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https://en.wikipedia.org/wiki/Bessel%E2%80%93Clifford_function | # Bessel–Clifford function
In mathematical analysis, the Bessel–Clifford function, named after Friedrich Bessel and William Kingdon Clifford, is an entire function of two complex variables that can be used to provide an alternative development of the theory of Bessel functions. If
${\displaystyle \pi (x)={\frac {1}{\Pi (x)}}={\frac {1}{\Gamma (x+1)}}}$
is the entire function defined by means of the reciprocal Gamma function, then the Bessel–Clifford function is defined by the series
${\displaystyle {\mathcal {C}}_{n}(z)=\sum _{k=0}^{\infty }\pi (k+n){\frac {z^{k}}{k!}}}$
The ratio of successive terms is z/k(n + k), which for all values of z and n tends to zero with increasing k. By the ratio test, this series converges absolutely for all z and n, and uniformly for all regions with bounded |z|, and hence the Bessel–Clifford function is an entire function of the two complex variables n and z.
## Differential equation of the Bessel–Clifford function
It follows from the above series on differentiating with respect to x that ${\displaystyle {\mathcal {C}}_{n}(x)}$ satisfies the linear second-order homogeneous differential equation
${\displaystyle xy''+(n+1)y'=y.\qquad }$
This equation is of generalized hypergeometric type, and in fact the Bessel–Clifford function is up to a scaling factor a Pochhammer–Barnes hypergeometric function; we have
${\displaystyle {\mathcal {C}}_{n}(z)=\pi (n)\ _{0}F_{1}(;n+1;z).}$
Unless n is a negative integer, in which case the right hand side is undefined, the two definitions are essentially equivalent; the hypergeometric function being normalized so that its value at z = 0 is one.
## Relation to Bessel functions
The Bessel function of the first kind can be defined in terms of the Bessel–Clifford function as
${\displaystyle J_{n}(z)=\left({\frac {z}{2}}\right)^{n}{\mathcal {C}}_{n}\left(-{\frac {z^{2}}{4}}\right);}$
when n is not an integer we can see from this that the Bessel function is not entire. Similarly, the modified Bessel function of the first kind can be defined as
${\displaystyle I_{n}(z)=\left({\frac {z}{2}}\right)^{n}{\mathcal {C}}_{n}\left({\frac {z^{2}}{4}}\right).}$
The procedure can of course be reversed, so that we may define the Bessel–Clifford function as
${\displaystyle {\mathcal {C}}_{n}(z)=z^{-n/2}I_{n}(2{\sqrt {z}});}$
but from this starting point we would then need to show ${\displaystyle {\mathcal {C}}}$ was entire.
## Recurrence relation
From the defining series, it follows immediately that ${\displaystyle {\frac {d}{dx}}{\mathcal {C}}_{n}(x)={\mathcal {C}}_{n+1}(x).}$
Using this, we may rewrite the differential equation for ${\displaystyle {\mathcal {C}}}$ as
${\displaystyle x{\mathcal {C}}_{n+2}(x)+(n+1){\mathcal {C}}_{n+1}(x)={\mathcal {C}}_{n}(x),}$
which defines the recurrence relationship for the Bessel–Clifford function. This is equivalent to a similar relation for 0F1. We have, as a special case of Gauss's continued fraction
${\displaystyle {\frac {{\mathcal {C}}_{n+1}(x)}{{\mathcal {C}}_{n}(x)}}={\cfrac {1}{n+1+{\cfrac {x}{n+2+{\cfrac {x}{n+3+{\cfrac {x}{\ddots }}}}}}}}.}$
It can be shown that this continued fraction converges in all cases.
## The Bessel–Clifford function of the second kind
The Bessel–Clifford differential equation
${\displaystyle xy''+(n+1)y'=y\qquad }$
has two linearly independent solutions. Since the origin is a regular singular point of the differential equation, and since ${\displaystyle {\mathcal {C}}}$ is entire, the second solution must be singular at the origin.
If we set
${\displaystyle {\mathcal {K}}_{n}(x)={\frac {1}{2}}\int _{0}^{\infty }\exp \left(-t-{\frac {x}{t}}\right){\frac {dt}{t^{n+1}}}}$
which converges for ${\displaystyle \Re (x)>0}$, and analytically continue it, we obtain a second linearly independent solution to the differential equation.
The factor of 1/2 is inserted in order to make ${\displaystyle {\mathcal {K}}}$ correspond to the Bessel functions of the second kind. We have
${\displaystyle K_{n}(x)=\left({\frac {x}{2}}\right)^{n}{\mathcal {K}}_{n}\left({\frac {x^{2}}{4}}\right).}$
and
${\displaystyle Y_{n}(x)=\left({\frac {x}{2}}\right)^{n}{\mathcal {K}}_{n}\left(-{\frac {x^{2}}{4}}\right).}$
In terms of K, we have
${\displaystyle {\mathcal {K}}_{n}(x)=x^{-n/2}K_{n}(2{\sqrt {x}}).}$
Hence just as the Bessel function and modified Bessel function of the first kind can both be expressed in terms of ${\displaystyle {\mathcal {C}}}$, those of the second kind can both be expressed in terms of ${\displaystyle {\mathcal {K}}}$.
## Generating function
If we multiply the absolutely convergent series for exp(t) and exp(z/t) together, we get (when t is not zero) an absolutely convergent series for exp(t + z/t). Collecting terms in t, we find on comparison with the power series definition for ${\displaystyle {\mathcal {C}}_{n}}$ that we have
${\displaystyle \exp \left(t+{\frac {z}{t}}\right)=\sum _{n=-\infty }^{\infty }t^{n}{\mathcal {C}}_{n}(z).}$
This generating function can then be used to obtain further formulas, in particular we may use Cauchy's integral formula and obtain ${\displaystyle {\mathcal {C}}_{n}}$ for integer n as
${\displaystyle {\mathcal {C}}_{n}(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {\exp(t+z/t)}{t^{n+1}}}\,dt={\frac {1}{2\pi }}\int _{0}^{2\pi }\exp(z\exp(-i\theta )+\exp(i\theta )-ni\theta )\,d\theta .}$
## References
• Clifford, William Kingdon (1882), "On Bessel's Functions", Mathematical Papers, London: 346–349.
• Greenhill, A. George (1919), "The Bessel–Clifford function, and its applications", Philosophical Magazine, Sixth Series: 501–528.
• Legendre, Adrien-Marie (1802), Éléments de Géometrie, Note IV, Paris.
• Schläfli, Ludwig (1868), "Sulla relazioni tra diversi integrali definiti che giovano ad esprimere la soluzione generale della equazzione di Riccati", Annali di Matematica Pura ed Applicata, 2 (I): 232–242.
• Watson, G. N. (1944), A Treatise on the Theory of Bessel Functions (Second ed.), Cambridge: Cambridge University Press.
• Wallisser, Rolf (2000), "On Lambert's proof of the irrationality of π", in Halter-Koch, Franz; Tichy, Robert F., Algebraic Number Theory and Diophantine Analysis, Berlin: Walter de Gruyer, ISBN 3-11-016304-7. | 2018-12-11T20:12:50 | {
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https://suchanutter.net/ItCanBeShown/moments-and-moment-generating-functions.html | 28 Moments and Moment Generating Functions
28.1 Definitions of Moments
28.1.1 Definition: General Definition of Moments
The $$k^{th}$$ moment of a random variable $$X$$ about some point $$c$$ is defined to be $$E[(X-c)^k]$$.
There are two moments that are of particular use in statistics. First, the moment of $$X$$ about the origin; second, the moment of $$X$$ about the mean.
28.1.2 Definition: Ordinary Moments
The $$k^{th}$$ moment of a random variable $$X$$ about the origin is defined to be $$E[(X-0)^k] = E(X^k)$$.
28.1.3 Definition: Central Moments
The $$k^{th}$$ moment of a random variable $$X$$ about the mean $$\mu$$ is defined to be $$E[(X-\mu)^k]$$.
Using these definitions we can derive the first three central moments as follows:
\begin{aligned} E[(X-\mu)^1] &= E(X - \mu) \\ &= E(X) - \mu \\ &= E(X) - E(X) \\ \\ E[(X-\mu)^2] &= E[(X-\mu)(X-\mu)] \\ &= E(X^2-\mu X-\mu X+\mu^2) \\ &= E(X^2-2\mu X+\mu^2) \\ &= E(X^2) - E(2\mu X) + E(\mu^2) \\ &= E(X^2) - 2\mu E(X) + \mu^2 \\ &= E(X^2) - 2\mu\cdot\mu + \mu^2 \\ &= E(X^2) - 2\mu^2 + \mu^2 \\ &= E(X^2) - \mu^2 \\ &= E(X^2) - E(X)^2 \\ \\ \\ E[(X-\mu)^3] &= E[(X-\mu)(X-\mu)(X-\mu)] \\ &= E[(X^2-2\mu X+\mu^2)(X-\mu)] \\ &= E(X^3-\mu X^2-2\mu X^2+2\mu^2X+\mu^2X+\mu^3) \\ &= E(X^3-3\mu X^2+3\mu^2X-\mu^3) \\ &= E(X^3) - E(3\mu X^2) + E(3\mu^2X) - E(\mu^3) \\ &= E(X^3) - 3\mu E(X^2) + 3\mu^2E(X) - \mu^3 \\ &= E(X^3) - 3\mu E(X^2) + 3\mu^3 - \mu^3 \\ &= E(X^3) - 3\mu E(X^2) + 2\mu^3 \end{aligned}
It should be noticed that with all of these results, the moment about the mean can be evaluated by finding the ordinary moments. Thus, if we can find a consistent way to generate ordinary moments , we may use these results to find various parameters of a distribution.
28.2 Moment Generating Functions
28.2.1 Definition: Moment Generating Function
The moment generating function of a random variable, denoted $$M_X(t)$$, is defined to be:
$M_X(t) = E(e^{tX})$
The moment generating function of $$X$$ is said to exist if for any positive constant $$c,\ M_X(t)$$ is finite for $$|t|<c$$. The definition can be expanded to
\begin{aligned} M_X(t) &= E(e^{tX}) \\ &= \sum\limits_{i=1}^{\infty}e^{tx_i}p(x_i) \\ ^{[1]} &= \sum\limits_{i=1}^{\infty}[\frac{(tx_i)^0}{0!}+\frac{(tx_i)^1}{1!} + \frac{(tx_i)^2}{2!}+\frac{(tx_i)^3}{3!}+\cdots]p(x_i) \\ &= \sum\limits_{i=1}^{\infty}[1+tx_i+\frac{(tx_i)^2}{2!} + \frac{(tx_i)^3}{3!}+\cdots]p(x_i) \\ &= \sum\limits_{i=1}^{\infty}[p(x_i)+tx_ip(x_i) + \frac{(tx_i)^2}{2!}p(x_i)+\frac{(tx_i)^3}{3!}p(x_i) + \cdots] \\ &= \sum\limits_{i=1}^{\infty}p(x_i)+\sum\limits_{i=1}^{\infty}tx_ip(x_i) + \sum\limits_{i=1}^{\infty}\frac{(tx_i)^2}{2!}p(x_i) + \sum\limits_{i=1}^{\infty}\frac{(tx_i)^3}{3!}p(x_i) + \cdots \\ &= \sum\limits_{i=1}^{\infty}p(x_i)+t\sum\limits_{i=1}^{\infty}x_ip(x_i) + \frac{t^2}{2!}\sum\limits_{i=1}^{\infty}x_i^2p(x_i) + \frac{t^3}{3!}\sum\limits_{i=1}^{\infty}x_i^3p(x_i)+\cdots \\ &= 1 + tE(X) + \frac{t^2}{2!}E(X^2) + \frac{t^3}{3!} E(X^3) + \cdots \end{aligned}
1. Taylor Series Expansion: $$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}\cdots = 1+x+\frac{x^2}{2!}+\cdots$$
28.2.2 Theorem: Extraction of Moments from Moment Generating Functions
Let $$M_X^{(k)}(t)$$ denote the $$k^{th}$$ derivative of $$M_X(t)$$ with respect to $$t$$. Then $$M_X^{(k)}(0)=E(X^k)$$.
Proof:
\begin{aligned} M_X(t) &= 1 + tE(X) \\ &= \frac{t^2}{2!}E(X^2) + \frac{t^3}{3!} + \cdots \\ \\ \\ M_X^{(1)}(t) &= 0 + E(X) + \frac{2t}{2!}E(X^2) + \frac{3t^2}{3!}E(X^3) + \cdots \\ &= E(X) + tE(X^2) + \frac{t^2}{2!}E(X^3) + \cdots \\ \\ \\ M_X^{(2)}(t) &= 0 + E(X^2) + \frac{2t}{2!}E(X^3) + \frac{3t^2}{3!}E(X^4) + \cdots \\ &= E(X^2) + tE(X^3) + \frac{t^2}{2!}E(X^4) + \cdots \\ \vdots \\ \\ \\ M_X^{(k)}(t) &= 0 + E(X^k) + \frac{2t}{2!}E(X^{k+1}) + \frac{3t^2}{3!}E(X^{k+2}) + \cdots \\ &= E(X^k) + tE(X^{k+1}) + \frac{t^2}{2!}E(X^{k+2}) + \cdots \\ \\ \\ M_X^{(1)}(0) &= 0 + E(X) + \frac{2\cdot 0}{2!}E(X^2) + \frac{3\cdot 0t^2}{3!}E(X^3) + \cdots \\ &= E(X)\\ \\ \\ M_X^{(2)}(0) &= 0 + E(X^2) + \frac{2\cdot 0}{2!}E(X^3) + \frac{3\cdot 0^2}{3!}E(X^4) + \cdots \\ &= E(X^2) \\ \\ \\ \vdots \\ \\ \\ M_X^{(0)}(t) &= 0 + E(X^k) + \frac{2\cdot 0}{2!}E(X^{k+1}) + \frac{3\cdot 0^2}{3!}E(X^{k+2}) + \cdots \\ &= E(X^k) \end{aligned} | 2019-12-10T23:47:07 | {
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https://www.studyadda.com/solved-papers/jee-main-paper-held-on-10-4-2019-morning_q71/817/374675 | • # question_answer Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is : [JEE Main 10-4-2019 Morning] A) $\frac{1}{11}$ B) $\frac{1}{17}$ C) $\frac{1}{10}$ D) $\frac{1}{12}$
$P\left( B \right)=P\left( G \right)=1/2$ Required Proballity = $\frac{\text{all 4girls}}{\left( \text{all 4girls} \right)\text{ (exactly+3girls+1boy)+(exactly2girls+2boys)}}$ $=\frac{{{\left( \frac{1}{2} \right)}^{4}}}{{{\left( \frac{1}{2} \right)}^{4}}{{+}^{4}}{{C}_{3}}{{\left( \frac{1}{2} \right)}^{4}}{{+}^{4}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{4}}}=\frac{1}{11}$ | 2021-01-24T19:52:01 | {
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https://en.m.wikibooks.org/wiki/Calculus_Course/Differentiation | # Calculus Course/Differentiation
## DerivativeEdit
A derivative is a mathematical operation to find the rate of change of a function.
## FormulaEdit
For a non linear function ${\displaystyle f(x)=y}$ . The rate of change of ${\displaystyle f(x)}$ correspond to change of ${\displaystyle x}$ is equal to the ratio of change in ${\displaystyle f(x)}$ over change in ${\displaystyle x}$
${\displaystyle {\frac {\Delta f(x)}{\Delta x}}={\frac {\Delta y}{\Delta x}}}$
Then the Derivative of the function is defined as
${\displaystyle {\frac {d}{dx}}f(x)=\lim _{\Delta x\to 0}\sum {\frac {\Delta f(x)}{\Delta x}}=\lim _{\Delta x\to 0}\sum {\frac {y}{x}}}$
but the derivative must exist uniquely at the point x. Seemingly well-behaved functions might not have derivatives at certain points. As examples, ${\displaystyle f(x)={\frac {1}{x}}}$ has no derivative at ${\displaystyle x=0}$ ; ${\displaystyle F(x)=|x|}$ has two possible results at ${\displaystyle x=0}$ (-1 for any value for which ${\displaystyle x<0}$ and 1 for any value for which ${\displaystyle x>0}$) On the other side, a function might have no value at ${\displaystyle x}$ but a derivative of ${\displaystyle x}$ , for example ${\displaystyle f(x)={\frac {x}{x}}}$ at ${\displaystyle x=0}$ . The function is undefined at ${\displaystyle x=0}$ , but the derivative is 0 at ${\displaystyle x=0}$ as for any other value of ${\displaystyle x}$ .
Practically all rules result, directly or indirectly, from a generalized treatment of the function.
## Table of DerivativeEdit
### General RulesEdit
${\displaystyle {\frac {d}{dx}}(f+g)={\frac {df}{dx}}+{\frac {dg}{dx}}}$
${\displaystyle {\frac {d}{dx}}(c\cdot f)=c\cdot {\frac {df}{dx}}}$
${\displaystyle {\frac {d}{dx}}(f\cdot g)=f\cdot {\frac {dg}{dx}}+g\cdot {\frac {df}{dx}}}$
${\displaystyle {\frac {d}{dx}}\left({\frac {f}{g}}\right)={\frac {g\cdot {\frac {df}{dx}}-f\cdot {\frac {dg}{dx}}}{g^{2}}}}$
### Powers and PolynomialsEdit
${\displaystyle {\frac {d}{dx}}(c)=0}$
${\displaystyle {\frac {d}{dx}}x=1}$
${\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}}$
${\displaystyle {\frac {d}{dx}}{\sqrt {x}}={\frac {1}{2{\sqrt {x}}}}}$
${\displaystyle {\frac {d}{dx}}{\frac {1}{x}}=-{\frac {1}{x^{2}}}}$
${\displaystyle {{\frac {d}{dx}}(c_{n}x^{n}+c_{n-1}x^{n-1}+c_{n-2}x^{n-2}+\cdots +c_{2}x^{2}+c_{1}x+c_{0})=nc_{n}x^{n-1}+(n-1)c_{n-1}x^{n-2}+(n-2)c_{n-2}x^{n-3}+\cdots +2c_{2}x+c_{1}}}$
### Trigonometric FunctionsEdit
${\displaystyle {\frac {d}{dx}}\sin(x)=\cos(x)}$
${\displaystyle {\frac {d}{dx}}\cos(x)=-\sin(x)}$
${\displaystyle {\frac {d}{dx}}\tan(x)=\sec ^{2}(x)}$
${\displaystyle {\frac {d}{dx}}\cot(x)=-\csc ^{2}(x)}$
${\displaystyle {\frac {d}{dx}}\sec(x)=\sec(x)\tan(x)}$
${\displaystyle {\frac {d}{dx}}\csc(x)=-\csc(x)\cot(x)}$
### Exponential and Logarithmic FunctionsEdit
${\displaystyle {\frac {d}{dx}}e^{x}=e^{x}}$
${\displaystyle {\frac {d}{dx}}a^{x}=a^{x}\ln(a)\qquad {\mbox{if }}a>0}$
${\displaystyle {\frac {d}{dx}}\ln(x)={\frac {1}{x}}}$
${\displaystyle {\frac {d}{dx}}\log _{a}(x)={\frac {1}{\ln(a)x}}\qquad {\mbox{if }}a>0,a\neq 1}$
${\displaystyle {\frac {d}{dx}}(f^{g})={\frac {d}{dx}}\left(e^{g\ln(f)}\right)=f^{g}\left({\frac {df}{dx}}\cdot {\frac {g}{f}}+{\frac {dg}{dx}}\cdot \ln(f)\right),\qquad f>0}$
${\displaystyle {\frac {d}{dx}}(c^{f})={\frac {d}{dx}}\left(e^{f\ln(c)}\right)={\frac {df}{dx}}\cdot c^{f}\ln(c)}$
### Inverse Trigonometric FunctionsEdit
${\displaystyle {\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}$
${\displaystyle {\frac {d}{dx}}\arccos(x)=-{\frac {1}{\sqrt {1-x^{2}}}}}$
${\displaystyle {\frac {d}{dx}}\arctan(x)={\frac {1}{1+x^{2}}}}$
${\displaystyle {\frac {d}{dx}}\operatorname {arcsec}(x)={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
${\displaystyle {\frac {d}{dx}}\operatorname {arccot}(x)=-{\frac {1}{1+x^{2}}}}$
${\displaystyle {\frac {d}{dx}}\operatorname {arccsc}(x)=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
### Hyperbolic and Inverse Hyperbolic FunctionsEdit
${\displaystyle {\frac {d}{dx}}\sinh(x)=\cosh(x)}$
${\displaystyle {\frac {d}{dx}}\cosh(x)=\sinh(x)}$
${\displaystyle {\frac {d}{dx}}\tanh(x)={\rm {sech}}^{2}(x)}$
${\displaystyle {\frac {d}{dx}}{\rm {sech}}(x)=-\tanh(x){\rm {sech}}(x)}$
${\displaystyle {\frac {d}{dx}}\coth(x)=-{\rm {csch}}^{2}(x)}$
${\displaystyle {\frac {d}{dx}}{\rm {csch}}(x)=-\coth(x){\rm {csch}}(x)}$
${\displaystyle {\frac {d}{dx}}{\rm {arcsinh}}(x)={\frac {1}{\sqrt {x^{2}+1}}}}$
${\displaystyle {\frac {d}{dx}}{\rm {arccosh}}(x)=-{\frac {1}{\sqrt {x^{2}-1}}}}$
${\displaystyle {\frac {d}{dx}}{\rm {arctanh}}(x)={\frac {1}{1-x^{2}}}}$
${\displaystyle {\frac {d}{dx}}{\rm {arcsech}}(x)={\frac {1}{x{\sqrt {1-x^{2}}}}}}$
${\displaystyle {\frac {d}{dx}}{\rm {arccoth}}(x)=-{\frac {1}{1-x^{2}}}}$
${\displaystyle {\frac {d}{dx}}{\rm {arccsch}}(x)=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}}$ | 2016-09-29T13:38:56 | {
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http://people.math.sfu.ca/~alspach/guide/node2.html | # Definitions, Notation and Results
Definition 2.1 The cardinality of a set is the number of elements in the set. If a set has cardinality n, we often write n-set.
Notation 2.2 We use n! to denote the product . It is read as n factorial. The numbers n! grow very rapidly. Some small values are 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, and 7! = 5,040.
We adopt the convention that 0! = 1. This is done in order to simplify the statements of certain theorems. One useful fact to note about n!is that n! = n(n-1)!. Similarly, n! = n(n-1)(n-2)!. This observation allows some useful cancellation to take place when computing formulas involving factorials.
Notation 2.3 We use to denote the number of ways of selecting k distinct objects from n objects. Another common notation for this number is C(n,k). One standard way of reading it is to say n choose k''. It is also called a binomial coefficient.
The numbers n choose k'' occur all the time in counting problems because we frequently are choosing k objects in such a way that a given object may occur at most once (distinctness), and the order in which the objects are chosen is irrelevant. For example, when being dealt a hand of 5 cards, our main concern is with the final hand and not in the order in which we receive the cards. It is true that we may develop some anxiety as a hand develops if we watch each card as it arrives in our hand, but in analyzing our chances of winning the particular game, we start with the composition of the hand and do not consider the order in which they arrived.
Definition 2.4 A partition of a set A is a collection of non-empty subsets of A such that
and
for all , . Each of the subsets in the collection is called a part of the partition.
The word partition makes sense for the previous concept because the set is being broken into non-empty pieces which have no overlap. This is one of the colloquial uses of the word partition as well.
Notation 2.5 We use to denote the number of ways of partitioning a given n-set into parts whose respective cardinalities are . In the case all mparts have the same cardinality k, we write to denote the number of partitions of an n-set into m parts, where each part has cardinality k.
Brian &
2000-01-31 | 2013-12-11T16:22:19 | {
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https://www.physics.uoguelph.ca/trigonometry-tutorial-2 | # Trigonometry Tutorial 2
We should now investigate the rules for these functions in other quadrants as shown in Panel 7. In the first case, we see that the angle to be dealt with is $\alpha = \theta - 180^\circ$. Now both $r$ and $p$ are negative. So we have the $\sin(\theta) = (-r)/q = -\sin(\alpha) = -\sin (\alpha) = -\sin (\theta - 180^\circ)$. Similarly $\cos(\theta) = (-p)/q = -\cos(\alpha) = -\cos (\theta - 180^\circ)$. And $\tan (\theta) = (-r)/(-p) = r/p = \tan (\alpha) = \tan(\theta - 180^\circ)$.
Notice in this case only the tangent is positive. Using your calculator determine the sine, cosine and tangent of $190^\circ$.
The angle we are concerned with is $190^\circ - 180^\circ = 10^\circ$.
$\sin(10^\circ) = 0.1736$
$\cos(10^\circ) = 0.9848$
$\tan(10^\circ) = 0.1763$.
Therefore,
$\sin(190^\circ) = - 0.1736$
$\cos(190^\circ) = - 0.9848$
$\tan(190^\circ) = 0.1763$
Finally, in the bottom of Panel 7 is the last case. The angle of concern $\theta$ is now $(360^\circ - \alpha)$, and $p$ and $q$ are positive while $r$ is negative. As before, we get the following relations given below. Notice only the cosine is positive.
$\sin(\theta) = -\frac{r}{q} = \sin(\alpha) = -\sin(360^\circ - \theta)$
$\cos (\theta) = \frac{p}{q} = \cos (\alpha) = \cos (360^\circ - \theta)$
$\tan (\theta) = - \frac{r}{p} = - \tan(\alpha) = - \tan (360^\circ - \theta)$
There is a simple rule by which you can remember all of these results, summarized in Panel 8. Notice in the first quadrant, all the functions are positive; in the second quadrant, only the sine is positive; in the third, only the tangent is positive; and in the fourth quadrant, only the cosine is positive. The little mnemonic at the right, called the CAST Rule, tells you which function is positive in each quadrant. And the c, a, s and t stand for cosine, all, sine and tangent
Let us now look at a graph of these functions. But before we examine it in detail, let's see what we can learn just from inspection.
Look at Panel 9. From our definition, $\sin(q) = r/q$, you can see that sine of $0$ is $0$ since $r$ will be 0. The value of sine will rise as $\theta$ increases and reach a value of $1$ when $q = 90^\circ$, since then $r$ will be equal to $q$. Obviously, then, the sine function is one which increases from $0$ to a maximum value of $1$ as $\theta$ increases from $0^\circ$ to $90^\circ$. You can see this plotted in Panel 10a
From $90^\circ$ to $180^\circ$, it decreases back to $0$ but remains positive as we saw earlier.
From $180^\circ$ to $360^\circ$, you remember, it was always negative, and you can readily see, it has a minimum value of $-1$ at $270^\circ$. Beyond $360^\circ$, of course, it just repeats. The function is what we call an oscillatory function, and in your studies in physics, you will find it most important to appreciate this property, particularly in the study of waves and alternating current in electricity and electronics.
How does the cosine function behave? Panel 10 again shows you. At $0^\circ$, $q = p$ and so the value is $1$. At $90^\circ$, $p = 0$, and so $\cos(90^\circ) = 0$. This is shown in Panel 10b. You can work out the rest of it for yourself. You see that the cosine function is exactly the same as the sine function if you slide the sine function graph back $90^\circ$. This says that $\sin(90^\circ + q) = \cos (\theta)$.
The tangent curve looks quite different. Panel 11 shows that $\tan(0) = r/p = 0$, but $\tan(90^\circ) = r/p = \infty$ since $r$ is finite but $p$ has gone to $0$. Panel 11 shows this function. The tangent function is a repeating one but not oscillatory.
One last point which you will find used many times are the approximate values of these functions when $\theta$ is very small. Look at Panel 12. This is like our other figures except that we have put the triangle into the sector of a circle of radius $q$. You can readily see that as $\theta$ gets very small, $p$ and $q$ become very nearly equal and so the cosine approaches the value $1$ as we already know. But in this approximation, $r/q$ and $r/p$ are very nearly equal. These, of course, are the sine and the tangent of $\theta$. So we may say that for very small values of $\theta$, $\sin(\theta)$ is very nearly equal to $\tan(\theta)$. Indeed, if you look back below Panel 7, you will see we looked up the functions for $\theta = 10^\circ$. Even for an angle of this size, you can see that $\cos(10^\circ)$ is within $1.5\%$ of $1$; and $\sin(10^\circ)$ and $\tan(10^\circ)$ are equal to within about $2$ parts in $170$, or about $1.2\%$
We can even go further. You will remember the relation that the arc length $s$ is equal to the radius $q$ multiplied by the angle measured in radians $\theta$; that is, $s = q \theta$. Now in the small angle limit I have been talking about, $s$ and $r$ are nearly equal. Therefore, for the ratio $r/q$, we can use $s/q$ and we immediately see that, for small angles, the sine, the tangent, and the angle $\theta$ itself measured in radians are all nearly equal. We need to stress that this is only true if $\theta$ is measured in radians, where you will recall that $2\pi$ radians was equal to $360^\circ$.
You have now learned the simple rules of trigonometry which, with practice, should equip you for your introductory science courses. Try this Trigonometry Quiz to check your understanding? | 2022-01-18T03:29:42 | {
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http://mathhelpforum.com/calculus/164085-interesting-limit.html | # Math Help - An interesting limit
1. ## An interesting limit
$\lim \{n!e\}= \:?$
where {.} is the symbol of the fractinal part
2. I think that the important observation is that $\displaystyle e=\sum_{k=1}^\infty \frac{1}{k!}$, so $\displaystyle n!\sum_{k=1}^\infty \frac{1}{k!}=\sum_{k=1}^\infty \frac{n!}{k!}$ will be very close an integer when $n$ is large. So, $\{n!e\}\to 0$.
I haven't thought it all the way through, but I think that you can make it rigorous by looking at the partial sums of the series and using Taylor's theorem. Give it a try.
$\lim \{n!e\}= \:?$
$\displaystyle{e=\sum\limits^\infty_{k=0}\frac{1}{k !}\Longrightarrow n!e=n!\left(1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}+\ sum\limits^\infty_{k=n+1}\frac{1}{k!}\right)}$ $\Longrightarrow\displaystyle{\{n!e\}=\sum\limits_{ k=n+1}^\infty\frac{n!}{k!}}$ | 2014-07-23T19:50:41 | {
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https://drexel28.wordpress.com/2012/01/12/universal-arrows-and-universal-elements-pt-ii/ | # Abstract Nonsense
## Universal Arrows and Universal Elements (Pt. II)
Point of Post: This is a continuation of this post.
$\text{ }$
To get a richer class of examples we consider a more specific type of universal arrow. Suppose that we had some functor $F:\mathcal{C}\to\mathbf{Set}$ and some singleton set $\ast$. What would a universal arrow from $\ast$ to $F$ look like? Well, such a universal arrow would consist of $\mathcal{C}$-object $c$ and an arrow $\ast\to F(c)$ such that given any other $\mathcal{C}$-object $x$ with a specified arrow $\ast\to F(x)$ there exists a unique arrow $c\xrightarrow{j}x$ whose image under $F$ commutatively completes the triangle given by $F(c)\longleftarrow\ast\longrightarrow F(x)$. Ok, fine, we literally just wrote down the definition of a universal arrow from $\ast$ to $F$, so how can we make things simpler? The key observation is recognizing (really) what an arrow $\ast\to X$ (where $X$ is any set) really is. Truly, yes, it’s a function, but for all practical purposes we can think about such an arrow as “picking out” some element $x_0$ of $X$ (the one corresponding to the image of the arrow). So, let’s recast our discussion in this language. We see then what we are really when we are talking about a universal arrow from $\ast$ to $F$ is some $\mathcal{C}$-element $c$ along with some distinguished element $a_0\in F(c)$ with the property that given any other $\mathcal{C}$-element $x$ and a distinguished element $b_0\in F(x)$ there exists a unique arrow $c\xrightarrow{j}x$ such that $F(j)$ respects the distinguished objects (i.e. $F(j)(a_0)=b_0$). Such a pair $(c,a_0)$ is called a universal element for $F$.
$\text{ }$
There are two classic examples of universal elements (both of which can be found in [1]) which I can’t help but discuss.
$\text{ }$
Fix some group $G$ and some normal subgroup $N\unlhd G$. Suppose that we had another group, $H$ and some homomorphism $f:G\to H$. A natural question that comes up is when can we factor $f$ through the canonical projection $\pi:G\to G/N$? In other words, under which conditions can we find some homomorphism $g:G/N\to H$ such that $f=g\circ\pi$? Well, this question is quickly answered in the first homomorphism theorem, a ‘lemma’ (really the whole proof, or the important parts of it) to the first isomorphism theorem, which says that such a factorization is possible precisely when $N\subseteq\ker f$. But, this tells us that $\pi$ is a universal element of a certain functor. Indeed, let $F$ be the functor $\mathbf{Grp}\to\mathbf{Set}$ given by sending $H$ to
$\text{ }$
$F(H)=\left\{f\in\text{Hom}_\mathbf{Grp}(G,H):N\subseteq\ker f\right\}$
$\text{ }$
and sending $g:G\to H$ to $g^\ast$ where, as in the case of the covariant Hom functor, we define $g^\ast(h)=g\circ h$ if $h:H\to H'$. We claim that $(G/N,\pi)$ is a universal element for $F$. Indeed, suppose we are given some group $H$ and some distinguished $f\in F(H)$. Since $f$ annihilates $N$ we have by the first homomorphism theorem that $f=g\circ \pi$ for some $g:G/N\to H$. Thus, $g$ is an arrow $G/N\to H$ and $F(g)(\pi)=g\circ\pi=f$. Moreover, it’s clear that $g$ is unique, since $\pi$ is epic (being surjective). This proves that $(G/N,\pi)$ is a universal element for $F$. The cool thing is that (as we have remarked before) this is really the characterization of being a quotient group–it’s what really matters. In particular, all quotient group related theorems are recoverable from the fact that $(G/N,\pi)$ is a universal element for $F$. For example, the clever proof of the first isomorphism theorem shows that if $f:G\to H$ is a surjective group homomorphism (an epimorphism in $\mathbf{Grp}$) with kernel $N$ then $H\cong G/N$ by (implicitly) showing that $(H,f)$ is a universal element for $F$ showing that $H$ and $G/N$ must be isomorphic (via a unique isomorphism!).
$\text{ }$
The other classic example is something we have recently been discussing. Namely, fix some commutative ring $R$ and consider two left $R$-modules $M$ and $N$. We can then consider the set-valued functor $B:R\text{-}\mathbf{Mod}\to\mathbf{Set}$ given by $B(L)=\text{Bil}(M,N;L)$ on objects (where, as we have previously discussed, $\text{Bil}(M,N;L)$ is the set of bilinear maps $M\times N\to L$) and given an arrow $g:L\to L'$ we define $B(g)=g^\ast$ where (as usual) $g^\ast(h)=g\circ h$. So, what does a universal element for $B$ look like? Well, if you haven’t guessed it by now (hard to guess, yes I know) a universal element is given by $(M\otimes_R N,\otimes)$ where $M\otimes_R N$ is the tensor product and $\otimes$ is the tensor map. Why is this true? Well, probably because it’s how we (implicitly) defined the tensor product! Indeed, we defined the tensor product to be an initial object in the comma category $(\ast\downarrow B)$, but by the above discussion this is the same thing as defining the tensor product to be a universal arrow from $\ast$ to $B$, or as we are now calling such an object, a universal element for $B$. Neat, huh?
$\text{ }$
As a last remark we define the “dual” concept to a universal arrow, a couniversal arrow. More descriptively we can think of couniveral arrows as being universal arrows away from a given functor. To be more specific, if we have some functor $F:\mathcal{C}\to\mathcal{D}$ and some object $d$ of $\mathcal{D}$ we define a couniversal arrow from $d$ to $F$ or a universal arrow $F$ to $d$ to be an ordered pair $(c,u)$ where $c$ is some $\mathcal{C}$-object and $u$ is an arrow $F(c)\to d$ such that given any other $\mathcal{C}$-object $x$ and an arrow $F(x)\xrightarrow{f}d$ there exists a unique arrow $x\xrightarrow{j}c$ such that $u\circ F(j)=f$.
