M_1^2 \leq 4 M_0 M_2.$$
Hint: If $h > 0$, Taylor's theorem shows that $$f^\prime(x) = \frac{1}{2h} \left[ f(x+2h) - f(x) \right] - h f^{\prime\prime}(\zeta)$$ for some $\zeta \in (x, x+2h)$. Hence $$\left| f^\prime(x) \right| \leq h M_2 + \frac{M_0}{h}.$$
To show that $M_1^2 = 4M_0M_2$ can actually happen, take $a = -1$, define $$f(x) = \begin{cases} 2x^2 - 1 \ & \ (-1 < x < 0), \\ \frac{x^2 - 1 }{ x^2 + 1} \ & \ (0 \leq x < \infty), \end{cases}$$ and show that $M_0 = 1$, $M_1 = 4$, $M_2 = 4$.
Does $M_1^2 \leq 4 M_0 M_2$ hold for vector-valued functions too?
My Attempt:
Let $x > a$ and $h > 0$. Then by Taylor's theorem (i.e. Theorem 5.15 in Baby Rudin, 3rd edition), we can find a real number $\zeta \in (x, x+2h)$ such that $$f(x+2h) = f(x) + 2h f^\prime(x) + \frac{(2h)^2}{2!} f^{\prime\prime}(\zeta) = f(x) + 2h f^\prime(x) + 2h^2 f^{\prime\prime}(\zeta),$$ and solving for $f^\prime(x)$ we obtain $$f^\prime(x) = \frac{1}{2h} \left[ f(x+2h) - f(x) \right] - h f^{\prime\prime}(\zeta),$$ and hence \begin{align} \left| f^\prime(x) \right| &\leq \frac{1}{2h} \left[ \left| f(x+2h) \right| + \left| f(x) \right| \right] + h \left| f^{\prime\prime}(\zeta) \right| \\ &\leq \frac{1}{2h} \left( M_0 + M_0 \right) + h M_2 \\ &= \frac{ M_0}{h} + h M_2. \end{align} Thus for all $x \in (a, \infty)$ and for all $h > 0$, we see that $$\left| f^\prime(x) \right| \leq M_2 h + \frac{M_0}{h}.$$ Now keeping $h$ fixed and taking the supremum of the left-hand side over all $x \in (a, \infty)$, we obtain $$M_1 \leq h M_2 + \frac{M_0}{h} \ \mbox{ for all h > 0. } \tag{1}$$ Now if $M_2 = 0$, then (1) yields $$0 \leq M_1 \leq \frac{M_0}{h} \ \mbox{ for all h > 0. } \tag{2}$$ But $$\lim_{h \to \infty} \frac{M_0}{h} = 0;$$ so from (2) we can conclude that $M_1 = 0$ also and hence $$M_1^2 = 0 = 4M_0 M_2,$$ as required. Am I right?
So let us suppose that $M_2 > 0$, and let us define a function $\varphi$ on $(0, \infty)$ by the formula $$\varphi(h) \colon= h M_2 + \frac{M_0}{h} \ \mbox{ for all h > 0. }$$ Then $$\varphi^\prime(h) = M_2 - \frac{M_0}{h^2} = \frac{M_2}{h^2} \left( h^2 - \frac{M_0}{M_2} \right) \ \mbox{ for all h > 0. }$$ Thus, $$\varphi^\prime(h) \begin{cases} > 0 \ & \ \mbox{ if } 0 < h < \sqrt{\frac{M_0}{M_2}}, \\ = 0 \ & \ \mbox{ if } h = \sqrt{\frac{M_0}{M_2}}, \\ < 0 \ & \ \mbox{ if } h > \sqrt{\frac{M_0}{M_2}}. \end{cases}$$ Thus by Theorem 5.11 in Baby Rudin, our $\varphi$ is monotonically (in fact, strictly) decreasing for $0 < h \leq \sqrt{\frac{M_0}{M_2}}$ and monotonically (in fact strictly) increasing for $h \geq \sqrt{\frac{M_0}{M_2}}$. Therefore $\varphi$ has a local minimum at $h = \frac{M_0}{M_2}$. Moreover, since $$\varphi \left(\sqrt{\frac{M_0}{M_2}} \right) = M_2 \sqrt{\frac{M_0}{M_2}} + \frac{M_0}{\sqrt{\frac{M_0}{M_2}}} = 2 \sqrt{M_0 M_2}$$ is the only local extreme value that $\varphi$ has, so it is in fact the absolute minimum value of $\varphi$. Therefore we can conclude that $$\varphi \left( \sqrt{ \frac{M_0}{M_2} } \right) = 2 \sqrt{M_0 M_2} \leq \varphi(h) \ \mbox{ for all } h > 0. \tag{3}$$ And, since $\sqrt{\frac{M_0}{M_2}} > 0$, therefore from (1) and (3) we can also conclude that $$0 \leq M_1 \leq 2 \sqrt{M_0 M_2},$$ which (upon squaring both sides) yields $$M_1^2 \leq 4 M_0 M_2,$$ as required. Am I right?
Now let $f$ be defined on $(-1, \infty)$ by the formula $$f(x) = \begin{cases} 2x^2-1 \ & \ \mbox{ if } -1 < x < 0, \\ \frac{x^2-1}{x^2+1} = 1 - \frac{2}{x^2+1} \ & \ \mbox{ if } 0 \leq x < \infty. \end{cases}$$ Then $$f^\prime(x) = \begin{cases} 4x \ & \ \mbox{ if } -1 < x < 0, \\ 0 \ & \ \mbox{ if } x = 0, \\ \frac{4x}{(x^2+1)^2} \ & \ \mbox{ if } 0 < x < \infty. \end{cases} \tag{*}$$ At $x= 0$, we see that $$\lim_{h \to 0+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0+} \frac{ \frac{h^2-1}{h^2+1} - (-1) }{h} = \lim_{h \to 0+} \frac{2h}{h^2+1} = 0,$$ and $$\lim_{h \to 0-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0-} \frac{ 2h^2-1 - (-1)}{h} = \lim_{h \to 0-} 2h = 0.$$ Therefore $f^\prime(0) = 0$.
As $f^\prime$ is negative in $(-1, 0)$, zero at $x = 0$, and positive on $(0, \infty)$, so $f$ is (strictly) decreasing on $(-1, 0]$ and (strictly) increasing on $[0, \infty)$. Therefore $f$ has a local minimum value at $x=0$, and $f(0) = -1$, being the unique extreme value, is also the absolute minimum value of $f$. Thus $$f(x) \geq -1 = f(0) \ \mbox{ for all } x \in (-1, \infty). \tag{4}$$
On the other hand, we note that $f$ is strictly increasing on $[0, \infty)$, and $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2-1}{x^2+1} = \lim_{x \to \infty} \frac{ 1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} = \frac{1-0}{1+0} = 1.$$ So we can conclude that $$f(x) < 1 \ \mbox{ for all } x \in [ 0, \infty) \tag{5}$$ And, as $f$ is strictly decreasing on $(-1, 0]$ and as $$\lim_{x \to -1+} f(x) = \lim_{x \to -1+} \left( 2x^2-1 \right) = 1,$$ so we can conclude that $$f(x) < 1 \ \mbox{ for all } x \in (-1, 0]. \tag{6}$$ From (4), (5), and (6), we can conclude that $$\left| f(x) \right\vert \leq 1 = \left| f(0) \right| = \left| f(0) \right| \ \mbox{ for all } x \in (-1, \infty). \tag{7}$$ Thus we have $M_0 = 1$.
From (*) above, we find that $$f^{\prime\prime}(x) = \begin{cases} 4 \ & \ \mbox{ if } -1 < x \leq 0, \\ \frac{4}{(x^2+1)^2} - \frac{16x^2}{(x^2+1)^3} = \frac{4\left( 1-3x^2 \right) }{(x^2+1)^3} \ & \ \mbox{ if } 0 < x < \infty. \end{cases} \tag{**}$$ At $x = 0$, we find that $$\lim_{h \to 0+} \frac{ f^\prime(h) - f^\prime(0)}{h} = \lim_{h \to 0+} \frac{4}{(h^2+1)^2} = 4,$$ and $$\lim_{h \to 0-} \frac{ f^\prime(h) - f^\prime(0) }{h} = \lim_{h\to 0-} \frac{4h}{h} = 4,$$ and hence $f^{\prime\prime}(0) = 4$.
Now from (**) we see that $$f^{\prime\prime}(x) \mbox{ is } \begin{cases} \mbox{ positive } \ & \mbox{ if } -1 < x < \frac{1}{\sqrt{3}}, \\ \mbox{ zero } \ & \mbox{ if } x = \frac{1}{ \sqrt{3}}, \\ \mbox{ negative } \ & \mbox{ if } \frac{1}{\sqrt{3}} < x < \infty. \end{cases}$$ Thus, $f^\prime$ is strictly increasing on $\left(-1, \frac{1}{\sqrt{3}} \right]$ and strictly decreasing on $\left[ \frac{1}{\sqrt{3}}, \infty \right)$. So $f^\prime$ has a local maximum value at $x = \frac{1}{\sqrt{3}}$, which, being the only extreme value of $f^\prime$, is also the absolute maximum value. Thus, $$f^\prime \left( \frac{1}{\sqrt{3}}\right) = \frac{ \frac{4}{\sqrt{3}} }{ \left( \frac{1}{3} + 1 \right)^2 } = \frac{ \frac{4}{\sqrt{3}} }{ \frac{16}{9} } = \frac{3 \sqrt{3}}{4},$$ and $$f^\prime(x) \leq f^\prime \left( \frac{1}{\sqrt{3}}\right) = \frac{3 \sqrt{3}}{4} \ \mbox{ for all } x \in (-1, \infty). \tag{8}$$ From (*) we also note that $f^\prime(x) \to -4$ as $x \to -1+$, which together with (8) implies that $$-4 < f^\prime(x) \leq \frac{3\sqrt{3}}{4} \ \mbox{ for all } x \in (-1, \infty),$$ and so $\left| f^\prime(x) \right| \to 4$ as $x \to -1+$, which implies that our $M_1 = 4$. Am I right?
Now form (**) we see that $$f^{(3)}(x) = \begin{cases} 0 \ & \ \mbox{ if } -1 < x \leq 0, \\ -\frac{24x}{(x^2+1)^3} + \frac{24x(3x^2-1) }{((x^2+1)^4 } = \frac{ 48x \left( x^2-1 \right)}{(x^2+1)^4} \ & \mbox{ if } 0 < x < \infty. \end{cases} \tag{***}$$ At $x = 0$, we find that $$\lim_{h \to 0-} \frac{ f^{\prime\prime}(h) - f^{\prime\prime}(0)}{h} = \lim_{h \to 0-} \frac{4-4}{h} = 0,$$ and $$\lim_{h \to 0+} \frac{ f^{\prime\prime}(h) - f^{\prime\prime}(0) }{h} = 4 \lim_{h \to 0+} \frac{ 1- 3h^2 - (h^2+1)^3 }{ h (h^2+1)^3} = 4 \lim_{h \to 0+} \frac{h^5-3 h^3- 6h }{ (h^2+1)^3} = 0,$$ showing that $f^{(3)}(0) = 0$.
Thus we find that $$f^{(3)}(x) \ \mbox{ is } \begin{cases} \mbox{ negative } \ & \mbox{ if } 0 < x < 1, \\ \mbox{ zero } \ & \mbox{ if } -1 < x \leq 0 \ \mbox{ or if } x =1, \\ \mbox{ positive } \ & \mbox{ if } 1 < x < \infty. \end{cases}$$ Thus $f^{\prime\prime}$ is strictly decreasing on $[0, 1]$ and strictly increasing on $[1, \infty)$. So $f^{\prime\prime}$ has a local minimum at $x=1$, and $f^{\prime\prime}(1) = -1$, and moreover $f^{\prime\prime}(x) \to 0$ as $x \to \infty$. So $$0 \leq \left| f^{\prime\prime}(x) \right| \leq 4,$$ and $4 = f^{\prime\prime}(0)$. So our $M_2 = 4$, as required. Am I right?
Is whatever I've done so far correct? Is my approach correct? And if so, then is my solution correct and lucid enough in its presentation?
What is the situation for vector-valued functions?
• could be streamlined a bit but otherwise this looks good (+1) – tired May 14 '17 at 22:30
• Wow you typed a lot – user217285 May 14 '17 at 23:02
• For vector valued function, you might consider the function defined by $g(x)=f(x)\cdot c$, where $c$ is a constant vector, then apply the result to $g$, and use Cauchy-Schwartz inequality. – Li Chun Min May 14 '17 at 23:35
• @tired can you please point out exactly where my proof is in need of streamlining? I'd be grateful. – Saaqib Mahmood May 15 '17 at 13:30 | 2019-06-17T09:28:48 | {
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http://followingtherules.info/getting-to-the-root.html | ## 9.17 Getting to the Root
We are used to using a slide rule to perform simple operations on numbers, such as obtaining values of trigonometric functions, reciprocals and squares and cubes of numbers, and performing relevant multiplication and division calculations and so on. Here we give a standard, but often overlooked technique for solving a general quadratic equation through the use of a slide rule.
Consider the general quadratic equation of variable $$x$$, expressed as
$a x^2 + b x + c = 0$
where the coefficients $$a$$, $$b$$, and $$c$$ are real numbers. It can be shown that for the case where the combination (called the discriminant) $$b^2 - 4ac > 0$$, then two real solutions exist that satisfy the above equation. We will look at a couple of examples of where such solutions can be found using the slide rule.
Square Root with C/D Scales
One often obtains the square root of a number using a Mannheim slide rule, for example, by locating the number on the A scale and reading its square root on the D scale. But what if for some reason there were no A/B scales. We can still solve the problem using the C and D scales alone. So let’s solve for the positive root of $$x^2 = c$$, or $$x=\sqrt{c}$$, using the C and D scales:
• Move the cursor to $$c$$ on the D scale.
• Next, move the slide until the number on C under the cursor is the same as the number on D under the C index. As the rule is now set to multiply a number by the same number, this “same number” is $$x$$!
Here is a calculation of the square root of 29.4:
By computer, the square root of 29.4 is 5.422.
$$~$$
Square Root with D and CI Scales
Next, note that $x^2 = c \longrightarrow x = c/x$ and hence $\log x = \log c + \log(1/x), ~~~ or ~~~ \log x - \log(1/x) = \log c.$
So, if there happens to be a CI scale on the slide of the rule, then:
• Align the index of CI with the number $$c$$ on the D scale. Or, equivalently, align the number $$c$$ on the CI scale with the index on the D scale. The two operations should give the same setting of the slide.
• Visually scan the two scales D and CI using the cursor to find where the same number on each lines up. This “same number” is, again, $$x$$. In this case, the rule is set to divide $$x$$ by $$1/x$$ to produce the value $$c$$! In our previous example, note that if 29.4 is on CI at the D index, then 29.4 will also be on the D scale at the CI index.
If there is no CI scale on your slide rule, you can perhaps remove and re-insert the C scale upside down to produce the same effect.
$$~$$
Now take the general quadratic equation we started with at the beginning of this story. It can always be reduced to a form where the coefficient of $$x$$ is equal to 1 (i.e., by dividing through by $$a$$) and hence another form of the quadratic equation would be
$x^2 + gx +h = 0$
where $$g=b/a$$ and $$h=c/a$$. The solutions to this equation will be the same as the solutions to our original equation.
To solve this new equation on a slide rule, we will use the D and CI scales. (Again, the C scale can be reversed if there is no CI scale on the rule.) First, we note that this equation can also be written in terms of its solutions $$u$$ and $$v$$ in the following way, $(x-u)(x-v) = 0$ from which $x^2 -(u+v)x + uv = 0.$ Comparing to our original form, we see that $$u+v = -g$$ and $$u\times v = h$$. So the strategy is to search for two numbers $$u$$ and $$v$$ that satisfy these two conditions:
• Move the slide to align the number $$h$$ on the CI scale with the index on the D scale.
• Move the cursor to where the numbers on the CI and D scales add or subtract to give $$g$$.
• These two numbers are $$u$$ and $$v$$, where a possible minus sign may need to be inferred.
The technique may not be so “automatic,” but with a little practice and patience (as it may take a few iterations to find this special combination) the solution can be found on the rule.
For example, let’s solve the following: $$2x^2 + 3x - 10 = 0$$.
• First, check the discriminant: $$b^2 - 4ac$$ = $$3^2 -4\times 2\times(-10)$$ = $$9+80 >0.~~~~$$ OK. We can do this!.
• Next, we divide everything by 2 to get: $$x^2 + 1.5x - 5 = 0$$.
• Now move the slide to align 5 on the CI scale with the index on the D scale.
• Since $$h$$ in our equation is negative, then one of our two roots will be positive and one will be negative. Thus, we will be looking for two numbers that subtract to make our value of $$-g$$ = 1.5.
• Sliding the cursor, we look for numbers on the CI and D scales at the cursor that add or subtract to 1.5. As expected, we don’t see any numbers adding to 1.5, but a close possibility of a difference of 1.5 is found near the combination of 1.6 and 3.1.
• These two numbers do not quite line up, but a slight tweak of the cursor finds that we can see a more accurate combination that appears to line up: 1.61 and 3.11.
• Since $$g = -(u+v)$$ is $$+1.5$$ in our equation, then the larger of our two numbers must be the negative one, and so our two solutions must be $$x$$ = -3.11 and +1.61.
Here are the roots to our original equation, via computer,
# coeff's for this function are listed in order "c, b, a"
Re(polyroot( c(-10, 3, 2) ))
## [1] 1.608495 -3.108495
Similar techniques using A, K, and BI scales can even be used to solve certain classes of cubic and higher-order equations. Discussions of such techniques can be found through standard on-line searches. | 2022-09-27T11:31:30 | {
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https://www.toktol.com/notes/context/2056/maths/bijections-and-compositions/composition-is-not-commutative | Use adaptive quiz-based learning to study this topic faster and more effectively.
# Composition is not commutative
Contrary to the properties of sums and products, order matters in composition. In general, $$f\circ g\ne g\circ f.$$ We say that composition is not commutative.
Setting $f(x) = x^2$ and $g(x) = x+1$, we see that, for $x\ne0$, $$f\circ g(x) = (x+1)^2\ne x^2+1 = g\circ f(x).$$
However, for the inverse function, we have $$f\circ f^{-1}(y) =y,\qquad f^{-1}\circ f(x) =x$$ This comes from $y=f(x)\Longleftrightarrow x=f^{-1}(y)$.
The composition of monotonic functions is monotonic. If both functions have same monotonicity (both increasing or decreasing), the composite function is increasing; otherwise it is decreasing.
To see this, assume that both $f$ and $g$ are decreasing. Then, for $x\le y$, we have $g(x)\ge g(y)$, hence $f\left(g\left(x\right)\right)\le f\left(g\left(x\right)\right)$. This means that means that $f\circ g$ is increasing. | 2018-01-20T21:27:38 | {
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http://zbmath.org/?q=an:0739.05007 | # zbMATH — the first resource for mathematics
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An algorithmic proof theory for hypergeometric (ordinary and “$q$”) multisum/integral identities. (English) Zbl 0739.05007
It is shown that every ‘proper-hypergeometric’ multisum/integral identity, or $q$-identity, with fixed number of summation and/or integration signs, possesses a short, computer-constructible proof. We give a fast algorithm for finding such proofs. Most of the identities that involve the classical special functions of mathematical physics are readily reducible to the kind of identities treated here. We give many examples of the method, including computer-generated proofs of identities of Mehta-Dyson, Selberg, Hille-Hardy, $q$-Saalschütz, and others. The prospect of using the method for proving multivariate identities that involve an arbitrary number of summations/integrations is discussed.
Reviewer: H.S.Wilf
##### MSC:
05A19 Combinatorial identities, bijective combinatorics 05A10 Combinatorial functions 11B65 Binomial coefficients, etc. 05A30 $q$-calculus and related topics 33C99 Hypergeometric functions 39A10 Additive difference equations | 2014-04-17T06:51:15 | {
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https://www.coursehero.com/file/49457385/02-PPT-63-Riemann-Sums-and-Definite-Integralspptx/ | 02 - PPT - 6.3 Riemann Sums and Definite Integralsx - MATH 12 6.3 Riemann Sums and Definite Integrals � ∗ � � lim ∑ � � ∆ �=� �
# 02 - PPT - 6.3 Riemann Sums and Definite Integralsx -...
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6.3 Riemann Sums and Definite Integrals lim ? → ∞ ? = 1 ? ? ( ? ? ) ? = ? ? ? ( ? ) ?? 10/29/19 MATH 12
Lesson Objective By the end of the lesson, I will be able to… Relate Riemann sums and definite integrals
Riemann Sum The Riemann sum helps us find the area under a curve using n rectangles Height Width Where is Right hand endpoint Left hand endpoint Midpoint rule Right hand endpoint Left hand endpoint Midpoint rule # of rectangles Sum of…
Ex1. Evaluate Riemann sum for…a)taking sample points to be right hand endpoints andwith 4 rectanglesb)taking sample points to be left hand endpoints and with3 rectangles
is…
Note on integrals Area above -axis = + Area below -axis = - + - + ? 1 ? 2 ? 3
Comparing… Riemann Sum Integral If , the Riemann sum is the sum of the areas of n rectangles. If , is the area under the curve from to (using an infinite number of rectangles) .