$\text{ }$
$\text{ }$
References:
[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.
[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.
[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.
[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.
[5] Rotman, Joseph J. Introduction to Homological Algebra. Springer-Verlag. Print.
[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print. | 2018-01-20T08:54:22 | {
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http://mathhelpforum.com/trigonometry/85467-trigonometry-2-a.html | 1. Trigonometry 2
$sin(x+y) = m$ and $sin(x-y) = n$, $\frac{tgx}{tgy}$ is ?
2. Originally Posted by Apprentice123
$sin(x+y) = m$ and $sin(x-y) = n$, $\frac{tgx}{tgy}$ is ?
note that $\frac {\tan x}{\tan y} = \frac {\sin x \cos y}{\sin y \cos x}$
now, use the addition formula to expand $\sin (x + y)$ and $\sin (x - y)$ and see if you can form an expression equivalent to the one above
3. thank you | 2016-10-28T22:27:18 | {
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https://stacks.math.columbia.edu/tag/044G | Lemma 77.18.2. Let $B \to S$ as in Section 77.3. Let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $B$.
1. If $s$ and $t$ are open, then for every open $W \subset U$ the open $s(t^{-1}(W))$ is $R$-invariant.
2. If $s$ and $t$ are open and quasi-compact, then $U$ has an open covering consisting of $R$-invariant quasi-compact open subspaces.
Proof. Assume $s$ and $t$ open and $W \subset U$ open. Since $s$ is open we see that $W' = s(t^{-1}(W))$ is an open subspace of $U$. Now it is quite easy to using the functorial point of view that this is an $R$-invariant open subset of $U$, but we are going to argue this directly by some diagrams, since we think it is instructive. Note that $t^{-1}(W')$ is the image of the morphism
$A := t^{-1}(W) \times _{s|_{t^{-1}(W)}, U, t} R \xrightarrow {\text{pr}_1} R$
and that $s^{-1}(W')$ is the image of the morphism
$B := R \times _{s, U, s|_{t^{-1}(W)}} t^{-1}(W) \xrightarrow {\text{pr}_0} R.$
The algebraic spaces $A$, $B$ on the left of the arrows above are open subspaces of $R \times _{s, U, t} R$ and $R \times _{s, U, s} R$ respectively. By Lemma 77.11.4 the diagram
$\xymatrix{ R \times _{s, U, t} R \ar[rd]_{\text{pr}_1} \ar[rr]_{(\text{pr}_1, c)} & & R \times _{s, U, s} R \ar[ld]^{\text{pr}_0} \\ & R & }$
is commutative, and the horizontal arrow is an isomorphism. Moreover, it is clear that $(\text{pr}_1, c)(A) = B$. Hence we conclude $s^{-1}(W') = t^{-1}(W')$, and $W'$ is $R$-invariant. This proves (1).
Assume now that $s$, $t$ are both open and quasi-compact. Then, if $W \subset U$ is a quasi-compact open, then also $W' = s(t^{-1}(W))$ is a quasi-compact open, and invariant by the discussion above. Letting $W$ range over images of affines étale over $U$ we see (2). $\square$
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https://math.stackexchange.com/questions/2305193/a-generalization-of-the-euler-mascheroni-constant | # A generalization of the Euler-Mascheroni constant
Let $f:[1,+\infty)\rightarrow \mathbb{R}$ be a differentiable function. We are dealing with the limit of the sequence $$f(n)-\sum_{k=1}^nf'(k).$$ If $f=\log$, then it is convergent to $-\gamma$ (where $\gamma$ is the Euler-Mascheroni constant). Now,
(a) Are there some criteria for its convergence (by putting some conditions on $f$)?
(b) Does anyone know some references (paper, book, etc.) about it?
• Are you familiar with the Euler-Maclaurin summation formula? – Steven Stadnicki Jun 1 '17 at 6:05
• Yes, I know it. – M.H.Hooshmand Jun 1 '17 at 6:07
• You may be able to build a set of convergence criteria from the error bounds that are part of that formula. – Steven Stadnicki Jun 1 '17 at 6:09 | 2019-07-16T19:01:52 | {
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http://bollu.github.io/elementary-uses-of-sheaves-in-complex-analysis.html | ## § Elementary uses of Sheaves in complex analysis
I always wanted to see sheaves in the wild in a setting that was both elementary but 'correct': In that, it's not some perverse example created to show sheaves (DaTaBaSeS arE ShEAvEs). Ahlfors has a great example of this which I'm condensing here for future reference.
#### § Sheafs: Trial 1
• We have function elements $(f: \Omega \rightarrow \mathbb C, \Omega \subseteq \mathbb C)$.$f$ is complex analytic, $\Omega$ is an open subset of $\mathbb C$.
• Two function elements $(f_1, \Omega_1), (f_2, \Omega_2)$ are said to be analyticcontinuations of each other iff $\Omega_1 \cap \Omega_2 \neq \emptyset$, and$f_1 = f_2$ on the set $\Omega_1 \cap \Omega_2)$.
• $(f_2, \Omega_2)$ can be called as the continuation of $(f_1, \Omega_1)$ toregion $\Omega_2$.
• We will have that the analytic continuation of $f_1$ to $\Omega_2$ is unique.If there exists a function element $(g_2, \Omega_2)$, $(h_2, \Omega_2)$ such that$g_2 = f_1 = h_2$ in the region $\Omega_1 \cap \Omega_2$, then by analyticity,this agreement will extend to all of $\Omega_2$.
• Analytic continuation is therefore an equivalence relation (prove this!)
• A chain of analytic continuations is a sequence of $(f_i, \Omega_i)$ such thatthe adjacent elements of this sequence are analytic continuations of each other.$(f_i, \Omega_i)$ analytically continues $(f_{i+1}, \Omega_{i+1})$.
• Every equivalence class of this equivalence relation is called as a globalanalytic function. Put differently, it's a family of function elements$(f, U)$ and $(g, V)$ such that we can start from $(f, U)$ and buildanalytic continuations to get to $(g, V)$.
#### § Sheafs: Trial 2
• We can take a different view, with $(f, z \in \mathbb C)$ such that $f$is analytic at some open set $\Omega$ which contains $z$. So we shouldpicture an $f$ sitting analytically on some open set $\Omega$ which contains $z$.
• Two pairs $(f, z)$, $(f', z')$ are considered equivalent if $z = z'$ and$f = f'$ is some neighbourhood of $z (= z')$.
• This is clearly an equivalence relation. The equivalence classes are called as germs.
• Each germ $(f, z)$ has a unique projection $z$. We denote a germ of $f$ with projection $z$as $f_z$.
• A function element $(f, \Omega)$ gives rise to germs $(f, z)$ for each $z \in \Omega$.
• Conversely, every germ $(f, z)$ is determined by some function element $(f, \Omega)$since we needed $f$ to be analytic around some open neighbourhood of $z$: Callthis neighbourhood $\Omega$.
• Let $D \subseteq \mathbb C$ be an open set. The set of all germs $\{ f_z : z \in D \}$ is called as a sheaf over $D$. If we are considering analytic $f$ thenthis will be known as the sheaf of germs of analytic functions over $D$. Thissheaf will be denoted as $Sh(D)$.
• There is a projection $\pi: Sh(D) \rightarrow D; (f, z) \mapsto z$. For a fixed $z0 \in D$,the inverse-image $\pi^{-1}(z0)$ is called as the stalk over $z0$. It isdenoted by $Sh(z)$.
• $Sh$ carries both topological and algebraic structure. We can give the sheafa topology to talk about about continuous mappings in and out of $Sh$.It also carries a pointwise algebraic structure at each stalk: we canadd and subtract functions at each stalk; This makes it an abelain group.
#### § Sheaf: Trial 3
A sheaf over $D$ is a topological space $Sh$ and a mapping $\pi: Sh \rightarrow D$ with the properties:
• $\pi$ is a local homeomorphism. Each $s \in S$ has an open neighbourhood $D$such that $\pi(D)$ is open, and the restriction of $\pi$ to $D$ is a homeomorphism.
• For each point $z \in D$, the stalk $\pi^{-1}(z) \equiv S_z$ has the structre of an abeliangroup.
• The group operations are continuous with respect to the topology of $Sh$.
We will pick $D$ to be an open set in the complex plane; Really, $D$ can be arbitrary. | 2021-01-24T02:41:35 | {
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https://socratic.org/questions/what-is-the-derivative-of-cos-a-3-x-3 | # What is the derivative of cos(a^3+x^3)?
Jun 20, 2016
$- \left(3 {x}^{2}\right) \left(\sin \left({a}^{3} + {x}^{3}\right)\right)$
#### Explanation:
Use chain rule for derivatives.
Consider this as $\frac{d}{\mathrm{dx}} \left(\cos \left(f \left(x\right)\right)\right)$ where $f \left(x\right)$ = ${a}^{3} + {x}^{3}$
The answer would be composition of derivatives of $\cos \left(x\right)$ (and putting x as f(x) after differentiating and f(x). Let me demonstrate this in this question.
Derivative of $\cos \left(x\right)$ is $- \sin \left(x\right)$. Now, let's substitute x with f(x)
So the answer is $\left(- \sin \left(f \left(x\right)\right) \times \left(\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right)\right)$.
Now, $f \left(x\right) = {a}^{3} + {x}^{3}$. Assuming a to be a constant, ${a}^{3}$ is a constant and derivative of a constant is $0$. Derivative of ${x}^{3}$ is $3 {x}^{2}$. I won't explain this because you need to learn this yourself if you can't already figure it out.
Ans: $\left(- \sin \left({a}^{3} + {x}^{3}\right) \times \left(\frac{d}{\mathrm{dx}} \left({a}^{3} + {x}^{3}\right)\right)\right)$
Final Ans: $- 3 {x}^{2} \times \sin \left({a}^{3} + {x}^{3}\right)$
Jun 20, 2016
Just another way of saying the same thing
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} \sin \left({a}^{3} + {x}^{3}\right)$
#### Explanation:
Let $u = {a}^{3} + {x}^{3} \text{ "->" } \frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$
Let $y = \cos \left(u\right) \text{ "->" } \frac{\mathrm{dy}}{\mathrm{du}} = - \sin \left(u\right)$
But $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \times \frac{\mathrm{dy}}{\mathrm{du}}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} \sin \left({a}^{3} + {x}^{3}\right)$
Jun 20, 2016
$\frac{d}{\mathrm{dx}} = - \sin \left({a}^{3} + {x}^{3}\right) 3 {x}^{2}$
the main function is $\cos \left(x\right)$
the sub function is ${a}^{3} + {x}^{3}$
$\frac{d}{\mathrm{dx}} = - \sin \left({a}^{3} + {x}^{3}\right) 3 {x}^{2}$ | 2019-06-25T18:25:11 | {
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http://fr.mathworks.com/help/signal/ug/cross-correlation-of-phase-lagged-sine-wave.html?nocookie=true | Accelerating the pace of engineering and science
Documentation
Cross-Correlation of Phase-Lagged Sine Wave
This example shows how to use the cross-correlation sequence to estimate the phase lag between two sine waves. The theoretical cross-correlation sequence of two sine waves at the same frequency also oscillates at that frequency. Because the sample cross-correlation sequence uses fewer and fewer samples at larger lags, the sample cross-correlation sequence also oscillates at the same frequency, but the amplitude decays as the lag increases.
Create two sine waves with frequencies of rad/sample. The starting phase of one sine wave is 0, while the starting phase of the other sine wave is radians. Add white noise to the sine wave with the phase lag of radians. Set the random number generator to the default settings for reproducible results.
rng default
t = 0:99;
x = cos(2*pi*1/10*t);
y = cos(2*pi*1/10*t-pi)+0.25*randn(size(t));
Obtain the sample cross-correlation sequence for two periods of the sine wave (10 samples). Plot the cross-correlation sequence and mark the known lag between the two sine waves (5 samples).
[xc,lags] = xcorr(y,x,20,'coeff');
stem(lags(21:end),xc(21:end),'filled')
hold on
plot([5 5],[-1 1])
ax = gca;
ax.XTick = 0:5:20;
You see that the cross-correlation sequence peaks at lag 5 as expected and oscillates with a period of 10 samples. | 2015-03-02T23:43:46 | {
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http://planetmath.org/DiscreteValuation | # discrete valuation
A on a field $K$ is a valuation $|\cdot|:K\to\mathbb{R}$ whose image is a discrete subset of $\mathbb{R}$.
For any field $K$ with a discrete valuation $|\cdot|$, the set
$R:=\{x\in K:|x|\leq 1\}$
is a subring of $K$ with sole maximal ideal
$M:=\{x\in K:|x|<1\},$
and hence $R$ is a discrete valuation ring. Conversely, given any discrete valuation ring $R$, the field of fractions $K$ of $R$ admits a discrete valuation sending each element $x\in R$ to $c^{n}$, where $0 is some arbitrary fixed constant and $n$ is the order of $x$, and extending multiplicatively to $K$.
Note: Discrete valuations are often written additively instead of multiplicatively; under this alternate viewpoint, the element $x$ maps to $\log_{c}|x|$ (in the above notation) instead of just $|x|$. This transformation reverses the order of the absolute values (since $c<1$), and sends the element $0\in K$ to $\infty$. It has the advantage that every valuation can be normalized by a suitable scalar multiple to take values in the integers.
Title discrete valuation DiscreteValuation 2013-03-22 13:59:14 2013-03-22 13:59:14 djao (24) djao (24) 6 djao (24) Definition msc 13F30 msc 12J20 rank one valuations rank-one valuations DiscreteValuationRing Valuation | 2018-03-17T22:15:52 | {
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https://www.projectrhea.org/rhea/index.php/ECE_PhD_QE_CNSIP_2007_Problem1.3 | Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2007
3. (25 Points)
Let $\mathbf{X}\left(t\right)$ be a real Gaussian random process with mean function $\mu\left(t\right)$ and autocovariance function $C_{\mathbf{XX}}\left(t_{1},t_{2}\right)$ .
(a)
Write the expression for the $n$ -th order characteristic function of $\mathbf{X}\left(t\right)$ in terms of $\mu\left(t\right)$ and $C_{\mathbf{XX}}\left(t_{1},t_{2}\right)$ .
ref.
There are the note about the n-th order characteristic function of Gaussians random process . The only difference between the note and this problem is that this problem use the $\mu\left(t\right)$ rather than $\eta_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right]$ .
Solution
$\Phi_{\mathbf{X}\left(t_{1}\right)\cdots\mathbf{X}\left(t_{n}\right)}\left(\omega_{1},\cdots,\omega_{n}\right)=\exp\left\{ i\sum_{k=1}^{n}\mu_{\mathbf{X}}\left(t_{k}\right)\omega_{k}-\frac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}C_{\mathbf{XX}}\left(t_{j},t_{k}\right)\omega_{j}\omega_{k}\right\}$ .
(b)
Show that the probabilistic description of $\mathbf{X}\left(t\right)$ is completely characterized by $\mu\left(t\right)$ and autocovariance function $C_{\mathbf{XX}}\left(t_{1},t_{2}\right)$ .
Solution
From (a), the characteristic function of $\mathbf{X}\left(t\right)$ is specified completely in terms of $\mu_{\mathbf{X}}\left(t\right)$ and $C_{\mathbf{XX}}\left(t_{1},t_{2}\right)$ . Thus, probabilistic description of $\mathbf{X}\left(t\right)$ is completely characterized by the characteristic function.
Note
$f_{\mathbf{X}}\left(x\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega x}\Phi_{\mathbf{X}}\left(\omega\right)d\omega.$
(c)
Show that if $\mathbf{X}\left(t\right)$ is wide-sense stationary then it is also strict-sense stationary.
Note
You can use the theorem and its proof for solving this problem.
## Alumni Liaison
To all math majors: "Mathematics is a wonderfully rich subject."
Dr. Paul Garrett | 2019-01-21T13:45:00 | {
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https://learn.careers360.com/ncert/question-find-the-distance-of-the-point-minus-1-minus-5-minus-10-from-the-point-of-intersection-of-the-line-r-vector-equal-to-2-i-caret-minus-j-caret-plus-2-k-caret-plus-lambda-3-i-caret-plus-4-j-caret-plus-2-k-caret-and-the-plane/ | Q
# Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line r vector = 2 i caret - j caret + 2 k caret + lambda 3 i caret + 4 j caret + 2 k caret and the plane
18 Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$ and the plane $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$.
Views
Given,
Equation of a line :
$\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$
Equation of the plane
$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$
Let's first find out the point of intersection of line and plane.
putting the value of $\vec r$ into the equation of a plane from the equation from line
$\left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5$
$(2+3\lambda)-(4\lambda -1)+(2+2\lambda)=5$
$\lambda+5=5$
$\lambda=0$
Now, from the equation, any point p in line is
$P=(2+3\lambda,4\lambda-1,2+2\lambda)$
So the point of intersection is
$P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)$
SO, Now,
The distance between the points (-1,-5,-10) and (2,-1,2) is
$d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}$
$d=\sqrt{169}=13$
Hence the required distance is 13.
Exams
Articles
Questions | 2020-02-29T03:46:30 | {
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https://en.m.wikipedia.org/wiki/Bayesian_hierarchical_modeling | # Bayesian hierarchical modeling
Bayesian hierarchical modelling is a statistical model written in multiple levels (hierarchical form) that estimates the parameters of the posterior distribution using the Bayesian method.[1] The sub-models combine to form the hierarchical model, and Bayes' theorem is used to integrate them with the observed data and account for all the uncertainty that is present. The result of this integration is the posterior distribution, also known as the updated probability estimate, as additional evidence on the prior distribution is acquired.
Frequentist statistics, the more popular foundation of statistics[citation needed], may yield conclusions seemingly incompatible with those offered by Bayesian statistics due to the Bayesian treatment of the parameters as random variables and its use of subjective information in establishing assumptions on these parameters.[2] As the approaches answer different questions the formal results aren't technically contradictory but the two approaches disagree over which answer is relevant to particular applications. Bayesians argue that relevant information regarding decision making and updating beliefs cannot be ignored and that hierarchical modeling has the potential to overrule classical methods in applications where respondents give multiple observational data. Moreover, the model has proven to be robust, with the posterior distribution less sensitive to the more flexible hierarchical priors.
Hierarchical modeling is used when information is available on several different levels of observational units. The hierarchical form of analysis and organization helps in the understanding of multiparameter problems and also plays an important role in developing computational strategies.[3]
## Philosophy
Numerous statistical applications involve multiple parameters that can be regarded as related or connected in such a way that the problem implies dependence of the joint probability model for these parameters.[4] Individual degrees of belief, expressed in the form of probabilities, come with uncertainty.[5] Amidst this is the change of the degrees of belief over time. As was stated by Professor José M. Bernardo and Professor Adrian F. Smith, “The actuality of the learning process consists in the evolution of individual and subjective beliefs about the reality.” These subjective probabilities are more directly involved in the mind rather than the physical probabilities.[6] Hence, it is with this need of updating beliefs that Bayesians have formulated an alternative statistical model which takes into account the prior occurrence of a particular event.[7]
## Bayes' theorem
The assumed occurrence of a real-world event will typically modify preferences between certain options. This is done by modifying the degrees of belief attached, by an individual, to the events defining the options.[8]
Suppose in a study of the effectiveness of cardiac treatments, with the patients in hospital j having survival probability ${\displaystyle \theta _{j}}$ , the survival probability will be updated with the occurrence of y, the event in which a hypothetical controversial serum is created which, as believed by some, increases survival in cardiac patients.
In order to make updated probability statements about ${\displaystyle \theta _{j}}$ , given the occurrence of event y, we must begin with a model providing a joint probability distribution for ${\displaystyle \theta _{j}}$ and y. This can be written as a product of the two distributions that are often referred to as the prior distribution ${\displaystyle P(\theta )}$ and the sampling distribution ${\displaystyle P(y\mid \theta )}$ respectively:
${\displaystyle P(\theta ,y)=P(\theta )P(y\mid \theta )}$
Using the basic property of conditional probability, the posterior distribution will yield:
${\displaystyle P(\theta \mid y)={\frac {P(\theta ,y)}{P(y)}}={\frac {P(y\mid \theta )P(\theta )}{P(y)}}}$
This equation, showing the relationship between the conditional probability and the individual events, is known as Bayes' theorem. This simple expression encapsulates the technical core of Bayesian inference which aims to incorporate the updated belief, ${\displaystyle P(\theta \mid y)}$ , in appropriate and solvable ways.[8]
## Exchangeability
The usual starting point of a statistical analysis is the assumption that the n values ${\displaystyle y_{n}}$ are exchangeable. If no information – other than data y – is available to distinguish any of the ${\displaystyle \theta _{j}}$ ’s from any others, and no ordering or grouping of the parameters can be made, one must assume symmetry among the parameters in their prior distribution.[9] This symmetry is represented probabilistically by exchangeability. Generally, it is useful and appropriate to model data from an exchangeable distribution, as independently and identically distributed, given some unknown parameter vector ${\displaystyle \theta }$ , with distribution ${\displaystyle P(\theta )}$ .
### Finite exchangeability
For a fixed number n, the set ${\displaystyle y_{1},y_{2},\ldots ,y_{n}}$ is exchangeable if the joint probability ${\displaystyle P(y_{1},y_{2},\ldots ,y_{n})}$ is invariant under permutations of the indices. That is, for every permutation ${\displaystyle \pi }$ or ${\displaystyle (\pi _{1},\pi _{2},\ldots ,\pi _{n})}$ of (1, 2, …, n), ${\displaystyle P(y_{1},y_{2},\ldots ,y_{n})=P(y_{\pi _{1}},y_{\pi _{2}},\ldots ,y_{\pi _{n}}).}$ [10]
Following is an exchangeable, but not independent and identical (iid), example: Consider an urn with red balls and blue balls inside, with probability ${\displaystyle {\frac {1}{2}}}$ of drawing either. Balls are drawn without replacement, i.e. after one ball is drawn from the n balls, there will be n − 1 remaining balls left for the next draw.
${\displaystyle {\text{Let }}Y_{i}={\begin{cases}1,&{\text{if the }}i{\text{th ball is red}},\\0,&{\text{otherwise}}.\end{cases}}}$
Since the probability of selecting a red ball in the first draw and a blue ball in the second draw is equal to the probability of selecting a blue ball on the first draw and a red on the second draw, both of which are equal to 1/2 (i.e. ${\displaystyle [P(y_{1}=1,y_{2}=0)=P(y_{1}=0,y_{2}=1)={\frac {1}{2}}]}$ ), then ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are exchangeable.
But the probability of selecting a red ball on the second draw given that the red ball has already been selected in the first draw is 0, and is not equal to the probability that the red ball is selected in the second draw which is equal to 1/2 (i.e. ${\displaystyle [P(y_{2}=1\mid y_{1}=1)=0\neq P(y_{2}=1)={\frac {1}{2}}]}$ ). Thus, ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are not independent.
If ${\displaystyle x_{1},\ldots ,x_{n}}$ are independent and identically distributed, then they are exchangeable, but the converse is not necessarily true.[11]
### Infinite exchangeability
Infinite exchangeability is the property that every finite subset of an infinite sequence ${\displaystyle y_{1}}$ , ${\displaystyle y_{2},\ldots }$ is exchangeable. That is, for any n, the sequence ${\displaystyle y_{1},y_{2},\ldots ,y_{n}}$ is exchangeable.[11]
## Hierarchical models
### Components
Bayesian hierarchical modeling makes use of two important concepts in deriving the posterior distribution,[1] namely:
1. Hyperparameters: parameters of the prior distribution
2. Hyperpriors: distributions of Hyperparameters
Suppose a random variable Y follows a normal distribution with parameter θ as the mean and 1 as the variance, that is ${\displaystyle Y\mid \theta \sim N(\theta ,1)}$ . Suppose also that the parameter ${\displaystyle \theta }$ has a distribution given by a normal distribution with mean ${\displaystyle \mu }$ and variance 1, i.e. ${\displaystyle \theta \mid \mu \sim N(\mu ,1)}$ . Furthermore, ${\displaystyle \mu }$ follows another distribution given, for example, by the standard normal distribution, ${\displaystyle {\text{N}}(0,1)}$ . The parameter ${\displaystyle \mu }$ is called the hyperparameter, while its distribution given by ${\displaystyle {\text{N}}(0,1)}$ is an example of a hyperprior distribution. The notation of the distribution of Y changes as another parameter is added, i.e. ${\displaystyle Y\mid \theta ,\mu \sim N(\theta ,1)}$ . If there is another stage, say, ${\displaystyle \mu }$ follows another normal distribution with mean ${\displaystyle \beta }$ and variance ${\displaystyle \epsilon }$ , meaning ${\displaystyle \mu \sim N(\beta ,\epsilon )}$ , ${\displaystyle {\mbox{ }}}$ ${\displaystyle \beta }$ and ${\displaystyle \epsilon }$ can also be called hyperparameters while their distributions are hyperprior distributions as well.[4]
### Framework
Let ${\displaystyle y_{j}}$ be an observation and ${\displaystyle \theta _{j}}$ a parameter governing the data generating process for ${\displaystyle y_{j}}$ . Assume further that the parameters ${\displaystyle \theta _{1},\theta _{2},\ldots ,\theta _{j}}$ are generated exchangeably from a common population, with distribution governed by a hyperparameter ${\displaystyle \phi }$ .
The Bayesian hierarchical model contains the following stages:
${\displaystyle {\text{Stage I: }}y_{j}\mid \theta _{j},\phi \sim P(y_{j}\mid \theta _{j},\phi )}$
${\displaystyle {\text{Stage II: }}\theta _{j}\mid \phi \sim P(\theta _{j}\mid \phi )}$
${\displaystyle {\text{Stage III: }}\phi \sim P(\phi )}$
The likelihood, as seen in stage I is ${\displaystyle P(y_{j}\mid \theta _{j},\phi )}$ , with ${\displaystyle P(\theta _{j},\phi )}$ as its prior distribution. Note that the likelihood depends on ${\displaystyle \phi }$ only through ${\displaystyle \theta _{j}}$ .
The prior distribution from stage I can be broken down into:
${\displaystyle P(\theta _{j},\phi )=P(\theta _{j}\mid \phi )P(\phi )}$ [from the definition of conditional probability]
With ${\displaystyle \phi }$ as its hyperparameter with hyperprior distribution, ${\displaystyle P(\phi )}$ .
Thus, the posterior distribution is proportional to:
${\displaystyle P(\phi ,\theta _{j}\mid y)\propto P(y_{j}\mid \theta _{j},\phi )P(\theta _{j},\phi )}$ [using Bayes' Theorem]
${\displaystyle P(\phi ,\theta _{j}\mid y)\propto P(y_{j}\mid \theta _{j})P(\theta _{j}\mid \phi )P(\phi )}$ [12]
### Example
To further illustrate this, consider the example: A teacher wants to estimate how well a male student did in his SAT. He uses information on the student’s high school grades and his current grade point average (GPA) to come up with an estimate. His current GPA, denoted by ${\displaystyle Y}$ , has a likelihood given by some probability function with parameter ${\displaystyle \theta }$ , i.e. ${\displaystyle Y\mid \theta \sim P(Y\mid \theta )}$ . This parameter ${\displaystyle \theta }$ is the SAT score of the student. The SAT score is viewed as a sample coming from a common population distribution indexed by another parameter ${\displaystyle \phi }$ , which is the high school grade of the student.[13] That is, ${\displaystyle \theta \mid \phi \sim P(\theta \mid \phi )}$ . Moreover, the hyperparameter ${\displaystyle \phi }$ follows its own distribution given by ${\displaystyle P(\phi )}$ , a hyperprior. To solve for the SAT score given information on the GPA,
${\displaystyle P(\theta ,\phi \mid Y)\propto P(Y\mid \theta ,\phi )P(\theta ,\phi )}$
${\displaystyle P(\theta ,\phi \mid Y)\propto P(Y\mid \theta )P(\theta \mid \phi )P(\phi )}$
All information in the problem will be used to solve for the posterior distribution. Instead of solving only using the prior distribution and the likelihood function, the use of hyperpriors gives more information to make more accurate beliefs in the behavior of a parameter.[14]
### 2-stage hierarchical model
In general, the joint posterior distribution of interest in 2-stage hierarchical models is:
${\displaystyle P(\theta ,\phi \mid Y)={P(Y\mid \theta ,\phi )P(\theta ,\phi ) \over P(Y)}={P(Y\mid \theta )P(\theta \mid \phi )P(\phi ) \over P(Y)}}$
${\displaystyle P(\theta ,\phi \mid Y)\propto P(Y\mid \theta )P(\theta \mid \phi )P(\phi )}$ [14]
### 3-stage hierarchical model
For 3-stage hierarchical models, the posterior distribution is given by:
${\displaystyle P(\theta ,\phi ,X\mid Y)={P(Y\mid \theta )P(\theta \mid \phi )P(\phi \mid X)P(X) \over P(Y)}}$
${\displaystyle P(\theta ,\phi ,X\mid Y)\propto P(Y\mid \theta )P(\theta \mid \phi )P(\phi \mid X)P(X)}$ [14]
## References
1. ^ a b Allenby, Rossi, McCulloch (January 2005). "Hierarchical Bayes Model: A Practitioner’s Guide". Journal of Bayesian Applications in Marketing, pp. 1–4. Retrieved 26 April 2014, p. 3
2. ^ Gelman, Andrew; Carlin, John B.; Stern, Hal S. & Rubin, Donald B. (2004). Bayesian Data Analysis (second ed.). Boca Raton, Florida: CRC Press. pp. 4–5. ISBN 1-58488-388-X.
3. ^ Gelman et al. 2004, p. 6.
4. ^ a b Gelman et al. 2004, p. 117.
5. ^ Good, I.J. (February 1980). “Some history of the hierarchical Bayesian methodology”. Trabajos de Estadistica Y de Investigacion Operativa Volume 31 Issue 1. Springer – Verlag, p. 480
6. ^ Good, I.J. (February 1980). “Some history of the hierarchical Bayesian methodology”. Trabajos de Estadistica Y de Investigacion Operativa Volume 31 Issue 1. Springer – Verlag, pp. 489–490
7. ^ Bernardo, Smith(1994). Bayesian Theory. Chichester, England: John Wiley & Sons, ISBN 0-471-92416-4, p. 23
8. ^ a b Gelman et al. 2004, pp. 6–8.
9. ^ Bernardo, Degroot, Lindley (September 1983). “Proceedings of the Second Valencia International Meeting”. Bayesian Statistics 2. Amsterdam: Elsevier Science Publishers B.V, ISBN 0-444-87746-0, pp. 167–168
10. ^ Gelman et al. 2004, pp. 121–125.
11. ^ a b Diaconis, Freedman (1980). “Finite exchangeable sequences”. Annals of Probability, pp. 745–747
12. ^ Bernardo, Degroot, Lindley (September 1983). “Proceedings of the Second Valencia International Meeting”. Bayesian Statistics 2. Amsterdam: Elsevier Science Publishers B.V, ISBN 0-444-87746-0, pp. 371–372
13. ^ Gelman et al. 2004, pp. 120–121.
14. ^ a b c Box G. E. P., Tiao G. C. (1965). "Multiparameter problem from a bayesian point of view". Multiparameter Problems From A Bayesian Point of View Volume 36 Number 5. New York City: John Wiley & Sons, ISBN 0-471-57428-7 | 2019-01-24T06:18:40 | {
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https://math.stackexchange.com/questions/1484411/providing-initial-conditions-for-differential-equation-phase-portrait/1484418#1484418 | # Providing initial conditions for differential equation? - Phase portrait
I'm given the system of differential equations \begin{align*} \begin{cases} x'(t) &= y(1 + x^2 + y^2) \\ y'(t) &= x (1 + x^2 + y^2) \end{cases} \end{align*} I need to plot (with Maple) the phase portrait of this, and also the direction field. From this I need to determine the nature and stability of the critical point $(0, 0)$. Now, for this critical point we have the linearized system (which results when you set all the non-linear terms zero) \begin{align*} \begin{pmatrix} \tilde{x}'(t) \\ \tilde{y}'(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \tilde{x}(t) \\ \tilde{y}(t) \end{pmatrix} \end{align*} I used $\tilde{x}$ notation to make clear this is an approximation and not the same system as the original one. The eigenvalues of this matrix are $\lambda = \pm 1$, which means this critical point is a saddle point, and this is always unstable.
Now, I was wondering if there some trick to provide 'nice' initial conditions, such that the phase portrait shows clearly the solution curves? I entered this in maple:
DEplot([diff(x(t), t) = y(t)*(1+x(t)^2+y(t)^2), diff(y(t), t) = x(t)*(1+x(t)^2+y(t)^2)],
[x(t), y(t)], t = 0 .. 5, [[x(0) = -2, y(0) = -2], [x(0) = 3, y(0) = 0]],
x = -3 .. 3, y = -5 .. 5, arrows = medium, stepsize = 0.05, linecolor = blue)
The result I got was:
But I'm not satisfied with this, because the solution curves are not nicely shown. How does one specify the initial conditions such that this happens? Is it just by trial and error?
Eg: Something like $x(0)=-3$, $y(0)=2$ should get you a good picture of a solution curve on the left of the origin. | 2021-09-24T13:13:19 | {
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http://mathhelpforum.com/differential-geometry/117644-using-creeping-lemma-prove-ivt.html | ## Using creeping lemma to prove IVT
The quote below is an outline of an intermediate value theorem proof that uses the creeping lemma.
"Assume that $f(x)-L$ is zero for no $c\in[a,b]$. Then, prove that $f(a)-L$ and $f(b)-L$ are of the same sign by applying the creeping lemma with $\rho$ the relation defined by requiring $u\rho\\v$ to be true if and only if $f(u)-L$ and $f(v)-L$ are of the same sign"
Where the creeping lemma is: Let $\rho$ be a transitive relation on the interval $[a, b]$. If each $x\in[a, b]$ has a neighborhood $N_{x}$ such that $u\rho\\v$ whenever $u\in[a, x]\cap\\N_{x}$ and $v\in[x, b]\cap\\N_{x}$ , then $a\rho\\b$.
So, with the above information I need to prove the intermediate value theorem, but I am not sure I understand the outline of the proof entirely and what parts of the lemma I am free to use without proof.
proof: Suppose $f$ is continuous on $[a,b]$ and there is a real number $L$ satisfying $f(a). Now, we can consider this statement in terms of the function $\varphi(x)=f(x)-L$. Suppose there is no $c\in[a,b]$ such that $\varphi(x)=f(x)-L=0$; then, without loss of generality $\varphi(a)<0$ and $\varphi(b)>0$. So, we want to show, at some point $c\in[a,b]$, $\varphi(c)=0$. I do not know how to use the relation $\rho$ given in the above quote to produce the rest of this proof.
If anyone knows about the creeping lemma and how to apply it, I'd really appreciate some advice on how to move forward.
Thanks | 2017-06-28T11:00:56 | {
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https://www.statisticsassignmenthelp.com/labor-cost-optimization | Using Statistics to Optimize Cost and Labor
Statistics has been used in the optimization of cost and laborfor many years. It provides methodologies and principles that help business managers to minimize variances in operations. The process usually requires automation, and thanks to the various statistical programs we have today, data analysts can be able to explore and analyze different sets of company data to identify patterns and trends that may aid in making informed decisions regarding labor and cost optimization.
Combination of labor and capital
The optimal combination of Labor (L) and Capital (K) (represented by machines) for fixed output, is achieved through cost minimization. This optimal point is found where the slope of the Iso-cost is equal to the slope of the Isoquant curve. An iso-cost reflects the possible combination of production factors that hold the same total cost, while the isoquant is the possible combination of production factors that generate the same level of output (q).
The isoquant is represented by:
q=F(L,K)Where the output (q) is a function of the production factors Labour (L) and Capital (K). And the iso-cost is represented by
C=w*L+r*K
Where C is the Total fixed cost, L and K are the units of labor and machines, and w and r represent the price per unit of labor and capital respectively.
With all this information, and assuming that the given combinations of L and K are all the existing combinations to achieve the desired level of output, the optimal combination can be found where the C value is the lowest, knowing that:
w=2
r=3
so;
With a higher output of 20, both L and K are higher, and the iso-cost is also higher. The optimal input combination is 21 units of labor and 8 units of capital.
Q=10 Q=20 Q=30 Q=40 Labor (L) Capital (K) Cost (C ) Labor (L) Capital (K) Cost (C ) Labor (L) Capital (K) Cost (C ) Labor (L) Capital (K) Cost (C ) 1 14 44 9 14 60 21 14 84 39 14 120 3 14 48 11 14 64 25 14 92 45 14 132 8 14 58 15 14 72 30 14 102 53 14 148 12 14 66 21 14 84 33 14 108 60 14 162 15 14 72 28 14 98 39 14 120 66 14 174 21 14 84 39 14 120 45 14 132 72 14 186 39 14 120 57 14 156 56 14 154 83 14 208 87 14 216
Quantity (q) Fixed Cost (FC) Variable Cost (VC) Total Cost (TC) Average Variable Cost (AVC) Marginal Cost (MC) 10 42 2 44 0.2 - 20 42 18 60 0.9 1.6 30 42 42 84 1.4 2.4 40 42 78 120 1.95 3.6
Applied statistics
a) The marginal product of labor is the change in output resulting from each new unit of labor added:
Units of Labor (L) Output (q) Marginal Product (MPL) Marginal revenue Product (MRPL) Fixed Cost (FC) Variable Cost (VC) Total Cost (TC) Marginal Cost (MC) revenue (r ) Marginal Revenue (MR) Profit (u) 1 5 $50.00$ 10.00 $60.00$ 10.00 $50.00 2 15 10 20$ 50.00 $20.00$ 70.00 $1.00$ 30.00 $2.00$40.00 3 30 15 30 $50.00$ 30.00 $80.00$ 0.67 $60.00$ 2.00 $20.00 4 50 20 40$ 50.00 $40.00$ 90.00 $0.50$ 100.00 $2.00$10.00 5 65 15 30 $50.00$ 50.00 $100.00$ 0.67 $130.00$ 2.00 $30.00 6 77 12 24$ 50.00 $60.00$ 110.00 $0.83$ 154.00 $2.00$44.00 7 86 9 18 $50.00$ 70.00 $120.00$ 1.11 $172.00$ 2.00 $52.00 8 94 8 16$ 50.00 $80.00$ 130.00 $1.25$ 188.00 $2.00$58.00 9 98 4 8 $50.00$ 90.00 $140.00$ 2.50 $196.00$ 2.00 $56.00 10 96 -2 -4$ 50.00 $100.00$ 150.00 $(5.00)$ 192.00 $2.00$42.00
MPL increases with the first 4 employees, but it starts decreasing after the 5th. By the 10th employee, the MPL is negative, meaning that this new employee is causing the total output to decrease.
VC increases with each unit of labor added, and the marginal cost, like the marginal product of labor, gets worse with the 5th unit of labor added.
This firm, with its cost and production function as it is, sees positive profits when producing 50 units of output.
b) Maximization of profit through optimization of inputs comes directly from the relationship between iso-cost and isoquant. The maximization point is where the slopes of both curves are the same:
c) Either through labor or output, the maximization point is the same: 8 employees that produce 94 units of output. So, for this firm, it doesn´t matter if it chooses to maximize labor or output, because the firm is within a perfectly competitive market, with no control of either the output´s price or the cost of labor.
To get professional assistance with this area, connect with our providers of applied statistics assignment help.
Econometrics
If both firms collude, they will be behaving effectively as a monopoly, so the output that maximizes profit will be a function of: | 2021-03-07T02:16:33 | {
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https://plainmath.net/46459/what-did-you-learn-from-binomial-and-hyper-geometric-distributions | # What did you learn from binomial and Hyper- geometric distributions?
What did you learn from binomial and Hyper- geometric distributions? Write a brief note of five lines on these distributions.
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Foreckije
Step 1
BINOMIAL DISTRIBUTION
$$\displaystyle\Rightarrow$$ It is a discrete probability distribution which gives the theoretical probabilities.
$$\displaystyle\Rightarrow$$ Binomial distribution allow us to deal with circumstances in which the outcomes belong to two relevant categories such as success or failure.
$$\displaystyle\Rightarrow$$ It depends on the parameter p or q i.e. the probability of success or failure and n (i.e. the number of trials). The parameter n is always a positive integer.
$$\displaystyle\Rightarrow$$ The distribution will be symmetrical if $$\displaystyle{p}={q}$$. It is skew-symmetric or asymmetric if $$\displaystyle{p}\ne{q}$$.
$$\displaystyle\Rightarrow$$ The possibility of outcome of any trial does not change and is independent of the results of previous trials.
Formula
$$\displaystyle{P}{\left({x}\right)}=^{{{n}}}{C}_{{{x}}}\cdot{p}^{{{x}}}\cdot{q}^{{{n}-{x}}}$$ where $$\displaystyle{x}\Rightarrow$$ number of successes
Step 2
HYPER GEOMMETRIC DISTRIBUTION
$$\displaystyle\Rightarrow$$ A hypergeometric distribution has a specified number of dependent trials having two possible outcomes, success or failure.
$$\displaystyle\Rightarrow$$ The random variable is the number of successful outcomes in the specified number of trials.
$$\displaystyle\Rightarrow$$ In statistics and probability theory, the hypergeometric distribution is a discrete probability distribution that describes the probability of successes in draws without replacement.
$$\displaystyle\Rightarrow$$ The probability of a success is not the same on each trial without replacement, thus events are not independent.
Formula
$$\displaystyle{P}{\left({x}\right)}={\frac{{{s}{C}_{{{k}}}\cdot^{{{N}-{S}{C}_{{{N}-{S}}}}}}}{{{N}{C}_{{{n}}}}}}$$
where $$\displaystyle{S}\Rightarrow$$ successes from population
$$\displaystyle{N}\Rightarrow$$ population size
$$\displaystyle{n}\Rightarrow$$ sample size
$$\displaystyle{k}\Rightarrow$$ successes from sample
###### Not exactly what you’re looking for?
ramirezhereva
Step 1
Consider a random experiment involving n independent trials, such that the outcome of each trial can be classified as either a “success” or a “failure”. The numerical value “1” is assigned to each success and “0” is assigned to each failure.
Moreover, the probability of getting a success in each trial, p, remains a constant for all the n trials. Denote the probability of failure as q. As success and failure are mutually exclusive, $$\displaystyle{q}={1}-{p}$$.
Let the random variable X denote the number of successes obtained from the n trials. Thus, X can take any of the values $$\displaystyle{0},\ {1},\ {2},с\dot{{s}},\ {n}.$$
Then, the probability distribution of X is a Binomial distribution with parameters (n, p) and the probability mass function (pmf) of X, that is, of a Binomial random variable, is given as:
$f(x)=\begin{cases}(\begin{array}{c}n\\ x\end{array})p^{x}q^{n-x} & x=0,\ 1,\dots\end{cases} Step 2 Hyper geometric distribution: A hyper geometric distribution is a discrete probability distribution that determines the probability of getting k successes in n draws (without replacement) from a finite population of size N that contains exactly K success states. Denote the total number of successes as k, Denote the total number of objects that are drawn without replacement as n, Denote the population size as N, Denote the total number of success states in the population as K. The probability distribution of k is a hyper geometric distribution with parameters (N, K, n) and the probability mass function (pmf) of k is given as: \[P(k)=\frac{(\begin{array}{c}K\\ k\end{array})\times(\begin{array}{c}N-K\\ n-k\end{array})}{(\begin{array}{c}N\\ n\end{array})}$
nick1337
Binomial distribution :
1. Trials are independent
2. Occurences are classified into 2 categories namely success and failure
3. Probability of success is a constant in every single trial
4. Trials are WR
5. Population is finite .
Hypergeometric distribution :
1. Trials are WOR
2. Trials are dependent .
3. Probability of success changesw in every trial.
4. When population size is large , binomial becomes approximately equal to hypergeometric .
5. Population is infinite | 2022-01-20T08:33:46 | {
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https://socratic.org/questions/how-do-you-find-the-common-ratio-of-3-6-12-24-1 | # How do you find the common ratio of -3,6,-12,24,...?
Common ratio is $- 2$
Here we have the series $\left\{- 3 , 6 , - 12 , 24 , \ldots \ldots \ldots \ldots \ldots . .\right\}$
and hence common ratio is $\frac{6}{-} 3 = - \frac{12}{6} = \frac{24}{-} 12 = - 2$ | 2021-10-26T12:10:38 | {
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http://www.ck12.org/algebra/Applications-of-Linear-Graphs/lesson/Applications-of-Linear-Functions-Honors/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Applications of Linear Graphs
## Linear equations in word problems
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Practice Applications of Linear Graphs
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Applications of Linear Functions
Joe’s Warehouse has banquet facilities to accommodate a maximum of 250 people. When the manager quotes a price for a banquet she is including the cost of renting the room plus the cost of the meal. A banquet for 70 people costs $1300. For 120 people, the price is$2200.
(a) Plot a graph of cost versus the number of people.
(b) From the graph, estimate the cost of a banquet for 150 people.
(c) Determine the slope of the line. What quantity does the slope of the line represent?
(d) Write an equation to model this real-life situation.
### Guidance
Linear relationships are often used to model real-life situations. In order to create an equation and graph to model the real-life situation, you need at least two data values related to the real-life situation. When the data values have been represented graphically and the equation of the line has been determined, questions relating to the real-life situation can be presented and answered.
When equations and graphs are used to model real-life situations, the domain of the graph is sometimes xϵN\begin{align*}x \epsilon N\end{align*}. However, it is often more convenient to sketch the graph as though xϵR\begin{align*}x \epsilon R\end{align*} instead of showing the function as a series of points in the plane.
#### Example A
A cab company charges $2.00 for the first 0.6 miles and$0.50 for each additional 0.2 miles.
(a) Draw the graph of cost versus distance.
(b) Determine the equations that model this situation.
(c) What is the cost to travel 16 miles by cab?
Solution: This example demonstrates a real-life situation that cannot be modeled with just one equation.
(a) On the x\begin{align*}x\end{align*}-axis is the distance in miles and on the y\begin{align*}y\end{align*}-axis is the cost in dollars. The first line from A\begin{align*}A\end{align*} to B\begin{align*}B\end{align*} extends horizontally across the distance from 0 to 0.6 miles. The cost is constant at 2.00. The equation for this constant function is y=2.00\begin{align*}y=2.00\end{align*} or c=2.00\begin{align*}c=2.00\end{align*}. The second line from B\begin{align*}B\end{align*} to C\begin{align*}C\end{align*} and upward is not constant. (b) The equation that models the second graph can be determined by using the data points (0.6, 2.00) and (1, 3.00) mmmmmyy1y2y2y2+2yc=y2y1x2x1=3.002.0010.6=3.002.0010.6=1.000.4=2.5=m(xx1)=2.5(x0.6)=2.5x1.5=2.5x1.5+2=2.5x+0.5=2.5d+0.5Use the data points to calculate the slope.Use the slope and one point to determine the equation. Therefore, the equations that model this situation are: c={2.002.5d+0.50<d0.6d>0.6} (c) The cost to travel 16 miles in the cab is: The distance is greater than 0.6 miles. The cost must be calculated using the equation c=2.5d+0.5\begin{align*}c=2.5d+0.5\end{align*}. Substitute 16 in for ‘d\begin{align*}d\end{align*}’. c=2.5d+0.5c=2.5(16)+0.5c=40+0.5c=40.50
#### Example B
When a 40 gram mass was suspended from a coil spring, the length of the spring was 24 inches. When an 80 gram mass was suspended from the same coil spring, the length of the spring was 36 inches.
(a) Plot a graph of length versus mass.
(b) From the graph, estimate the length of the spring for a mass of 70 grams.
(c) Determine an equation that models this situation. Write the equation in slope-intercept form.
(d) Use the equation to determine the length of the spring for a mass of 60 grams.
(e) What is the y\begin{align*}y\end{align*}-intercept? What does the y\begin{align*}y\end{align*}-intercept represent?
Solution:
(a) On the x\begin{align*}x\end{align*}-axis is the mass in grams and on the y\begin{align*}y\end{align*}-axis is the length of the spring in inches.
(b) The length of the coil spring for a mass of 70 grams is approximately 33 inches.
(c) The equation of the line can be determined by using the two data values (40, 24) and (80, 36).
mmm my242424241212=y2y1x2x1=36248040=1240=310=mx+b=310(40)+b=310(404)+b=12+b=1212+b=b
The y\begin{align*}y\end{align*}-intercept is (0, 12). The equation that models the situation is
y=310x+12\begin{align*}y=\frac{3}{10}x+12\end{align*}
l=310m+12
where ‘l\begin{align*}l\end{align*}’ is the length of the spring in inches and ‘m\begin{align*}m\end{align*}’ is the mass in grams.
(d)
l=310m+12l=310(60)+12l=310(606)+12l=18+12l=30 inchesUse the equation and substitute 60 in for m.
(e) The y\begin{align*}y\end{align*}-intercept is (0, 12). The y\begin{align*}y\end{align*}-intercept represents the length of the coil spring before a mass was suspended from it. The length of the coil spring was 12 inches.
#### Example C
Juan drove from his mother’s home to his sister’s home. After driving for 20 minutes he was 62 miles away from his sister’s home and after driving for 32 minutes he was only 38 miles away. The time driving and the distance away from his sister’s home form a linear relationship.
(a) What is the independent variable? What is the dependent variable?
(b) What are the two data values?
(c) Draw a graph to represent this problem. Label the axis appropriately.
(d) Write an equation expressing distance in terms of time driving.
(e) What is the slope and what is its meaning in this problem?
(f) What is the time-intercept and what does it represent?
(g) What is the distance-intercept and what does it represent?
(h) How far is Juan from his sister’s home after he had been driving for 35 minutes?
Solution:
(a) The independent variable is the time driving. The dependent variable is the distance.
(b) The two data values are (20, 62) and (32, 38).
(c) On the x\begin{align*}x\end{align*}-axis is the time in minutes and on the y\begin{align*}y\end{align*}-axis is the distance in miles.
(d) (20, 62) and (32, 38) are the coordinates that will be used to calculate the slope of the line.
mmmmy626262+40102=y2y1x2x1=38623220=2412=2=mx+b=2(20)+b=40+b=40+40+b=bThe y-intercept is (0,102)
yyd=mx+b=2x+102=2t+102
(e) The slope is 2=21=2(miles)1(minute)\begin{align*}-2=\frac{-2}{1}=\frac{-2(miles)}{1(minute)}\end{align*}. The slope means that for each minute of driving, the distance that Juan has to drive to his sister’s home is reduced by 2 miles.
(f) The time-intercept is actually the x\begin{align*}x\end{align*}-intercept. This value is:
d00+2t2t2t22t2t=2t+102=2t+102=2t+2t+102=102=1022=1022=51 minutesSet d=0 and solve for t.
The time-intercept is 51 minutes and this represents the time it took Juan to drive from his mother’s home to his sister’s home.
(g) The distance-intercept is the y\begin{align*}y\end{align*}-intercept. This value has been calculated as (0, 102). The distance-intercept represents the distance between his mother’s home and his sister’s home. The distance is 102 miles.
(h)
dddd=2t+102=2(35)+102=70+102=32 milesSubstitute 35 into the equation for t and solve for d.
After driving for 35 minutes, Juan is 32 miles from his sister’s home.
#### Concept Problem Revisited
Joe’s Warehouse has banquet facilities to accommodate a maximum of 250 people. When the manager quotes a price for a banquet she is including the cost of renting the room plus the cost of the meal. A banquet for 70 people costs $1300. For 120 people, the price is$2200.
(a) Plot a graph of cost versus the number of people.
(b) From the graph, estimate the cost of a banquet for 150 people.
(c) Determine the slope of the line. What quantity does the slope of the line represent?
(d) Write an equation to model this real-life situation.
Solution:
(a) On the \begin{align*}x\end{align*}-axis is the number of people and on the \begin{align*}y\end{align*}-axis is the cost of the banquet.
(b) The approximate cost of a banquet for 150 people is $2700. (c) The two data points (70, 1300) and (120, 2200) will be used to calculate the slope of the line. The slope represents the cost of the banquet for each person. The cost is$18 per person.
When a linear function is used to model the real life situation, the equation can be written in the form or in the form \begin{align*}y=mx+b\end{align*} or in the form \begin{align*}Ax+By+C=0\end{align*}.
(d)
The \begin{align*}y\end{align*}-intercept is (0, 40)
The equation to model the real-life situation is \begin{align*}y=18x+40\end{align*}. The variables should be changed to match the labels on the axes. The equation that best models the situation is \begin{align*}c=18n+40\end{align*} where ‘\begin{align*}c\end{align*}’ represents the cost and ‘\begin{align*}n\end{align*}’ represents the number of people.
### Guided Practice
1. Some college students who plan on becoming math teachers decide to set up a tutoring service for high school math students. One student was charged $25 for 3 hours of tutoring. Another student was charged$55 for 7 hours of tutoring. The relationship between the cost and time is linear.
(a) What is the independent variable?
(b) What is the dependent variable?
(c) What are two data values for this relationship?
(d) Draw a graph of cost versus time.
(e) Determine an equation to model the situation.
(f) What is the significance of the slope?
(g) What is the cost-intercept? What does the cost-intercept represent?
2. A Glace Bay developer has produced a new handheld computer called the Blueberry. He sold 10 computers in one location for $1950 and 15 in another for$2850. The number of computers and cost forms a linear relationship
(a) State the dependent and independent variables.
(b) Sketch a graph.
(c) Find an equation expressing cost in terms of the number of computers.
(d) State the slope of the line and tell what the slope means to the problem.
(e) State the cost-intercept and tell what it means to this problem.
(f) Using your equation, calculate the number of computers you could get for $6000. 3. Handy Andy sells one quart can of paint thinner for$7.65 and a two quart can for 13.95. Assume there is a linear relationship between the volume of paint thinner and the price. (a) What is the independent variable? (b) What is the dependent variable? (c) Write two data values for this relationship. (d) Draw a graph to represent this relationship. (e) What is the slope of the line? (f) What does the slope represent in this problem? (g) Write an equation to model this problem. (h) What is the cost-intercept? (i) What does the cost-intercept represent in this problem? (j) How much would you pay for 6 quarts of paint thinner? Answers: 1. (a) The cost for tutoring depends upon the amount of time. The independent variable is the time. (b) The dependent variable is the cost. (c) Two data values for this relationship are (3, 25) and (7, 55). (d) On the \begin{align*}x\end{align*}-axis is the time in hours and on the \begin{align*}y\end{align*}-axis is the cost in dollars. (e) Use the two data values (3, 25) and (7, 55) to calculate the slope of the line. \begin{align*}m = \frac{15}{2}\end{align*}. Determine the \begin{align*}y\end{align*}-intercept of the graph. The equation to model the relationship is \begin{align*}y=\frac{15}{2}x+\frac{5}{2}\end{align*}. To match the variables of the equation with the graph the equation is . The relationship is cost in dollars versus time in hours. The equation could also be written as . (f) The slope of \begin{align*}\frac{15}{2}\end{align*} means that it costs15.00 for 2 hours of tutoring. If the slope is expressed as a decimal, it means that it costs 7.50 for 1 hour of tutoring. (g) The cost-intercept is the \begin{align*}y\end{align*}-intercept. The \begin{align*}y\end{align*}-intercept is (0, 2.50). This value could represent the cost of having a scheduled time or the cost that must be paid for cancelling the appointment. In a problem like this, the \begin{align*}y\end{align*}-intercept must represent a meaningful quantity for the problem. 2. (a) The number of dollars in sales from the computers depends upon the number of computers sold. The dependent variable is the dollars in sales and the independent variable is the number of computers sold. (b) On the \begin{align*}x\end{align*}-axis is the number of computers and on the \begin{align*}y\end{align*}-axis is the cost of the computers. (c) Use the data values (10, 1950) and (15, 2850) to calculate the slope of the line. \begin{align*}m=180\end{align*}. Next determine the \begin{align*}y\end{align*}-intercept of the graph. The equation of the line that models the relationship is . To make the equation match the variables of the graph the equation is . (d) The slope is \begin{align*}\frac{180}{1}\end{align*}. This means that the cost of one computer is180.00.
(e) The cost intercept is the \begin{align*}y\end{align*}-intercept. The \begin{align*}y\end{align*}-intercept is (0, 150). This could represent the cost of renting the location where the sales are being made or perhaps the salary for the sales person.
(f)
With 6000 you could get 32 computers. 3. (a) The independent variable is the volume of paint thinner. (b) The dependent variable is the cost of the paint thinner. (c) Two data values are (1, 7.65) and (2, 13.95). (d) On the \begin{align*}x\end{align*}-axis is the volume in quarts and on the \begin{align*}y\end{align*}-axis is the cost in dollars. (e) Use the two data values (1, 7.65) and (2, 13.95) to calculate the slope of the line. The slope is \begin{align*}m=6.30\end{align*}. (f) The slope represents the cost of one quart of paint thinner. The cost is6.30.
(g)
The equation to model the relationship is \begin{align*}y=6.30x+1.35\end{align*}. The equation that matches the variables of the graph is
.
(h) The cost-intercept is (0, 1.35).
(i) This could represent the cost of the can that holds the paint thinner.
(j)
The cost of 6 quarts of paint thinner is $39.15. ### Explore More Players on the school soccer team are selling candles to raise money for an upcoming trip. Each player has 24 candles to sell. If a player sells 4 candles a profit of$30 is made. If he sells 12 candles a profit of $70 is made. The profit and the number of candles sold form a linear relation. 1. State the dependent and the independent variables. 2. What are the two data values for this relation? 3. Draw a graph and label the axis. 4. Determine an equation to model this situation. 5. What is the slope and what does it mean in this problem? 6. Find the profit-intercept and explain what it represents. 7. Calculate the maximum profit that a player can make. 8. Write a suitable domain and range. 9. If a player makes a profit of$90, how many candles did he sell?
Jacob leaves his summer cottage and drives home. After driving for 5 hours, he is 112 km from home, and after 7 hours, he is 15 km from home. Assume that the distance from home and the number of hours driving form a linear relationship.
1. State the dependent and the independent variables.
2. What are the two data values for this relationship?
3. Represent this linear relationship graphically.
4. Determine the equation to model this situation.
5. What is the slope and what does it represent?
6. Find the distance-intercept and its real-life meaning in this problem.
7. How long did it take Jacob to drive from his summer cottage to home?
8. Write a suitable domain and range.
9. How far was Jacob from home after driving 4 hours?
10. How long had Jacob been driving when he was 209 km from home?
### Vocabulary Language: English
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed.
Slope
Slope
Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$ | 2015-10-05T13:45:33 | {
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https://proxies-free.com/values-p-4p21-and-6p21-are-prime-numbers-determine-p/ | # Values \$p\$, \$4p^2+1\$ and \$6p^2+1\$ are prime numbers. Determine \$p\$.
Values $$p$$, $$4p^2+1$$ and $$6p^2+1$$ are prime numbers. Determine $$p$$.
Quickly testing the first primes it’s clear that $$5$$ is a solution. However, how can I show that this is the only possible solution? | 2020-09-23T09:36:47 | {
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https://math.stackexchange.com/questions/3028090/if-y-is-a-hausdorff-space-and-x-which-is-a-subspace-of-that-space-is-limit-p | If $Y$ is a Hausdorff space and $X$ which is a subspace of that space is limit point compact space . Then $X$ is closed.
If $$Y$$ is a Hausdorff space and $$X$$ which is a subspace of that space is limit point compact space . Then $$X$$ is closed.
Can anyone give me a trivial counter example?
• Can you define the term 'limit point compact'? – Kavi Rama Murthy Dec 6 at 5:51
• If $X$ is a limit point compact space then every infinite set has a limit point. E.g - $\mathbb R$ is not limit point compact space where so is any of it's closed interval.@KaviRamaMurthy – cmi Dec 6 at 5:59
Let $$Y$$ be $$\omega_1 + 1$$ and $$X$$ is $$\omega_1$$, in the order topology. Here $$\omega_1$$ is the first uncountable ordinal, and $$Y$$ is its successor (one extra point).
Another example: let $$Y$$ be $$[0,1]^\mathbb{R}$$ in the product topology, and $$X$$ the limit point compact subset of all elements that are $$0$$ except for at most countably many coordinates. ( A $$\Sigma$$-product, this is called). This $$X$$ is dense and not closed in $$Y$$.
• @cmi Munkres calls it $W$. Look it up – Henno Brandsma Dec 6 at 7:02 | 2018-12-13T02:57:40 | {
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https://jmanton.wordpress.com/2013/10/17/long-versus-short-proofs/ | Home > Education, Uncategorized > Long versus Short Proofs
## Long versus Short Proofs
Proofs are very similar to computer programs. And just like computer programs, there are often many different ways of writing a proof. Sometimes the difference between proofs is great; one might be based on geometry while another on analysis. The purpose of this short note is to emphasise that different proofs are possible at the “simple” level too. While this may appear mundane, it is actually an important part of learning mathematics: once you have proved a result, spend a few minutes trying to simplify the proof. It will make the next proof you do easier. The two key messages are as follows.
• There is no substitute for experience when it comes to deriving proofs.
• Practicing on simple examples to find the “best” proof is training for being able even just to find a proof of a harder result.
For lack of a better on-the-spot example, consider the following problem. Let $f\colon\mathbb{C} \rightarrow \mathbb{C}$ be an analytic function having a double zero at the origin and whose first derivative has at most linear growth: $| f'(x) | \leq \alpha |x|$ for some $\alpha \in \mathbb{R}$. What is the most general form of $f$?
## First Approach
A double zero at the origin means $f(x) = x^2 h(x)$ for some analytic function $h$. Therefore $f'(x) = 2xh(x) + x^2h'(x)$ and $\frac{f'(x)}{x} = 2h(x) + xh'(x)$. Since $latex 2h(x) + xh'(x)$ is both analytic and bounded, it must be constant, by Liouville’s theorem. (Here, all analytic functions are entire because they are defined on the whole of the complex plane.) Solving the differential equation $latex 2h(x) + xh'(x) = c$ by substituting in the power series expansion $h(x) = \sum a_i x^i$ shows that $h(x) = \frac{c}2$. [The general solution of $latex 2h(x) + xh'(x) = 0$ is $h(x) = \beta x^{-2}$ where $\beta$ is an arbitrary scalar. The only general solution that is analytic is $h(x)=0$.] The conclusion is that the most general form of $f$ is $f(x) = ax^2$ for some complex scalar $a$.
## Second Approach
The first approach is unattractive and unnecessarily complicated. Instead of starting with the double zero, start instead with the first derivative having a linear bound. (A standard generalisation of Liouville’s theorem is that an entire function with a linear bound must itself be linear. We will pretend here we do not know this fact.) If $f(x)$ has a double zero at the origin then $f'(x)$ has a zero at the origin, therefore $f'(x) = x g(x)$ for some analytic $g(x)$. The linear bound together with Liouville’s theorem means $g(x)$ is constant, that is $f'(x) = 2 a x$ for some scalar $a$. Therefore $f(x)$ must equal $ax^2$ if it is to satisfy both $f'(x) = 2 a x$ and $f(0)=0$.
## What was the Difference?
The first approach expressed $f$ as $f(x) = x^2 h(x)$ while the second approach expressed $f'$ as $f'(x) = x g(x)$. Both approaches resulted in a differential equation, but the second approach resulted in the simpler differential equation $f'(x) = 2ax$. Underlying this example is that a “change of coordinates” can simplify a differential equation. Although both approaches could be made to work in this simple example, there are situations where some approaches are too difficult to follow through to completion.
One could argue that because the linear bound constraint is “harder” than the double-zero constraint, one should start with the linear bound constraint and not the double-zero constraint, and therefore be led to the simpler differential equation. Yet the real messages are as stated at the start:
• There is no substitute for experience when it comes to deriving proofs.
• Practicing on simple examples to find the “best” proof is training for being able even just to find a proof of a harder result. | 2017-09-24T21:12:29 | {
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https://socratic.org/questions/what-is-the-area-of-a-circle-with-radius-of-10-cm | # What is the area of a circle with radius of 10 cm?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Write a one sentence answer...
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
Describe your changes (optional) 200
13
Dec 23, 2015
314.16 $c {m}^{2}$
#### Explanation:
The formula for the area of a circle is given by:
Area = $\pi$${r}^{2}$ , r = radius
Therefore, you only need to substitute radius into the equation and calculate the area using calculator.
Area = $\pi$${r}^{2}$, r = 10 cm
Area = $\pi$${10}^{2}$
= 314.16 $c {m}^{2}$
However, sometimes the question will give you the $\pi$ as 3.142 or $\frac{22}{7}$, so your answer might varies a bit with the actual calculation. Let's proof it for you! :D
Area = 3.142 x ${10}^{2}$
= 3.142 x 100
= 314.2 $c {m}^{2}$
Area = $\frac{22}{7}$ x ${10}^{2}$
= $\frac{22}{7}$ x 100
= $\frac{2200}{7}$ [You can leave your answer in this form]
= 314$\frac{2}{7}$ $c {m}^{2}$
##### Just asked! See more
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http://math.stackexchange.com/questions/254544/normal-operators-in-hilbert-spaces | # Normal operators in Hilbert spaces
Let $H$ be a separable Hilbert space and let $T:H\to H$ be a continues linear map such that there exists an orthonormal basis of $H$ that consists of the eigenvectors of $T$. Show that $T$ is normal. That is $T^*T = TT^*$
Any hints would be appriciated!
-
Fix $\{v_j\}$ an orthonormal basis of eigenvectors of $T$. We have $Tv_k=\lambda_kv_k$, where $\lambda_k$ is the corresponding eigenvalue. This gives for each $j$, $$\langle T^*v_j,v_k\rangle=\langle v_j,Tv_k\rangle =\bar \lambda_k\delta_{j,k}.$$ As the sequence $\{v_j\}$ spans a dense subspace, we have that $T^*$ is completely determined. By boundedness and linearity, we just need to show the relationship $$\forall k, TT^*v_k=T^*Tv_k.$$
-
$TT^*v_k = T\bar{\lambda_k}v_k = \|\lambda\|^2 v_k = T^*{\lambda_k}v_k =T^*Tv_k$ So then we can use the boundedness and linearity of the composition of maps? – Johan Dec 9 '12 at 16:24
Yes, in other to establish the equality for linear combinations, then for all the vectors. – Davide Giraudo Dec 9 '12 at 16:29
I made the post CW, as the ideas were in @Norbert's answer. – Davide Giraudo Dec 16 '12 at 14:53 | 2015-01-29T16:20:37 | {
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http://mathhelpforum.com/calculus/90786-average-value.html | # Math Help - Average Value
1. ## Average Value
The value, V of a Tiffany lamp, worth $225 in 1965, increases at 15% per year. Its value in dollars t years after 1965 is given by $V = 225(1.15)^t$. Find the average value of the lamp over the period 1965-2000 -Can some1 please explain to me how to do this? I understand average value is 1 / b-a. But how do i set up the integral? from 0 to 45? or from 1965 - 2000 ? 2. Originally Posted by VkL The value, V of a Tiffany lamp, worth$225 in 1965, increases at 15% per year. Its value in dollars t years after 1965 is given by $V = 225(1.15)^t$. Find the average value of the lamp over the period 1965-2000
-Can some1 please explain to me how to do this? I understand average value is 1 / b-a. But how do i set up the integral? from 0 to 45? or from 1965 - 2000 ?
Neither!
The problem gives the value "t years after 1965" so the value of t starts at 1965-1965= 0 and ends at 2000- 1965= 35, not 45.
The average value is $\frac{225\int_0^{35} (1.15)^t dt}{35}$. | 2014-08-27T15:59:34 | {
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http://math.tutornext.com/calculus/calculus-problems.html | # Calculus Problems
Sub Topics
Introduction: The topic Calculus is a vast topic in Math which will help us to find the volume of a solid generated by a curve, area between the curves and various topics in physics which involve in rate of change of certain quantities. To learn all these we should be familiar with certain topics like, Relations, Functions, Types of functions, Limits and Derivatives of functions, Applications of Derivatives in different fields, Integration, Properties of Integration, Application of integration in various fields etc. Tutor Next is the organisation which conducts one on one tutoring which help you you to master these topics. In this section let us discuss with some of the problems in calculus and the various methods of solving them.
## Calculus Problem Solver
Let us discuss certain problems on calculus from the basics.
1. Which of the following relations are functions.
R1 = { ( 1, 2), ( 2, 3), ( 3, 4), ( 4, 5) }
Every element of the domain has its image in the co-domain, therefore the above relation is a function
R2 = { ( 1, -1 ), ( 4, -2 ) , (0 , 0 ), ( 1, 1 ), ( 4, 2 ) }
The element 1 has two images -1 and 1 and the element 4 has its 2 images -2 and 2 in the co-domain.
Therefore the above relation is not a function.
#### f(x) = 3x + 5, find the range of { -4, 1, 4 }
Solution: We have f ( x) = 3x + 5
f( 1) = 3 ( 1 ) + 5 = 3 + 5 = 8, which belongs to Natural numbers N
f( -4 ) = 3 ( -4 ) + 5 = - 12 + 5 = - 7 which is not a natural number.
f( 4 ) = 3 ( 4 ) + 5 = 12 + 5 = 17 , which is a natural number
Therefore Range = { 8, 17 }
3. Let f : R ---> R and f ( x ) = $\frac{ 2 } { x + 3 }$, find the domain and range of the function. Express your answer in interval notation.
Solution: We have f ( x ) = $\frac{ 2 } { x + 3 }$
The above function is a rational function.
The function is not defined when the denominator x + 3 = 0
(i. e ) x = -3
Therefore the Domain consists of all real numbers except x = -3
Domain = { x : x $\epsilon$ R - {3} }
As x tends to infinity, y = -2/3
and for no value of x, y = 0, therefore y is not equal to 0.
Therefore, the Range of the function = { y : y $\epsilon$ R - { 0 } }
4. If f and g are functions such that f : A ---> B and g : B------> C.
If A = { 1,2,3 }, B = { 1, 2, ,3 4, 5,6 } and C = { 10, 20, 30 }
such that f ( 1 ) =2, f( 2) =3, f ( 3 ) = 4, g(2)=20, g(3)= 30 and g(4) = 40.
Find g o f(1), (g o f)(2), (g o f )(3)
Solution: We are given that, f : A ---> B and g : B------> C.