Ex1. Express the following as integrals a) on the interval b) on the interval ? ? ? ( ? ) ?? = lim ? →∞ ? = 1 ? ? ( ? ? ) ? 2 1 1 lim [ ] 2 n i i n i x x x 2 1 1 lim [ 3 ] n i n i i x x x
Ex3. Use the midpoint rule with… a)to approximateb)to approximate
Properties of Definite Integrals
Properties of Definite Integrals e.g.
e.g.e.g.
Properties of Definite Integrals e.g. Given 5 8 8 5 ( ) 6, ( ) 3, ( ) f x dx f x dx find f x dx
Isaac Barrow “Differentiation and integration are inverse processes”
Lesson Objective By the end of the lesson, I will be able to… Relate Riemann sums and definite integrals
Suggested Practice Page 428 # 39 – 43 (odd #s) Riemann sum and Definite Integrals worksheet #5, 6, 7, 13 Draw it first! Vocabulary: definite integral, Riemann sums, integration | 2021-07-25T13:15:30 | {
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https://www.math.tolaso.com.gr/?author=1 | Home » Articles posted by Tolaso
Author Archives: Tolaso
Convergence of series
For we define and . For which does the series
converge?
Solution
We may assume that is not the zero vector and otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.
(a) If the above condition is satisfied then, without loss of generality, let be the component of maximal absolute value and let . Hence , as
and the given series is convergent because .
(b) If the above condition is not satisfied, then there are at least components of maximal absolute value and therefore
and the given series is not convergent because .
Inequality
Let be endowed with the usual product and the usual norm. If then we define . Prove that
Huygen’s Inequality
Let . Prove that
Solution
Let . is twice differentiable with
Hence is convex. The tangent at has equation . The result follows.
Sigma divisor sum
Let denote the divisor function. Prove that
Solution
We have successively:
Derivative at 0
Let
Evaluate .
Solution
The domain of is . Let us consider the logarithmic of
and differentiate it; Hence,
For we have
Thus, since . | 2020-10-28T23:23:36 | {
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http://en.wikibooks.org/wiki/Measure_Theory/Basic_Structures_And_Definitions/Measures | # Measure Theory/Basic Structures And Definitions/Measures
In this section, we study measure spaces and measures.
## Measure Spaces
Let $X$ be a set and $\mathcal{M}$ be a collection of subsets of $X$ such that $\mathcal{M}$ is a σ-ring.
We call the pair $\left\langle X,\mathcal{M}\right\rangle$ a measure space. Members of $\mathcal{M}$ are called measurable sets.
A positive real valued function $\mu$ defined on $\mathcal{M}$ is said to be a measure if and only if,
(i)$\mu (\varnothing)=0$ and
(i)"Countable additivity": $\mu\left(\bigcup_{i=1}^\infty E_i\right) = \sum_{i=1}^\infty \mu(E_i)$, where $E_i\in\mathcal{M}$ are pairwise disjoint sets.
we call the triplet $\left\langle X,\mathcal{M},\mu\right\rangle$ a measurable space
A probability measure is a measure with total measure one (i.e., μ(X)=1); a probability space is a measure space with a probability measure.
## Properties
Several further properties can be derived from the definition of a countably additive measure.
### Monotonicity
$\mu$ is monotonic: If $E_1$ and $E_2$ are measurable sets with $E_1\subseteq E_2$ then $\mu(E_1) \leq \mu(E_2)$.
### Measures of infinite unions of measurable sets
$\mu$ is subadditive: If $E_1$, $E_2$, $E_3$, ... is a countable sequence of sets in $\Sigma$, not necessarily disjoint, then
$\mu\left( \bigcup_{i=1}^\infty E_i\right) \le \sum_{i=1}^\infty \mu(E_i)$.
$\mu$ is continuous from below: If $E_1$, $E_2$, $E_3$, ... are measurable sets and $E_n$ is a subset of $E_{n+1}$ for all n, then the union of the sets $E_n$ is measurable, and
$\mu\left(\bigcup_{i=1}^\infty E_i\right) = \lim_{i\to\infty} \mu(E_i)$.
### Measures of infinite intersections of measurable sets
$\mu$ is continuous from above: If $E_1$, $E_2$, $E_3$, ... are measurable sets and $E_{n+1}$ is a subset of $E_n$ for all n, then the intersection of the sets $E_n$ is measurable; furthermore, if at least one of the $E_n$ has finite measure, then
$\mu\left(\bigcap_{i=1}^\infty E_i\right) = \lim_{i\to\infty} \mu(E_i)$.
This property is false without the assumption that at least one of the $E_n$ has finite measure. For instance, for each nN, let
$E_n = [n, \infty) \subseteq \mathbb{R}$
which all have infinite measure, but the intersection is empty.
## Counting Measure
Start with a set Ω and consider the sigma algebra X on Ω consisting of all subsets of Ω. Define a measure μ on this sigma algebra by setting μ(A) = |A| if A is a finite subset of Ω and μ(A) = ∞ if A is an infinite subset of Ω, where |A| denotes the cardinality of set A. Then (Ω, X, μ) is a measure space. μ is called the counting measure.
## Lebesgue Measure
For any subset B of Rn, we can define an outer measure $\lambda^*$ by:
$\lambda^*(B) = \inf \{\operatorname{vol}(M) : M \supseteq B \}$, and $M \$ is a countable union of products of intervals .
Here, vol(M) is sum of the product of the lengths of the involved intervals. We then define the set A to be Lebesgue measurable if
$\lambda^*(B) = \lambda^*(A \cap B) + \lambda^*(B - A)$
for all sets B. These Lebesgue measurable sets form a σ-algebra, and the Lebesgue measure is defined by λ(A) = λ*(A) for any Lebesgue measurable set A. | 2014-09-23T18:25:23 | {
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https://franklin.dyer.me/post/83 | ## Roulettes, non-Russian
2017 July 3
As stated in the title, I'm not talking about a game of chance that you play with old-fashioned pistols. Nor am I talking about the casino game. In the mathematics of curves, a roulette is what you get when you roll one curve along another, and trace a point on or inside of the curve as it is rolled. One example of a roulette is a cycloid, which is obtained by rolling a circle along a straight line and tracing a point on its circumference. This is what a cycloid looks like:
We will be deriving the parametric equation for the cycloid, as well as the parametric equations for many other interesting roulettes. But first, something a little more basic. We will start with the roulettes formed by rolling a regular polygon along a line. These are a good place to start, as they are much easier to work with than some other roulettes. The problem to which we will be turning is the length of the path traced by a vertex of a polygon in one "cycle" of its roulette.
The way that we would generate such a roulette is by placing a regular polygon of side length $l$ into the coordinate plane so that the point which we are tracing is on top of the origin and one side of the polygon containing the tracing point is laid over the $x$-axis, covering the length from $x=0$ to $x=l$. Then, to roll the polygon, we would "pivot" it about its rightmost point resting on the $x$-axis until another point touches the $x$-axis, and then we would pivot about that one, and so on, until all vertices of the polygon have been "pivoted about". Here are the pictures of the cycloids formed by an equilateral triangle, a square, and a regular pentagon:
This is more of a geometry problem than anything else, other than a partial sum involving trigonometric functions that we will have to evaluate. As you can see from the pictures, each $n$-gonal cycloid is basically composed of $n-1$ circle sectors attached to each other. A bit of further observation reveals that each circle sector measures $\frac{2\pi}{n}$ radians.
First let's just calculate the length for the triangular cycloid as an easy example. We can see that the angle of each sector is $\frac{2\pi}{3}$ radians, and the radius of each one is equal to the side length of the triangle, the length of the total path is
Now for the square. This time we have three sectors. The radii of the two smaller ones are equal to the side length $l$ of the square, and the radius of the larger one is equal to the diagonal length of the square, or $l\sqrt 2$. The angle of each sector is $\frac{\pi}{2}$, and so the total length is
Now we are ready to generalize. Since each circle is formed by rotating one vertex of the polygon (namely, the tracing point) about another, then the radii of the circles are equal to the lengths of the diagonals of the polygon. Thus the total length is equal to We can extract the constant factor from the summation and simplify it to Where $d_x$ is the length of the xth diagonal counterclockwise from the tracing point. If we can find $d_x$ as a function of $x$ and $l$, then we may be able to simplify our sum even further.
Using a bit of geometry, we can find that the angle formed by joining the tracing point to the center of the polygon and the xth vertex counterclockwise measures $\frac{2\pi x}{n}$ radians. Furthermore, if we join the tracing point to that vertex to form the xth diagonal, we then have a isocscles triangle with angle $\frac{2\pi x}{n}$ opposite its base and two congruent sides that join the center of the polygon to two different vertices of the polygon. Again, using a bit of geometry, we can find that the length of the line segment joining the center of a regular $n$-gon with side length $l$ is given by and so now we have an isoscles triangle measuring as shown, and we need to find the base length:
I'll spare you the details of the rest of the geometry. It turns out that the diagonal length turns out the be
and that means that we can now transform our sum into
Luckily, we learned how to evaluate this type of sum in my post regarding non-telescoping sums. The value of the summation is equal to the imaginary part of And now we can readily sum these terms using the formula used for the sum of the terms of a geometric sequence: now we just need to find the imaginary part. and so the imaginary part is We can now simplify this down a little bit:
And so the length of the path traced is
Now onto something a bit more interesting. We will now derive the parametric equation of a cycloid.
A cycloid is typically formed by a circle with radius $r$ starting with its center at $(0,r)$ and the tracing point at the origin. There is one further given that we must accept - the idea that the length of the segment of the $x$-axis rolled over by the circle at some time $t$ is always equal to the arclength of the circle that has been rolled on. This is fundamental to the concept of roulettes in general, and this law can be generalized by replacing the statement "segment of the $x$-axis" with "segment of the curve rolled on" and "circle" with "rolling curve".
Okay, time to derive the parametric equation! Let $r$ be radius of the circle and let the circle turn $1$ radian per unit of time, so that the total angle that it has turned is $t$. Thus we can construct the following diagram for any time $t$:
Before we derive the parametric equation for the cycloid using this diagram, I would like to say a word about a method that I call "relative coordinates". Often, when dealing with tranformations and parametric equations, this strategy can be very useful. Here is an example of how it is used.