If A = { 1,2,3 },
B = { 1, 2, ,3 4, 5,6 }
and C = { 10, 20, 30, 40, 50 }
such that f ( 1 ) =2, f( 2) =3, f ( 3 ) = 4,
g(2)=20, g(3)= 30 and g(4) = 40.
g o f ( 1 ) = g [ f ( 1 ) ] = g ( 2 ) = 20
g o f ( 2 ) = g [ f ( 2 ) ] = g ( 3 ) = 30
g o f ( 3 ) = g [ f ( 3 ) ] = g ( 4 ) = 40
Therefore, g o f(1) = 20 , (g o f)(2) = 30, (g o f )(3) = 40
#### 5. If f : R --> R and g : R----> R such that f ( x) = 3x and g ( x ) = x2 find f o g ( x) and g o f ( x ).
Solution: We are given that, f ( x) = 3x and g ( x ) = x2
f o g ( x ) = f [ g ( x ) ] =f ( x2 ) = 3 ( x2 ) = 3 x2
g o f ( x ) = g [ f ( x ) ] = g ( 3 x ) = ( 3 x ) 2 = 9 x2
6. Find the limit of the following functions.
a. $\lim_{x -> 2}$ ( x2 - 3x + 4 )
Solution:
$\lim_{x -> 2}$ ( x2 - 3x + 4 ) = 22 - 3 ( 2 ) + 4
= 4 - 6 + 4
= 2
b. $\lim_{ x -> 2 }$ $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$
Solution:
$\lim_{ x -> 2 }$ $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$
= $\frac{(6-6)(8+5)}{(2-2)}$
= $\frac{0}{0}$
Therefore, $\lim_{ x -> 2 }$ $\frac{( 3 x-6 )( 4 x + 5 )}{( x - 2 ) }$
= $\lim_{ x -> 2 }$ $\frac{3(x-2)(4x+5)}{(x-2)}$
= $\lim_{ x -> 2 }$ $\frac{3(4x+5)}{1}$
= 3[ 4(2) + 5 ]
= 3 [ 8 + 5 ] = 3 [13 ] = 39
c. $\lim_{x->-3}$ $\frac{(x^{2}-9)}{(x+3)}$
Solution: $\lim_{x->-3}$ $\frac{(x^{2}-9)}{(x+3)}$
= $\frac{9-9}{-3+3}$ = $\frac{0}{0}$
Therefore, $\lim_{x->-3}$ $\frac{(x^{2}-9)}{(x+3)}$
= $\lim_{x->-3}$ $\frac{(x^{2}-3^{2})}{(x+3)}$
= $\lim_{x->-3}$ $\frac{(x+3)(x-3)}{(x+3)}$
= $\lim_{x->-3}$ ( x- 3 )
= - 3 - 3 = - 6
d. $\lim_{x->\pi/2}$ $\frac{sin x + cos x}{sin x}$
Solution:
$\lim_{x->\pi/2}$ $\frac{sin x + cos x}{sin x}$
= $\frac{sin(\pi /2)+cos(\pi /2)}{sin(\pi /2)}$
= $\frac{1+0}{1}$ = 1
## Free Calculus Problem Solver
1. Find the Derivatives of the following function using derivative rules
If f(x) = 2x + 5, find f ' ( x ) at x = 2.
Solution: We have f (x ) = 2 x + 5
f ' ( x ) = $\lim_{h->0}$ $\frac{f(x+h)-f(x)}{h}$
f ' ( 2 ) = $\lim_{h->0}$ $\frac{f(2+h)-f(2)}{h}$
= $\lim_{h->0}$ $\frac{2(2+h)+5-(2(2)+5)}{h}$
= $\lim_{h->0}$ $\frac{4+2h+5-4-5}{h}$
= $\frac{2h}{h}$ = 2
2. A man 6 feet high walks at a uniform rate of 4 miles per hour away from a lamp 20 feet high. Find the rate at which the length of the shadow increases.
Solution: Let us Study the following diagram.
Let L be the lamp and AB the position of the man at t seconds. The length of his shadow is AC and let it be x feet.
The rate at which he walks is given.
(i. e ) the rate at which AM = y is changing is given
i. e $\frac{dy}{dx}$ = 4 miles per hour
= $\frac{88}{15}$ feet per second
$\Delta$ ABC and $\Delta$ LMC are similar.
Therefore, $\frac{CA}{CM}$ = $\frac{6}{20}$
i.e $\frac{x}{x+y}$ = $\frac{3}{10}$
i.e 7x = 3y
The rate at which the shadow is increasing is $\frac{dx}{dt}$
Differentiating 7x = 3y with respect to t, we get,
7 $\frac{dx}{dt}$ = 3 $\frac{dy}{dt}$
= 3 x $\frac{88}{15}$
$\frac{dx}{dt}$
= 3 x $\frac{88}{15}$ x $\frac{1}{7}$
= $\frac{3\times 88}{15\times 7}$
= $\frac{88}{35}$ feet per second
* * * * * | 2017-10-19T08:56:34 | {
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https://math.stackexchange.com/questions/3170837/prove-that-ip-perp-cq | # Prove that $IP \perp CQ$.
Consider $$I$$ being the incentre of $$\triangle ABC$$. $$IF \perp AB$$ ($$F \in AB$$). $$AI$$ extended intersects the circumcircle of $$\triangle ABC$$ at $$D$$ ($$D \not\equiv A$$) and $$AD \cap BC = \{K\}$$. $$DF$$ intersects the circumcircle of $$\triangle ABC$$ and $$\triangle BKD$$ respectively at $$Q$$ and $$P$$. Prove that $$IP \perp CQ$$.
I have proven that $$KP \parallel CQ$$.
Let $$BC \cap QD = \{E\}$$. We have that $$ED \cdot EP = EB \cdot EK$$ and $$ED \cdot EQ = EB \cdot EC$$.
$$\implies \dfrac{EP}{EQ} = \dfrac{ED \cdot EP}{ED \cdot EQ} = \dfrac{EB \cdot EK}{EB \cdot EC} = \dfrac{EK}{EC}$$
Using the intercept theorem for $$\triangle ECQ$$ and $$\dfrac{EP}{EQ} = \dfrac{EK}{EC}$$, we have that $$KP \parallel CQ$$.
I have tried to prove that $$KP \perp PI$$ by proving $$\widehat{KHB} = \widehat{PIF}$$ where $$KP \cap AB = \{H\}$$. But it didn't work.
I would be grateful if you could solve the problem.
Let's prove that $$IP \perp PK$$, since indeed $$PK || CQ$$. (because $$\angle DPK= \angle DBK =$$ half the measure of arc CD $$= \angle DQC$$)
Consider an inversion centered at D with radius DB. I will denote by $$T'$$ the image of point $$T$$ under that inversion.
$$B'=B$$, $$C'=C$$, $$I'=I$$ (the latter is due to so-called "trillium lemma", that is, $$DI=DB=DC$$)
The line BC after the inversion becomes the circumscribed circle of $$ABC$$. So, $$K' = A$$.
The circle $$BPKD$$ becomes a line passing through $$B'=B$$ and $$K' = A$$ So, $$P' = F$$.
Now, we know that $$F$$ lies on the circle with diameter $$AI$$. The image of that circle under our inversion is the circle with diameter $$KI$$. We're done, since $$F' = P$$. | 2021-10-28T02:20:28 | {
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https://proofwiki.org/wiki/Identity_of_Monoid_is_Cancellable | # Identity of Monoid is Cancellable
## Theorem
The identity of a monoid is cancellable.
## Proof
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$.
Let $x, y \in S$ such that $x \circ e = y \circ e$
Then, by the definition of the identity:
$x = x \circ e = y \circ e = y$
... thus $x = y$ and the result is proved.
$\blacksquare$ | 2020-01-26T09:40:32 | {
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https://eccc.weizmann.ac.il/report/2020/155/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Revision(s):
Revision #1 to TR20-155 | 22nd October 2020 19:10
#### Log-rank and lifting for AND-functions
Revision #1
Authors: Alexander Knop, Shachar Lovett, Sam McGuire, Weiqiang Yuan
Accepted on: 22nd October 2020 19:10
Keywords:
Abstract:
Let $f: \{0,1\}^n \to \{0, 1\}$ be a boolean function, and let $f_\land (x, y) = f(x \land y)$ denote the AND-function of $f$, where $x \land y$ denotes bit-wise AND. We study the deterministic communication complexity of $f_\land$ and show that, up to a $\log n$ factor, it is bounded by a polynomial in the logarithm of the real rank of the communication matrix of $f_\land$. This comes within a $\log n$ factor of establishing the log-rank conjecture for AND-functions with no assumptions on $f$. Our result stands in contrast with previous results on special cases of the log-rank
conjecture, which needed significant restrictions on $f$ such as monotonicity or low $\mathbb{F}_2$-degree. Our techniques can also be used to prove (within a $\log n$ factor) a lifting theorem for AND-functions, stating that the deterministic communication complexity of $f_\land$ is polynomially-related to the AND-decision tree complexity of $f$.
The results rely on a new structural result regarding boolean functions $f:\{0, 1\}^n \to \{0, 1\}$ with a sparse polynomial representation, which may be of independent interest. We show that if the polynomial computing $f$ has few monomials then the set system of the monomials has a small hitting set, of size poly-logarithmic in its sparsity. We also establish extensions of this result to multi-linear polynomials $f:\{0,1\}^n \to \mathbb{R}$ with a larger range.
Changes to previous version:
Fixed author order.
### Paper:
TR20-155 | 18th October 2020 17:09
#### Log-rank and lifting for AND-functions
TR20-155
Authors: Alexander Knop, Shachar Lovett, Sam McGuire, Weiqiang Yuan
Publication: 18th October 2020 17:44
Let $f: \{0,1\}^n \to \{0, 1\}$ be a boolean function, and let $f_\land (x, y) = f(x \land y)$ denote the AND-function of $f$, where $x \land y$ denotes bit-wise AND. We study the deterministic communication complexity of $f_\land$ and show that, up to a $\log n$ factor, it is bounded by a polynomial in the logarithm of the real rank of the communication matrix of $f_\land$. This comes within a $\log n$ factor of establishing the log-rank conjecture for AND-functions with no assumptions on $f$. Our result stands in contrast with previous results on special cases of the log-rank
conjecture, which needed significant restrictions on $f$ such as monotonicity or low $\mathbb{F}_2$-degree. Our techniques can also be used to prove (within a $\log n$ factor) a lifting theorem for AND-functions, stating that the deterministic communication complexity of $f_\land$ is polynomially-related to the AND-decision tree complexity of $f$.
The results rely on a new structural result regarding boolean functions $f:\{0, 1\}^n \to \{0, 1\}$ with a sparse polynomial representation, which may be of independent interest. We show that if the polynomial computing $f$ has few monomials then the set system of the monomials has a small hitting set, of size poly-logarithmic in its sparsity. We also establish extensions of this result to multi-linear polynomials $f:\{0,1\}^n \to \mathbb{R}$ with a larger range. | 2021-05-14T23:07:10 | {
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http://mathhelpforum.com/advanced-algebra/200698-finding-eigenvalues-matrix.html | # Thread: Finding the eigenvalues of a matrix
1. ## Finding the eigenvalues of a matrix
Let A be a square matrix of real numbers whose eigenvalues are positive integers.
It is given that |adj(adj(A))| = 81.
What is the characteristic polynomial of the matrix?
--------------------------------------------------
What I did was this:
|adj(adj(A))| = |adj(A)|^(n-1) = |A|^(n-1)^2 = 81
But I don't know how to proceed.
Any suggestions? Or if there's any other possible way...
Thanks!
2. ## Re: Finding the eigenvalues of a matrix
The determinant of A^{n-1} is either -9 or 9.
3. ## Re: Finding the eigenvalues of a matrix
How do I conclude that?
4. ## Re: Finding the eigenvalues of a matrix
Its square is 81. | 2016-10-26T19:30:02 | {
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https://encyclopediaofmath.org/wiki/Borel_field_of_sets | # Borel field of sets
family of Borel sets, generated by a system of sets $M$
The smallest system of sets containing $M$ and closed with respect to the operations of countable union and taking complements (see also $\sigma$-algebra). | 2022-07-05T09:17:55 | {
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https://math.stackexchange.com/questions/1161556/compute-lim-n-to-infty-int-01-n-fracn1n2x2n6x8dx/1161562 | # Compute $\lim_{n\to \infty}\int_0^{1/n}\frac{n}{1+n^2x^2+n^6x^8}dx$
The integrand decreases when $n\ge 1$ but the statement of the monotone convergence theorem requires that the sequence is increasing. Fatou's lemma doesn't apply since it will only give a lower bound of $0$. That leaves the dominated convergence theorem (DCT) , but to use it I'd have to find an integrable upper bound for $\frac{n}{1+n^2x^2+n^6x^8}$ but I can't find a better upper bound than $f(x)=n$ and I'm not sure if such a function is allowed as an upper bound when using DCT since it depends on $n$.
It seems clear that the integrand and the region of integration $[0,1/n]$ tend to $0$ but I don't see how to show it formally.
Let $u=nx$ then the given integral becomes
$$\int_0^1\frac{du}{1+u^2+\frac1{n^2}u^8}$$ and the function which dominates is $\phi(u)=\frac1{1+u^2}$. Can you take it from here using the DCT?
Hint. You may perfom the change of variable $u=nx$, giving $ndx=du$ to get $$\int_0^{1/n}\frac{n}{1+n^2x^2+n^6x^8}dx=\int_0^{1}\frac{1}{1+u^2+\dfrac{u^8}{n^2}}du\leq\int_0^{1}\frac{1}{1+u^2}du$$ and then easily use the dominated convergence theorem. | 2021-07-23T16:24:40 | {
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https://math.stackexchange.com/questions/1588357/k-subalgebra-of-finite-extension-of-k-is-a-field | # $k$-subalgebra of finite extension of $k$ is a field
I am working on a proof which has
Let $\mathfrak m,\mathfrak n$ be maximal ideals and $A$ Noetherian. Given that, $A[T_1,\dots,T_n]/\mathfrak n$ is a finite extension of $A/\mathfrak m$, if $\mathfrak n_1 = \mathfrak n \cap A[T_1,\dots,T_{n-1}]$ then $A[T_1,\dots,T_{n-1}]/\mathfrak n_1$ is an $A/\mathfrak m$-subalgebra of $A[T_1, \dots, T_n]/\mathfrak n$ and therefore is a field.
Is this a special case of a more general result such as
Let $K/k$ be a finite extension. If $L$ is a $k$-subalgebra of $K$, then $L$ is a field.
If the above result is true, how can one justify it?
If the above result is false, how should one justify the original proof?
A $k$-subalgebra $A$ of a finite extension $K/k$ is (a subring of a field, hence) an integral domain, so multiplication by any nonzero $a \in A$ is injective on $A$. But since $A$ is (a $k$-subalgebra of a finite-dimensional algebra, hence) finite-dimensional, this means multiplication is also surjective. So $A$ is a field. | 2020-12-02T15:37:02 | {
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http://clay6.com/qa/12265/planet-neptune-travels-around-the-sun-with-a-period-of-165-years-find-the-r | # Planet Neptune travels around the sun with a period of 165 years. Find the ratio of $\large\frac{R_2}{R_1}$ where $R_2$ radius of Neptune orbit and $R_1$ radius of earth , both being considerd circular.
$(a)\;20 \quad (b)\;30 \quad (c)\; 25 \quad (d)\;35$
$T_1=T_{earth}=1 year$
$T_2=T_{neptune}=165 years$
$\large\frac{{T_1}^2}{{T_2}^2}=\frac{{R_1}^3}{{R_2}^3}$
=> ${R_2}^3=\large\frac{{R_1}^3{T_2}^2}{{T_1}^2}$
=>${R_2}^3=165^2{R_1}^3$
$R_2 \approx 30 \;R_1$
hence b is the correct answer.
edited Feb 17, 2014 by meena.p | 2018-02-23T04:48:11 | {
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https://math.stackexchange.com/questions/1169816/another-infinite-telescoping-series-question-high-school-calc | # Another infinite (telescoping?) series question (high school calc)
$$\sum^\infty_{n=0} \frac{n}{(n+1)!}$$
I don't even know where to begin. I posted another question about a topic like this, but with the factorial thrown into the mix, it is dubious if the same methodology would work.
How do I evaluate this?
Thanks.
• Write out the terms, list them out one by one. – IAmNoOne Mar 1 '15 at 2:22
• Indeed telescoping. The $n$-th term is $\frac{n+1-1}{(n+1)!}$, which is $\frac{1}{n!}-\frac{1}{(n+1)!}$. – André Nicolas Mar 1 '15 at 2:23
• How did you get to that last conclusion? EDIT: figured it out. Thanks for the hint! – louie mcconnell Mar 1 '15 at 2:24
• Add the terms, modified to look like above, and observe the mass cancellations. Added: good! – André Nicolas Mar 1 '15 at 2:25
• Why don't you post that as an answer, @AndréNicolas? – hjhjhj57 Mar 1 '15 at 3:46
Write $n/(n+1)!=(n+1-1)/(n+1)!=1/n!-1/(n+1)!$
Then the sum from 0 to infinity is trivially $1$! | 2019-06-16T06:50:53 | {
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https://astarmathsandphysics.com/ib-physics-notes/measurements-units-uncertainty-and-principles/1337-choosing-variables-for-a-straight-line.html | Choosing Variables For a Straight Line
Not every equation naturally gives rise to a straight line. If we have variablesandrelated byorandrelated bythen graphs ofagainstoragainst will not result in a straight line. We are however not restricted to plotagainstor againstWe can often rearrange an equation into a form which will give a straight line if we plot a suitable function of one variable against a suitable function of the other. In order to do this we need to:
• Identify which symbols in the equation are variables and which are constants.
• The symbols that correspond toandmust be variables and the symbols that correspond toandmust be constants.
• If a variable is cubed, square rooted or the reciprocal, log or exponential is taken, the result is still a variable and may still be used to label one of the axes.
• Any function of the readings may be used to label the axes, since the result is still a variable.
• Sometimes the physical quantities use the same symbols as in our notation e.g.is used to denote the speed of light. Do not get these confused.
For the equationabove, taking natural logs results in a straight line.
For the equationabove, plottingagainstresults in a straight line with gradient -1 andinterceptsince | 2022-05-27T13:38:52 | {
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https://math.stackexchange.com/questions/954321/change-of-variable-pdf-inverse-function | # Change of variable pdf inverse function
I've been given the following problem: $f(x,y) = e^{-(x+y)}$ on intervals $x \ge 0$ and $y \ge 0$, and $f(x,y) = 0$ otherwise. I'm also given that $Φ_1(x,y) = \frac{x}{y} = U$ and $Φ_2(x,y) = x + y = V$. I've proven that $f(x,y)$ is a pdf as asked by the problem but then the problem asks me to find the inverse functions of $Ψ_1(U,V) = x$ and $Ψ_2(U,V) = y$. I don't have a clue how to go about this: I certainly know how to find the inverse of a function but this looks like nothing I've seen in my textbook thus far. | 2019-07-17T00:33:15 | {
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https://proofwiki.org/wiki/Relation_on_Empty_Set_is_Equivalence | Relation on Empty Set is Equivalence
Theorem
Let $S = \varnothing$, that is, the empty set.
Let $\mathcal R \subseteq S \times S$ be a relation on $S$.
Then $\mathcal R$ is the null relation and is an equivalence relation.
Proof
As $S = \varnothing$, we have from Cartesian Product is Empty iff Factor is Empty that $S \times S = \varnothing$.
Then it follows that $\mathcal R \subseteq S \times S = \varnothing$.
Reflexivity
From the definition:
$\mathcal R = \varnothing \implies \forall x \in S: \left({x, x}\right) \notin \mathcal R$
But as $\neg \, \exists x \in S$ it follows vacuously that $\mathcal R$ is reflexive.
$\Box$
Symmetry
It follows vacuously that:
$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R$
and so $\mathcal R$ is symmetric.
$\Box$
Transitivity
It follows vacuously that:
$\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$
and so $\mathcal R$ is transitive.
$\Box$
It follows from the definition that $\mathcal R$ is an equivalence relation.
$\blacksquare$ | 2019-10-18T23:58:17 | {
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http://mathhelpforum.com/differential-equations/130210-solving-de.html | 1. ## Solving a DE
I must solve $(1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}$.
I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
So far I've rewritten the equation as $\frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)} y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}$.
I found the integrating factor to be $2 \ln (1+x^2)$.
So multiplying the DE by it and integrating with respect to x, I found out that $2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx$.
I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of $\ln (1+x^2)$, as stupid as it may sound.
Am I in the right direction? If so, could you provide a tip to find a primitive of $\ln (1+x^2)$?
2. Originally Posted by arbolis
I must solve $(1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}$.
I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
So far I've rewritten the equation as $\frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)} y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}$.
I found the integrating factor to be $2 \ln (1+x^2)$.
So multiplying the DE by it and integrating with respect to x, I found out that $2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx$.
I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of $\ln (1+x^2)$, as stupid as it may sound.
Am I in the right direction? If so, could you provide a tip to find a primitive of $\ln (1+x^2)$?
You are definitely on the right track, but you have the wrong integrating factor.
The integrating factor is actually
$e^{\int{\frac{4x}{1+x^2}\,dx}} = e^{2\ln{(1 + x^2)}} = e^{\ln{(1 + x^2)^2}} = (1 + x^2)^2$.
So multiplying through by the integrating factor:
$(1 + x^2)^2\frac{dy}{dx} + 4x(1 + x^2)y = \frac{1}{1 + x^2}$
$\frac{d}{dx}[(1 + x^2)^2y] = \frac{1}{1 + x^2}$
$(1 + x^2)^2y = \int{\frac{1}{1 + x^2}\,dx}$
$(1 + x^2)^2y = \arctan{x} + C$
$y = \frac{\arctan{x} + C}{(1 + x^2)^2}$.
3. Oops, I forgot to take the exponential of "my" integrating factor.
Thanks for the rest. | 2016-08-27T11:44:53 | {
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https://ccssmathanswers.com/eureka-math-precalculus-module-2-end-of-module-assessment/ | # Eureka Math Precalculus Module 2 End of Module Assessment Answer Key
## Engage NY Eureka Math Precalculus Module 2 End of Module Assessment Answer Key
### Eureka Math Precalculus Module 2 End of Module Assessment Task Answer Key
Question 1.
Find values for a, b, c, d, and e so that the following matrix product equals the 3×3 identity matrix. Explain how you obtained these values.
The row 1, column 1 entry of the product is a – 3 + 10 . This should equal 1 , so a = -6.
The row 2, column 1 entry of the product is c + c + 2 . This should equal zero, so c = -1.
The row 3, column 1 entry of the product is 5 + b – 8 . This should equal zero, so b = 3.
The row 1, column 3 entry of the product is ad – 3e + 5b = –6d – 3e + 15 , and this should equal zero. So, we should have 6d + 3e = 15.
The row 2, column 3 entry of the product is cd + ce + b = –d- e + 3 , and this should equal zero. So, we need d + e = 3.
Solving the two linear equations d and e gives d = 2 and e = 1.
So, in summary, we have a = –6 , b = 3 , c = –1 , d = 2 , and e = 1.
b. Represent the following system of linear equations as a single matrix equation of the form Ax=b, where A is a 3×3 matrix and x and b are 3×1 column matrices.
x+3y+2z=8
x-y+z=-2
2x+3y+3z=7
We have Ax = b with A = $$\left[\begin{array}{ccc} 1 & 3 & 2 \\ 1 & -1 & 1 \\ 2 & 3 & 3 \end{array}\right]$$ and x = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ and b = $$\left[\begin{array}{c} 8 \\ -2 \\ 7 \end{array}\right]$$.
c. Solve the system of three linear equations given in part (b).
The solution is given by x = A-1b , if the matrix inverse exists. But part (a) shows that if we set
,
then we have BA = I . We could check that AB = I , as well (in which case B is the matrix inverse of A), but even without knowing this, from
we get
that is, that x = –7 , y = 1 , z = 6 is a solution to the system of equations. (In fact, it can be the only solution to the system.)
Question 2.
The following diagram shows two two-dimensional vectors v and w in the place positioned to each have an endpoint at point P.
a. On the diagram, make reasonably accurate sketches of the following vectors, again each with endpoint at P. Be sure to label your vectors on the diagram.
i. 2v
ii. -w
iii. v+3w
iv. w-2v
v. $$\frac{1}{2}$$ v
Vector v has magnitude 5 units, w has magnitude 3, and the acute angle between them is 45°.
b. What is the magnitude of the scalar multiple -5v?
We have ||–5v || = 5 ||v || = 5 × 5 = 25 .
c. What is the measure of the smallest angle between -5v and 3w if these two vectors are placed to have a common endpoint?
The two vectors in question have an angle of smallest measure 135° between them.
Question 3.
Consider the two-dimensional vectors and .
a. What are the components of each of the vectors v+w and v-w?
b. On the following diagram, draw representatives of each of the vectors v, w, and v+w, each with an endpoint at the origin.
c. The representatives for the vectors v and w you drew form two sides of a parallelogram, with the vector v+w corresponding to one diagonal of the parallelogram. What vector, directed from the third quadrant to the first quadrant, is represented by the other diagonal of the parallelogram? Express your answer solely in terms of v and w, and also give the coordinates of this vector.
Label the points A and B as shown. The vector we seek is $$\overrightarrow{A B} .$$ To move from A to B we need to follow –w and then v. Thus, the vector we seek is –w + v, which is the same as v – w.
Also, we see that to move from A to B we need to move 4 units to the right and 4 units upward. This is consistent with v – w =
d. Show that the magnitude of the vector v+w does not equal the sum of the magnitudes of v and of w.
We have ||v|| = $$\sqrt{2^{2}+3^{2}}$$ = $$\sqrt{13}$$ and ‖w ‖ = $$\sqrt{(-2)^{2}+(-1)^{2}}$$ = $$\sqrt{5}$$, so ‖v ‖ + ‖w ‖ = $$\sqrt{13}$$ + $$\sqrt{5}$$.
Now, ‖v + w ‖ = $$\sqrt{o^{2}+2^{2}}$$ = 2 . This does not equal $$\sqrt{13}$$ + $$\sqrt{5}$$.
e. Give an example of a non-zero vector u such that ‖v+u‖ does equal ‖v‖+‖u‖.
Choosing u to be the vector v works.
‖v + u ‖ = ‖v + v ‖ = ‖2v ‖ = 2 ‖v ‖ = ‖v ‖ + ‖v ‖
(In fact, u = kv for any positive real number k works.)
f. Which of the following three vectors has the greatest magnitude: v+(-w), w-v, or (-v)-(-w)?
(-v) – (–w) = –v + w = w – v and
v + (–w) = v – w = –(w – v).
So, each of these vectors is either w – v or the scalar multiple (–1)(w – v) , which is the same vector but with opposite direction. They all have the same magnitude.
g. Give the components of a vector one-quarter the magnitude of vector v and with direction opposite the direction of v.
Question 4.
Vector a points true north and has a magnitude of 7 units. Vector b points 30° east of true north. What should the magnitude of b be so that b-a points directly east?
a. State the magnitude and direction of b-a.
We hope to have the following vector diagram incorporating a right triangle.
We have ‖a ‖ = 7 , and for this 30 – 60 – 90 triangle we need
||b|| = 2||b-a|| and 7 = ||a|| = $$\sqrt{3}$$||b-a|| This shows the magnitude of b should be ‖b ‖ = $$\frac{14}{\sqrt{3}}$$.
The vector b – a points east and has magnitude $$\frac{7}{\sqrt{3}}$$.
b. Write b-a in magnitude and direction form.
has a magnitude of $$\frac{7}{\sqrt{3}}$$ and a direction of 0° measured from the horizontal.
Question 5.
Consider the three points A=(10,-3,5), B=(0,2,4), and C=(2,1,0) in three-dimensional space.
Let M be the midpoint of $$\overline{A B}$$ and N be the midpoint of $$\overline{A C}$$.
a. Write down the components of the three vectors $$\overrightarrow{A B}$$, $$\overrightarrow{B C}$$, and $$\overrightarrow{C A}$$, and verify through arithmetic that their sum is zero. Also, explain why geometrically we expect this to be the case.
We have the following:
This is to be expected as the three points A, B, and C are vertices of a triangle (even in three-dimensional space), and the vectors $$\overrightarrow{A B}$$, $$\overrightarrow{B C}$$, and $$\overrightarrow{C A}$$, when added geometrically, traverse the sides of the triangle and have the sum effect of “returning to start.” That is, the cumulative effect of the three vectors is no vectorial shift at all.
b. Write down the components of the vector $$\overrightarrow{M N}$$. Show that it is parallel to the vector $$\overrightarrow{B C}$$ and half its magnitude.
We have M = (5,- $$\frac{1}{2}$$, $$\frac{9}{2}$$) and N = (6,-1, $$\frac{5}{2}$$). Thus, $$\overrightarrow{M N}$$ = .
We see that $$\overrightarrow{M N}$$ = $$\frac{1}{2}$$ $$\frac{1}{2}$$$$\overrightarrow{B C}$$, which shows that $$\overrightarrow{M N}$$ has the same direction as $$\overrightarrow{B C}$$ (and hence is parallel to it) and half the magnitude.
Let G=(4,0,3).
c. What is the value of the ratio ?
d. Show that the point G lies on the line connecting M and C. Show that G also lies on the line connecting N and B.
That $$\overrightarrow{M G}$$= $$\frac{1}{3}$$$$\overrightarrow{M C}$$ means that the point G lies a third of the way along $$\overline{M C}$$.
So, G also lies on $$\overline{N B}$$ (and one-third of the way along, too).
Question 6.
A section of a river, with parallel banks 95 ft. apart, runs true north with a current of 2 ft/sec. Lashana, an expert swimmer, wishes to swim from point A on the west bank to the point B directly opposite it. In still water she swims at an average speed of 3 ft/sec.
The diagram to the right illustrates the situation.
To counteract the current, Lashana realizes that she is to swim at some angle θ to the east/west direction as shown.
With the simplifying assumptions that Lashana’s swimming speed will be a constant 3 ft/sec and that the current of the water is a uniform 2 ft/sec flow northward throughout all regions of the river (e.g., we ignore the effects of drag at the river banks), at what angle θ to east/west direction should Lashana swim in order to reach the opposite bank precisely at point B? How long will her swim take?
a. What is the shape of Lashana’s swimming path according to an observer standing on the bank watching her swim? Explain your answer in terms of vectors.
Lashana’s velocity vector v has magnitude 3 and resolves into two components as shown, a component in the east direction vE and a component in the south direction vS.
We see
||vs|| =||v ||sin (θ) = 3sin (θ).
Lashana needs this component of her velocity vector to counteract the northward current of the water. This will ensure that Lashana will swim directly toward point B with no sideways deviation.
Since the current is 2 ft/sec , we need 3sinθ = 2 , showing that θ = sin –1 ($$\frac{2}{3}$$) ≈ 41.8° .
Lashana will then swim at a speed of ||vE|| = 3cosθ ft/sec toward the opposite bank.
Since sin (θ) = $$\frac{2}{3}$$, θ is part of a 2 – $$\sqrt{5}$$ – 3 right triangle, so cos (θ) = $$\frac{\sqrt{5}}{3}$$. Thus,
‖vE ‖ = $$\sqrt{5}$$ ft/sec .
She needs to swim an east/west distance of 95 ft. at this speed. It will take her
$$\frac{95 \mathrm{ft}}{\sqrt{5} \mathrm{ft} / \mathrm{sec}}$$ = $$19 \sqrt{5}$$ seconds ≈ 42 seconds to do this.
b. If the current near the banks of the river is significantly less than 2 ft/sec, and Lashana swims at a constant speed of 3 ft/sec at the constant angle θ to the east/west direction as calculated in part (a), will Lashana reach a point different from B on the opposite bank? If so, will she land just north or just south of B? Explain your answer.
As noted in the previous solution, Lashana will have no sideways motion in her swim. She will swim a straight-line path from A to B .
If the current is slower than 2 ft/sec at any region of the river surface, Lashana’s velocity vector component vS, which has magnitude 2 ft/sec , will be larger in magnitude than the magnitude of the current. Thus, she will swim slightly southward during these periods. Consequently, she will land at a point on the opposite bank south of B .
Question 7.
A 5 kg ball experiences a force due to gravity $$\vec{F}$$ of magnitude 49 newtons directed vertically downward. If this ball is placed on a ramp tilted at an angle of 45°, what is the magnitude of the component of this force, in newtons, on the ball directed 45° toward the bottom of the ramp? (Assume the ball has a small enough radius that all forces are acting at the point of contact of the ball with the ramp.)
The force vector can be resolved into two components as shown: Framp and Fperp.
We are interested in the component Framp.
We see a 45 – 90 – 45 triangle in this diagram, with hypotenuse of magnitude 49 N . This means that the magnitude of Framp is $$\frac{49 \mathrm{~N}}{\sqrt{2}}$$ ≈ 35 N .
Question 8.
Let A be the point (1,1,-3) and B be the point (-2,1,-1) in three-dimensional space.
A particle moves along the straight line through A and B at uniform speed in such a way that at time
t=0 seconds the particle is at A, and at t=1 second the particle is at B. Let P(t) be the location of the particle at time t (so, P(0)=A and P(1)=B).
a. Find the coordinates of the point P(t) each in terms of t.
We will write the coordinates of points as 3 × 1 column matrices, as is consistent for work with matrix notation.
The velocity vector of the particle is $$\overrightarrow{A B}$$ = So, its position at time t is
b. Give a geometric interpretation of the point P(0.5).
Since P(0) = A and P(1) = B , P(0.5) is the midpoint of $$\overline{A B}$$.
Let L be the linear transformation represented by the 3×3 matrix $$\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 3 & 0 \\ 0 & 1 & 1 \end{array}\right]$$, and let A’=LA and B’=LB be the images of the points A and B, respectively, under L.
c. Find the coordinates of A’ and B’.
A second particle moves through three-dimensional space. Its position at time t is given by L(P(t)), the image of the location of the first particle under the transformation L.
d. Where is the second particle at times t=0 and t=1? Briefly explain your reasoning.
L (P (t )) = A’ + t$$\overrightarrow{\mathrm{A}^{\prime} B^{\prime}} .$$
Thus, the second particle is moving along the straight line through A’ and B’ at a uniform velocity given by the vector $$\overrightarrow{\mathrm{A}^{\prime} B^{\prime}}$$ = | 2021-11-27T23:16:57 | {
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http://hyperspacewiki.org/index.php?title=Tietze_Extension_Theorem&oldid=1525 | Tietze Extension Theorem
The Tietze Extension Theorem deals with the problem of extending a continuous real-valued function that is defined on a subspace of a topological space \$X\$ to a continuous function defined on all of \$X\$.
Definition
Let \$X\$ be a normal space and let \$A\$ be a closed subspace of \$X\$.
1. Any continuous map of \$A\$ into the closed interval \$[a,b]\$ of \$\mathbb{R}\$ may be extended to a continuous map of all of \$X\$ onto \$[a,b]\$.
2. Any continuous map of \$A\$ into \$\mathbb{R}\$ may be extended to a continuous map of all of \$X\$ into \$\mathbb{R}\$.[1] | 2020-12-05T11:03:54 | {
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https://mathoverflow.net/questions/342627/how-are-characteristic-classes-morphisms-of-infinite-loop-spaces-if-they-are/342643 | # How are characteristic classes morphisms of infinite loop spaces? (if they are)
The direct sum of real vector bundles endows $$BO=\mathrm{colim} BO(n)$$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $$O$$, i.e., one has a natural structure of abelian group up to homotopy on $$BSO$$, $$BSpin$$, $$BString$$, etc.
As $$w(E\oplus F)=w(E)\cup w(F)$$ one sees that $$w_1\colon BO \to K(\mathbb{Z}/2\mathbb{Z},1)$$ is a morphism of abelian groups up to homotopy, and that similarly $$w_2\colon BSO \to K(\mathbb{Z}/2\mathbb{Z},2)$$ is a morphism of abelian groups up to homotopy. One can even make a step further and see $$BSO \to BO\xrightarrow{w_1} K(\mathbb{Z}/2\mathbb{Z},1)$$ and $$BSpin \to BSO\xrightarrow{w_2} K(\mathbb{Z}/2\mathbb{Z},2)$$ as short exact sequences of abelian groups up to homotopy''.
One can be more ambitious here. Not only $$BO$$ is an abelian group up to homotopy, but it is an $$\infty$$-loop space, i.e. $$BO=\Omega^\infty bo$$ for a certain connective spectrum $$bo$$. The same applies to $$BSO$$, $$BSpin$$,etc., and it also applies to $$K(\mathbb{Z}/2\mathbb{Z},n)$$ as $$K(\mathbb{Z}/2\mathbb{Z},n)=\Omega^\infty\Sigma^n H\mathbb{Z}/2$$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations $$bso \to bo\xrightarrow{\Omega^{-\infty}w_1} \Sigma H\mathbb{Z}/2$$ and $$bspin \to bso\xrightarrow{\Omega^{-\infty}w_2} \Sigma^2 H\mathbb{Z}/2$$ of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$$BSO \to BO\xrightarrow{w_1} K(\mathbb{Z}/2\mathbb{Z},1)$$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $$BSpin^c$$ participates in a fibration of infinite loop spaces $$K(\mathbb{Z},2)\to BSpin^c\to BSO\xrightarrow{bw_2}K(\mathbb{Z},3)$$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF.
My question is:
• Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $$\Omega^\infty$$?
• Where can I find a rigorous proof of this statement?
Your sequences are all arise in the following standard way. Suppose $$x$$ is an $$(n-1)$$--connected spectrum and let $$X = \Omega^{\infty} x$$. One always has a fibration sequence $$y \rightarrow x \rightarrow \Sigma^n H\pi_n(X)$$ and applying $$\Omega^\infty$$ to this yields a fibration sequence of spaces $$Y \rightarrow X \rightarrow K(\pi_n(X),n).$$
$$Y$$ is the $$n$$--connected cover of $$X$$.
(In your situation, one has successive covers $$bspin \rightarrow bso \rightarrow bo$$.)
• Ah, sure! For any spectrum $x$ one has the fiber sequence $x_{>n} \to x \to x_{\leq n}$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_{\leq n}=\Sigma^n H\pi_n(x) = \Sigma^n H\pi_n(\Omega^\infty x)$. Then one applies $\Omega^\infty$. I was missing the obvious here. – domenico fiorenza Sep 28 '19 at 5:52
Yes it is true. You have correctly interpreted the intended meaning of the phrase fibration of infinite loop spaces''. One early reference is chapter I of $$E_{\infty}$$ ring spaces and $$E_{\infty}$$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf | 2020-02-17T21:10:48 | {
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https://www.physicsforums.com/threads/volume-between-a-circular-paraboloid-and-a-plane.549623/ | # Volume Between A Circular Paraboloid and a Plane
## Homework Statement
Find the volume of the solid E bounded by z = 3+x2 +y2 and z = 6.
## The Attempt at a Solution
I'm going to use cylindrical coordinates. So, I have,
z = 3 + r2
Clearly, my bounds on z are 3 and 6. If I project the intersection of the paraboloid and the plan onto the xy-plane, I have 3 = r^2. My bounds on r are then 0 and ±√3, but I reject the negative boundary. I will use the bounds 0 and 2*pi for θ, though I don't know how these bounds are known to be appropriate. So, if I perform a triple integral of r in the order dz dr dθ, I get an answer of 9*pi.
Does that sound legitimate? | 2022-07-02T11:36:36 | {
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http://mathoverflow.net/questions/67252/norm-of-tridiagonal-operator | # Norm of tridiagonal operator
Recently, I needed to estimate the operator norm of the tridiagonal operator, but I am sure answers much more refined than my simple observations must be known.
Let $T$ be the linear operator that maps a square matrix to its tridiagonal part. Thus, the action of $T$ on a matrix $X$ can be defined by the Hadamard product $M \circ X$, where $m_{ij}=1$ if $|i-j| \le 1$, and $m_{ij}=0$ otherwise.
What is the operator 2-norm of $T$? (or what is a good approximation thereof?)
The observation $\|M \circ X\| \le \|M\| \|X\|$ shows that $\|T\| \le 3$. A more refined estimate follows from Theorem 5.5.3 of Horn and Johnson's Topics in Matrix Analysis, which says that $\|M\circ X\| \le r_1(M)c_1(X)$, where $r_1$ is maximum row-length (Euclidean norm) and $c_1$ is max column length. This result then implies that $\|T\| \le \sqrt{3}$.
I am sure that significantly more refined estimates of $\|T\|$ are available, and will be thankful if you can provide me a reference, or maybe a short proof itself.
-
If you consider the operator $T_r$ that retains the diagonals defined by $|i-j|\le r$ (yours is $T_1$), its norm is accurately bounded by $$L_r=\frac{1}{2\pi}\int_{-\pi}^\pi |D_r(\theta)|d\theta,$$ where $D_r$ is the Dirichlet kernel.
I like it! I teach the case $r=0$ (diagonal extraction) here. – Denis Serre Jun 8 '11 at 15:35 | 2015-12-01T23:48:28 | {
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https://homes.cs.washington.edu/~jrl/teaching/cseP521wi17/homework/hw5.html | ## CSE P521: HW #5
Out: Saturday, Feb 18
Due: Saturday, Feb 25 at midnight (in the dropbox)
### Assignment
You should use the guidelines for HW#3, stated in slide 2 here. In particular, you may work with a partner on this homework.
#### Part I: Theory questions
1. See the proof of Theorem 2 (analysis of the power method) in these notes. Now suppose that the matrix $$A$$ has eigenvalues $$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$$, and that $$\lambda_1=\lambda_2=\cdots=\lambda_k$$, but $$\lambda_{k+1} < \lambda_k$$. Give a formal analysis of the power method in this setting, analogous to the statement of Theorem 2. Be careful to specify what replaces the vector $$\mathbf{v}_1$$. One cannot specify a single vector ahead of time. Make sure you explain why.
2. Suppose that $$A=X^T X$$ is an $$n\times n$$ symmetric matrix with real entries formed from an $$m \times n$$-dimensional matrix of data points $$X$$. Let $$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$$ denote the eigenvalues of $$A$$.
• Show that all the eigenvalues of $$A$$ are non-negative.
• In class, we argued that if we have an algorithm to find the top component $$\mathbf{v}_1$$ of $$A$$, then we can use this recursively to find the top $$k$$ components. State formally an algorithm that achieves this (assuming a function that computes the top component).
Under the assumption that all the eigenvalues of $$A$$ are distinct, use induction to prove that your algorithm works.
#### Part II: Analysis
You should do only Part I from this homework assignment.
If you attempt the bonus questions, I will also assign some extra credit value to your answers.
Resources: The data set from the 1000 genomes project. | 2018-01-22T14:32:23 | {
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https://web2.0calc.com/questions/helllppp_2 | +0
# Helllppp
0
312
4
Find $\angle BPC$ below. [asy] pair A = (0,0); pair B = (1,0); pair P = (1/2, sqrt(3)/2); pair D = (0,-1); pair C = (1,-1); draw(A--B--C--D--cycle); draw(A--P--B); draw(rightanglemark(B,A,D, 2)); draw(rightanglemark(C,B,A, 2)); draw(rightanglemark(D,C,B, 2)); draw(rightanglemark(A,D,C, 2)); add(pathticks(P--A, 1, .5, 1, 2)); add(pathticks(P--B, 1, .5, 1, 2)); add(pathticks(A--B, 1, .5, 1, 2)); add(pathticks(B--C, 1, .5, 1, 2)); add(pathticks(C--D, 1, .5, 1, 2)); add(pathticks(A--D, 1, .5, 1, 2)); label("$P$", P, N); label("$A$", A, W); label("$B$", B, E); label("$C$", C, SE); label("$D$", D, SW); [/asy]
Sep 21, 2019
#1
+195
+3
Triangle BPC is an isossilies triangle because sides BP and BC are the same length, so angles BPC and BCP are congruent. To find angle BPC you first need to find angle PBC.
Triangle APB is equilateral so all the angles measure 60 degrees, so angle APB measures 60 degrees. All the angles in a square measure 90 degrees. The measure of angle PBC is the sum of the angles ABC=90 and ABP=60 so PBC=150 degrees.
$$2 \cdot m\angle BPC+m\angle PBC=180$$
150 can be substituted for angle PBC
$$2 \cdot m\angle BPC+150=180$$
subtract 150
$$2 \cdot m\angle BPC=30$$
divide by 2
$$\boxed{m\angle BPC=15}$$
.
Sep 21, 2019
#3
+2847
+4
Good job Power27!
How were able to decipher the picture code?
CalculatorUser Sep 22, 2019
#4
+195
+3
http://artofproblemsolving.com/texer/
AoPS's TeXeR supports bbCode, LaTeX and Asymptote (what was used to create the drawing).
Asymptote is pretty cool, I've been able to make some fun stuff with it.
power27 Sep 22, 2019 | 2020-04-01T12:07:06 | {
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https://courses.wikinana.org/math214/01-27 | math214:01-27
# 2020-01-27, Monday
Today, we talk about the linear approximation of a smooth manifold at a point $p \in M$: the tangent space $T_p M$. There are several approaches to define the tangent space, one possible way is to define it in each coordinate patch, and then show the compatibilities in coordinate patch; another way is to characterize its action on a function by directional derivatives, and characterize the tangent space as such.
In class, I followed basically Lee's Ch 3, first two sections. Spivak also did a very good job explaning what is a tangent bundle, and gives an example of why the tangent bundle of a sphere is not trivial (you cannot comb the hair on a sphere).
Also, I suggested reading about examples of smooth manifolds. Lee's Ch 1, Ex 1.30, 1.31, 1.33 are all worth reading. | 2022-06-25T14:39:12 | {
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https://www.iitianacademy.com/ib-dp-physics-2-3-work-energy-and-power-question-bank-sl-paper-2-part-a/ | # IBDP Physics 2.3 – Work, energy, and power: IB style Question Bank SL Paper 2
Question
This question is in two parts. Part 1 is about the motion of a ship. Part 2 is about melting ice.
Part 1 Motion of a ship
Some cargo ships use kites working together with the ship’s engines to move the vessel.
The tension in the cable that connects the kite to the ship is 250 kN. The kite is pulling the ship at an angle of 39° to the horizontal. The ship travels at a steady speed of $${\text{8.5 m}}\,{{\text{s}}^{ – 1}}$$ when the ship’s engines operate with a power output of 2.7 MW.
The ship’s engines are switched off and the ship comes to rest from a speed of $${\text{7 m}}\,{{\text{s}}^{ – 1}}$$ in a time of 650 s.
Part 2 Melting ice
A container of negligible mass, isolated from its surroundings, contains 0.150 kg of ice at a temperature of –18.7 °C. An electric heater supplies energy at a rate of 125 W.
a. Outline the meaning of work. [2]
b.i. Calculate the work done on the ship by the kite when the ship travels a distance of 1.0 km. [2]
b.ii. Show that, when the ship is travelling at a speed of $${\text{8.5 m}}\,{{\text{s}}^{ – 1}}$$, the kite provides about 40% of the total power required by the ship. [4]
c. The kite is taken down and no longer produces a force on the ship. The resistive force $$F$$ that opposes the motion of the ship is related to the speed $$v$$ of the ship by
$F = k{v^2}$
where $$k$$ is a constant.
Show that, if the power output of the engines remains at 2.7 MW, the speed of the ship will decrease to about $${\text{7 m}}\,{{\text{s}}^{ – 1}}$$. Assume that $$k$$ is independent of whether the kite is in use or not. [3]
d.i. Estimate the distance that the ship takes to stop. Assume that the acceleration is uniform. [2]
d.ii. It is unlikely that the acceleration of the ship will be uniform given that the resistive force acting on the ship depends on the speed of the ship. Using the axes, sketch a graph to show how the speed $$v$$ varies with time $$t$$ after the ship’s engines are switched off.
[2]
e. Describe, with reference to molecular behaviour, the process of melting ice. [2]
f.i. After a time interval of 45.0 s all of the ice has reached a temperature of 0 °C without any melting. Calculate the specific heat capacity of ice. [2]
f.ii. The following data are available.
Specific heat capacity of water $$= 4200{\text{ J}}\,{\text{k}}{{\text{g}}^{ – 1}}{{\text{K}}^{ – 1}}$$
Specific latent heat of fusion of ice $$= 3.30 \times {10^5}{\text{ J}}\,{\text{k}}{{\text{g}}^{ – 1}}$$
Determine the final temperature of the water when the heater supplies energy for a further 600 s. [3]
g. The whole of the experiment in (f)(i) and (f)(ii) is repeated with a container of negligible mass that is not isolated from the surroundings. The temperature of the surroundings is 18 °C. Comment on the final temperature of the water in (f)(ii). [3]
## Markscheme
a. $${\text{work done}} = {\text{force}} \times {\text{distance moved}}$$;
(distance moved) in direction of force;
or
energy transferred;
from one location to another;
or
$${\text{work done}} = Fs\cos \theta$$;
with each symbol defined;
b.i.
$${\text{horizontal force}} = 250\,000 \times \cos 39^\circ {\text{ }}( = 1.94 \times {10^5}{\text{ N}})$$;
$${\text{work done}} = 1.9 \times {10^8}{\text{ J}}$$;
b.ii.
$${\text{power provided by kite}} = (1.94 \times {10^5} \times 8.5 = ){\text{ }}1.7 \times {10^6}{\text{ W}}$$;
$${\text{total power}} = (2.7 + 1.7) \times {10^6}{\text{ W }}( = 4.4 \times {10^6}{\text{ W}})$$;
$${\text{fraction provided by kite}} = \frac{{1.7}}{{2.7 + 1.7}}$$;
38% or 0.38; (must see answer to 2+ sig figs as answer is given)
Allow answers in the range of 37 to 39% due to early rounding.
or
Award [3 max] for a reverse argument such as:
if 2.7 MW is 60%;
then kite power is $$\frac{2}{3} \times 2.7{\text{ MW}} = 1.8{\text{ MW}}$$;
shows that kite power is actually 1.7 MW; (QED)
c.
$$P = (k{v^2}) \times v = k{v^3}$$;
$$\frac{{{v_1}}}{{{v_2}}} = \left( {\sqrt[3]{{\left( {\frac{{{P_1}}}{{{P_2}}}} \right)}} = } \right){\text{ }}\sqrt[3]{{\left( {\frac{{7.7}}{{4.4}}} \right)}}$$;
$${\text{final speed of ship}} = 7.2{\text{ m}}\,{{\text{s}}^{ – 1}}$$; (at least 2 sig figs required).
Approximate answer given, marks are for working only.
d.i. correct substitution of 7 or 7.2 into appropriate kinematic equation;
an answer in the range of 2200 to 2400 m;
d.ii.
starts at $$7.0/7.2{\text{ m}}\,{{\text{s}}^{ – 1}}$$; (allow ECF from (d)(i))
correct shape;
e.
in ice, molecules vibrate about a fixed point;
as their total energy increases, the molecules (partly) overcome the attractive force between them;
in liquid water the molecules are able to migrate/change position;
f.
$$(Q = ){\text{ }}45.0 \times 125{\text{ }}( = 5625{\text{ J}})$$;
$$c = \left( {\frac{Q}{{m\Delta \theta }} = } \right){\text{ }}2.01 \times {10^3}{\text{ J}}\,{\text{k}}{{\text{g}}^{ – 1}}{{\text{K}}^{ – 1}}$$;
f.i.
$${\text{energy available}} = 125 \times 600{\text{ }}( = 75000{\text{ J}})$$;
$${\text{energy available to warm the water}} = 75000 – [0.15 \times 3.3 \times {10^5}]{\text{ }}( = 25500{\text{ J}})$$;
$${\text{temperature}} = \left( {\frac{{25500}}{{0.15 \times 4200}} = } \right){\rm{ 40.5 ^\circ C}}$$;
f.ii.
ice/water spends more time below 18 °C;
so net energy transfer is in to the system;
so final water temperature is higher;
or
ice/water spends less time below 18 °C;
so net energy transfer is out of the system;
so final water temperature is lower;
## Examiners report
a.[N/A]
b.i.[N/A]
b.ii. This is a “show that” question which means that the candidate is obliged to show their line of reasoning. Very few SL candidates did this.
c. This was easy using proportionality, but most candidates at SL attempted to calculate k unnecessarily. Even so there were many correct answers.
d.i.[N/A]
d.ii.[N/A]
e. A minority of candidates knew that molecules made a transition from being localised to being free to migrate, but had difficulty expressing their answers coherently. Candidates are so used to commenting on the energy transformations when ice melts, that many completely misread the question.
f.i.
Many good answers, although those that did not get the correct answers presented their working in such a way that part marks (ECF) were not able to be given.
g.
The significance of the temperature of the surroundings was ignored by nearly all candidates, but most were able to obtain 2 marks for suggesting that thermal energy would be lost to the surroundings causing a lower final temperature.
Question
This question is in two parts. Part 1 is about the motion of a car. Part 2 is about electricity.
Part 1 Motion of a car
A car is travelling along the straight horizontal road at its maximum speed of $${\text{56 m}}\,{{\text{s}}^{ – 1}}$$. The power output required at the wheels is 0.13 MW.
A driver moves the car in a horizontal circular path of radius 200 m. Each of the four tyres will not grip the road if the frictional force between a tyre and the road becomes less than 1500 N.
Part 2 Electricity
A lemon can be used to make an electric cell by pushing a copper rod and a zinc rod into the lemon.
A student constructs a lemon cell and connects it in an electrical circuit with a variable resistor. The student measures the potential difference V across the lemon and the current I in the lemon.
a. A car accelerates uniformly along a straight horizontal road from an initial speed of $${\text{12 m}}\,{{\text{s}}^{ – 1}}$$ to a final speed of $${\text{28 m}}\,{{\text{s}}^{ – 1}}$$ in a distance of 250 m. The mass of the car is 1200 kg. Determine the rate at which the engine is supplying kinetic energy to the car as it accelerates. [4]
b. A car is travelling along a straight horizontal road at its maximum speed of $${\text{56 m}}\,{{\text{s}}^{ – 1}}$$. The power output required at the wheels is 0.13 MW.
(i) Calculate the total resistive force acting on the car when it is travelling at a constant speed of $${\text{56 m}}\,{{\text{s}}^{ – 1}}$$.
(ii) The mass of the car is 1200 kg. The resistive force $$F$$ is related to the speed $$v$$ by $$F \propto {v^2}$$. Using your answer to (b)(i), determine the maximum theoretical acceleration of the car at a speed of $${\text{28 m}}\,{{\text{s}}^{ – 1}}$$. [5]
c.
(i) Calculate the maximum speed of the car at which it can continue to move in the circular path. Assume that the radius of the path is the same for each tyre.
(ii) While the car is travelling around the circle, the people in the car have the sensation that they are being thrown outwards. Outline how Newton’s first law of motion accounts for this sensation. [6]
d.
(i) Draw a circuit diagram of the experimental arrangement that will enable the student to collect the data for the graph.
(ii) Show that the potential difference $$V$$ across the lemon is given by
$V = E – Ir$
where $$E$$ is the emf of the lemon cell and $$r$$ is the internal resistance of the lemon cell.
(iii) The graph shows how $$V$$ varies with $$I$$.
Using the graph, estimate the emf of the lemon cell.
(iv) Determine the internal resistance of the lemon cell.
(v) The lemon cell is used to supply energy to a digital clock that requires a current of $${\text{6.0 }}\mu {\text{A}}$$. The clock runs for 16 hours. Calculate the charge that flows through the clock in this time.
[10]
d.
## Markscheme
a. use of a kinematic equation to determine motion time $$( = 12.5{\text{ s)}}$$;
change in kinetic energy $$= \frac{1}{2} \times 1200 \times \left[ {{{28}^2} – {{12}^2}} \right]{\text{ }}( = 384{\text{ kJ)}}$$;
rate of change in kinetic energy $$= \frac{{384000}}{{12.5}}$$; } (allow ECF of 162 from (28 – 12)2 for this mark)
31 (kW);
or
use of a kinematic equation to determine motion time $$( = 12.5{\text{ s)}}$$;
use of a kinematic equation to determine acceleration $$( = {\text{1.28 m}}\,{{\text{s}}^{ – 2}}{\text{)}}$$;
work done $$= \frac{{F \times s}}{{{\text{time}}}} = \frac{{1536 \times 250}}{{12.5}}$$;
31 (kW);
b.
(i) $${\text{force}} = \frac{{{\text{power}}}}{{{\text{speed}}}}$$;
2300 or 2.3k (N);
Award [2] for a bald correct answer.
(ii) resistive force $$= \frac{{2300}}{4}$$$$\,\,\,$$or$$\,\,\,$$$$\frac{{2321}}{4}{\text{ }}( = 575)$$; (allow ECF)
so accelerating force = $$(2300 – 580 = ){\text{ }}1725{\text{ (N)}}$$$$\,\,\,$$or$$\,\,\,$$1741 (N);
$$a = \frac{{1725}}{{1200}} = 1.44{\text{ (m}}{{\text{s}}^{ – 2}}{\text{)}}$$$$\,\,\,$$or$$\,\,\,$$$$a = \frac{{1741}}{{1200}} = 1.45{\text{ (m}}\,{{\text{s}}^{ – 2}}{\text{)}}$$;
Award [2 max] for an answer of 0.49 (m$$\,$$s–2) (omits 2300 N).
c.
(i) centripetal force must be $$< {\text{6000 (N)}}$$; (allow force = 6000 N)
$${v^2} = F \times \frac{r}{m}$$;
$${\text{31.6 (m}}\,{{\text{s}}^{ – 1}}{\text{)}}$$;
Allow [3] for a bald correct answer.
Allow [2 max] if 4$$\times$$ is omitted, giving 15.8 (m$$\,$$s–1).
(ii) statement of Newton’s first law;
(hence) without car wall/restraint/friction at seat, the people in the car would move in a straight line/at a tangent to circle;
(hence) seat/seat belt/door exerts centripetal force;
(in frame of reference of the people) straight ahead movement is interpreted as “outwards”;
d.
(i) voltmeter in parallel with cell; (allow ammeter within voltmeter leads)
ammeter in series with variable resistor; } (must draw as variable arrangement or as potential divider)
Allow cell symbol for lemon/cell/box labelled “lemon cell”.
Award [1 max] if additional cell appears in the circuit.
(ii) $$E = I(R + r)$$ and $$V = IR$$ used; (must state both explicitly)
re-arrangement correct ie $$E = V + Ir$$; } (accept any other correct re-arrangement eg. involving energy conversion)
(iii) line correctly extrapolated to y-axis; (judge by eye)
1.6 or 1.60 (V); (allow ECF from incorrect extrapolation)
(iv) correct read-offs from large triangle greater than half line length;
290 to 310 $${\text{(}}\Omega {\text{)}}$$;
Award [2 max] for the use of one point on line and equation.
(v) 0.35 (C);
## Examiners report
a. There were at least two routes to tackle this problem. Some solutions were so confused that it was difficult to decide which method had been used. Common errors included: forgetting that the initial speed was $${\text{12 m}}\,{{\text{s}}^{ – 1}}$$ not zero, power of ten errors, and simple mistakes in the use of the kinematic equations, or failure to evaluate work done = force $$\times$$ distance correctly. However, many candidates scored partial credit. Scores of two or three out of the maximum four were common showing that many are persevering to get as far as they can.
b.
(i) Many correct solutions were seen. Candidates are clearly comfortable with the use of the equation force = power/speed.
(ii) The method to be used here was obvious to many. What was missing was a clear appreciation of what was happening in terms of resistive force in the system. Many scored two out of three because they indicated a sensible method but did not use the correct value for the force. Scoring two marks does require that the explanation of the method is at least competent. Those candidates who give limited explanations of their method leading to a wrong answer will generally accumulate little credit. A suggestion (never seen in answers) is that candidates should have begun from a free-body force diagram which would have revealed the relationship of all the forces.
c.
(i) The major problem here was that most candidates did not recognise that 1500 N of force acting at each of four wheels will imply a total force of 6 kN. Again, partial credit was available only if it was clear what the candidate was doing and what the error was.
(ii) Statements of Newton’s first law were surprisingly poor. As in previous examinations, few candidates appear to have learnt this essential rule by heart and they produce a garbled and incomplete version under examination pressure. The first law was then only loosely connected to the particular context of the question. Candidates have apparently not learnt to relate the physics they learn to everyday contexts.
d.
(i) Circuit diagrams continue to be a particular issue for many candidates. Neat, well-drawn diagrams are rarely seen. Some diagrams had two cells, the lemon cell and another. Variable resistors were sometimes absent (or were drawn as fixed). Potential dividers were often attempted usually unsuccessfully. Generally candidates gained an average one mark for what should have been a familiar task.
(ii) Those who quoted the data booklet equation and the definition of resistance were generally able to show the final expression. Some however could not convince the examiners that they knew what they were doing.
(iii) Candidates were expected to understand the physical point that the emf can be determined when the current in the cell is zero. For many, an extrapolation of the obvious straight line to the emf axis and a correct read-off gave an easy couple of marks. Some however did not understand the physics of the circuit and gave poorly described solutions.
(iv) The internal resistance was best obtained from a large triangle drawn on the graph. Many however gained two of the three marks because they engendered power of ten errors or because they used only one point, or because their triangle was too small.
(v) Only a minority were able to use the data to calculate the charge transferred correctly.
Question
Impulse and momentum
The diagram shows an arrangement used to test golf club heads.
The shaft of a club is pivoted and the centre of mass of the club head is raised by a height h before being released. On reaching the vertical position the club head strikes the ball.
a. (i) Describe the energy changes that take place in the club head from the instant the club is released until the club head and the ball separate.
(ii) Calculate the maximum speed of the club head achievable when h = 0.85 m. [4]
b. The diagram shows the deformation of a golf ball and club head as they collide during a test.
Explain how increasing the deformation of the club head may be expected to increase the speed at which the ball leaves the club. [2]
c.
In a different experimental arrangement, the club head is in contact with the ball for a time of 220 μs. The club head has mass 0.17 kg and the ball has mass 0.045 kg. At the moment of contact the ball is at rest and the club head is moving with a speed of 38 ms–1. The ball moves off with an initial speed of 63 ms–1.
(i) Calculate the average force acting on the ball while the club head is in contact with the ball.
(ii) State the average force acting on the club head while it is in contact with the ball.
(iii) Calculate the speed of the club head at the instant that it loses contact with the ball. [5]
## Markscheme
a. (i) (gravitational) potential energy (of club head) goes to kinetic energy (of club head);
some kinetic energy of club head goes to internal energy of club head/kinetic energy of ball;
(ii) equating mgh to $$\frac{1}{2}$$mv2 ;
v=4.1(ms-1);
Award [0] for answers using equation of motion – not uniform acceleration.
b.
deformation prolongs the contact time;
increased impulse => bigger change of momentum/velocity;
or
(club head) stores (elastic) potential energy on compression;
this energy is passed to the ball;
c.
(i) any value of $$\frac{{{\rm{mass}} \times {\rm{velocity}}}}{{{\rm{time}}}}$$;
1.3 x 104 (N);
(ii)-1.3 x 10-4 (N);
Accept statement that force is in the opposite direction to (c)(i).
Allow the negative of any value given in (c)(i).
(iii) clear use of conservation of momentum / impulse = change of momentum;
21(ms-1 );
or
a=$$\left( {\frac{{\rm{F}}}{{\rm{m}}} = \frac{{ – 13000}}{{0.17}} = } \right)\left( – \right)76500\left( {{\rm{m}}{{\rm{s}}^{ – 1}}} \right)$$;
v=(u+at=38-76500 x 0.00022=) 21(ms-1);
Award [2] for a bald correct answer.
## Examiners report
a. (i) Nearly all candidates gained a mark for recognising the change from kinetic to potential energy in this part. Fewer recognised that the club head would not transfer all of its energy to the ball and therefore retained a significant amount of energy.
(ii) This part was well done by many.
b.
A minority of candidates became bogged down by the deformation of the ball and club head idea and ventured into elastic potential energy ideas. This had a successful outcome in many cases when there was discussion of the compression providing further kinetic energy to the ball on recovering its shape. The most straightforward solution was to use to principle of impulse being equal to the change in momentum (as shown in the question heading) and simply to recognise that an increased contact time would be expected to give a greater change of momentum for a constant force.
c.
(i) This was well done with the only real problem being deciding which was the speed change of the ball.
(ii) Less candidates than anticipated recognised that the force on the club head was equal and opposite to that acting on the ball (applying Newton’s third law of motion).
(iii) Most made a good attempt at calculating the speed of the club head. | 2022-08-08T04:03:34 | {
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https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_11&diff=prev&oldid=75389 | # Difference between revisions of "2016 AMC 10A Problems/Problem 11"
## Problem
What is the area of the shaded region of the given $8 \times 5$ rectangle?
$[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("1",(1/2,5),dir(90)); label("7",(9/2,5),dir(90)); label("1",(8,1/2),dir(0)); label("4",(8,3),dir(0)); label("1",(15/2,0),dir(270)); label("7",(7/2,0),dir(270)); label("1",(0,9/2),dir(180)); label("4",(0,2),dir(180)); [/asy]$
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$
## Solution
First, split the rectangle into $4$ triangles: $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("1",(1/2,5),dir(90)); label("7",(9/2,5),dir(90)); label("1",(8,1/2),dir(0)); label("4",(8,3),dir(0)); label("1",(15/2,0),dir(270)); label("7",(7/2,0),dir(270)); label("1",(0,9/2),dir(180)); label("4",(0,2),dir(180)); draw((0,5)--(8,0)); [/asy]$
The bases of these triangles are all $1$, and their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$. | 2023-01-31T22:37:41 | {
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https://demo.formulasearchengine.com/wiki/Monomial | Monomial
In mathematics, a monomial is, roughly speaking, a polynomial which has only one term. Two different definitions of a monomial may be encountered:
In the context of Laurent polynomials and Laurent series, the exponents of a monomial may be negative, and in the context of Puiseux series, the exponents may be rational numbers.
Since the word "polynomial" comes from "poly-" plus the Greek word "νομός" (nomós, meaning part, portion), a monomial should theoretically be called a "mononomial". "Monomial" is a syncope of "mononomial".[1]
Comparison of the two definitions
With either definition, the set of monomials is a subset of all polynomials that is closed under multiplication.
Both uses of this notion can be found, and in many cases the distinction is simply ignored, see for instance examples for the first[2] and second[3] meaning, and an unclear definition. In informal discussions the distinction is seldom important, and tendency is towards the broader second meaning. When studying the structure of polynomials however, one often definitely needs a notion with the first meaning. This is for instance the case when considering a monomial basis of a polynomial ring, or a monomial ordering of that basis. An argument in favor of the first meaning is also that no obvious other notion is available to designate these values (the term power product is in use, in particular when monomial is used with the first meaning, but it does not make the absence of constants clear either), while the notion term of a polynomial unambiguously coincides with the second meaning of monomial.
As bases
The most obvious fact about monomials (first meaning) is that any polynomial is a linear combination of them, so they form a basis of the vector space of all polynomials - a fact of constant implicit use in mathematics.
Number
The number of monomials of degree d in n variables is the number of multicombinations of d elements chosen among the n variables (a variable can be chosen more than once, but order does not matter), which is given by the multiset coefficient ${\displaystyle \textstyle {\left(\!\!{n \choose d}\!\!\right)}}$. This expression can also be given in the form of a binomial coefficient, as a polynomial expression in d, or using a rising factorial power of d + 1:
${\displaystyle \left(\!\!{n \choose d}\!\!\right)={\binom {n+d-1}{d}}={\binom {d+(n-1)}{n-1}}={\frac {(d+1)\times (d+2)\times \cdots \times (d+n-1)}{1\times 2\times \cdots \times (n-1)}}={\frac {1}{(n-1)!}}(d+1)^{\overline {n-1}}.}$
The latter forms are particularly useful when one fixes the number of variables and lets the degree vary. From these expressions one sees that for fixed n, the number of monomials of degree d is a polynomial expression in d of degree ${\displaystyle n-1}$ with leading coefficient ${\displaystyle {\tfrac {1}{(n-1)!}}}$.
For example, the number of monomials in three variables (${\displaystyle n=3}$) of degree d is ${\displaystyle \textstyle {\frac {1}{2}}(d+1)^{\overline {2}}=\textstyle {\frac {1}{2}}(d+1)(d+2)}$; these numbers form the sequence 1, 3, 6, 10, 15, ... of triangular numbers.
The Hilbert series is a compact way to express the number of monomials of a given degree: the number of monomials of degree Template:Mvar in Template:Mvar variables is the coefficient of degree Template:Mvar of the formal power series expansion of
${\displaystyle {\frac {1}{(1-t)^{n}}}.}$
Notation
Notation for monomials is constantly required in fields like partial differential equations. If the variables being used form an indexed family like ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$, ${\displaystyle x_{3}}$, ..., then multi-index notation is helpful: if we write
${\displaystyle \alpha =(a,b,c)}$
we can define
${\displaystyle x^{\alpha }=x_{1}^{a}\,x_{2}^{b}\,x_{3}^{c}}$
and save a great deal of space.
Degree
The degree of a monomial is defined as the sum of all the exponents of the variables, including the implicit exponents of 1 for the variables which appear without exponent; e.g., in the example of the previous section, the degree is ${\displaystyle a+b+c}$. The degree of ${\displaystyle xyz^{2}}$ is 1+1+2=4. The degree of a nonzero constant is 0. For example, the degree of -7 is 0.
The degree of a monomial is sometimes called order, mainly in the context of series. It is also called total degree when it is needed to distinguish it from the degree in one of the variables.
Monomial degree is fundamental to the theory of univariate and multivariate polynomials. Explicitly, it is used to define the degree of a polynomial and the notion of homogeneous polynomial, as well as for graded monomial orderings used in formulating and computing Gröbner bases. Implicitly, it is used in grouping the terms of a Taylor series in several variables.
Geometry
In algebraic geometry the varieties defined by monomial equations ${\displaystyle x^{\alpha }=0}$ for some set of α have special properties of homogeneity. This can be phrased in the language of algebraic groups, in terms of the existence of a group action of an algebraic torus (equivalently by a multiplicative group of diagonal matrices). This area is studied under the name of torus embeddings. | 2020-11-29T01:37:45 | {
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https://math.stackexchange.com/questions/3600041/if-int-ab-fx-dx-geq-0-is-fx-geq0-forall-x-ina-b/3600043 | # If $\int_{a}^{b} f(x) dx \geq 0$ is $f(x)\geq0, \forall x \in[a,b]$?
Let $$f:[a,b]\rightarrow R$$ be a continiuous function. If $$\int_{a}^{b} f(x) dx \geq 0$$ is $$f(x)\geq0, \forall x \in[a,b]$$? I know that if $$\int_{a}^{b} f(x) dx > 0$$ then $$f(x)\geq0$$ is false, but what about the inquality that also has the equals? Is it true or false?
• If the integral is strictly positive it is also positive. Therefore just take the same counter-example for the statement which you know is false. Mar 29 '20 at 11:19
$$\int_{0}^{2\pi}\sin{x}dx=0\geq 0$$ but certainly not $$\sin{x}\geq 0$$. | 2021-09-27T02:04:42 | {
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https://www.shaalaa.com/textbook-solutions/c/rd-sharma-solutions-10-mathematics-chapter-10-circles_296 | Share
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# R.D. Sharma solutions for Class 10 Mathematics chapter 10 - Circles
## Chapter 10 - Circles
#### Page 0
Fill in the blank:
The common point of a tangent to a circle and the circle is called ____
Fill in the blank:
The common point of a tangent to a circle and the circle is called ____
Fill in the blank:
A circle can have __________ parallel tangents at the most.
Fill in the blank:
A circle can have __________ parallel tangents at the most.
Fill in the blank:
A tangent to a circle intersects it in _______ point (s).
Fill in the blank:
A tangent to a circle intersects it in _______ point (s).
Fill in the blank:
A line intersecting a circle in two points is called a __________.
Fill in the blank:
A line intersecting a circle in two points is called a __________.
Fill in the blank
The angle between tangent at a point on a circle and the radius through the point is ........
How many tangents can a circle have?
How many tangents can a circle have?
O is the center of a circle of radius 8cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB
If the tangent at point P to the circle with center O cuts a line through O at Q such that PQ= 24cm and OQ = 25 cm. Find the radius of circle
#### Page 0
If PT is a tangent at T to a circle whose center is O and OP = 17 cm, OT = 8 cm. Find the length of tangent segment PT.
Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle.
A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
If from any point on the common chord of two intersecting circles, tangents be drawn to circles, prove that they are equal.
If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair.
If AB, AC, PQ are tangents in Fig. and AB = 5cm find the perimeter of ΔAPQ.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at center.