Suppose we want to find the image of the point $(3,3)$ dilated about the point $(2,5)$ by a factor of $2$. It is very tiresome to dilate points about points, but if we were dilating about the origin, it would be easy - we would just have to multiply the coordinates by two! So that is what we will do. We will translate both points in such a way that $(2,5)$ maps onto the origin, and then we will proceed with the dilation, and then we will move $(2,5)$ back to its original position. Thus we will translate both points $2$ units left and $5$ units down, moving $(2,5)$ to $(0,0)$ and $(3,3)$ to $(1,-2)$. Now we can dilate, and now we need only double the abscissa and ordinate of $(1,-2)$ to get $(2,-4)$. Now we translate everything back to where it was before, by shifting both points right $2$ units and back up $5$ units. $(0,0)$ goes back to $(2,5)$ and our dilated point $(2,-4)$ becomes $(4,1)$. That wasn't so bad!
We can use the same process for parametric equations. It is often overcomplicated to go directly to finding the coordinates of the point that we wish to trace. We must often first choose some "reference point" that has coordinates that are simpler to find. Then if we find the coordinates of the reference point as a function of $t$ and then find the relationship between the reference point and the tracing point, we can "combine" these to get our final parametric equation.
Time to derive the parametric equation of the cycloid! Here's our beautiful diagram again:
From the diagram, we can see that an obvious reference point would be the center of the rolling circle. The center of the circle obviously has the coordinates Now we need to find the coordinates of the tracing point relative to the center of the circle - that is, what the coordinates of the tracing point would be if the center of the circle was always at the origin. If we suppose that the center of our circle is at the origin, then we get something like this:
This should be familiar to you - it's basically the unit circle dilated by a factor of $r$!
As you should know, the parametric equation for the normal circle (starting with a point at $(r,0)$ and then rotating counterclockwise) is but our circle is a little different. It starts at $(0,-r)$ and travels clockwise rather than counterclockwise. Thus the angle at which it starts is $\frac{\pi}{2}$ less than the starting angle of the normal circle. Furthermore, since it is rotating clockwise rather than counterclockwise, the $x$ coordinate of the point is reflected. Thus the parametric equation of our special circle is or When we came up with this, we were treating the center of our circle as if it was the origin. Now we must translate it back to $(rt,r)$, and, in the process, translate our tracing point to And this is it - this is the parametric equation for a cycloid! We did it!
Now we shall find the length of one revolution of the cycloid, from $t=0$ to $t=2\pi$.
But wait... that would require us to evaluate a messy integral of a parametric equation. There's a much easier way to do this using a formula that we have already derived: the formula for the length of the path traced by a polygon rather than a circle. We can also use the fact that as a polygon with apothem $r$ gains more and more sides, it approaches a circle with radius $r$. So suppose a polygon has apothem $r$ and $n$ sides. Then (using a little geometry) we can say that the side length of the polygon is and so we can use our previous formula to say that the length of a cycloid is If we make the substitution $n \to \frac{1}{n}$, we get And since $\lim_{n \to 0} \sec(n \pi)=1$, we have And with the substitution $n \to \frac{2}{\pi}n$, we have
And so the length of one revolution of a cycloid is $8$ times the radius of the circle that generates it. What a beautiful property!
Now it is time for something trickier - the hypocycloid. This is slightly different from the regular cycloid. It still features a rolling circle... but the circle now rolls along another larger circle, rather than along a line. In a hypocycloid, a smaller circle rolls along the inside of a larger one:
Let $t$ be the amount of arclength rolled on by the smaller circle, and let $\alpha$ be the angle formed by joining the origin to both the center of the rolling circle and the point at which its center started. Then the amount of arclength rolled over on the large one must be equal to the arclength rolled on by the circle, which means that and Now, as a reference point to the tracing point, we can use the center of the small circle. When traced, it forms a circle of radius $R-r$, which is easy to see, and since the angle that it makes with the origin and the starting position is $\alpha$, we have that its coordinates are or and, relative to the center of the small circle (if it was the origin), the relative coordinates of the tracing point are or and so, when we translate the origin back to the center of the small circle where it should be, the coordinates of the tracing point become
And this is it - the general parametric equation for a hypocycloid. Here are some pictures of hypocycloids:
The shape made by the first is called a deltoid, the shape made by the second is called an astroid, and the third is just a random cool-looking hypocycloid.
Just one more thing before I end this blog post.
We've been working with rolling circles so far, but now I'd like to show you what happens when we try to roll a line along another curve, and trace a point on it. The curve generated by this process is called the involute of the curve along which the line is rolled.
How would this work? Basically, if we were rolling a line along the curve $y=f(x)$, we would let $t$ be the $x$ coordinate at which the line touches the curve. Then the equation of the line would basically just be the tangent to the curve at that point, and the tracing point would appear some length $L$ away from the point of tangency along the line, where $L$ is the length of the curve from $x=0$ to $x=t$. Recall the formula for the arclength of a curve from $x=0$ to $x=t$:
and so we can construct the following diagram:
We can see that the slope of the tangent line is $f'(t)$ and so its angle of elevation is $\arctan f'(t)$. This means that, if we let the tracing point be $(x,y)$, we can construct a right triangle and say that and
And that's it! Our parametric equation is
Here are a few interesting curve involutes. The original curves are in red and their involutes are in blue. They are, respectively, a parabola, an exponential function, and a sine wave:
If you want to play around with this a little more, I have created, at desmos.com, a graph in which you can input any function whose involute you wish to see, and there you can watch the line roll along it, which should help with any trouble understanding what it means for a line to "roll" along a curve. You can find it at this link. Have fun!
Well, that concludes this post. I will definitely return to this topic later, though - there is much more to learn about the involutes of curves. For examples, our formula does not apply to parametric equations, who also have involutes. Furthermore, you don't just have to roll lines along curves - you can roll curves along other curves. But I guess I'll just have to put that off until later and end this post with a cliff hanger... | 2021-05-09T14:22:35 | {
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https://encyclopediaofmath.org/wiki/First_fundamental_form | # First fundamental form
metric form, of a surface
A quadratic form in the differentials of the coordinates on the surface that determines the intrinsic geometry of the surface in a neighbourhood of a given point.
Let the surface be defined by the equation
$$\mathbf r = \mathbf r ( u , v),$$
where $u$ and $v$ are coordinates on the surface, while
$$d \mathbf r = \mathbf r _ {u} du + \mathbf r _ {v} dv$$
is the differential of the position vector $\mathbf r$ along the chosen direction $du :dv$ of displacement from a point $M$ to an infinitesimally close point $M ^ \prime$( see Fig. a).
Figure: f040450a
The square of the principal linear part of the increment of the length of the arc $MM ^ \prime$ can be expressed in terms of the square of the differential $d \mathbf r$:
$$\textrm{ I } = ds ^ {2} = d \mathbf r ^ {2} = \mathbf r _ {u} ^ {2} du ^ {2} + 2 \mathbf r _ {u} \mathbf r _ {v} du dv + \mathbf r _ {v} ^ {2} dv ^ {2} ,$$
and is called the first fundamental form of the surface. The coefficients in the first fundamental form are usually denoted by
$$E = \mathbf r _ {u} ^ {2} ,\ \ F = ( \mathbf r _ {u} , \mathbf r _ {v} ) ,\ \ G = \mathbf r _ {v} ^ {2} ,$$
or, in tensor symbols,
$$d \mathbf r ^ {2} = g _ {11} du ^ {2} + 2g _ {12} du dv + g _ {22} dv ^ {2} .$$
The tensor $g _ {ij}$ is called the first fundamental, or metric, tensor of the surface. The first fundamental form is a positive-definite form at regular points on the surface:
$$EG - F ^ { 2 } > 0.$$
The first fundamental form characterizes the metric properties of the surface: Knowledge of the first fundamental form enables one to calculate arc lengths on the surface:
$$s = \int\limits _ {t _ {0} } ^ { t } \sqrt {E \left ( \frac{du}{dt} \right ) ^ {2} + 2F \frac{du}{dt} \frac{dv}{dt} + G \left ( \frac{dv}{dt} \right ) ^ {2} } dt,$$
where $t$ is a parameter on the curve; angles between curves on the surface:
$$\cos ( \widehat{ {dr \delta r }} ) =$$
$$= \ \frac{E du \delta u + F( du \delta v+ dv \delta u)+ G dv \delta v }{\sqrt {E du ^ {2} + 2F du dv+ G dv ^ {2} } \sqrt {E \delta u ^ {2} + 2F \delta u \delta v+ G \delta v ^ {2} } } ,$$
where $du : dv$ and $\delta u : \delta v$ are the directions of the tangent vectors to the curves (see Fig. b); and areas of regions on the surface:
$$\sigma = \int\limits \int\limits \sqrt {EG - F ^ { 2 } } du dv .$$
Figure: f040450b
The form of the coefficients of the first fundamental form substantially depends on the choice of the coordinates. The first fundamental form has so-called orthogonal form
$$E( u, v) du ^ {2} + G( u, v) dv ^ {2}$$
in orthogonal coordinates, canonical form
$$du ^ {2} + G ^ {2} dv ^ {2}$$
in semi-geodesic coordinates, and isothermal (isometric) form in isothermal coordinates:
$$ds ^ {2} = \lambda ^ {2} ( u, v)( du ^ {2} + dv ^ {2} ).$$
Sometimes a surface is characterized by special forms of the first fundamental form. For example, a Liouville surface is characterized by:
$$[ \phi ( u) + \psi ( v)]( du ^ {2} + dv ^ {2} ).$$
The first fundamental form is a bending invariant for the surface: The Gaussian curvature at a given point can be calculated from the coefficients of the first fundamental form and their derivatives only (Gauss' theorem).
See Fundamental forms of a surface for the relation between the first fundamental form and other quadratic forms, as well as for literature.
In more current terminology the first fundamental form of an imbedded surface $f: U \rightarrow \mathbf R ^ {3}$, $U \subset \mathbf R ^ {2}$, is defined as follows. The inclusion $( Tf ) _ {u} ( \mathbf R _ {u} ^ {2} ) \subset T _ {f ( u) } \mathbf R ^ {3}$ defines a quadratic form on the tangent space $( Tf ) _ {u} ( \mathbf R _ {u} ^ {2} )$ to the surface at $f ( u)$ by restricting the canonical scalar product on $T _ {f ( u) } \mathbf R ^ {3} \simeq \mathbf R ^ {3}$. Thus the first fundamental form of the surface $f ( U) \subset \mathbf R ^ {3}$ simply assigns to a pair of tangent vectors $X, Y$ to $f ( U)$ at $z \in f ( U)$ the inner product of $X$ and $Y$ in $\mathbf R ^ {3}$. | 2023-03-27T20:52:04 | {
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https://upscprelims.online/online-course-csat-paper-ii-basic-numeracy-average/ | # (Online Course) CSAT Paper – II : Basic Numeracy: Average
## Average
The average of a given number of quantities of the same kind is expressed as
Average = Sum of the quantities/Number of the quantities
Average is also called the Arithmetic Mean.