In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°. Find
∠PRS.
If PA and PB are tangents from an outside point P. such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.
From an external point P, tangents PA and PB are drawn to the circle with centre O. If CD is the tangent to the circle at point E and PA = 14 cm. Find the perimeter of ABCD.
In the fig. ABC is right triangle right angled at B such that BC = 6cm and AB = 8cm. Find the radius of its in circle.
From a point P, two tangents PA and PB are drawn to a circle with center O. If OP =
diameter of the circle shows that ΔAPB is equilateral.
Two tangent segments PA and PB are drawn to a circle with center O such that ∠APB =120°. Prove that OP = 2AP
If ΔABC is isosceles with AB = AC and C (0, 2) is the in circle of the ΔABC touching BC at L, prove that L, bisects BC.
In fig. a circle touches all the four sides of quadrilateral ABCD with AB = 6cm, BC = 7cm, CD = 4cm. Find AD.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre
In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that ∠BAC +
∠ACD = 90°
Two circles touch externally at a point P. from a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and E respectively. Prove that TQ = TR.
In the fig. a circle is inscribed in a quadrilateral ABCD in which ∠B = 90° if AD = 23cm,
AB = 29cm and DS = 5cm, find the radius of the circle.
In fig. there are two concentric circles with Centre O of radii 5cm and 3cm. From an
external point P, tangents PA and PB are drawn to these circles if AP = 12cm, find the
tangent length of BP.
In the given figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA.
In figure PA and PB are tangents from an external point P to the circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN
In the given figure, PO⊥QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisector of each other.
In the fig two tangents AB and AC are drawn to a circle O such that ∠BAC = 120°. Prove that OA = 2AB.
In the given figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ΔAEO~Δ ABC.
The lengths of three consecutive sides of a quadrilateral circumscribing a circle are 4cm,5cm and 7cm respectively. Determine the length of fourth side.
In fig common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.
In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of (DO')/(CO')
In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of (DO')/(CO')
In figure OQ : PQ = 3 : 4 and perimeter of ΔPDQ = 60cm. determine PQ, QR and OP.
## R.D. Sharma solutions for Class 10 Mathematics chapter 10 - Circles
R.D. Sharma solutions for Class 10 Mathematics chapter 10 (Circles) include all questions with solution and detail explanation from 10 Mathematics. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has created the CBSE 10 Mathematics solutions in a manner that help students grasp basic concepts better and faster.
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Concepts covered in Class 10 Mathematics chapter 10 Circles are Theorem of External Division of Chords, Theorem of Internal Division of Chords, Converse of Theorem of the Angle Between Tangent and Secant, Theorem of Angle Between Tangent and Secant, Converse of Cyclic Quadrilateral Theorem, Corollary of Cyclic Quadrilateral Theorem, Theorem of Cyclic Quadrilateral, Corollaries of Inscribed Angle Theorem, Inscribed Angle Theorem, Intercepted Arc, Inscribed Angle, Property of Sum of Measures of Arcs, Tangent Segment Theorem, Converse of Tangent Theorem, Circles passing through one, two, three points, Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers, Cyclic Properties, Tangent - Secant Theorem, Cyclic Quadrilateral, Angle Subtended by the Arc to the Point on the Circle, Angle Subtended by the Arc to the Centre, Introduction to an Arc, Touching Circles, Number of Tangents from a Point on a Circle, Tangent to a Circle, Tangents and Its Properties, Theorem - Converse of Tangent at Any Point to the Circle is Perpendicular to the Radius, Number of Tangents from a Point to a Circle, Tangent to a Circle, Number of Tangents from a Point on a Circle, Circles Examples and Solutions, Introduction to Circles.
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S | 2019-02-20T18:34:41 | {
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https://math.stackexchange.com/questions/2248751/what-is-the-boundary-map-in-the-long-exact-sequence-in-homotopy-groups-especia | What is the boundary map in the long exact sequence in homotopy groups? (Especially for $SO(n)$)
Suppose that $E \to B$ is a fibration with fibers $F$. There is a long exact sequence in homotopy groups. In particular, we have boundary maps $\partial: \pi_n(B) \to \pi_{n-1} (F)$. Is there a good, geometric interpretation for this boundary map?
In particular, I want to understand the long exact sequence arising from $SO(2) \to SO(3) \to S^2$. How does an element $a \in \pi_2(S^2)$ induce a loop in $SO(2) \cong S^1$ via the boundary map. In particular, is it is visibly clear that the generator for $\pi_2(S^2)$ induces $[2] \in \pi_1(SO(1)) \cong \mathbb{Z}$?
Yes. Let's chase through the proof that this long exact sequence exists. First, the construction of the relative long exact sequence of a pair is reasonably elementary; and there is an isomorphism $p: \pi_n(E,F_b) \to \pi_n(B,b)$ given by projection commuting with the obvious map from $\pi_n(E,*)$ to both. And the boundary map is constructed very geometrically for the relative pair: you send a (homotopy class of) map $(D^n,S^{n-1}) \to (E,F_b)$ to the map $S^{n-1} \to F_b$. So the construction of $p^{-1}$ is the geometric content of the theorem.
Suppose you have a sphere in $B$. Think of the sphere as a homotopy $f_t: S^{n-1} \times [0,1] \to B$ with $f_0(x) = b_0$ and $f_1(x) = b$. Picking an arbitrary lift of $b_0$, applying the homotopy lifting property we get a disc in $E$ with boundary (corresponding to points of the form $(x,1)$) having having values in $F_b$. Another application of homotopy lifting shows this map is well-defined. That the two are mutually inverse is also simple.
For $S^1 \to S^3 \to S^2$, start with the identity map $S^2 \to S^2$. Thinking of $b_0$ as the north pole and $b$ as the south, and run the above procedure. In more generality, on the total space of the circle bundle with Euler number $n$, the construction will show that the boundary map in the corresponding long exact sequence $\pi_2(S^2) \to \pi_1(S^1)$ is multiplication by $n$. (As a hint, use that you can trivialize the bundles over the top and bottom hemispheres, while keeping in mind the data of the transition function over the center circle. It's easy to construct the lift over the trivial bundle.) | 2019-08-26T02:42:58 | {
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https://math.stackexchange.com/questions/967780/ellipse-1-e-1-fracx2a2-fracy2b2-1-a-b-another-ellipse-2 | # Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru…
Problem :
Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru extremities of major axis of $E_1$ and has its foci at ends of its minor axis. If eccentricity of both the ellipses are same, then find their eccentricity.
Since $E_2$ passes through extremities of major axis of $E_1$ therefore its length of minor axis is 2a; and coordinates of focus of $E_2$ = $2b$ but I am not getting further idea to solve this please help.
## 2 Answers
Let $e_1,e_2$ is the eccentricity of $E_1,E_2$ respectively.
First, we have $$e_1=\frac{\sqrt{a^2-b^2}}{a}.\tag1$$
Let $$E_2\ :\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1.$$ Here, note that $B\gt A.$
Since we have $$A=a,\ \ \ \sqrt{B^2-A^2}=b\Rightarrow B=\sqrt{a^2+b^2},$$ we have $$e_2=\frac{\sqrt{B^2-A^2}}{B}=\frac{b}{\sqrt{a^2+b^2}}.\tag2$$ Hence, from $(1),(2)$, we have $$\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{\sqrt{a^2+b^2}}\Rightarrow a^2b^2=a^4-b^4$$$$\Rightarrow \left(\frac ba\right)^4+\left(\frac ba\right)^2-1=0\Rightarrow \left(\frac ba\right)^2=\frac{\sqrt 5-1}{2}.$$ Hence, the answer is $$\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-\left(\frac ba\right)^2}=\sqrt{1-\frac{\sqrt 5-1}{2}}=\sqrt{\frac{3-\sqrt 5}{2}}=\frac{\sqrt 5-1}{2}.$$
So, the equation of $E_2$ $$\frac{x^2}{p^2}+\frac{y^2}{a^2}=1$$ where $p$ is the major axis
So we have $b^2=a^2(1-e^2)\ \ \ \ (0)$ and $a^2=p^2(1-e^2)\ \ \ \ (1)$
As the coordinates of the foci are $(0\pm pe),b=pe$ where $e$ is the common eccentricity
So,from $(0), a^2(1-e^2)=(pe)^2\ \ \ \ (2)$
Divide $(1)$ by $(2)$ | 2020-05-27T04:10:28 | {
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https://math.stackexchange.com/questions/3323056/what-is-the-a-matrix-decomposition-to-swap-out-an-eigenvalue | # What is the a matrix decomposition to swap out an eigenvalue?
I have the following problem: let $$A$$ be an $$m \times m$$ matrix that is not symmetric and real. Then, I am interested in its eigenspaces that correspond to the largest eigenvalue, $$\lambda_{\text{max}}$$. Without affecting the column span of the eigenspace associated with the largest root, say $$\text{sp}\,r_{\text{max}}$$, what can I do to $$A$$ to effect the following change: 1. remove $$\lambda_{\text{max}}$$, 2. insert $$\mu$$ in its stead, 3. obtain the relation $$A\, r_{\text{max}} = \mu r_{\text{max}}$$ so that the eigenvalue associated with $$\text{sp}\,r_{\text{max}}$$ is now $$\mu$$. I was thinking of possibly using the real Schur decomposition, thanks to its numerical stability. Any help is much appreciated!
Follow-up: suppose that the eigenvalue is potentially complex but the matrix is real. Then, as suggested, calculating $$A + I \left(\mu -\lambda_{\text{max}} \right)$$ would cause a problem because $$\lambda_{\text{max}}$$ comes in a conjugate pair. We (re-)define the notion of 'max' to be maximal by modulus, so $$\lambda_{\text{max}}$$ is such that $$||\lambda_{\text{max}}||$$ is maximal. (This arises from the fact that I am estimating $$A$$ using least squares. A few references, avenues I have considered is applying the method of Gershgorin discs or the Bauer-Fike theorem just to bound $$||\lambda_{\text{max}}||$$ using the (real) entries of $$A$$ but have not made much progress.
Moreover, it would be interesting to know if one could use the derivative of $$A$$ with respect to its eigenvalues to 'slide it' towards a matrix that would have the desired root, very much one uses the first derivative of a function to attain a linear approximation to.
The "obvious" thing to do is to change each $$\lambda_{max}$$ on the diagonal of the Schur decomposition to $$\mu$$, but that may change the eigenspace. For example, if $$A = \pmatrix{1 & 1\cr 0 & 2\cr}$$ (already upper triangular) the eigenspace for $$2$$ is spanned by $$\pmatrix{1\cr 1\cr}$$, but if you change the $$2$$ to $$\mu$$ the eigenspace for $$\mu$$ becomes $$\pmatrix{1/(\mu-1)\cr 1\cr}$$.
Instead, why not just add $$(\mu - \lambda_{max}) I$$ to the matrix? The eigenspaces stay the same, just that all eigenvalues are shifted by $$\mu - \lambda_{max}$$.
EDIT: Another possibility is to use the functional calculus: for any polynomial $$p(X)$$, $$p(A)$$ is a matrix whose eigenvalues are $$p(\lambda)$$ for eigenvalues $$\lambda$$ of $$A$$. If $$v$$ is an eigenvector of $$A$$ for $$\lambda$$, then $$p(A) v = p(\lambda) v$$. Moreover, if $$p$$ has real coefficients, $$p(A)$$ will still be real. Choose a polynomial such that $$p(\lambda_{max}) = \mu$$.
• Thanks so much, but just to make sure I understand, did you mean to write if $v$ is an eigenvector of $A$? Aug 15 '19 at 16:17
• If $Av = \lambda v$ then by induction $A^n v = \lambda^n v$ for all nonnegative integers $n$ (including $n=0$ where $A^0 = I$), and so if $p(x) = \sum_{k=0}^n c_k x^k$ is a polynomial, $p(A) v = \sum_{k=0}^n c_k A^k v = \sum_{k=0}^n c_k \lambda^k v = p(\lambda) v$. Aug 15 '19 at 17:42 | 2021-09-18T19:29:02 | {
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https://cs.stackexchange.com/questions/136902/finding-closest-line-segment-intersecting-rays | # Finding closest line segment intersecting rays
Assuming we have a fixed set of line segments $$S$$ such that any two segments are either disjoint or have a common endpoint.
A query would look like this; given a query point $$q$$ shoot 4 rays from $$q$$ (both sides horizontally and both sides vertically) and report the line segment in $$S$$ closest to $$q$$ intersecting any on these rays.
Is there a way to do this faster than $$O(n)$$ by preprocessing the data and storing it in some data structure? Or do we have to check for intersections with all segments for each ray and then compute the distances from the intersected line segments to $$q$$?
• – D.W.
Mar 21 at 22:52
## 1 Answer
You'll need two data structures - for vertical rays and for horizontal rays. I'll describe a data structure for vertical rays. Draw a vertical line through each segment end - you'll get a set of vertical slabs. Each of these slabs will be divided by pieces of original segments into trapezoids (possibly infinite ones).
Number of these slabs will not exceed $$2n$$, and number of trapezoids in each slab will not exceed $$n$$. In order to find a segment, closest to the query point along a vertical ray (going up or down), you need to use binary search to find a slab at first, and then - another binary search to find a trapezoid inside the found slab. Upper and lower boundary of this trapezoid will give you the answer. These two binary searches will take $$O(log(n))$$ time.
The data structure for horizontal rays is similar - the only difference is that slabs should be horizontal. | 2021-10-28T18:49:09 | {
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https://www.mathdept.okstate.edu/announce/?showtalk=1157 | OSU Mathematics Seminars and Colloquia
Upcoming By Week By Month All series Algebra and Geometry SeminarAlgebra SeminarAnalysis SeminarApplied Mathematics SeminarArthur Reading SeminarAwards CeremonyColloquiumCombinatorial and Commutative Algebra SeminarCreative Component PresentationDistinguished Colloquium SeriesDoctoral Thesis DefenseEnumerative Geometry / Intersection Theory SeminarGraduate Student SeminarLie Groups SeminarMasters Report PresentationMasters Thesis DefenseMath Club MeetingMathematics Education SeminarMGSS Special LectureNumber Theory SeminarNumerical Analysis SeminarOU/OSU Automorphic Forms SeminarQualifying Exam PresentationSenior Honors Thesis DefenseSpecial LectureTopology SeminarOther Calendar
Tue, Oct 20, 2020
Number Theory Seminar
8:30 AM
Online (Zoom)
On the uniformity of unlikely intersections
Hang Fu, National Taiwan University
Host: Paul Fili
NOTE THE SPECIAL TIME! Contact Paul Fili for the Zoom link.
[Abstract] [PDF]
Abstract: In this talk, we will discuss some recent progress in the uniformity of unlikely intersections. We will introduce some applications of Arakelov-Zhang pairing and take the following result as a concrete example. Fix $d\geq2$ and let $f_{t}(z)=z^{d}+t$ be the family of polynomials parameterized by $t\in\mathbb{C}$. Then there exists a constant $C(d)$ such that for any $a,b\in\mathbb{C}$ with $a^{d}\neq b^{d}$, the number of $t\in\mathbb{C}$ such that $a$ and $b$ are both preperiodic for $f_{t}$ is at most $C(d)$. This result can be seen as an analog of DeMarco-Krieger-Ye and as a uniform version of Baker-DeMarco. | 2020-10-26T01:57:03 | {
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https://www.embeddedrelated.com/showarticle/671.php | # Second-Order Systems, Part I: Boing!!
October 29, 2014
I’ve already written about the unexciting (but useful) 1st-order system, and about slew-rate limiting. So now it’s time to cover second-order systems.
The most common second-order systems are RLC circuits and spring-mass-damper systems.
Spring-mass-damper systems are fairly common; you’ve seen these before, whether you realize it or not. One household example of these is the spring doorstop (BOING!!):
(For what it’s worth: the spring doorstop appears to have been invented by Frank H. Chase during World War I.)
The mechanical systems may be more fun. But the electrical circuits are more relevant to embedded systems, so let’s start there.
## Differential equations
Consider a series LRC circuit:
The equations for this system are
\begin{aligned} C\frac{dV_C}{dt}&=I, \cr L\frac{dI}{dt}&=V_{in}-V_C-IR \end{aligned}
(Incidentally, I made that diagram with a short snippet of circuitikz. LaTeX and tikz are kind of quirky but once you get past the initial learning curve, you can get some really nice results.)
If you solve these equations for $V_C$, you get
$$LC\frac{d^2V_C}{dt}+RC\frac{dV_C}{dt}+V_C=V_{in}$$
The time-domain derivative operator $\frac{d}{dt}$ is isomorphic to multiplication by the Laplace domain variable s, so in the Laplace domain, we have
$$H(s) = \frac{V_C}{V_{in}} = \frac{1}{LCs^2 + RCs + 1}$$
Hey, look! It’s a second-order system with a DC gain of 1. What else can we figure out about this system?
There are two standard forms of this type of second-order system, depending on whether you want to think of the timescale in terms of a time constant or a natural frequency:
$$H(s) = \frac{1}{\tau^2s^2 + 2\zeta\tau s + 1} = \frac{{\omega_n}^2}{s^2 + 2\zeta\omega_n s + {\omega_n}^2}$$
With our LRC circuit, this gives us $\tau = \sqrt{LC}$, $\omega_n = \frac{1}{\sqrt{LC}}$ and $\zeta = \frac{R}{2}\sqrt{\frac{C}{L}}$.
The reason we put the transfer function into this form is because we can normalize the system response and scale the Laplace frequency to define $\bar{s} = \tau s = s/\omega_n$:
$$H(\bar{s}) = \frac{1}{\bar{s}^2 + 2\zeta\bar{s} + 1}$$
This leaves only one parameter ζ. Second order systems with a constant numerator in the transfer function (no zeros) have a behavior that is completely determined by a timescale (natural frequency $\omega_n$ or time constant τ) and damping factor ($\zeta$). Let’s forget about the damping factor for a moment, and just take a look at how the system response varies with $\omega_n$:
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import lti, lsim, step
zeta = 0.456
t = np.arange(0,10,0.01)
wn_range = [1,2,5]
for wn in wn_range:
H = lti([wn*wn],[1,2*zeta*wn,wn**2])
t,y = H.step(T=t)
plt.plot(t,y)
plt.xlabel('t')
plt.ylabel('step response')
plt.legend(['$\\omega_n = %.1f$' % wn for wn in wn_range],
'lower right', fontsize=16)
Same shape, different time scale. If we normalize the time scale $\bar{t} = \omega_n t$, we get this:
zeta = 0.456
wn_range = [1,2,5]
for wn in wn_range:
H = lti([wn*wn],[1,2*zeta*wn,wn**2])
t,y = H.step()
plt.plot(wn*t,y)
plt.xlabel('$\\bar{t} = \\omega_n t$',fontsize=16)
plt.ylabel('step response')
plt.legend(['$\\omega_n = %.1f$' % wn for wn in wn_range],
'lower right', fontsize=16)
Hey, great! It’s just like first-order systems: the natural frequency $\omega_n$ is not an interesting parameter. All systems with the same damping factor but different values of $\omega_n$ are exactly the same, just with different time scales.
## Damping factor
So let’s focus on the damping factor ζ and choose a fixed value for the natural frequency. We’ll choose $\omega_n = 2\pi$ and look what happens for different values of ζ:
zeta_range = [0.1, 0.2, 0.3, 0.5, 0.7, 1.000001, 1.2, 1.5, 2.0]
wn = 2*np.pi
t = np.arange(0,4,0.005)
n = len(t)
plt.figure(figsize=(8,6))
hlines = []
for k, zeta in enumerate(zeta_range):
H = lti([wn*wn],[1,2*zeta*wn,wn**2])
_,y = H.step(T=t)
hlines.append(plt.plot(t,y))
if zeta >= 1:
klbl = int(n * 0.12)
tl = t[klbl]
yl = y[klbl]
else:
zcomp = np.sqrt(1-zeta**2)
wd = zcomp*wn
tl = np.pi/wd
yl = 1+np.exp(-np.pi*zeta/zcomp)
plt.annotate('$\\zeta=%.3f$' % zeta, xy=(tl,yl), xytext=(-0.45,yl),
xycoords='data', textcoords='data',
arrowprops=dict(arrowstyle="-", shrinkB=0),
verticalalignment='top',
fontsize=13
)
for k, zeta in enumerate ([z for z in zeta_range if z < 0.5]):
A = 1/np.sqrt(1-zeta**2)
hl2 = plt.plot(t,1+A*np.exp(-zeta*wn*t), '--',
t,1-A*np.exp(-zeta*wn*t), '--')
for hl in hl2:
hl.set_color(hlines[k][0].get_color())
hl.set_linewidth(hlines[k][0].get_linewidth()*0.4)
plt.xlim([-0.5,4])
plt.ylim([0,1.8])
plt.xlabel('t')
plt.title('Second-order system step response for '
+'$\\omega_n = 2\\pi$, '
+'different damping factors'
)
Values of ζ that are less than 1.0 lead to underdamped systems, which have an overshoot. Values of ζ that are greater than 1.0 lead to overdamped systems, which do not have an overshoot, and which settle more slowly. If ζ = 1.0 then the system is critically damped; this is the minimum value for ζ that does not have an overshoot.
How do we figure out the overshoot as a function of ζ ?
## Poles and Overshoot
Let’s look again at the denominator of the transfer function, $s^2 + 2\zeta\omega_n s + {\omega_n}^2$. If we factor it, we can find the poles of the transfer function. So solve for $s^2 + 2\zeta\omega_n s + {\omega_n}^2 = 0$ using the quadratic formula:
$$s = \frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2{\omega_n}^2 - 4{\omega_n}^2} }{2} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)$$
If $\zeta > 1$, both poles are real and the equation tells us how to compute them. (For example, if $\zeta = 1.6$, then the poles are at $s=-0.351\omega_n$ and $s=-2.849\omega_n$.)
If $\zeta = 1$, then both poles are at $s=-\omega_n$.
If $\zeta < 1$ — the underdamped case — then the poles are a complex conjugate pair at $s = \omega_n\left(-\zeta \pm j\sqrt{1-\zeta^2}\right) = -\alpha \pm j\omega_d$ where $\alpha = \zeta\omega_n$ is the decay rate and $\omega_d = \omega_n\sqrt{1-\zeta^2}$ is the damped natural frequency.
The underdamped case is the most interesting one. We can solve for the step response $h_1(t)$ and impulse response $h(t) = \frac{d}{dt}h_1(t)$ exactly; if you go through the grungy math, it turns out that
\begin{aligned} h(t) &= u(t) \frac{\omega_n}{\sqrt{1-\zeta^2}} e^{-\alpha t} \sin \omega_d t \cr h_1(t) &= u(t) \left(1-\frac{\zeta}{\sqrt{1-\zeta^2}} e^{-\alpha t} \sin \omega_d t - e^{-\alpha t}\cos\omega_d t\right) \cr &= u(t) \left(1+\frac{1}{\sqrt{1-\zeta^2}} e^{-\alpha t} \sin (\omega_d t + \phi)\right) \end{aligned}
which is not too complicated. The nice thing is that we can make a couple of conclusions here:
1. At t = 0, the slope of the step response is zero. (In our LCR circuit, that’s because the rate of change in output voltage $\frac{dV_C}{dt} = I/C$, and the inductor current starts off at zero.)
2. The local extrema (the points that are a local minimum or maximum) of the step response are the instants where its derivative (= the impulse response) is zero, which occurs when $\sin \omega_d t = 0$, namely at integer multiples of $\frac{T}{2} = \frac{\pi}{\omega_d}$.
3. The point of maximum overshoot is the first extremum after $t=0$, and occurs at $t=\frac{\pi}{\omega_d}$. For our example with $\omega_n=2\pi$, at low values of ζ this occurs at t=0.5; as ζ increases, the damped natural frequency slows down a little bit.
4. The value of maximum overshoot is pretty easy to calculate; we just plug in $t=\frac{\pi}{\omega_d}$, and since $\sin\omega_d t = 0$ there, we are left with
$$h_1(t) = 1- e^{-\alpha t}\cos\omega_d t = 1+e^{-\alpha \frac{\pi}{\omega_d}}$$
and therefore the overshoot is the amount this exceeds 1.0, which is
$$OV = e^{-\alpha \frac{\pi}{\omega_d}} = e^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}}$$
zeta = np.arange(0.001,0.999,0.001)
plt.plot(zeta,np.exp(-np.pi*zeta/np.sqrt(1-zeta**2)))
plt.xlabel('$\zeta$',fontsize=15)
plt.ylabel('overshoot',fontsize=12)
plt.grid('on')
zeta = np.arange(0.7,0.999,0.001)
plt.figure()
plt.plot(zeta,np.exp(-np.pi*zeta/np.sqrt(1-zeta**2)))
plt.xlim(0.7,1)
plt.xlabel('$\zeta$',fontsize=15)
plt.ylabel('overshoot',fontsize=12)
plt.grid('on')
If we want to solve this equation for the damping factor, we get $\zeta = \sqrt{\frac{(\ln OV)^2}{\pi^2 + (\ln OV)^2}}$
Note that the overshoot decreases rapidly with increasing ζ, and once you reach about ζ = 0.7, the overshoot is small and decreases more slowly. That’s an important factor for system design. Let’s say you are tuning a system and need to trade off the value of ζ against other design choices. If you can tolerate a small amount of overshoot, and your design tradeoffs would benefit from a smaller value of ζ, you don’t need an overdamped system; you can generally tolerate a ζ in the range of 0.7 - 0.9.
## Summary and what’s next
We looked at second order systems of the form
$$H(s) = \frac{1}{\tau^2s^2 + 2\zeta\tau s + 1} = \frac{{\omega_n}^2}{s^2 + 2\zeta\omega_n s + {\omega_n}^2}$$
and examined features of the step response. This transfer function has a DC gain of 1, two poles, and no zeroes. Its timescale is determined by the natural frequency $\omega_n$ and its shape is determined by the damping factor ζ. Systems with $\zeta < 1$ are underdamped and have two conjugate poles. Systems with $\zeta > 1$ are overdamped and have two real poles.
Other characteristics of the step response:
• The poles are located at $s=\omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)$ for overdamped systems and $s=\omega_n\left(-\zeta \pm j\sqrt{1-\zeta^2}\right)$ for underdamped systems.
• For underdamped systems:
• the step response looks like $1 + \frac{1}{\sqrt{1-\zeta^2}}e^{-\alpha t} \sin(\omega_d t + \phi)$
• the envelope of the step response decays with a rate of $\alpha = \zeta\omega_n$
• the resonance occurs at $\omega_d = \omega_n\sqrt{1-\zeta^2}$
• the overshoot is equal to $OV = e^{-\alpha \frac{\pi}{\omega_d}} = e^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}}$
In Part II we’ll look at how the presence of a zero affects the behavior of second-order systems.
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Important Programming Concepts (Even on Embedded Systems) Part IV: Singletons | 2017-04-24T21:04:45 | {
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https://www.talkstats.com/threads/chi-squared-hypothesis-test-of-homogeneity-using-r.50406/ | # chi-squared hypothesis test of homogeneity using R
#### Cynderella
##### New Member
A survey of drivers was taken to see if they had been in an accident during the previous year, and if so was it a minor or major accident. The results are tabulated by age group:
$$\begin{array}{c|lcr} \text{Age} & \text{None} & \text{Minor} & \text{Major} \\ \hline \text{under} 18 &67 & 10 & 5 \\ 18-25 & 42 & 6 & 5 \\ 26-40 & 75 &8 & 4 \\ 40-65 &56 &4 & 6\\ 65+ &57 &15 &1 \\ \end{array}$$
Do a chi-squared hypothesis test of homogeneity using R to see if there is difference in distributions based on age and draw a column wise normal qqplot of these data by age.
I used the function "chisq.test" but do not understanding how does it interpret whether there is difference in distributions based on age ?
Also If i was asked to check the independence, i would use the same function "chisq.test". Am i wrong to use the function "chisq.test" for chi-squared hypothesis test of "homogeneity"?
#### gianmarco
##### TS Contributor
Hi!
I am little confused from your terminology (i.e., "homogeneity"), but since you have been given a cross-tabulation, I guess you are to evaluate if there is a correlation between the levels (i.e., categories) of the two categorical variables being compared (age class and type of accident). This is where chisq test comes into play.
The test indicates that there is not a significant correlation between age classes and type of accident:
Code:
Pearson's Chi-squared test
data: mydata
X-squared = 12.5862, df = 8, p-value = 0.1269
It can be seen that the expected frequencies are not far from the observed ones:
Code:
none minor major
under18 67.46260 9.767313 4.770083
18-25 43.60388 6.313019 3.083102
26-40 71.57618 10.362881 5.060942
40-65 54.29917 7.861496 3.839335
over65 60.05817 8.695291 4.246537
Hope this helps
regards
Gm | 2022-11-30T23:17:44 | {
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https://www.shaalaa.com/question-bank-solutions/the-number-values-x-0-2-that-satisfy-equation-sin-2-x-cos-x-1-4-trigonometric-equations_58835 | # The Number of Values of X in [0, 2π] that Satisfy the Equation Sin 2 X − Cos X = 1 4 - Mathematics
MCQ
Sum
The number of values of x in [0, 2π] that satisfy the equation $\sin^2 x - \cos x = \frac{1}{4}$
• 1
• 2
• 3
• 4
#### Solution
2
$\sin^2 x - \cos x = \frac{1}{4}$
$\Rightarrow (1 - \cos^2 x) - \cos x = \frac{1}{4}$
$\Rightarrow 4 - 4 \cos^2 x - 4 \cos x = 1$
$\Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0$
$\Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0$
$\Rightarrow 2 \cos x ( 2 \cos x + 3) - 1 ( 2 \cos x + 3) = 0$
$\Rightarrow (2 \cos x + 3 ) (2 \cos x - 1) = 0$
$\Rightarrow 2 \cos x + 3 = 0$ or, $2 \cos x - 1 = 0$
$\Rightarrow \cos x = - \frac{3}{2}$ or $\cos x = \frac{1}{2}$
Here,
$\cos x = - \frac{3}{2}$ is not possible.
$\cos x = \frac{1}{2}$
$\Rightarrow \cos x = \cos \frac{\pi}{3}$
$\Rightarrow x = 2n\pi \pm \frac{\pi}{3}$
Now for n = 0 and 1, the values of $x are \frac{\pi}{3}, \frac{5\pi}{3}\text{ and }\frac{7\pi}{3},\text{ but }\frac{7\pi}{3} \text{ is not in }$ $\left[ 0, 2\pi \right]$
Hence, there are two solutions in $\left[ 0, 2\pi \right]$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Q 14 | Page 27 | 2021-04-22T17:01:04 | {
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https://math.stackexchange.com/questions/2570116/proving-that-strictly-monotonic-curvature-implies-no-self-intersections-more-sp | # Proving that strictly monotonic curvature implies no self intersections (more specifically, using the following inequalities)
Let $a(s)$ be a regular curve that is parametrized by arclength. Prove that, if the curvature $k(s)$ is a strictly monotonic function, then $a(s)$ has no self intersections. Suggestions:
a) [will be ommited since I managed to prove it using this first suggestion so there's no point putting it here]
b) Consider the curve $b(s) = a(s) + \frac{n(s)}{k(s)}$, with $k'(s) > 0$. Verify that for all $s > s_0$, the following are true, and then conclude that $a(s)$ has no self intersections:
$|b(s) - b(s_0)| <\int_{s_0}^{s} |b'(s)| ds = \frac{1}{k(s_0)} - \frac{1}{k(s)}$
$|b(s) - b(s_0)| < \frac{1}{k(s_0)}$
I managed to compute the first integral, but I'm lost on how to prove the other two inequalities. I would appreciate any ideas. I know there are other ways to prove the statement, but I want to do it specifically using the second suggestion. Also, I don't know where the contradiction in b) is, I'm not seeing how both being true is not possible. | 2019-08-23T03:14:04 | {
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https://stats.stackexchange.com/questions/386750/probability-that-exactly-12-buses-will-arrive-within-3-hours | # Probability that exactly 12 buses will arrive within 3 hours
Let's suppose there are two buses $$A$$ and $$B$$. They draw up at the bus stop under the Poisson distribution with intensities $$3$$ and $$5$$ times per hour. (a) What's the expected length of time after the $$15$$th bus will arrive?, (b) What's the probability that exactly $$12$$ buses will arrive within $$3$$ hours?
If bus $$A$$ arrives on average $$3$$ times per hour and bus $$B$$ arrives on average $$5$$ times per hour, then a bus comes to the bus stop on average $$8$$ times per hour.
Poisson distribution: $$P(N(t)=j)=\frac{(\lambda t)^j}{j!}e^{-\lambda t}$$.
(b) $$P(N(t)=j)=\frac{(8*3)^{12}}{12!}e^{-8*3}\approx 0,00288$$
(a) $$P(N(t)=j)=\frac{8t^{15}}{15!}e^{-8t}$$, but I don't think it's a proper approach. I suppose the proper answer is $$\frac{15}{8} \approx 1,9$$, but I'm not sure how to show it.
I'll be thankful for any tips and help.
• Are you familiar with the relationship between Poisson processes and Gamma waiting times? If not, review some of the posts at stats.stackexchange.com/…. – whuber Jan 11 at 17:53
• The arrival rate per 3 hrs is $\lambda = 24.$ So why isn't the probability of exactly 12, found by using the Poisson PDF: 0.00288? – BruceET Jan 12 at 4:58 | 2019-01-19T13:04:21 | {
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https://spc.unige.ch/en/teaching/courses/algorithmes-probabilistes/chaotic-behaviour-random-numbersb/ | # From Chaotic behaviour to Random Numbers
Fig.1 - The Relative number of appearances of the word "random" in Google Books database.
## What is randomness?
The definition of "random" given by the Collins dictionary is:
"If you describe events as random, you mean that they do not seem to follow a definite plan or pattern."
The word "seem" in this definition for our purposes will differentiate pseudo-random (do not seem) from the random (do not) definition. The word "seem" has a crucial implication in our vision of nature. In fact, by adopting this definition, If the world followed classical physics, randomness would not exist. The intrinsically deterministic and time-symmetric principles of the classical physic prevent the arising of truly random processes. True physical randomness is conceived uniquely by the quantum theory (in its probabilistic interpretation). Nevertheless, the extraordinarily high number of degrees of freedom of our universe and the non-linearity of the governing laws allows the emergence of the (pseudo) random processes even when the quantum phenomena are irrelevant.
Summing up:
1. Quantum-like processes are theoretically accepted to be subject to real randomness $\Rightarrow$ non-deterministic behaviour (click here to see how to create random numbers in this way);
2. Deterministic calculations cannot yield to random numbers because by definition they follow a rule;
3. Non-linear deterministic maps can yield effectively unpredictable sequences of pseudo-random numbers $\Rightarrow$ deterministic chaotic behaviour (pseudorandom-generators follow this way).
## The Logistic Map - deterministic chaos
A logistic map is one of the simplest non-linear discrete dynamic system representation. A continuous dynamical systems is a system that can be described by an equation of the type: $$\frac{{\rm d} s} {{\rm d} t} = f(s)$$ Where s is the state variable and t is the time. We can write the discrete time equivalent system as: $$s_{t+1}=f(s_t)$$ The easiest possible time-discrete dynamic equation is the linear map $$s_{t+1}=\gamma s_t$$ The logistic map is a non-linear time-discrete map defined as: $$s_{t+1} = F\left(x_s\right) = \gamma s_t \cdot (1-s_t)$$ as you can see the state of the system s at the time step $t+1$ depends in a non linear way on the system state at time $t$. Depending on the value of $\gamma$ the behaviour of the system changes:
• If $0 < \gamma < 3.5$ the system is bounded in the interval $[0,1]$
• if $\gamma < 3$ the system is stable and has a fixed point $S = \dfrac{\gamma-1}{\gamma}$
• if $3 < \gamma < 1 + \sqrt{6}$ the system starts to oscillate between two fixed points $\Rightarrow$ bifurcation $$S_{1,2} = \left[ \gamma +1 \pm ((\gamma +1)(\gamma -3))^{1/2} \right] / 2\gamma$$ where $S_1 = F(S_2) \quad S_2 = F(S_1)$
• at $\gamma > 1+\sqrt{6}$ there is a so called period doubling and the $s_t$ flips around between 4 points.
• $3 < \gamma < 3.57$ infinite number of period doubling occurs as $\gamma \uparrow$
• $a > 3.57$ the behavior becomes aperiodic and chaotic, nevertheless there still exist islands of tranquility in the sea of chaos in which the periodic behavior is recovered.
Fig. - Bifurcation diagram for the logistic map. The attractor for any value of the parameter r is shown on the vertical line at that r(Wikipedia)
### Exercise
• In your opinion what is the difference between a chaotic and a random behaviour?
• Implement the logistic map in c++ and, using high values of gamma, compare the generated numbers with the random number generators that you implemented in TP1.
• Discuss the stability of 2-flip points condition.
• Do you think that the output of the logistic map can be seen as a random number generator or there are some missing proprieties?
## random number generators
The random number generator can be of two types:
1. hardware random-number generators;
2. pseudo-random number generators (PRNG).
The first class takes advantage of quantum processes to generate real random number sequences. The second class instead, consist of algorithms that starting from an initial value (seed) generate a sequence of numbers that follow a chaotic behaviour, but that must have some other important features:
• uniformly distributed $\Rightarrow$ from a uniformly distributed random sequence is easy to do a mapping to a non-uniform distribution of random numbers
• the generated numbers should be (almost) uncorrelated $\Rightarrow$ the sequence should pass all the random test (these tests are usually derived from the $\chi^2$ Pearson test);
• long period: with finite precision arithmetic a sequence must repeat itself, its period must be much longer of the random numbers needed by the application
• insensitivity to seed: the above properties should not depend on the initial seed. | 2020-02-17T09:23:34 | {
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http://mathoverflow.net/feeds/user/12261 | User russell may - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:14:53Z http://mathoverflow.net/feeds/user/12261 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108066/sequences-recurrence-relation/108081#108081 Answer by Russell May for sequences - recurrence relation Russell May 2012-09-25T17:09:36Z 2012-09-25T17:09:36Z <p>One standard way to solve recurrence relations is with generating functions. In this case, let $f$ and $g$ be the ordinary generating functions of the sequences $y$ and $z$. Then the generating function equivalent of your recurrence relations would be $$\frac{g(x)-g(0)}x=d\cdot g(x)+\frac e{1-x}$$ and $$\frac{f(x)-f(0)}x=a\cdot f(x)+b\cdot g(x)+\frac c{1-x}.$$ You can then solve these relations for the generating functions $$g(x)=\left(\frac{e\cdot x}{1-x}+g(0)\right)\cdot \frac 1{1-xd}$$and $$f(x)=\left(x\cdot b\cdot g(x)+\frac{x\cdot c}{1-x}+f(0)\right)\frac 1{1-ax}.$$ Lastly, you need to find the partial fraction decomposition of $f$. Using geometric series and its derivatives, you can then read off the coefficients of the partial fraction decomposition to get an explicit solution for the terms in your sequences. </p> <p>There are lots of examples along these lines in Wilf's book <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">Generatingfunctionology</a>, chapter 2 sections 1-2.</p> http://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical/108067#108067 Answer by Russell May for Which popular games are the most mathematical? Russell May 2012-09-25T15:32:02Z 2012-09-25T15:32:02Z <p>The game of <a href="http://en.wikipedia.org/wiki/Cootie_%28game%29" rel="nofollow">Cootie</a>, where players roll dice to collect parts of an insect (cootie), is a variant of the coupon collector's problem.</p> <p><img src="http://upload.wikimedia.org/wikipedia/en/thumb/a/a6/Original_Cootie_box_cover_and_components.jpg/250px-Original_Cootie_box_cover_and_components.jpg" alt="alt text"></p> <p>Instead of collecting a single instance of each coupon, players must collect multiple copies (6 legs, 2 eyes, 1 head, etc.) to win. It turns out you can compute the expected number of rolls to win at Cootie (even with a weighted die) with a <em>finite</em> sum. </p> <p>In particular, if you have $L$ objects to collect and for each object <code>$\ell<L$</code> you need $q_\ell$ copies and the probability of getting the object is $p_\ell$, then the expected number of rolls to get all of the needed objects is</p> <p><code>$\displaystyle\sum_{\ell\in L} \frac{{p_\ell}^{q_\ell}}{(q_\ell -1)!}\int_0^\infty x^{q_\ell}\exp(-x) \prod_{k\in L-\{\ell\}}(\exp(p_k x)-\exp_{<q_k}(p_k x))dx.$</code></p> <p>If you're interested, check out my <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v15i1n31" rel="nofollow">paper</a> out for the full computation.</p> http://mathoverflow.net/questions/107443/probability-of-first-collision-with-replacement/108020#108020 Answer by Russell May for Probability of first collision with replacement Russell May 2012-09-25T03:47:32Z 2012-09-25T04:20:38Z <p>There's no closed form for this expectation. However, you can get good approximations. One common way to do this is with generating functions. Herb Wilf's <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">book</a> is an excellent reference.</p> <p>As noted, the probability that the first collision occurs on the $n$th draw with $m$ balls is $p_n=d_{n-1}\frac{n-1}m$, where $d_n=\frac{m!}{m^n (m-n)!}$. Consider the ordinary generating function $D(x)=\sum_{n=0}^{m}d_nx^n$. Then the expected number of draws to get a collision would be $\langle p\rangle=\sum_{n=1}^{m+1} n\cdot p_n$ or, in terms of the generating function,<br> $\langle p\rangle=\frac{d^2}{dx^2}(xD(x))|_{x=1}$. </p> <p>The generating function $D(x)$ is not summable. However, its exponential counterpart is: $E(x)=\sum_{n=0}^{m}\frac{d_n}{ n!}x^n=(1+x/m)^m$. The ordinary and exponential generating functions are related by the Laplace transform, $D(x)=\frac 1x\int_0^\infty e^{-t/x}E(t)dt$. Differentiating under the integral twice and then evaluating at $x=1$, we get $\langle p\rangle=\frac 1m\int_0^\infty e^{-t}(t^2-2t)(1+t/m)^mdt$. The dominant part in this integral comes from the $t^2$ term. So, $\langle p\rangle\approx\frac 1m\int_0^\infty e^{-t}t^2(1+t/m)^m dt$, and substituting $u=t/m$, we get $\langle p\rangle\approx m^2\int_0^\infty e^{m(-u+\log(1+u))}u^2 du$. This integral can be nicely approximated with Laplace's method, where the exponent $-u+\log(1+u)$ is replaced with its second order Taylor series about its maximum (at $u=0$), which turns out to be just $-u^2/2$. So, $\langle p\rangle\approx m^2\int_0^\infty e^{-u^2 m/2}u^2du=\sqrt{{\pi m}/2}$.</p> <p>If you need greater accuracy or if you want to consider higher moments of the distribution, you can always consider the other terms in the first integral representation of $\langle p\rangle$ and higher order terms in the Laplace's method. Another avenue to analyze this expectation, as suggested in the wikipedia article on the birthday problem, is to learn about the Ramanujan $Q$-function.</p> http://mathoverflow.net/questions/52370/eigenvalues-of-an-oblique-diagonal-matrix Eigenvalues of an "oblique diagonal" matrix Russell May 2011-01-18T02:25:59Z 2012-07-26T07:22:00Z <p>I am looking for guidance about the behavior of powers of a particular matrix (call it $A_n$ for $n\ge2$), which has come up in a counting problem about quantum knot mosaics (a good reference for quantum knot mosaics is <a href="http://www.csee.umbc.edu/~lomonaco/pubs/Final-Version-QIP-Quantum-Knots.pdf" rel="nofollow">here</a> ). Here's a description of the matrix. It has $4^n$ rows and columns. Instead of a traditional diagonal matrix with its non-zero entries on the main diagonal, the non-zero entries of $A_n$ are on an oblique diagonal" of slope $4$, modulo the size of the matrix. More precisely, the non-zero entries occur where $\left\lfloor\frac{\text{column}}4\right\rfloor=\text{row}\text{, mod}\;4^{n-1}$. The matrix has sixteen possibly non-zero values $a_1,\ldots,a_{16}$, arranged as follows (with boxes for visual clarity):</p> <p><code>$\begin{array}{l|lll|lll|} \text{row}\backslash\text{column}&0&4&\ldots&4^n/2&4^n/2+4&\ldots\\ \hline 0&\boxed{a_1\,a_2\,a_1\,a_2}\\ 1&&\boxed{a_1\,a_2\,a_1\,a_2}\\ \vdots&&&\ddots\\ \hline 4^n/8&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\ 4^n/8+1&&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\ \vdots&&&&&&\ddots\\ \hline 2\cdot4^n/8&\boxed{a_5\,a_6\,a_5\,a_6}\\ 2\cdot4^n/8+1&&\boxed{a_5\,a_6\,a_5\,a_6}\\ \vdots&&&\ddots\\ \hline 3\cdot4^n/8&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\ 3\cdot4^n/8+1&&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\ \vdots&&&&&&\ddots\\ \hline 4\cdot4^n/8&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\ 4\cdot4^n/8+1&&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\ \vdots&&&\ddots\\ \hline 5\cdot4^n/8&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\ 5\cdot4^n/8+1&&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\ \vdots&&&&&&\ddots\\ \hline 6\cdot4^n/8&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\ 6\cdot4^n/8+1&&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\ \vdots&&&\ddots\\ \hline 7\cdot4^n/8&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\ 7\cdot4^n/8+1&&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\ \vdots&&&&&&\ddots\\ \hline \end{array}$</code></p> <p>Since $A_n$ has a straightforward geometrical description with non-zero entries only on a diagonal (albeit an oblique one), the following question seems reasonable:</p> <blockquote> <p>Is there an elementary way to compute the eigenvalues of this matrix in terms of $a_1,\ldots,a_{16}$ and $n$? </p> </blockquote> <p>I'm no expert in tensor algebra, but it seems that $A_n$ might be expressed as the tensor product of $A_2$ with $n-1$ copies of another transformation. Even an approximation of the largest eigenvalue would be useful, but it would be best to avoid the power method with the Rayleigh quotient since I'm trying to analyze the powers of the matrix in terms of the eigenvalues, <em>not</em> vice versa. Any insight into the computation of the eigenvalues of $A_n$ would be greatly appreciated and would go a long way in answering a question in the paper referenced above.</p> http://mathoverflow.net/questions/98876/a-probability-question-about-removing-stones-from-piles/98943#98943 Answer by Russell May for A probability question about removing stones from piles Russell May 2012-06-06T11:24:27Z 2012-06-06T11:24:27Z <p>As Brendan noted, the literature on the coupon collector's problem is large. I suspect a close fit to your problem would be <a href="http://www-irma.u-strasbg.fr/~foata/paper/pub89brother.pdf" rel="nofollow">The Collector’s Brotherhood Problem Using the Newman-Shepp Symbolic Method</a> by Foata and Zeilberger. For the case of unequal probabilities, you might consider the methods in my paper <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v15i1n31/pdf" rel="nofollow">Coupon Collecting with Quotas</a>. It turns out that the case of unequal probabilities is not too much "hairier" than the uniform case, so it seems feasible to mesh these two papers to obtain the result you're looking for.</p> http://mathoverflow.net/questions/93744/estimating-a-partial-sum-of-weighted-binomial-coefficients/93750#93750 Answer by Russell May for Estimating a partial sum of weighted binomial coefficients Russell May 2012-04-11T10:31:51Z 2012-04-11T10:37:12Z <p>Your sum can also be thought of as the first $\alpha n$ terms in a binomial distribution with probability of success $p=1-\frac1{\lambda+1}$. So, it is closely approximated by a normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$, i.e., $$\sum_{k=0}^{\alpha n} \binom nk \lambda^k\approx (1-p)^{-n}\Phi\left((\alpha-p)\sqrt{\frac n{p(1-p)}}\right),$$ where $\Phi$ is the cumulative standard normal distribution.</p> http://mathoverflow.net/questions/91390/generating-function-for-regular-tournaments generating function for regular tournaments Russell May 2012-03-16T17:01:33Z 2012-03-16T21:10:10Z <p>At the beginning of a <a href="http://cs.anu.edu.au/~bdm/papers/euler.pdf" rel="nofollow">paper</a> by McKay and Robinson on enumerating eulerian circuits, the authors state that the number of regular tournaments containing a directed rooted tree $T$ on vertices $v_1,\dots,v_n$ with root $v_n$ coincides with the constant term in the generating function <code>$$\prod_{1\le j<k\le n}(x_j^{-1}x_k+x_jx_k^{-1})\,\prod_{jk\in\textrm{ edges of }T}\frac{x_jx_k^{-1}}{x_j^{-1}x_k+x_jx_k^{-1}}\,.$$</code> Unlike most everything else in the paper, this statement is made without justification, which makes me think that it's either a well-known result or obvious, i.e., except to me.</p> <p>Could someone provide a reference or a few words to justify this claim?</p> http://mathoverflow.net/questions/79051/theorems-proved-with-ad-whose-proof-is-also-known-in-the-zf-world/79092#79092 Answer by Russell May for Theorems proved with AD whose proof is also known in the ZF world Russell May 2011-10-25T15:55:54Z 2011-10-25T15:55:54Z <p>Perhaps, this is along the lines of what you're looking for. This <a href="http://digital.library.unt.edu/ark%3A/67531/metadc2789/m1/1/high_res_d/dissertation.pdf" rel="nofollow">thesis</a> gives a proof of the stong partition relation on $\omega_1$ from AD, and then "relativizes" the proof to $V$ to show, assuming the existence of Woodin cardinals, a collapsing result, namely, that some regular cardinal $<\aleph_{\omega_2}$ in $L[\mathbb{R}]$ must collapse in $V$.</p> http://mathoverflow.net/questions/29137/good-combinatorics-textbooks-for-teaching-undergraduates/73661#73661 Answer by Russell May for Good combinatorics textbooks for teaching undergraduates? Russell May 2011-08-25T13:54:22Z 2011-08-25T13:54:22Z <p>It's obviously slanted towards the generating-function view of enumeration, but I enthusiastically recommend <em>Generatingfunctionology</em> by Herb Wilf. It covers all the topics you mentioned, written mainly in the style of examples, rather than theory---something that usually appeals to undergraduates. To me what makes the book a great introduction for a newcomer to combinatorics is Wilf's obvious enthusiasm and easy-going (yet firmly exacting) writing style. The mileage he gets out of changing a recurrence relation into a generating function is truly amazing. I think most undergraduates would be amazed that their skills in calculus can help them enumerate discrete objects, and this book does exactly that over and over again. If price matters, this one is tough to beat---the second edition is free at Wilf's <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">website</a>.</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59872#59872 Answer by Russell May for Use of traces in physics Russell May 2011-03-28T17:49:31Z 2011-03-28T17:49:31Z <p>There was a similar question recently posed (and often answered) here:</p> <p><a href="http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace" rel="nofollow">http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace</a></p> http://mathoverflow.net/questions/56210/reference-for-a-edge-matching-problem reference for a edge-matching problem Russell May 2011-02-21T19:54:19Z 2011-02-22T04:49:44Z <p>Perhaps not surprisingly, a variation of a recreational math puzzle (a so-called edge-matching puzzle or scramble square) is equivalent to a combinatorics question of interest (in this case, about <a href="http://www.csee.umbc.edu/~lomonaco/pubs/Final-Version-QIP-Quantum-Knots.pdf" rel="nofollow">quantum knot mosaics</a>, question #9). In a traditional edge-matching puzzle, you are given $n^2$ tiles, each tile square in shape and bearing a design, with the goal of arranging the tiles in an $n\times n$ grid so that the designs on the side of adjacent tiles "match". For instance, here's a puzzle with 24 possible tiles (4 sides, 3 colors, 2 halves) of which at most 9 actually appear (link <a href="http://puzzles.guidestobuy.com/scramble-squares-butterflies" rel="nofollow">here</a> if the image is broken): <img src="http://puzzles.guidestobuy.com/scramble-squares-butterflies/71CVKT0J0BL.gif" alt="alt text"></p> <p>For what it's worth, solving a general edge-matching puzzle is NP-complete (see <a href="http://erikdemaine.org/papers/Jigsaw_GC/" rel="nofollow">article</a> of Demaine). The combinatorics problem is phrased in a slightly different fashion. You begin with a finite collection of designs (such as quadruples of colored halves of butterflies) and for each design an ample supply of square tiles bearing that design. The problem is to calculate the number of arrangements of these tiles in an $n\times n$ grid so that, as in the game described above, the designs on sides of adjacent tiles match. The number of arrangements should be in terms of the size of the grid and the collection of designs on the tiles.</p> <blockquote> <p>Is anyone aware of results along these lines or, even better, able to provide a quick calculation of the number of arrangements of tiles into an edge-matched grid?</p> </blockquote> <p>My suspicion is that the number of arrangements goes like $\lambda^{n^2}$ where $\lambda$ is determined from the collection of designs on the tiles.</p> http://mathoverflow.net/questions/107443/probability-of-first-collision-with-replacement Comment by Russell May Russell May 2012-09-18T13:46:52Z 2012-09-18T13:46:52Z This is a variant of the birthday problem. It looks like even the wikipedia article on this has the information you're asking for. http://mathoverflow.net/questions/52370/eigenvalues-of-an-oblique-diagonal-matrix/93173#93173 Comment by Russell May Russell May 2012-04-06T18:44:32Z 2012-04-06T18:44:32Z I agree that there's a permutation matrix P and a block diagonal matrix A' so that the oblique diagonal matrix A is PA'. However, it's not clear how to get the eigenvalues of a product, given the eigenvalues of the factors. http://mathoverflow.net/questions/52370/eigenvalues-of-an-oblique-diagonal-matrix Comment by Russell May Russell May 2012-04-05T12:32:24Z 2012-04-05T12:32:24Z Well, as Donald Knuth said, TeX is intended for the creation of beautiful mathematics. http://mathoverflow.net/questions/91390/generating-function-for-regular-tournaments/91413#91413 Comment by Russell May Russell May 2012-03-16T21:26:04Z 2012-03-16T21:26:04Z Thank you, Ira. That helps very much. http://mathoverflow.net/questions/61664/number-of-required-trials-to-sample-all-possible-states-of-a-d-sided-loaded-die/61668#61668 Comment by Russell May Russell May 2011-12-08T13:17:10Z 2011-12-08T13:17:10Z For what it's worth, this result goes back to: von Schelling, H., (1934). Auf der Spur des Zufalls. Deutsches Statistisches Zentralblatt, 26, 137-146. An English translation appeared later (in a much more accessible location): von Schelling, H., (1954). Coupon Collecting for Unequal Probabilities, Amer. Math. Monthly, 61, no. 5, 306-311. http://mathoverflow.net/questions/79051/theorems-proved-with-ad-whose-proof-is-also-known-in-the-zf-world/79092#79092 Comment by Russell May Russell May 2011-10-25T18:07:26Z 2011-10-25T18:07:26Z @Asaf Kargila: I thought about doing that, but I think it's fairer to say that all the major ideas there were from my advisor, Steve Jackson. http://mathoverflow.net/questions/56210/reference-for-a-edge-matching-problem/56244#56244 Comment by Russell May Russell May 2011-02-22T18:03:51Z 2011-02-22T18:03:51Z These references look very useful---many thanks. Now that you mention it, this puzzle does look like a slight generalization of an alternating sign matrix. I'll definitely try to see if Zeilberger's or Kuperberg's proofs of the alternating sign matrix conjecture generalize. http://mathoverflow.net/questions/56210/reference-for-a-edge-matching-problem/56216#56216 Comment by Russell May Russell May 2011-02-22T05:03:02Z 2011-02-22T05:03:02Z The "red lines" come from the referenced paper on quantum knot mosaics by Lomonaco and Kauffman, in which the tiles bear designs of zero, one or two strands of rope (which are drawn in red). | 2013-05-18T14:14:53 | {
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https://stacks.math.columbia.edu/tag/0FUV | Lemma 50.17.3. With notation as in Lemma 50.17.1 and denoting $f : X \to S$ the structure morphism there is a canonical distinguished triangle
$\Omega ^\bullet _{X/S} \to Rb_*(\Omega ^\bullet _{X'/S}) \oplus i_*\Omega ^\bullet _{Z/S} \to i_*Rp_*(\Omega ^\bullet _{E/S}) \to \Omega ^\bullet _{X/S}[1]$
in $D(X, f^{-1}\mathcal{O}_ S)$ where the four maps
$\begin{matrix} \Omega ^\bullet _{X/S} & \to & Rb_*(\Omega ^\bullet _{X'/S}), \\ \Omega ^\bullet _{X/S} & \to & i_*\Omega ^\bullet _{Z/S}, \\ Rb_*(\Omega ^\bullet _{X'/S}) & \to & i_*Rp_*(\Omega ^\bullet _{E/S}), \\ i_*\Omega ^\bullet _{Z/S} & \to & i_*Rp_*(\Omega ^\bullet _{E/S}) \end{matrix}$
are the canonical ones (Section 50.2), except with sign reversed for one of them.
Proof. Choose a distinguished triangle
$C \to Rb_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \to i_*Rp_*\Omega ^\bullet _{E/S} \to C[1]$
in $D(X, f^{-1}\mathcal{O}_ S)$. It suffices to show that $\Omega ^\bullet _{X/S}$ is isomorphic to $C$ in a manner compatible with the canonical maps. By the axioms of triangulated categories there exists a map of distinguished triangles
$\xymatrix{ C' \ar[r] \ar[d] & b_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \ar[r] \ar[d] & i_*p_*\Omega ^\bullet _{E/S} \ar[r] \ar[d] & C'[1] \ar[d] \\ C \ar[r] & Rb_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \ar[r] & i_*Rp_*\Omega ^\bullet _{E/S} \ar[r] & C[1] }$
By Lemma 50.17.2 part (3) and Derived Categories, Proposition 13.4.23 we conclude that $C' \to C$ is an isomorphism. By Lemma 50.17.2 part (2) the map $i_*\Omega ^\bullet _{Z/S} \to i_*p_*\Omega ^\bullet _{E/S}$ is an isomorphism. Thus $C' = b_*\Omega ^\bullet _{X'/S}$ in the derived category. Finally we use Lemma 50.17.2 part (1) tells us this is equal to $\Omega ^\bullet _{X/S}$. We omit the verification this is compatible with the canonical maps. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-06-25T17:28:10 | {
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https://proxieslive.com/tag/expz/ | ## Prove conjugation law for Exp(z) using only Exp(x + y) = Exp(x)Exp(y)
Starting with just the property $$E(x + y) = E(x)E(y)$$, one can prove quite a lot of the main properties of the exponential function on real numbers. For example, $$E(0) = 1$$, and $$E'(X) = E(X)$$, and $$E(nx) = E(x)^n$$ all follow from a straight-forward application of definitions.
To move this in to complex analysis and Euler’s formula, it seems to me that the key property that needs to be proved is $$E(\overline{z})$$ = $$\overline{E(z)}$$.
Is there a nice way to prove this that is in the spirit of this particular line of reasoning?
Edit: For example, Chapter 8 of Baby Rudin follows this reasoning, but resorts to the definition $$E(x) = \Sigma_{n\ge0}(\frac{z^n}{n!})$$ to prove the conjugation property. | 2019-09-23T18:42:00 | {
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https://www.shaalaa.com/question-bank-solutions/nature-roots-if-roots-equations-ax2-2bx-c-0-bx-2-2sqrt-ac-x-b-0-are-simultaneously-real-then-prove-that-b2-ac_23625 | Share
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# If the Roots of the Equations Ax2 + 2bx + C = 0 and Bx^2-2sqrt(Ac)X+B = 0 Are Simultaneously Real, Then Prove That B2 = Ac. - CBSE Class 10 - Mathematics
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#### Question
If the roots of the equations ax2 + 2bx + c = 0 and bx^2-2sqrt(ac)x+b = 0 are simultaneously real, then prove that b2 = ac.
#### Solution
The given equations are
ax2 + 2bx + c = 0 ............ (1)
bx^2-2sqrt(ac)x+b = 0 ............. (2)
Roots are simultaneously real
Then prove that b2 = ac
Let D1 and D2 be the discriminants of equation (1) and (2) respectively,
Then,
D1 = (2b)2 - 4ac
= 4b2 - 4ac
And
D_2=(-2sqrt(ac))^2-4xxbxxb
= 4ac - 4b2
Both the given equation will have real roots, if D1 ≥ 0 and D2 ≥ 0
4b2 - 4ac ≥ 0
4b2 ≥ 4ac
b2 ≥ ac ............... (3)
4ac - 4b2 ≥ 0
4ac ≥ 4b2
ac ≥ b2 ................... (4)
From equations (3) and (4) we get
b2 = ac
Hence, b2 = ac
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [7]
Solution If the Roots of the Equations Ax2 + 2bx + C = 0 and Bx^2-2sqrt(Ac)X+B = 0 Are Simultaneously Real, Then Prove That B2 = Ac. Concept: Nature of Roots.
S | 2019-10-19T07:54:59 | {
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https://www.jobilize.com/online/course/2-5-geometric-mean-descriptive-statistics-by-openstax?qcr=www.quizover.com | # 2.5 Geometric mean
Page 1 / 1
The mean (Arithmetic), median and mode are all measures of the “center” of the data, the “average”. They are all in their own way trying to measure the “common” point within the data, that which is “normal”. In the case of the arithmetic mean this is solved by finding the value from which all points are equal linear distances. We can imagine that all the data values are combined through addition and then distributed back to each data point in equal amounts. The sum of all the values is what is redistributed in equal amounts such that the total sum remains the same.
The geometric mean redistributes not the sum of the values but the product of multiplying all the individual values and then redistributing them in equal portions such that the total product remains the same. This can be seen from the formula for the geometric mean, $\stackrel{~}{x}$ : (Pronounced x-tilde)
$\stackrel{~}{x}={\left(\prod _{i=1}^{n}{\pi }_{i}\right)}^{\frac{1}{n}}=\sqrt[n]{{x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}$
where $\pi$ is the mathematical symbol that tells us to multiply all the ${x}_{i}$ numbers in the same way capital Greek sigma tells us to add all the ${x}_{i}$ numbers. Remember that a fractional exponent is calling for the nth root of the number thus an exponent of 1/3 is the cube root of the number.
The geometric mean answers the question, "if all the quantities had the same value, what would that value have to be in order to achieve the same product?” The geometric mean gets its name from the fact that when redistributed in this way the sides form a geometric shape for which all sides have the same length. To see this, take the example of the numbers 10, 51.2 and 8. The geometric mean is the product of multiplying these three numbers together (4,096) and taking the cube root because there are three numbers among which this product is to be distributed. Thus the geometric mean of these three numbers is 16. This describes a cube 16x16x16 and has a volume of 4,096 units.
The geometric mean is relevant in Economics and Finance for dealing with growth: growth of markets, in investment, population and other variables the growth of which there is an interest. Imagine that our box of 4,096 units (perhaps dollars) is the value of an investment after three years and that the investment returns in percents were the three numbers in our example. The geometric mean will provide us with the answer to the question, what is the average rate of return: 16 percent. The arithmetic mean of these three numbers is 23.6 percent. The reason for this difference, 16 versus 23.6, is that the arithmetic mean is additive and thus does not account for the interest on the interest, compound interest, embedded in the investment growth process. The same issue arises when asking for the average rate of growth of a population or sales or market penetration, etc., knowing the annual rates of growth. The formula for the geometric mean rate of return, or any other growth rate, is:
${r}_{s}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}-1$
Manipulating the formula for the geometric mean can also provide a calculation of the average rate of growth between two periods knowing only the initial value ${a}_{0}$ and the ending value ${a}_{n}$ and the number of periods, $n$ . The following formula provides this information:
${\left(\frac{{a}_{n}}{{a}_{0}}\right)}^{\frac{1}{n}}=\stackrel{~}{x}$
Finally, we note that the formula for the geometric mean requires that all numbers be positive, greater than zero. The reason of course is that the root of a negative number is undefined for use outside of mathematical theory. There are ways to avoid this problem however. In the case of rates of return and other simple growth problems we can convert the negative values to meaningful positive equivalent values. Imagine that the annual returns for the past three years are +12%, -8%, and +2%. Using the decimal multiplier equivalents of 1.12, 0.92, and 1.02, allows us to compute a geometric mean of 1.0167. Subtracting 1 from this value gives the geometric mean of +1.67% as a net rate of population growth (or financial return). From this example we can see that the geometric mean provides us with this formula for calculating the geometric (mean) rate of return for a series of annual rates of return:
${r}_{s}=\stackrel{~}{x}-1$
where ${r}_{s}$ is average rate of return and $\stackrel{~}{x}$ is the geometric mean of the returns during some number of time periods. Note that the length of each time period must be the same.
As a general rule one should convert the percent values to its decimal equivalent multiplier. It is important to recognize that when dealing with percents, the geometric mean of percent values does not equal the geometric mean of the decimal multiplier equivalents and it is the decimal multiplier equivalent geometric mean that is relevant.
## Formula review
The Geometric Mean: $\stackrel{~}{x}={\left(\prod _{i=1}^{n}{\pi }_{i}\right)}^{\frac{1}{n}}=\sqrt[n]{{x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}}={\left({x}_{1}*{x}_{2}\bullet \bullet \bullet {x}_{n}\right)}^{\frac{1}{n}}$
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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http://www.sumsar.net/files/posts/2017-bayesian-tutorial-exercises/modeling_exercise1.html | # Exercise 1: Bayesian A testing for Swedish Fish Incorporated (B comes later)
Rasmus Bååth
Swedish Fish Incorporated is the largest Swedish company delivering fish by mail order. They are now trying to get into the lucrative Danish market by selling one year Salmon subscriptions. The marketing department have done a pilot study and tried the following marketing method:
A: Sending a mail with a colorful brochure that invites people to sign up for a one year salmon subscription.
The marketing department sent out 16 mails of type A. Six Danes that received a mail signed up for one year of salmon and marketing now wants to know, how good is method A?
At the bottom of this document you’ll find a solution. But try yourself first!
### Question I) Build a Bayesian model that answers the question: What would the rate of sign-up be if method A was used on a larger number of people?
Hint 1: The answer is not a single number but a distribution over probable rates of sign-up.
Hint 2: As part of you generative model you’ll want to use the binomial distribution, which you can sample from in R using the rbinom(n, size, prob). The binomial distribution simulates the following process n times: The number of “successes” when performing size trials, where the probability of “success” is prob.
Hint 3: A commonly used prior for the unknown probability of success in a binomial distribution is a uniform distribution from 0 to 1. You can draw from this distribution by running runif(1, min = 0, max = 1)
Hint 4: Here is a code scaffold that you can build upon.
# Number of random draws from the prior
n_draws <- 10000
prior <- ... # Here you sample n_draws draws from the prior
hist(prior) # It's always good to eyeball the prior to make sure it looks ok.
# Here you define the generative model
generative_model <- function(parameters) {
...
}
# Here you simulate data using the parameters from the prior and the
# generative model
sim_data <- rep(NA, n_draws)
for(i in 1:n_draws) {
sim_data[i] <- generative_model(prior[i])
}
# Here you filter off all draws that do not match the data.
posterior <- prior[sim_data == observed_data]
hist(posterior) # Eyeball the posterior
length(posterior) # See that we got enought draws left after the filtering.
# There are no rules here, but you probably want to aim
# for >1000 draws.
# Now you can summarize the posterior, where a common summary is to take the mean
# or the median posterior, and perhaps a 95% quantile interval.
median(posterior)
quantile(posterior, c(0.025, 0.975))
### Question II) What’s the probability that method A is better than telemarketing?
So marketing just told us that the rate of sign-up would be 20% if salmon subscribers were snared by a telemarketing campaign instead (to us it’s very unclear where marketing got this very precise number from). So given the model and the data that we developed in the last question, what’s the probability that method A has a higher rate of sign-up than telemarketing?
Hint 1: If you have a vector of samples representing a probability distribution, which you should have from the last question, calculating the amount of probability above a certain value is done by simply counting the number of samples above that value and dividing by the total number of samples.
Hint 2: The answer to this question is a one-liner.
### Question III) If method A was used on 100 people what would be number of sign-ups?
Hint 1: The answer is again not a single number but a distribution over probable number of sign-ups.
Hint 2: As before, the binomial distribution is a good candidate for how many people that sign up out of the 100 possible.
Hint 3: Make sure you don’t “throw away” uncertainty, for example by using a summary of the posterior distribution calculated in the first question. Use the full original posterior sample!
Hint 4: The general patter when calculating “derivatives” of posterior samples is to go through the values one-by-one, and perform a transformation (say, plugging in the value in a binomial distribution), and collect the new values in a vector.
## Solutions (but this can be done in many ways)
### Question I
n_draw <- 10000
# Defining and drawing from the prior distribution
prior_rate <- runif(n_draw, 0, 1)
# Defining the generative model
gen_model <- function(rate) {
subscribers <- rbinom(1, size = 16, prob = rate)
subscribers
}
# Simulating the data
subscribers <- rep(NA, n_draw)
for(i in 1:n_draw) {
subscribers[i] <- gen_model(prior_rate[i])
}
# Filtering out those parameter values that didn't result in the
# data that we actually observed
post_rate <- prior_rate[subscribers == 6]
# Checking that there enough samples left
length(post_rate)
## [1] 578
# Plotting and summarising the posterior.
hist(post_rate, xlim = c(0, 1))
mean(post_rate)
## [1] 0.3862927
quantile(post_rate, c(0.025, 0.975))
## 2.5% 97.5%
## 0.1956573 0.6189745
# Question II
sum(post_rate > 0.2) / length(post_rate)
## [1] 0.9705882
# Question III
# This can be done with a for loop
singnups <- rep(NA, length(post_rate))
for(i in 1:length(post_rate)) {
singnups[i] <- rbinom(n = 1, size = 100, prob = post_rate[i])
}
# But since rbinom is vectorized we can simply write it like this:
signups <- rbinom(n = length(post_rate), size = 100, prob = post_rate)
hist(signups, xlim = c(0, 100))
quantile(signups, c(0.025, 0.975))
## 2.5% 97.5%
## 18 62
# So a decent guess is that is would be between 20 and 60 sign-ups. | 2017-07-22T00:49:32 | {
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https://aakashsrv1.meritnation.com/cbse-class-8/maths/rd-sharma-2019-2020/mensuration-ii-volumes-and-surface-areas-of-a-cuboid-and-a-cube/textbook-solutions/10_1_3523_5480_21.8_46617 | Rd Sharma 2019 2020 Solutions for Class 8 Maths Chapter 21 Mensuration Ii Volumes And Surface Areas Of A Cuboid And A Cube are provided here with simple step-by-step explanations. These solutions for Mensuration Ii Volumes And Surface Areas Of A Cuboid And A Cube are extremely popular among Class 8 students for Maths Mensuration Ii Volumes And Surface Areas Of A Cuboid And A Cube Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Maths Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.
#### Question 1:
Find the volume in cubic metre (cu. m) of each of the cuboids whose dimensions are:
(i) length = 12 m, breadth = 10 m, height = 4.5 cm
(ii) length = 4 m, breadth = 2.5 m, height = 50 cm.
(iii) length = 10 m, breadth = 25 dm, height = 50 cm.
#### Question 2:
Find the volume in cubic decimetre of each of the cubes whose side is
(i) 1.5 m
(ii) 75 cm
(iii) 2 dm 5 cm
#### Question 3:
How much clay is dug out in digging a well measuring 3 m by 2 m by 5 m?
#### Question 4:
What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2?
#### Question 5:
A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain?
#### Question 6:
The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its height and length are 10 m and 2.5 m respectively.