Also, Sum of the quantities = Average × Number of the quantities
Number of quantities = Sum of the quantities/Average
• If all the given quantities have the same value, then the number itself is
the average.
• If all the given quantities are not all the same,
then the average of the given quantities is always greater, then the
smallest number and always less than the largest number. Equivalently,
atleast one of the numbers is less than the average and atleast one is
greater then the average.
• If each of the given quantities is increased by a constant p, then
their average is also increased by p.
• If each of the given quantities is decreased by a constant p, then
their average is also decreased by p.
• If each of the given quantities is multiplied by a constant p, then
their average is also multiplied by p.
• Whenever the given quantities form an arithmetic sequence and if the
given quantities has odd terms, then the average is the middle term in
the sequence and if the given quantities has even terms, then the
average of the sequence is the average of the middle two terms.
• In order to calculate the weighted average of a set
of numbers, multiply each number in the set by the number of times it
appears, add all the products and divide by the total number of numbers
in the set.
• If the speed of an object from A to B is x km/h and from B to A is y
km/h, then the average speed during the whole journey is 2xy/(x + y) km/h.
• If the average of N1 quantities is x and N2
quantities is y then the average of total (N1 + N2)
quantities is given by (N1x+N2y)/N1+N2
Example 1: What is the average of first five even numbers.
Solution. The first prime even numbers are 2, 4, 6, 8, 10
Average = (2+4+6+8+10)/5 = 30/5 = 6
Example 2: The average of five consecutive even numbers is 50. What is
the largest of these numbers?
Solution. Let the numbers be x – 4, x – 2, x, x + 2, x + 4.
Average = Sum of the quantities/Number of the quantities
= (x-4+ x-2+ x+x+2+ x+4)/5 = 50
5x/5=50
x=50
So, the numbers are 46, 48, 50, 52, 54.
The largest of these numbers is 54.
Example 3: Average weight of 32 students of a class is 30.5 kg. If
weight of a teacher is also included then average weight is increased by 500 g.
What is the weight of the teacher?
Solution. Total weight of 32 students = 30.5 × 32 = 976 kg
Average weight of (32 students + 1 teacher) = (30.5 + 0.5) = 31 kg
Total weight of (32 students + 1 teacher) = 31 × 33 = 1023 kg
Weight of teacher = (1023 – 976) kg = 47 kg
Example 4: The average salary per head of all the
employees of an institution is Rs 60. The average salary per head of 12 officers
is Rs 400 and average salary per head of the rest is Rs. 56. Find the total
number of employees in the institution.
Solution. Let the total number of employees be x.
Then,
60 = Total salary of all employees/x
60 = {12 × 400 (x-12) × 56}/x
60x = 12 × 400 + (x – 12) × 56 = 4800 + 56x – 672
60x – 56x = 4800 – 672
4x = 4128 >> x = 1032
Hence, the total number of employees is 1032.
Example 5: If the average of p and q is 58 and the
average of q and 5 is 64, what is the value of s – p?
Example 6: 12 men went to a restaurant. 11 of them spent Rs. 5 each
and the 12th person spent Rs. 11 more than the average expenditure of all. Find
the total money spent by them?
Solution. Let the average money spent by the 12 men = Rs. x
Money spent by the 12th man = Rs. (x + 11)
Money spent by the other 11 men = Rs. (11 × 5) = Rs. 55
Total money spent by the 12 men = Rs. (55 + x + 11) = Rs. (x + 66) | 2023-02-07T17:48:19 | {
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https://www.cheenta.com/west-bengal-rmo-2015-problem-3-solution/ | Select Page
The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 3 Solution has been written for RMO preparation series.
## Problem:
Show that there are infinitely many triples $(x,y,z)$ of positive integers, such that $x^3+y^4=z^{31}$.
## Discussion
Suppose we have found one such triplet (x, y, z). Then $x^3 + y^4 = z^{31}$. Multiply $a^{372}$ to both sides where a is an arbitrary integer.
Clearly we have $a^{372}x^3 + a^{372}y^4 = a^{372}z^{31}$
$\Rightarrow (a^{124}x)^3 + (a^{93}y)^4 = (a^{12}z)^{31}$
Hence if (x, y, z) is a triple then $(a^{124}x, a^{93}y, a^{12}z )$ is another such triple. Since a can be any arbitrary integer, hence we have found infinitely many such triplets provided we have found at least one
To find one such triple, we use the following intuition: set x, y, z as some powers of 2 such that $x^3 = y^4 = 2^{r}$. Then r must be of the form 12k. Finally, their sum must be $x^3 + y^4 = 2^{r} + 2^{r} = 2^{r+1}$. This r+1 must be divisible by 31.
Let $r = 12s$ and $r+1 = 31m$ we get $12s +1 = 31m$. Since 12 and 31 are coprime there is integer solution to this linear diophantine equation (by Bezoat’s theorem). We can solve this linear diophantine equation by euclidean algorithm.
$31 = 12 \times 2 + 7$
$12 = 7\times 1 + 5$
$7 = 5\times 1 + 2$
$5 = 2 \times 2 + 1$
$\Rightarrow 1 = 5 - 2 \times 2 = 5 - 2 \times (7 - 5 \times 1)$
$\Rightarrow 1 = 3 \times 5 - 2 \times 7 = 3 \times (12 - 7 \times 1) - 2 \times 7$
$\Rightarrow 1 = 3 \times 12 - 5 \times 7 = 3 \times 12 - 5 \times (31 - 12 \times 2)$
$\Rightarrow 1 = 13 \times 12 - 5 \times 31$
$\Rightarrow 1 = 13 \times 12 - 5 \times 31 + 12 \times 31 - 12 \times 31$
$\Rightarrow 1 = (13 -31)\times 12 +(12- 5 )\times 31$
$\Rightarrow 1 = -18 \times 12 + 7\times 31$
$\Rightarrow 1 + 18 \times 12 = 7\times 31$
Hence we use this to form an equation:
$2^{18 \times 12} + 2^{18 \times 12} = 2^{216} + 2^{216} = 2^{217}=2^{7\times 31}$
$(2^{72})^3 + (2^{54})^4 = (2^7)^{31}$
Hence we have found one such triple : $(2^{72}, 2^{54} ,2^7 )$ (from which we have shown earlier that infinitely more can be generated)
## Chatuspathi:
• What is this topic: Number Theory
• What are some of the associated concept: Linear Diophantine Equation, Bézout’s Theorem, Euclidean Algorithm, Sum of powers of two
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course, Cheenta Math Olympiad Program, discuss these topics in the ‘Number Theory’ module.
• Book Suggestions: Elementary Number Theory by David Burton | 2019-10-19T10:11:23 | {
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https://plainmath.net/80703/we-ve-been-given-a-triangle-a-b | # We’ve been given a triangle A B C with an area = 1. Now Marcus gets to choose a point
We’ve been given a triangle $ABC$ with an area = 1. Now Marcus gets to choose a point $X$ on the line $BC$, afterwards Ashley gets to choose a point $Y$ on line $AC$ and finally Marcus gets to choose a point $Z$ on line AB. They can choose every point on their given line (Marcus: $BC$; Ashley: $CA$; Marcus: $AB$) except of $A$, $B$ or $C$. Marcus tries to maximize the area of the new triangle $XYZ$ while Ashley wants to minimize the area of the new triangle. What is the final area of the triangle $XYZ$ if both people choose in the best possible way?
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Your guess in the comment is (almost) correct. The first two points are the midpoints, and the last move makes no difference; the resulting area is $\frac{1}{4}$.
To prove this, consider Ashley's move. If her move is further away from $AB$ than the first move, Marcus will make the last move arbitrarily close to $A$, whereas if it's closer to $AB$ than the first move, Marcus will make the last move arbitrarily close to $B$. In either case, Ashley would have been better off making her move at the same distance from $AB$ as the first move, so this is what she does. Then the last move makes no difference, and the area is given by
$\frac{1}{2}\lambda h\left(1-\lambda \right)c$
in terms of the length $c$ of $AB$, the height $h$ above $AB$ and the fraction $\lambda \in \left(0,1\right)$ at which the first move is placed along $BC$. The maximum is at $\lambda =\frac{1}{2}$; the corresponding area is $\frac{1}{8}hc$, and this is $\frac{1}{4}$ of the area $\frac{1}{2}hc$ of triangle $ABC$. | 2022-12-06T06:56:46 | {
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http://hippomath.blogspot.com/2012/04/ | ## 22 Apr 2012
### My Favourite Proof
This post is inspired and adapted from Greg Martin's excellent talk on prime numbers (found here).
As a student of mathematics, one of the first things I ever learned was the concept of a proof. In fact, the first proof I ever saw was Euclid's classic proof that there were an infinite number of primes.
If you have never seen Euclid's proof before, it goes something like this:
Suppose there were a finite number of primes. We may label them $p_1, p_2, \ldots, p_n$. Let $N=p_1 p_2 \ldots p_n + 1.$ Now for every prime $p_i,\,1\leq i\leq n$ that we just labelled, the remainder of $N$ divided by $p_i$ is 1. However, since $N$ is not prime, $N$ must be able to be factorized into a product of primes. What primes are in that factorization if it doesn't contain any of the primes we just listed out?
Hence, $N$ must have some new prime in its factorization or be a prime itself. This contradicts our assumption that there are a finite number of primes, and therefore an infinite number of primes must exist. $\blacksquare$
Recently, I learned that a variant of Euclid’s proof can be used to prove the infinitude of primes of the form 4k-1 and 6k-1. The proof for 4k-1 is below:
Suppose there are a finite number of primes that is of the form 4k-1 for some $k\in\mathbb{N}$. Again, we may label them $p_1, p_2, \ldots, p_n$. Let $N=4 p_1 p_2 \ldots p_n - 1.$ Again, for every prime $p_i,\,1\leq i\leq n$ that we just labelled, the remainder of $N$ divided by $p_i$ is 1. Note that $N$ is not prime since it is of the form 4k-1 but greater than any of our $p_i$. However, since $N$ is not a prime, $N$ must be able to be factorized into a product of primes.
Furthermore, the factorization of $N$ must only contain primes of the form 4k+1 since $N$ is odd and all primes of the form 4k-1 are used. This is a contradiction since two numbers which are of the form 4k+1 must have a product of the form 4k+1. Hence, if $N$ was entirely the product of primes of the form 4k+1, $N$ must be a number of the form 4k+1 also. But since $N$ is obviously not expressible as 4k+1, $N$ must itself be a prime that is of the form 4k-1 or have some new prime factor that is of the form 4k-1. Again, by contradiction, there must exist an infinite number of primes of the form 4k-1. The case for 6k-1 is similar. $\blacksquare$
Amazingly, there is a general theorem which states that a Euclid-style proof can only be used to prove the infinitude of primes of the form $p \equiv a\bmod{m}$, where $a^2 \equiv 1 \bmod{m}$ for any two natural numbers $a$ and $m$ (proven here by Murty, 1988). For example, this means that infinitude any primes of the form $p \equiv -1 \bmod 4$ may be proven (as we did above), but any primes of the form $p \equiv 3 \bmod 7$ cannot be proven by Euclid's proof.
I find Euclid’s classic proof of the infinitude of primes beautiful -- not just because it establishes an amazing result in an elegant way, but because of its vast reusability. For these reasons, it is easily my favourite proof.
Bonus: From the theorem above, we may also prove infinitude of primes of the form 4k+1 by a Euclid style proof. Try proving this statement! Be warned though, there is an extra trick to defining $N$ than just multiplying all primes of the form 4k+1 together. The solution can be found in the slides of Greg Martin, as referenced in the first sentence of this post. | 2017-08-22T18:29:01 | {
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http://eulerarchive.maa.org/pages/E610.html | ## E610 -- Novae demonstrationes circa divisores numerorum formae $$xx+nyy$$
(New demonstrations about the divisors of numbers of the form $$xx+nyy$$)
Summary:
First, Euler presents some familiar theorems about quadratic residues. Then he presents two main theorems:
• If $$n$$ is of the form $$4k+1$$ or $$4k+2$$, then the only prime divisors of $$xx+nyy$$ are of the form $$4ni+2n-1$$.
• If $$n$$ is of the form $$4k$$ or $$4k-1$$, then the only prime divisors of $$xx+nyy$$ are of the form $$4ni-2n+1$$.
Publication:
• Originally published in Nova Acta Academiae Scientarum Imperialis Petropolitinae 1, 1787, pp. 47-74
• Opera Omnia: Series 1, Volume 4, pp. 197 - 220
• Reprinted in Commentat. arithm. 2, 1849, pp. 159-173 [E610b]
Documents Available:
• Original Publication: E610
Return to the Euler Archive | 2020-01-27T20:11:33 | {
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https://astarmathsandphysics.com/university-maths-notes/abstract-algebra-and-group-theory/1691-cyclic-groups.html?tmpl=component&print=1&page= | ## Cyclic Groups
A group G is called cyclic if there exists an element g in G such that Since any group generated by an element in a group is a subgroup of that group, showing that the only subgroup of a groupthat containsisitself suffices to show thatis cyclic.
For example, ifis a group, thenand G is cyclic. In fact,is isomorphic towith additionFor example, corresponds toWe can use the isomorphismdefined by
For every positive integerthere is exactly one cyclic group (up to isomorphism) whose order isand there is exactly one infinite cyclic group (the integers under addition). Hence, the cyclic groups are the simplest groups and they are completely classified.
Since the cyclic groups are abelian, they are often written additively and denotedor or C-n where n is the order, equal to the number of elements.
inwhereas 3 + 4 = 2 in
Cyclic groups and all their subgroups are abelian. Every element is of the formthen so every element commutes with every other.
Ifthenfor allThis is because
Ifis a cyclic group of orderthen every subgroup ofis cyclic. The order of any subgroup ofis a divisor ofand for each positive divisorofthe grouphas exactly one subgroup of order
Ifis finite, then there are exactlyelements that generate the group on their own, whereis the number of numbers inthat are coprime toMore generally, ifdividesthen the number of elements inwhich have orderisThe order of the residue class of m is
Ifis prime, then the only group (up to isomorphism) with p elements is the cyclic groupor
The direct product of two cyclic groupsandis cyclic if and only ifand are coprime. Thusis the direct product ofandbut not the direct product ofand
A primary cyclic group is a group of the formwhereis a prime number. The fundamental theorem of abelian groups states that every finitely generated abelian group is the direct product of finitely many finite primary cyclic and infinite cyclic groups.
The elements ofcoprime toform a group under multiplication modulowith elements,Whenwe get
is cyclic if and only ifforandin which case every generator ofis called a primitive root moduloThus,is cyclic for but not forwhere it is instead isomorphic to the Klein four-group.
The group is cyclic withelements for every primeand is also written because it consists of the non-zero elements. More generally, every finite subgroup of the multiplicative group of any field is cyclic. | 2018-02-26T03:49:11 | {
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https://astarmathsandphysics.com/university-maths-notes/analysis/1754-rearrangements-of-series.html?tmpl=component&print=1&page= | ## Rearrangements of Series
Definition: Letbe an infinite series. Ifis any one to one function fromontothen the infinite seriesis called a rearrangement of
With absolutely convergent series all rearrangements converge to the same limit:
Proof: For eachletandconverges toChoose then there issuch that forsince converges absolutely.is a one to one mappingontohence there is K>=N such thatAssumethen
Thusconverges to
Conversely ifis an infinite series with the property that every rearrangement converges to the same limit, the series is absolutely convergent. Ifconverges conditionally then the sum of all the positive termsdiverges and the sum of all the negative terms diverges.
Theorem
Ifconverges conditionally then given any real number r there is a rearrangement ofthat converges to r.
Proof: Supposeandare unbounded sequences andconverges to zero. Letbe the first positive integer such thatThere is such ansinceis an unbounded increasing sequence. Next letbe the first positive integer such thatLetbe the least positive integer such thatContinuing in this way we obtain a rearrangement converging tosinceconverges to zero. | 2018-06-18T17:29:40 | {
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https://www.yaclass.in/p/mathematics-state-board/class-10/relations-and-functions-8849/cartesian-product-of-sets-8661/re-45f35afd-c7d1-4502-b7f0-c6b7e146d437 | ### Theory:
Illustration 1:
Consider the sets $$A$$ and $$B$$.
Let $$A$$ $$=$$ $$\{2, 3\}$$ and $$B$$ $$=$$ $$\{1, 2, 3\}$$.
We shall write the product $$A \times B$$ as follows:
$$A \times B$$ $$=$$ $$\{(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$$
The graphical illustration of $$A \times B$$ is given by:
Similarly, we will write the product $$B \times A$$ as follows:
$$B \times A$$ $$=$$ $$\{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)\}$$
The graphical illustration of $$B \times A$$ is given by:
From the graphs of $$A \times B$$ and $$B \times A$$, it is observed that $$A \times B$$ $$\neq$$ $$B \times A$$.
Important!
• In general, $$A \times B$$ $$\neq$$ $$B \times A$$ but $$n(A \times B)$$ $$=$$ $$n(B \times A)$$.
• $$A \times B = \phi$$ if and only if $$A = \phi$$ or $$B = \phi$$.
• If $$n(A) = x$$ and $$n(B) = y$$, then $$n(A \times B) = xy$$.
Illustration 2:
Consider the sets $$A$$ and $$B$$.
Let $$A$$ $$=$$ $$\{\text{Set of all numbers in the interval } [2, 5]\}$$ and $$B$$ $$=$$ $$\{\text{Set of all numbers in the interval } [4, 5]\}$$.
The product $$A \times B$$ corresponds to the intersecting region of the given intervals. In other words, it the set of all points $$(x,y)$$ lying within the common region.
We shall represent the product $$A \times B$$ graphically as follows:
The product $$A \times B$$ corresponds to the middle rectangular region. That is, it consists of all points $$(x, y)$$ within this rectangular region.
Important!
The product $$\mathbb{R} \times \mathbb{R}$$ is called the cartesian plane where it represents the set of all points $$(x, y)$$ where $$x$$, $$y$$ are real numbers. | 2021-06-21T18:24:31 | {
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https://www.ask-math.com/void-relation.html | # Void Relation
Void relation : Let A be a set, then $\Phi \subseteq$ A X A and so it is a relation on A. This relation is called the void-relation or empty relation on set A.
In other words, a relation R on set A is called empty relation, if no elements of A is related to any element of A.
For example : The relation R on the set A = {1,2,3,4} defined by
R = {(a,b):a + b = 10}
We observe that a + b $\neq$ 10 for any two elements of set A . Therefore,
(a,b) $\notin$ R for any a, b $\in$ A.
⇒ R does not contain any element A X A
⇒ R is an empty set.
⇒ R is the void relation on A.
## Examples on Void Relation
Example 1 : Let A = {1, 2, 5, 8}, then R = {(x, y), x, y $\in$ A, x * y = 7}
Check whether this relation is empty relation or not.
Solution : This is an empty relation in A, since no ordered pair (x, y) A x A $\in$ satisfies x * y = 7.
Example 2 : Let B = {1, 2, 5, 8}, then R = {(x, y), x, y $\in$ A, x - y = 1}
Check whether this relation is empty relation or not.
Solution : This is not an empty relation in B, since ordered pair (2, 1) satisfies x - y = 1. | 2019-09-15T20:53:49 | {
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https://brilliant.org/discussions/thread/the-most-difficult-yet-so-easy-to-understand/ | ×
# The most difficult - yet so easy to understand - problem in the world!
Take any natural number $$n$$. If $$n$$ is even, divide it by $$2$$ to get $$n / 2$$. If $$n$$ is odd, multiply it by $$3$$ and add $$1$$ to obtain $$3n + 1$$. Repeat the process indefinitely. The problem is to prove that no matter what number you start with, you will always eventually reach $$1$$.
Don't be fooled by the simplicity of this problem, great mathematicians like Erdős had great trouble proving it. The latter even said: "Mathematics may not be ready for such problems."
Note by حكيم الفيلسوف الضائع
2 years, 5 months ago
Sort by:
When do you predict that the Collatz conjecture will be solved? @Sharky Kesa @Daniel Liu @Daniel Chiu @Trevor B. · 2 years, 5 months ago
Possibly in the next decade given the number of young mathematicians rising up and are being seen on Brilliant. · 2 years, 5 months ago
I also thought this ....one day....Thanks for posting it · 2 years, 5 months ago
The Collatz Conjecture? or The Ulam's Conjecture? · 2 years, 5 months ago
It's commonly named after Collatz name but it is also named Ulam's conjecture because he used to talk about it in the lectures he gave. · 2 years, 5 months ago
I did not understand what were you saying about odd numbers. Even one is true but can you explain me the odd process ? · 2 years, 5 months ago
Start with any natural number. If the number is even, divide it by two. If it is odd, triple it and add one. Repeat the process with the new number so formed. Ultimately the number 1 is reached. · 2 years, 5 months ago
Even one is true. But what about odd ? Let the no. be 5. if we triple it and add one, we will get 16 and then 49 then 148 ......... Which I don't think would approach 1. Or am I not getting the process. Can u explain the process with examples. · 2 years, 5 months ago
When you got 16, you'll have to use the operation we restricted to even numbers, that is: $$n/2$$. And so you will get 8. Repeat it again you'll have 4, then 2, then 1! · 2 years, 5 months ago
OK. Now I got it very well. Thanks · 2 years, 5 months ago
You're welcome! · 2 years, 5 months ago
You have done the operation wrong. Lets call the operation C(x) where x is your starting number.
$$C(5) = 16, C(16) = 8, C(8) = 4, C(4) = 2, C(2) = 1$$. And thus the operation ends. · 2 years, 5 months ago
couldn't we just consider it true until proven wrong? =) · 2 years, 4 months ago
That wouldn't be helpful in any way. · 2 years, 4 months ago
Yep, just taking the lazy way out. · 2 years, 4 months ago
Le problème de Syracuse ;) · 2 years, 5 months ago
Wouldn't it be correct if I say it will also reach every time reach 2 . · 2 years, 5 months ago
No. If you start on 2 itself, then you will have to carry out the operation n/2 anyway. You have to COMPLETE the sequence and hence end with 1. · 2 years, 4 months ago | 2016-10-25T15:44:21 | {
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https://mathspace.co/textbooks/syllabuses/Syllabus-1082/topics/Topic-21062/subtopics/Subtopic-273148/?textbookIntroActiveTab=guide&activeTab=worksheet | iGCSE (2021 Edition)
Worksheet
Substitution method
1
Use the given substitution to solve the following equations:
a
3^{2x}-12 \left(3^x \right) +27=0 where m=3^x
b
2^{2x}-12 \left(2^x \right) +32=0 where m=2^x
c
3^{2x}-4 \left(3^x \right) +3=0 where m=3^x
d
5^{2x}-30 \left(5^x \right) +125=0 where m=5^x
e
2^{2x}+3 \left(2^x \right) -10=0 where m=2^x
f
3^{2x}-5 \left(3^x \right) -36=0 where m=3^x
g
2 \left( 2^{2x} \right) -9 \left(2^x \right) +4=0 where m=2^x
h
3 \left( 3^{2x} \right) -4 \left(3^x \right) +1=0 where m=3^x
i
4^{ 2 x} - 65 \times 4^{x} + 64 = 0 where m = 4^{x}
2
Solve the following equations using substitution:
a
2^{2x}-5 \left(2^x \right) +4=0
b
3^{2x}-26 \left(3^x \right) +27=0
c
4^{2x}-6 \left(4^x \right) +8=0
d
2^{2x}-5 \left(2^x \right) -24=0
e
4 \left( 2^{2x} \right)-5 \left(2^x \right) +1=0
f
3\left( 3^{2x} \right) +8 \left(3^x \right) -3=0
3
A circle has equation \left(x + 2\right)^{2} + \left(y - 4\right)^{2} = 41. Find the x-intercepts of the circle.
4
On Earth, the equation d = 4.9 t^{2} is used to find the distance (in meters) an object has fallen through the air after t seconds.
a
8 seconds after a skydiver jumps out of the plane, what distance has she fallen?
b
Solve for t, the time it will take her to fall 78.4 metres.
5
On the moon, the equation d = 0.8 t^{2} is used to approximate the distance an object has fallen after t seconds. (Assuming no air resistance or buoyancy). On Earth, the equation is d = 4.9 t^{2}.
a
A rock is thrown from a height of 60 metres on Earth. Solve for t, the time it will take to hit the ground to the nearest second.
b
If a rock is thrown from a height of 60 metres on the moon, solve for the time, t, it will take to reach the ground to the nearest second.
6
The Gateway Arch in St. Louis, Missouri, is 630\text{ ft} tall.
We can model the behaviour of objects falling from the arch using Galileo's formula for falling objects: d = 16 t^{2}, where d is distance fallen in feet and t is time in seconds since the object was dropped.
How long would it take an object dropped from the arch to fall to the ground? Give your answer to the nearest second.
7
The kinetic energy E of a moving object is given by E = \dfrac{1}{2} m v^{2}, where m is its mass in kilograms and v is its speed in metres/second.
a
If a vehicle travelling at 18\text{ m/s} has kinetic energy E = 307\,800, what is the mass of the car?
b
If a vehicle is travelling at 18\text{ m/s}, what is its speed in kilometres/hour to one decimal place?
c
If a vehicle weighing 1600 kilograms has kinetic energy E = 204\,800, at what speed is it moving?
8
The sum of the series 1 + 2 + 3 + \ldots + n is given by S = \dfrac{n \left(n + 1\right)}{2}.
a
Find the sum of the first 32 positive integers.
b
Find the number, n, of positive integers required for a sum of 15.
9
In a room of n people, if everyone shakes hands with everyone else, the total number of handshakes is given by \dfrac{n \left(n - 1\right)}{2}. If 780 handshakes are made, how many people are there in the room?
10
The product of 2 positive consecutive numbers is 56. Let the smaller number be x.
a
Solve for x.
b
Find the larger number.
11
A number increased by 3 is multiplied by the same number decreased by 2, and the result is 50. Using x to represent the number, write an equation that represents the situation and solve it for x.
12
Mae throws a stick vertically upwards. After t seconds, its height h metres above the ground is given by the formula h = 25 t - 5 t^{2}.
a
At what time(s) will the stick be 30 metres above the ground?
b
How long does the stick take to hit the ground?
c
Can the stick ever reach a height of 36 metres?
13
A t-shirt is fired straight up from a t-shirt cannon at ground level. After t seconds, its height above the ground is h metres, where h = - 16 t^{2} + 48 t.
a
For what values of t is the shirt 11 metres above the ground?
b
Hence, calculate how long the t-shirt was at least 11 metres above the ground.
c
For what values of t is the shirt 35 metres above the ground?
d
Hence, calculate how long the t-shirt was at least 35 metres above the ground.
14
The perimeter of a rectangle is 28\text{ cm}. Let the length be x\text{ cm} and the width be y\text{ cm}.
a
Find an equation for y in terms of x.
b
Given that the area of the rectangle is 48\text{ cm}^2, find all of the possible values of x.
c
Find all of the possible values for y.
15
The area of a rectangle is 160\text{ m}^2. The length of the rectangle is 6\text{ m} longer than its width.
Let w be the width of the rectangle in metres.
a
Find the width of the rectangle, w.
b
Find the rectangle's length.
16
At time t seconds, the distance, s, travelled by an object moving in a straight line is given by s = u t + \dfrac{1}{2} a t^{2}where u is its starting speed and a is its acceleration.
When u = 16 and a = 8, find how long it would take for the object to travel 128 metres.
17
The base of a triangle is 3 metres more than twice its height. The area of the triangle is 115 square metres. Let x be the height of the triangle.
a
Find the height by solving for x.
b
Find the length of the base.
18
The product of 2 consecutive, positive odd numbers is 399.
Let the smaller number be 2 x + 1.
a
Solve for x.
b
Find the first odd number.
c
Find the other odd number.
19
The square of a number is 70 less than 17 times the number. Let x be the number.
What is the largest possible number that satisfies this condition?
20
The sum of the areas of two circles is 79.56 \pi\text{ m}^2. One of the circles has a radius 1.1 times the length of the other. Let r be the length of the shorter radius.
a
Find the length of the shorter radius.
b
Find the length of the longer radius.
21
A rectangular swimming pool is 16\text{ m} long and 6\text{ m} wide. It is surrounded by a pebble path of uniform width x\text{ m}. The area of the path is 104\text{ m}^2 .
a
Find an expression for the area of the path in terms of x. Express your answer in simplest form.
b
Hence, form an equation and solve for x, the width of the path.
22
Georgia can break a code in 5 hours less time than her younger sister Tricia. If they work together, the code will take them 6 hours to break.
a
Find the number of hours, m, that it would take Georgia to break the code.
b
Hence, find how long it would take Tricia to crack the code.
23
The diameter of a semicircle shares its edge with the width of a rectangle whose length is twice its width. Their combined area is 13\text{ cm}^2 .
a
Find the width of the rectangle correct to two decimal places. Let w be the width.
b
Hence, find the perimeter of the entire shape correct to one decimal place.
24
The sum of two integers is 70.
a
If one of the integers is 29, what is the other integer?
b
Let one of the integers be x. Write an expression for the other integer.
c
Hence, write an expression for the product of the two integers. Give your answer in expanded form.
d
When x = 35, what is the product of the two integers?
e
When x = 35, does this result in the largest possible product?
25
We want to investigate which points have a y-coordinate 7 and are 5 units from \left(2, 3\right).
a
Find the x-coordinate(s) of the points.
b
Write the coordinates of the points.
26
In the diagram, \triangle ABC is similar to \triangle ADE, with BC parallel to ED.
Find the value of x.
27
The rectangular pool in the image has a length of x + 7 metres and a width of x metres. It is surrounded by a path 4 metres wide.
a
Find an expression for the area of the pool in terms x.
b
Find an expression for the total area of the pool and surrounding path, in terms of x.
c
Suppose the total area of the pool and the surrounding path is 488\text{ m}^2 more than the area of the pool alone. Find the value of x.
d
Hence, state the width and length of the pool.
### Outcomes
#### 0606C2.4A
Solve quadratic equations for real roots.
#### 0606C2.4B
Find the solution set for quadratic inequalities.
#### 0606C3.3
Use substitution to form and solve a quadratic equation in order to solve a related equation. | 2022-01-29T13:37:04 | {
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http://www.topicsinmaths.co.uk/cgi-bin/sews.py?CommonFactor | A "Common Factor" of two (or more) integers is any integer that will divide into each of them. Thus a common factor of $a$ and $b$ is a divisor of each of them. By default, and by convention, we usually ask for positive common factors, but negative numbers can also be common factors.
When we work with Complex Numbers there is not concept of positive and negative.
Trivially, 1 is always a common factor of any collection of integers.
Examples:
• A common factor of 15 and 21 is 3.
• A common factor 105 and 70 is 5
• Other common factors of 105 are 7, and 35.
• A common factor of 154 and 455 is 1.
• ... and that is the only positive common factor.
When cancelling fractions we try to find a common factor of both the numerator and denominator, and then divide each by that common factor.
For two given numbers there is also a Highest Common Factor (HCF), which can be found by using the Euclidean Algorithm.
This is related to:
EquivalentFractions
Euclid
EuclideanAlgorithm
LeastCommonMultiple
ReducingFractionsToLowestTerms
ComplexConjugate
ComplexNumber
DividingComplexNumbers
SubtractingComplexNumbers
WhatIsATopic
CancellingFractions
CommonMultiple
HighestCommonFactor
MultiplyingComplexNumbers
## You are here
CommonFactor
ComplexNumber
Denominator
Divisor
EuclideanAlgorithm
Integer
Numerator
LeastCommonMultiple
ReducingFractionsToLowestTerms
ArgandDiagram
CategoryMetaTopic
ComplexConjugate
ComplexPlane
ContinuedFraction
DividingComplexNumbers
Euclid
Euler
ImaginaryNumber
ModuloArithmetic
MultiplyingComplexNumbers
PolarRepresentationOfAComplexNumber | 2020-07-12T22:22:27 | {
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https://answerbun.com/mathematics/growth-of-limits-of-n-th-terms-in-series/ | # Growth of Limits of $n$-th Terms in Series
Mathematics Asked on November 29, 2020
I have two divergent series when summed up to infinite $$n$$: $$frac{1}{n}$$ and $$frac{2^n – 2}{2^n}$$. However, in the limit of increasingly large $$n$$, the $$n$$-th term of $$frac{1}{n}$$ is $$0$$ and the $$n$$-th term of the second series goes to $$1$$ in the limit.
Basically, my question is although both are divergent as infinite series, what is the speed with which the $$n$$-th term approaches $$0$$ in the first case and what is the speed with which the $$n$$-th term approaches $$1$$ in the second case? My guess is that in the first case the speed of decay is slower than exponential whereas the speed of the growth of the second case is slower than logarithmic, but what is the formal analysis way to express this? Can it be shown that one decays slower than the other grows?
Note that the second one can be expressed as $$1-frac{1}{2^{n-1}}$$. This means that the second approaches $$1$$ faster than the first approaches $$0$$, because $$2^{n-1}$$ grows faster than $$n$$.
Correct answer by Joshua Wang on November 29, 2020
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https://www.mathsqrt.com/en/law-of-sines | # Law of Sines
In trigonometry, the law of sines is an equation relating the lengths of the sides of any shape triangle to the sines of its angles:
$$\frac{a}{\sin\alpha }=\frac{b}{\sin\beta }=\frac{c}{\sin\gamma}=2R$$
where,
a, b, c— sides of a triangle;
α, β, γ— angles of a triangle;
R— radius of the triangle's circumcircle.
The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known — a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle. In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn) | 2022-08-08T01:58:50 | {
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https://k12.libretexts.org/Bookshelves/Mathematics/Trigonometry/01%3A_Right_Triangles_and_the_Pythagorean_Theorem/1.09%3A_Special_Right_Triangles_and_Ratios | # 1.9: Special Right Triangles and Ratios
Properties of 30-60-90 and 45-45-90 triangles.
The Pythagorean Theorem is great for finding the third side of a right triangle when you already know two other sides. There are some triangles like 30-60-90 and 45-45-90 triangles that are so common that it is useful to know the side ratios without doing the Pythagorean Theorem each time. Using these patterns also allows you to totally solve for the missing sides of these special triangles when you only know one side length.
Given a 45-45-90 right triangle with sides 6 inches, 6 inches and $$x$$ inches, what is the value of $$x$$?
### Special Right Triangles
There are three types of special right triangles, 30-60-90 triangles, 45-45-90 triangles, and Pythagorean triple triangles.
#### 30-60-90 Triangles
A 30-60-90 right triangle has side ratios $$x$$, $$x\sqrt{3}$$,$$2x$$.
Confirm with Pythagorean Theorem:
\begin{aligned} x^2+(x\sqrt{3})^2&=(2x)^2 \\ x^2+3x^2&=4x^2 \\ 4x^2&=4x^2\end{aligned}
#### 45-45-90 Triangles
A 45-45-90 right triangle has side ratios x,x,x\sqrt{2}.
Confirm with Pythagorean Theorem:
\begin{aligned} x^2+x^2 &=(x\sqrt{2})^2\\ 2x^2&=2x^2 \end{aligned}
Note that the order of the side ratios $$x,x\sqrt{3},2x$$ and $$x,x,x\sqrt{2}$$ is important because each side ratio has a corresponding angle. In all triangles, the smallest sides correspond to smallest angles and largest sides always correspond to the largest angles.
#### Pythagorean Triple Triangles
Pythagorean number triples are special right triangles with integer sides. While the angles are not integers, the side ratios are very useful to know because they show up everywhere. Knowing these number triples also saves a lot of time from doing the Pythagorean Theorem repeatedly. Here are some examples of Pythagorean number triples:
• 3, 4, 5
• 5, 12, 13
• 7, 24, 25
• 8, 15, 17
• 9, 40, 41
More Pythagorean number triples can be found by scaling any other Pythagorean number triple. For example:
$$3,4,5 \rightarrow 6,8,10$$ (scaled by a factor of 2)
Even more Pythagorean number triples can be found by taking any odd integer like 11, squaring it to get 121, halving the result to get 60.5. The original number 11 and the two numbers that are 0.5 above and below (60 and 61) will always be a Pythagorean number triple.
$$11^2+60^2=61^2$$
Example $$\PageIndex{1}$$
Earlier you were asked about a 45-45-90 right triangle with sides 6 inches, 6 inches and $$x$$ inches.
Solution
If you can recognize the pattern for 45-45-90 right triangles, a right triangle with legs 6 inches and 6 inches has a hypotenuse that is 6\sqrt{2} inches. x=6\sqrt{2}.
Example $$\PageIndex{2}$$
A 30-60-90 right triangle has hypotenuse of length 10. What are the lengths of the other two sides?
Solution
The hypotenuse is the side opposite 90. Sometimes it is helpful to draw a picture or make a table.
30 60 90 x $$x\sqrt{3}$$ $$2x$$ 10
From the table you can write very small subsequent equations to solve for the missing sides.
\begin{aligned} 10&=2x\\ x&=5 \\ x\sqrt{3} &=5\sqrt{3}\end{aligned}
The other sides are $$5$$ and $$5\sqrt{3}$$.
Example $$\PageIndex{3}$$
A 30-60-90 right triangle has a side length of 18 inches corresponding to 60 degrees. What are the lengths of the other two sides?
Solution
Make a table with the side ratios and the information given, then write equations and solve for the missing side lengths.
30 60 90 x $$x\sqrt{3}$$ $$2x$$ 18
\begin{aligned}18&=x\sqrt{3} \\ \dfrac{18}{\sqrt{3}}&=x \\ x&=18\sqrt{3}=18\sqrt{3} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}=18\dfrac{\sqrt{3}}{3}=6\sqrt{3}\\ x&=6\sqrt{3}\end{aligned}
Note that you need to rationalize denominators.
Now use the calculated $$x$$ value to solve for $$2x$$.
\begin{aligned} 2x&=2(6\sqrt{3}) \\ 2x&=12\sqrt{3} \end{aligned}
The other sides are 6\sqrt{3} and 12\sqrt{3}.
Example $$\PageIndex{4}$$
Using your knowledge of special right triangle ratios, solve for the missing sides of the right triangle.
Solution
The other sides are each $$\dfrac{5\sqrt{2}}{2}$$.
45 45 90 x x x\sqrt{2} 5
$$x\sqrt{2}&=5 \\ x&=\dfrac{5}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{2}$$
The other sides are each $$5\dfrac{\sqrt{2}}{2}$$.
Example $$\PageIndex{1}$$
Using your knowledge of special right triangle ratios, solve for the missing sides of the right triangle.
Solution
The other sides are $$9$$ and $$6\sqrt{3}$$.
30 60 90 $$x$$ $$x\sqrt{3}$$ $$2x$$ $$3\sqrt{3}$$
\begin{aligned} x&=3\sqrt{3} \\ 2x&=6\sqrt{3} \\ x\sqrt{3}&=3\sqrt{3} \cdot \sqrt{3}=9 \end{aligned}
The other sides are $$9$$ and $$6\sqrt{3}$$.
## Review
For 1-4, find the missing sides of the 45-45-90 triangle based on the information given in each row.
Problem Number Side Opposite $$45^{\circ}$$ Side Opposite $$45^{\circ}$$ Side Opposite $$90^{\circ}$$ 1. 3 2. 7.2 3. 16 4. $$5\sqrt{2}$$
For 5-8, find the missing sides of the 30-60-90 triangle based on the information given in each row.
Problem Number Side Opposite $$30^{\circ}$$ Side Opposite $$60^{\circ}$$ Side Opposite $$90^{\circ}$$ 5. $$3\sqrt{2}$$ 6. 4 7. 15 8. $$12\sqrt{3}$$
Use the picture below for 9-11.
1. Which angle corresponds to the side that is 12 units?
2. Which side corresponds to the right angle?
3. Which angle corresponds to the side that is 5 units?
4. A right triangle has an angle of $$\dfrac{\pi}{6}$$ radians and a hypotenuse of 20 inches. What are the lengths of the other two sides of the triangle?
5. A triangle has two angles that measure $$\dfrac{\pi}{4}$$ radians. The longest side is 3 inches long. What are the lengths of the other two sides?
For 14-19, verify the Pythagorean Number Triple using the Pythagorean Theorem.
1. 3, 4, 5
2. 5, 12, 13
3. 7, 24, 25
4. 8, 15, 17
5. 9, 40, 41
6. 6, 8, 10
7. Find another Pythagorean Number Triple using the method explained for finding “11, 60, 61”.
## Vocabulary
Term Definition
30-60-90 Triangle A 30-60-90 triangle is a special right triangle with angles of $$30^{\circ}$$, $$60^{\circ}$$, and $$90^{\circ}$$.
45-45-90 Triangle A 45-45-90 triangle is a special right triangle with angles of $$45^{\circ}$$, $$45^{\circ}$$, and $$90^{\circ}$$.
Pythagorean number triple A Pythagorean number triple is a set of three whole numbers a,b and c that satisfy the Pythagorean Theorem, $$a^2+b^2=c^2$$.
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $$a^2+b^2=c^2$$, where a and b are legs of the triangle and c is the hypotenuse of the triangle. | 2021-05-14T23:31:18 | {
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