#### Question 7:
A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold?
#### Question 8:
The length , breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains.
#### Question 9:
A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold?
#### Question 10:
How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick?
#### Question 11:
How many bricks each of size 25 cm × 10 cm × 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick, assuming that the volume of sand and cement used in the construction is negligible?
#### Question 12:
A village, having a population of 4000, requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days will the water of this tank last?
#### Question 13:
A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m × 8 m × 6 m is dug outside the field and the earth dug-out from this well is spread evenly on the field. How much will the earth level rise?
#### Question 14:
A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water.
#### Question 15:
A beam 5 m long and 40 cm wide contains 0.6 cubic metre of wood. How thick is the beam?
#### Question 16:
The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day?
#### Question 17:
An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge.
#### Question 18:
The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed?
#### Question 19:
A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 cm3 of iron weighs 8 gm.
#### Question 20:
Fill in the blanks in each of the following so as to make the statement true:
(i) 1 m3 = .........cm3
(ii) 1 litre = ....... cubic decimetre
(iii) 1 kl = ....... m3
(iv) The volume of a cube of side 8 cm is ........
(v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is ........ cm.
(vi) 1 cu.dm = ........ cu. mm
(vii) 1 cu. km = ........ cu. m
(viii) 1 litre = ........ cu. cm
(ix) 1 ml = ........ cu. cm
(x) 1 kl = ........ cu. dm = ........ cu. cm.
#### Question 1:
Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2 m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
#### Question 2:
Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
#### Question 3:
A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
#### Question 4:
Find the surface area of a cube whose volume is
(i) 343 m3
(ii) 216 dm3
#### Question 5:
Find the volume of a cube whose surface area is
(i) 96 cm2
(ii) 150 m2
#### Question 6:
The dimensions of a cuboid are in the ratio 5 : 3 : 1 and its total surface area is 414 m2. Find the dimensions.
#### Question 7:
Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.
#### Question 8:
Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm.
#### Question 9:
The dimensions of an oil tin are 26 cm × 26 cm × 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs 10, find the cost of tin sheet used for these 20 tins.
#### Question 10:
A cloassroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows, etc.)
#### Question 11:
A swimming pool is 20 m long 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs 25 per square metre.
#### Question 12:
The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room.
#### Question 13:
Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.
#### Question 14:
The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5 m, 3 m and 350 cm, respectively. Find the cost of plastering at the rate of Rs 8 per square metre.
#### Question 15:
A cuboid has total surface area of 50 m2 and lateral surface area is 30 m2. Find the area of its base.
#### Question 16:
A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m2. What is the cost of white-washing the walls at the rate of Rs 1.50 per m2.
#### Question 17:
The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m × 1.5 m and 10 windows each of size 1.5 m × 1 m. If the cost of white-washing the walls of the hall at the rate of Rs 1.20 per m2 is Rs 2385.60, fidn the breadth of the hall.
#### Question 1:
Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.
#### Question 2:
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
$\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
#### Question 3:
The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V2 = xyz.
#### Question 4:
A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.
#### Question 5:
Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.
#### Question 6:
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.
#### Question 7:
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.
#### Question 8:
A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m × 8 m × 6 m, find the cost of iron sheet at Rs 17.50 per metre.
#### Question 9:
Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
2
i.e.,
the ratio of the total surface area cuboid to the sum of the surface areas of the three cubes =
Hence, the ratio is 7:9.
#### Question 10:
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre.
#### Question 11:
A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much is the level of field raised?
#### Question 12:
Two cubes, each of volume 512 cm3 are joined end to end. Find the surface area of the resulting cuboid.
#### Question 13:
Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.
#### Question 14:
The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.
#### Question 15:
The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.
#### Question 16:
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
#### Question 17:
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?
#### Question 18:
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm × 3 cm × 0.75 cm can be put in this box?
#### Question 19:
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs. 1248. Find the dimensions of the box.
#### Question 1:
Find the volume of a cuboid whose
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length =1.2 m, breadth = 30 cm, height = 15 cm
(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm.
#### Question 2:
Find the volume of a cube whose side is
(i) 4 cm
(ii) 8 cm
(iii) 1.5 dm
(iv) 1.2 m
(v) 25 mm
#### Question 3:
Find the height of a cuboid of volume 100 cm3, whose length and breadth are 5 cm and 4 cm respectively.
#### Question 4:
A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm3 of a liquid?
#### Question 5:
A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk?
#### Question 6:
A cuboidal wooden block contains 36 cm3 wood. If it be 4 cm long and 3 cm wide, find its height.
#### Question 7:
What will happen to the volume of a cube, if its edge is
(i) halved
(ii) trebled?
#### Question 8:
What will happen to the volume of a cuboid if its:
(i) Length is doubled, height is same and breadth is halved?
(ii) Length is doubled, height is doubled and breadth is sama?
#### Question 9:
Three cuboids of dimensions 5 cm × 6 cm × 7cm, 4cm × 7cm × 8 cm and 2 cm × 3 cm × 13 cm are melted and a cube is made. Find the side of cube.
#### Question 10:
Find the weight of solid rectangular iron piece of size 50 cm × 40 cm × 10cm, if 1 cm3 of iron weighs 8 gm.
#### Question 11:
How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage?
#### Question 12:
A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm3 each are to be made. Find the number of beads that can be made from the block.
#### Question 13:
Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm and 24 cm.
#### Question 14:
A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from this block? Assume cutting causes no wastage.
#### Question 15:
A cube A has side thrice as long as that of cube B. What is the ratio of the volume of cube A to that of cube B?
#### Question 16:
An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm?
#### Question 17:
Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volumes V1 and V2 of the cubes and compare them.
#### Question 18:
A tea-packet measures 10 cm × 6 cm × 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm × 30 cm × 0.2 m?
#### Question 19:
The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm.
#### Question 20:
How many soap cakes can be placed in a box of size 56 cm × 0.4 m × 0.25 m, if the size of a soap cake is 7 cm × 5 cm × 2.5 cm? | 2022-08-11T23:06:43 | {
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https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-4-quadratic-functions-and-factoring-4-7-complete-the-square-4-7-exercises-problem-solving-page-290/64 | ## Algebra 2 (1st Edition)
Published by McDougal Littell
# Chapter 4 Quadratic Functions and Factoring - 4.7 Complete the Square - 4.7 Exercises - Problem Solving - Page 290: 64
decreasing the price by $\$ 10$#### Work Step by Step$y=(70-x)(50+x)=-x^2+20x+3500=-(x^2-20x)+3500=-((x-10)^2-100)+3500=-(x-10)^2+100+3500=-(x-10)^2+3600$Thus the vertex of the graph is at$(10,3600)$, and the shop can maximize its revenue by decreasing the price by$ \$10$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2021-04-18T18:21:14 | {
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https://math.stackexchange.com/questions/3259381/meromorphic-function-at-a-point | # Meromorphic function at a point.
i'm study meromorphic function at the complex plane extended, but i have a trouble. I know if $$f \colon D \subseteq \mathbb{C} \to \mathbb{C}$$ $$f$$ is meromorphic on D if the singularities of $$f$$ are poles in $$D$$ But whats mean $$f$$ is meromorphic at a point $$a \in \mathbb {C}$$?
Other quickly question, anyone knows why $$\infty$$ is branched point of $$log(z)$$?
Thanks
• In answer to your second question: $\infty$ is a branch point of $\log z$ because $0$ is a branch point of $\log \frac{1}{z} = -\log z$. – mjw Jun 12 '19 at 4:47
• We usually don't define holomorphicity, meromorphicity, etc. at single points because singletons are closed and we need open sets for almost all the nice theorems. I would thus assume that when someone writes that "$f$ is meromorphic at $a$" the author is saying "$f$ is meromorphic in a neighborhood of $a$". However, without knowing the context you found this it is hard to say – Brevan Ellefsen Jun 12 '19 at 5:14
• I've found this in "Complex functions: An algebraic and geometric viewpoint" - Singerman. The point of this is extend the notions of analytic and meromorphic functions at complex plane extend, autor says $f(z)$ are analytic, meromorphic, etc... at infinity if $f(1/z)$ are analytic, meromorphic, etc... At $0$. – Davis We Jun 12 '19 at 12:31
• @DavisWe Oh, then yes - the author means analytic in a neighborhood. This is very standard terminology. You are essentially looking at some small neighborhood of the north pole on the Riemann sphere and seeing if you get holo/mero-morphicity. (For some sort of reference, here's a post on this site from a quick google search. Let me know if you still have questions on this) – Brevan Ellefsen Jun 13 '19 at 5:14 | 2020-06-02T05:25:03 | {
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http://gradientdescending.com/gradient-descent-on-linear-regression/ | Gradient descent is one of the more important optimisation techniques. It is used in a wide variety of machine learning techniques due to it’s flexibility in being able to be applied to any differentiable objective function. With each iteration steps are taken in the direction of the negative gradient until converging to a local minimum. As the algorithm approaches the local minimum the jumps become smaller until a specified tolerance is met, or the maximum iterations. To understand how this works gradient descent is applied to a common method, simple linear regression.
The simple linear regression model is of the form:
where
The objective is to find the parameters (\boldsymbol{\beta}) such that they minimise the mean squared error.
This is a good problem to start with since we know the analytical solution is given by
and can check our results.
## Example
Set up:
The analytical solution can be found manually by
And just to convince ourselves this is correct
The objective is to achieve the same result using gradient descent. Gradient descent works by updating the parameters with each iteration in the direction of negative gradient i.e.
where is the learning rate and
The learning rate is to ensure we don’t jump too far with each iteration and rather some proportion of the gradient, otherwise we could end up overshooting the local minimum taking mauch longer to converge or not find the optimal solution at all. Apply this to the problem above, we’ll initialise our values for to something sensible e.g. . I have chosen with 1000 iterations which is a reasonable learning rate to start with for this problem. It’s worth trying different values of to see how it changes convergence. The algorithm is setup as
As expected we obtain the same result. The lines show the gradient and how the parameters converge to the optimal values.
Let’s try a different set of starting values to see how well it converges.
In this example we can see how robust this method is on a simple problem like linear regression. Even when the initial values are very far away from the true values it converges very quickly. | 2019-02-22T14:54:50 | {
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http://math.stackexchange.com/questions/348943/trigonometry-calculating-the-pyramid-volume | # Trigonometry - Calculating the pyramid volume
The problem:
There be the points $P_0(0,0,0)$, $P_1(1,1,1)$, $P_2(2,-1,2)$ and $P_3(3,0,1)$. Calculate the volume of the pyramid.
Now I assumed the base of the pyramid is a triangle, with points $P_1$, $P_2$ and $P_3$.
So I know $\underline u:=\overrightarrow{P_1P_2} = <1, -2, 1>$ and $\underline v:=\overrightarrow{P_1P_3}=<2,-1,0>$.
I calculated the angle between $\underline u$ and $\underline v$:
$$\cos\theta:=\frac{\underline{u}\cdot\underline{v}}{||\underline{u}||\,||\underline{v}||}=\frac{4}{\sqrt{30}}\Longrightarrow \theta=43.0887^\circ$$ $$S_\Delta=\frac{||\underline{u}||\,||\underline{v}||\sin\theta}{2}=1.87$$
and I thought the pyramid's height to be $\,2\,$ , so the volume is $\,\,\displaystyle{V=\frac{S_\Delta\cdot 2}{3}=1.246}$
I believe I'm somewhat wrong with this.
Help will be much appreciated!
-
This problem is relatively simple because one of your points is at the origin. The volume of your pyramid is then equal to $1/6$ of the determinant of the matrix formed by the other points:
$$V = \frac{1}{6} \left |\begin{array}\\1&1&1\\2&-1&2\\3&0&1 \end{array} \right |$$
Evaluating the determinant, I get $V=1$. | 2014-03-09T00:37:27 | {
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https://www.physicsforums.com/threads/homomorphism-problem-need-work-checked.511067/ | # Homomorphism Problem - Need work checked
1. Jun 30, 2011
### Samuelb88
1. The problem statement, all variables and given/known data
Let $f : G \rightarrow H$ be a surjective homomorphism. Prove that if G is cyclic, then H is cyclic, and if G is abelian, then H is abelian.
2. Relevant equations
Notation: Let it be understood that $f^n(x) = [f(x)]^n$.
3. The attempt at a solution
Just would like to see if I am right.
Proof. (of G cyclic $\Rightarrow$ H cyclic)Suppose that $f$ is a surjective homomorphism, and suppose that $G$ is cyclic. Let $G$ be generated by $x$. I want to show that there exists $y$ such that $y$ generates $H$.
First I will prove that $\forall x^n \in G$ $\Rightarrow$ $f(x^n) = f^n(x)$. As our base case for $n=1$ we cite that $f(x e_G) = f(x) f(e_G) = f(x)$. Next suppose that $f(x^n) = f^n(x)$ (Inductive hypothesis). Then $f(x^{n+1}) = f(x^n x) = f(x^n) f(x) = f^n(x) f(x) =f^{n+1}(x)$, which from our inductive hypothesis.
The proof that $\forall x^{-n} \in G$ $\Rightarrow$ $f(x^{-n}) = f^{-n}(x)$ is similar.
Now since $f$ is surjective $\Rightarrow$ $\mathrm{im}(f) = H$ and therefore $H$ is generated by $f(x)$ and is therefore cyclic.
Proof. (of G abelian $\Rightarrow$ H abelian) Suppose now that $G$ is abelian and let $x, y \in G$. I want to show that $\forall f(x) , f(y)$ $\Rightarrow$ $f(x)f(y) = f(y)f(x)$. Then $f(xy) = f(x)f(y)$ and $f(yx) = f(y)f(x)$, but $f(xy) = f(yx)$ $\Rightarrow$ $f(x)f(y) = f(y)f(x)$. Since $f$ is surjective $\mathrm{im}(f) = H$ $\Rightarrow$ $\forall f(x), f(y) \in H$ $f(x)f(y) = f(y)f(x)$ $\Rightarrow$ $H$ is abelian.
My intuition says I'm right, but it seems my intuition is wrong more than right this quarter.
2. Jun 30, 2011
### micromass
Staff Emeritus
Hi Samuelb88!
Could you explain in more detail why H is generated by f(x). I'm sure you know it, but it's a crucial detail that should be mentioned.
That proof is correct! But it's perhaps a bit chaotic, here's a proposal to clean it:
Take h and h' in H. We want to show that hh'=h'h. Because of surjectivity, we know that there are g and g' in G such that f(g)=h and f(g')=h'. Since G is abelian, we know that gg'=g'g. But then it follows that hh'=f(g)f(g')=f(gg')=f(g'g)=f(g')f(g)=h'h.
I'm not saying your proof is wrong, though.
3. Jun 30, 2011
### Samuelb88
Since $f$ is surjective $\Rightarrow$ $\mathrm{im}(f)=\{ ..., f^{-1}(x), e_H , f(x) , f^2(x) , ... \} = H$. Doesn't this imply that $H$ is generated by $f$?
4. Jun 30, 2011
### micromass
Staff Emeritus
Yes, of course, I just wanted some more explanation on that part
Well, it seems it is all correct!
5. Jun 30, 2011
### Samuelb88
Great to hear! Hehe. Thanks much, micromass! | 2017-11-18T10:37:00 | {
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https://worddisk.com/wiki/Infinite_sequence/ | # Sequence
In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, the same elements can appear multiple times at different positions in a sequence, and unlike a set, the order does matter. Formally, a sequence can be defined as a function from natural numbers (the positions of elements in the sequence) to the elements at each position. The notion of a sequence can be generalized to an indexed family, defined as a function from an index set that may not be numbers to another set of elements.
For example, (M, A, R, Y) is a sequence of letters with the letter 'M' first and 'Y' last. This sequence differs from (A, R, M, Y). Also, the sequence (1, 1, 2, 3, 5, 8), which contains the number 1 at two different positions, is a valid sequence. Sequences can be finite, as in these examples, or infinite, such as the sequence of all even positive integers (2, 4, 6, ...).
The position of an element in a sequence is its rank or index; it is the natural number for which the element is the image. The first element has index 0 or 1, depending on the context or a specific convention. In mathematical analysis, a sequence is often denoted by letters in the form of ${\displaystyle a_{n}}$, ${\displaystyle b_{n}}$ and ${\displaystyle c_{n}}$, where the subscript n refers to the nth element of the sequence; for example, the nth element of the Fibonacci sequence ${\displaystyle F}$ is generally denoted as ${\displaystyle F_{n}}$.
In computing and computer science, finite sequences are sometimes called strings, words or lists, the different names commonly corresponding to different ways to represent them in computer memory; infinite sequences are called streams. The empty sequence ( ) is included in most notions of sequence, but may be excluded depending on the context. | 2022-08-14T04:11:46 | {
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https://www.wyzant.com/resources/answers/59981/find_a_vector_in_r3 | Dalia S.
# Find a vector in R3
Let H be the set of all vectors of the form [-4s,-3s,5s].
Find a vector v in R3 such that H=Span{v}
v=?
By: | 2020-01-26T12:05:48 | {
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https://www.studyxapp.com/homework-help/before-a-major-exam-you-realize-that-the-probability-of-yougetting-a-perfect-ma-q401322656 | # Question Solved1 AnswerBefore a major exam, you realize that the probability of you getting a perfect mark is 80% if you are on time to class, but only 55% if you are late to class. Unfortunately, you come late to class 4 out of 10 times. What is the probability you will get perfect marks on this exam?
D9MT5I The Asker · Probability and Statistics
Before a major exam, you realize that the probability of you getting a perfect mark is 80% if you are on time to class, but only 55% if you are late to class. Unfortunately, you come late to class 4 out of 10 times.
What is the probability you will get perfect marks on this exam?
More | 2023-02-03T00:00:35 | {
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"lm_q1_score": 0.9817357200450139,
"lm_q2_score": 0.6654105653819836,
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https://math.stackexchange.com/questions/1799298/homotopy-group-of-the-conformal-group/1799299 | # Homotopy group of the conformal group
I would like to know which are the first three homotopy groups of the conformal group SO(4,2): $$\pi_n(SO(4,2))=? \quad n=1,2,3$$
According to ncatlab, the maximal compact subgroup of (the connected component of) $SO(4,2)$ is $SO(4)\times SO(2)$. Any connected Lie group retracts onto its maximal compact subgroup, so your question is about $SO(4)\times SO(2)$.
Since $SO(2)$ is just a circle, we have $\pi_n(SO(2))=\mathbb{Z},0,0,\ldots$.
$SO(4)$ is a semi-simple group, $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z_2}$, so it is easy to compute the first homotopy groups.
Indeed because the second homotopy of any simple group is trivial, and the third is $\mathbb Z$, we have $\pi_2(SO(4))=0$ and $\pi_3(SO(4))=\mathbb{Z}^2$ (for $n>4$, $\pi_3(SO(n))=\mathbb{Z}$, since then $SO(n)$ is actually simple).
The fundamental group of $SO(n)$, $n\geq 3$ is $\mathbb{Z}_2$, as can be seen in this specific case from the isomorphism above.
Thus we find finally (for the connected component) $$\pi_n(SO(4,2))=\mathbb{Z}_2\times \mathbb{Z},0,\mathbb{Z}^2,\mathbb{Z}_2^2,\mathbb{Z}_2^2,\mathbb{Z}_{12}^2,\ldots,$$ or more generally for $n>1$ $$\pi_n(SO(4,2))=\pi_n(SU(2))\times \pi_n(SU(2))=\pi_n(S^3)\times \pi_n(S^3),$$ where the homotopy groups of spheres can be found on Wikipedia. Note that we do not know all the homotopy groups of $S^3$ in full generality, as noted by Mariano in the comments.
Edit: fixed the 3rd homotopy group and added the general relation to $S^3$.
• It would be probably useful to emphasize the fact that ellipses in your last formula denote groups which we probably do not know. Jun 1, 2016 at 4:15
• @marRrR I have updated the answer, which previously contained a mistake. I am not a topology expert, so it would be nice if someone could check the current version. Jun 11, 2016 at 23:33
The group $SO(n) \subset SO(n+1)$ by an $n-1$-connected map. Consequently for $k < n-1$ $\pi_k(O(n+1)) = \pi_k(O(n))$. I am euclideanizing $SO(4,2) \rightarrow SO(6)$, and not considering for the time the hyperbolic aspects. So all we have to consider is the fundamental group $\pi_1(SO(4,2))$ The Serre fibration $$SO(n) \rightarrow SO(n+1) \rightarrow SO(n+1)/SO(n) \sim S^n$$ gives the sequence of homotopies $$\pi_k(SO(n)) \rightarrow \pi_k(SO(n+1)) \rightarrow \pi_k(SO(n+1)/SO(n))$$ has $\pi_k(S^n) = 0$ this demonstrates the equality. I will now state that it is known that the fundamental group of Lie algebras are abelian.
To continue this, sorry I had to post due to interruption, I now appeal to Bott periodicity. I now use the fact from Bott periodicity theorem that $\pi_k(Sp) = \pi_{k+4}(O)$. Now we can focus in on $\pi_1(sp(2))$ and the knowledge that $sp(2) \sim U(1)$. The homotopy is abelian, which means it is equal to its homology group, which for the circle is $\mathbb Z$
As for not going hyperbolic, it is the case with physics problems that one looks at the Euclidean case first.
• What is an $n-1$-connected map?
– Danu
May 24, 2016 at 22:13
• Why would "Euclideanizing" preserve the homotopy groups? It doesn't preserve other topological properties like compactness. Also, you didn't actually give the homotopy groups, stating they are Abelian does not uniquely identify them. May 24, 2016 at 22:24
• Do you mean that the fundamental group of Lie groups (not algebras: the algebra is of course the simply connected $\mathbb{R}^N$) is Abelian (as an aside: a property of general topological groups)? May 25, 2016 at 0:23
• @WetSavannaAnimalakaRodVance -- yes, the fundamental group of any topological group is abelian (the fundamental group functor takes group objects in the category of spaces to group objects in the category of groups because it preserves products). May 25, 2016 at 4:43
• @WillO Indeed. See Thm 14.6 and 14.7 on my page herre for an elegant and very simple proof of this equivalence of products. Unfortunately the originator of this proof hasn't published it, and the blog where first I read about it is gone. May 25, 2016 at 6:06 | 2022-08-09T08:19:39 | {
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https://toph.co/p/object-detection | # Object Detection
Criterion 2021 Round 14
Limits 1s, 512 MB
Object detection is a computer vision technique that is used to identify and locate objects in an image. An image can be considered as a 2D array containing $P$ rows and $Q$ columns. Each element at position $(i, j)$ in the image is called a ‘picture element’ or pixel. In this problem, we will only consider Grayscale Images, which means, each pixel can be described by a single value denoting the intensity (amount of light) of that pixel. Let’s denote a $N\times M$ sub-image (where $1\leq N \leq P$ and $1 \leq M \leq Q$) of the image to be a rectangle in the image containing $N$ rows and $M$ columns.
You will be given an image containing $R$ rows and $C$ columns, and an object image containing $X$ rows and $Y$ columns $(X \leq R, \, Y \leq C)$. Your task is to find out an $X\times Y$ sub-image of the given image that has the smallest distance from the object image. To calculate the distance, you need to sum up the squared difference between the intensity of each pixel of the object image and its corresponding pixel in the sub-image. For the sub-image positioned at $i$th row and $j$th column of the original image, we can calculate the distance using:
$\text{Distance}_{ij} = \sum\limits_{p=0}^{x-1}\sum\limits_{q=0}^{y-1}\left(A_{(i+p)(j+q)}-B_{(p+1)(q+1)}\right)^2$,
where $A$ is the original image and $B$ is the object that we are looking for.
## Input
The first line of the input contains a single integer $T (1 \leq T \leq 5)$, denoting the number of test cases.
The following line contains two space-separated integers $R$ and $C (1 \leq R, \, C \leq 500)$, denoting the number of rows and the number of columns in the original image, respectively. The following $R$ lines each contain $C$ space-separated integers, describing each pixel’s intensity in the original image.
The next line contains two space-separated integers $X (1 \leq X \leq R)$ and $Y (1 \leq Y \leq C)$, denoting the number of rows and the number of columns in the object image, respectively. The following $X$ lines each contain $Y$ space-separated integers, describing each pixel’s intensity in the object image.
The intensity value of each pixel will be in the range $[0, 50]$.
## Output
Print two space-separated integers $x \, \, y$ denoting the index of the top-left corner of the sub-image that meets the criteria. If there are multiple grids with the smallest distance, print the lexicographically smallest $x \, \, y$. That means if there are multiple grids that meet the criteria, print the one with the smallest $x$. If there are multiple grids with the smallest $x$ that meet the criteria, print the one with the smallest $y$.
Assume that the indices are 1-indexed.
## Sample
InputOutput
1
3 3
1 2 3
4 4 5
6 7 8
1 1
4
2 1 | 2022-08-11T18:37:44 | {
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https://web2.0calc.com/questions/quadrilateral_18 | +0
0
115
1
In quadrilateral ABCD, we have AB = 3, BC = 6, CD = $$9$$, and DA = $$12$$.
If the length of diagonal AC is an integer, what are all the possible values for AC? Explain your answer in complete sentences.
May 13, 2022
Triangle inequality implies $$\begin{cases} 3 + 6 > AC\\ 3 + AC > 6\\ 6 + AC > 3\\ 9 + AC > 12\\ 9 + 12 > AC\\ 12 + AC > 9 \end{cases}$$. If you list the integers satisfying all 6 inequalities, the possible values would be 4, 5, 6, 7, 8. | 2022-12-03T04:53:28 | {
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https://math.stackexchange.com/questions/3610391/is-the-divisor-topology-second-countable | # Is the divisor topology second countable?
Let $$X\neq\varnothing$$ and $$B\subseteq P(X)$$. We define the topology generated by $$B$$ as follows: $$T:=\bigcap\limits_{\tau\supseteq B}\tau$$ where $$\tau$$ is a topology containing $$B$$.
I’m trying to show that the topology on $$\mathbb{Z}_{\geq 2}=\{x\in \mathbb{Z},\,x\geq 2\}$$, generated by the sets $$U_n:=\{x\in \mathbb{Z}_{\geq 2},\,x|n\}$$ is second countable. My attempt was trying to give an explicit countable base but I know that the sets $$U_n$$ aren’t an option. Could anyone give me a hint?
• Any space with a countable generating set is second-countable (count the finite intersections). – Noah Schweber Apr 5 '20 at 4:54
• If I understood correctly, would the set of all finite intersections of the generating set produce an actual basis for the topological space? – pmorelli Apr 5 '20 at 4:58
• That's exactly right. – Noah Schweber Apr 5 '20 at 4:58
• I got it. Thanks a lot! – pmorelli Apr 5 '20 at 5:03
What you're describing is that $$B$$ is a subbase (i.e. generatin set) for $$\tau$$, and if we have a countable subbase, the set of all finite intersections from $$B$$ (including $$X$$ (the empty intersection) and all members of $$B$$ itself) is also countable (standard set theory) and forms a base for $$\tau$$, so yes, $$\tau$$ is then second countable. | 2021-06-13T08:47:14 | {
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http://web2.0calc.com/questions/use-the-formula-s-n2-to-find-the-sum-of-1-3-5 | +0
# Use the formula S = n2 to find the sum of 1 + 3 + 5 + ... +
0
37
1
+138
Use the formula S = n2 to find the sum of 1 + 3 + 5 + ... +
861 = . (Hint: To find n, add 1 to the last term and divide by 2.)
cassie19 Sep 6, 2017
Sort:
#1
+18564
+1
Use the formula S = n2 to find the sum of
1 + 3 + 5 + ... + 861 = ?
(Hint: To find n, add 1 to the last term and divide by 2.)
$$\begin{array}{|rcll|} \hline n &=& \frac{861+1}{2} \\ &=& \frac{862}{2} \\ &=& 431 \\ \hline \end{array}$$
$$\begin{array}{rcll} sum &=& n^2 \\ &=& 431^2 \\ &=& 185761 \\ \end{array}$$
$$\begin{array}{rcll} 1\ + 3\ + 5\ + 7\ + \ ... \ + 859\ + 861 &=& 185761 \\ \end{array}$$
heureka Sep 6, 2017
edited by heureka Sep 6, 2017
### 13 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 2017-09-24T15:46:07 | {
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https://math.stackexchange.com/questions/1256250/how-much-is-it-worth-to-participate-in-a-second-price-auction | # How much is it worth to participate in a second price auction?
You have a valuation for an object (say $v_a$), which you don't know yet but you know is distributed U[0,1]. You will be competing in a second price auction against a completely identical guy as you, who has a valuation $v_b$~U[0,1].
Assume you and him will both bid truthfully in the auction. How much are you willing to pay to enter the auction, i.e. what is the expected gain for a bidder who doesn't know his valuation?
Your expected payoff is $\frac{v_a^2}{2}$, because the probability your valuation is higher is simply $F(v_a)=v_a$. Given that you win, you get your valuation minus the expected bid, which is uniform on [0, $v_a$]. I have two answers and would like to know which one is right and why the other is wrong. It maybe be that both are wrong!
1. Take $\int_0^1$ $\frac{v_a^2}{2}$ d $v_a$ = $\frac{v_a^3}{6}$ = 0.166
2. Evaluate $\frac{v_a^2}{2}$ at the expected value of $v_a$, and get 0.125.
Thanks!
But I would not be prepared to pay this amount as it reduces my expected gain to $0$ with some risk while not participating gives the same expected gain with no risk. | 2019-12-13T16:54:25 | {
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https://spinningnumbers.org/a/source-transformation.html | Source transformation is a way to simplify a circuit. You change a voltage source into a current source, or the other way around. The method is based on Thévenin’s theorem and Norton’s theorem.
Two simple circuits have special names,
The Thévenin form is a voltage source in series with a resistor.
The Norton form is a current source in parallel with a resistor.
It is possible to convert between Thévenin and Norton forms.
Written by Willy McAllister.
### Where we’re headed
To convert Thévenin to Norton: set $\text I_\text N = \text V_\text T / \text R_\text T$.
To convert Norton to Thévenin: set $\text V_\text T = \text I_\text N \, \text R_\text N$.
The Thévenin and Norton resistors have the same value, $\text R_\text T = \text R_\text N$.
Thévenin and Norton forms are equivalent because they have the same $i$-$v$ behavior from the viewpoint of the output port.
We’ll draw a lot of $i$-$v$ graphs to visualize what’s going on. The idea sinks in better if you to do most of the work. Please follow along with pencil and paper.
We use source transformation in the proof of Thévenin’s and Norton’s theorem.
## $i$-$v$ graphs for V, I, and R
Let’s first review the $i$-$v$ graphs of sources and resistors by themselves. Prepare an $i$-$v$ graph and carefully sketch each of these as separate lines,
• voltage source, $v = 5\,\text V$
• current source, $i = 6\,\text{mA}$
• resistor, $\text R = 2\,\text k\Omega$
The resistor line has a tilt of $1/\text R = 1/2\,\text k\Omega$. It passes through the origin.
The voltage source line is a vertical line passing through $v = 5\,\text V$.
The current source line is a horizontal line passing through $i = 6\,\text{mA}$.
A resistor appears as a slanted line through the origin, $i = \dfrac{v}{\text R}$.
The slope (rise over run) of a resistor line is $\dfrac{1}{\text R}$.
A voltage source plots as a vertical line. It’s the same voltage for any current.
A current source plots as a horizontal line. It’s the same current for any voltage.
Voltage and current source lines do not pass through the origin.
## Thévenin form
Now make it more interesting: Find the $i$-$v$ behavior of this resistor and voltage source circuit.
The pair of little circles on the right represents the port of this simple network. You are allowed to touch the port with voltmeter probes. You can also measure current if you connect an ammeter to the port. This is what it means to "look into" the circuit from the "viewpoint" of the port.
Derive the $i$-$v$ graph for this circuit.
Hint: Start by deriving a symbolic expression for $i$ in terms of $v,\text R_\text T,\text V_\text T$.
Try to find something with the form $i(v) = f(v,\text R_\text T,\text V_\text T)$. Use what you know about the two components and Ohm’s Law.
$i = f(v,\text R_\text T,\text V_\text T) =$ _____________
Plot your function for $\text V_\text T = 5\,\text V$ and $\text R_\text T = 2\,\text k\Omega$.
Now that you’ve had a try, it’s my turn. I will derive an $i$-$v$ expression by traditional circuit analysis, and then again with some EE cleverness.
#### Traditional circuit analysis
Write KVL around the loop; start in the lower left,
$\text V_\text T - i\,\text R_\text T - v = 0$
Solve for $i$ in terms of $v,\text R_\text T$, and $\text V_\text T$,
$i = \dfrac{\text V_\text T - v}{\text R_\text T}$
Rearrange a little,
$i = -\dfrac{v}{\text R_\text T} + \dfrac{\text V_\text T}{\text R_\text T}$
Do you recognize this as the equation of a line? It’s what we should expect since it is the combination of lines from a voltage source and resistor, $\text V_{\text T}$ and $\text R_{\text T}$.
With the given values,
$i = -\dfrac{v}{2\,\text k\Omega} + \dfrac{5\,\text V}{2\,\text k\Omega} = -\dfrac{v}{2\,\text k\Omega} + 2.5\,\text{mA}$
And it plots like this,
The line is tilted $i$-$v$ like a resistor, with a negative slope because of the way we defined the direction of the $i$ arrow, but it does not pass through the origin like a resistor would.
The voltage source makes the line shift to the right. It crosses the voltage axis at $v = \text V_\text T$.
#### EE cleverness
It is reasonable to expect the $i$-$v$ curve to be a straight line, since it’s made from the sum of two lines. If you know two points you can create the equation of a line. Can we get the circuit to tell us two points?
Two easy points are where the line crosses the voltage axis and where it crosses the current axis. For this we need some equipment: a voltmeter, an ammeter, and a short length of wire.
Where does the $i$-$v$ line cross the voltage axis?
The line crosses the voltage axis when $i = 0$. How might we force $i$ to be $0$?
We could connect nothing across the port to create an open circuit,
With $i = 0$, measure the voltage with a voltmeter (or do it in your head).
$v_{oc} = \text V_{\text T} = 5\,\text V$
$v_{oc}$ stands for “open circuit voltage”.
The open circuit voltage is the same as the voltage source, $\text V_\text T$.
The line crosses the voltage axis at $v_{oc} = \text V_{\text T}$.
Where does the $i$-$v$ line cross the current axis?
The line crosses the current axis when $v = 0$. How might we force $v$ to be $0$?
We could connect a wire across the port to short it out,
With $v = 0$, measure the current in the shorting wire. Insert an ammeter into the wire (or do it in your head).
$i_{sc} = \dfrac{\text V_{\text T}}{\text R_{\text T}} = \dfrac{5\,\text V}{2000\,\Omega} = 2.5\,\text{mA}$
$i_{sc}$ stands for “short circuit current.”
The line crosses the current axis at $i_{sc}$.
Caution: DO NOT put a short across a real circuit unless you already know it can survive the abuse.
Create an $i$-$v$ equation based on the two points.
You can see the plot right away. Just mark the two points and draw a line. The points are $(v_{oc},0)$ and $(0,i_{sc})$.
Now find the equation of the line. Given two points the equation of the line is,
$(y - y_1) = m\,(x - x_1)\qquad m = \dfrac{(y_1 - y_2)}{(x_1 - x_2)}$</p> The two points we know are,
$x_1,y_1 = (v_{oc},0)\quad$ and $\quad x_2,y_2 = (0,i_{sc})$
$y - 0 = \dfrac{(0 - i_{sc})}{(v_{oc} - 0)}(x - v_{oc})$
$y = \dfrac{0 - (2.5\,\text{mA})}{5 - 0}(x - 5)$
$y = -\dfrac{2.5\,\text{mA}}{5}(x - 5)$
$y = -\dfrac{2.5\,\text{mA}}{5}\,x + \dfrac{2.5\,\text{mA}}{5}\cdot 5$
$y = -\dfrac{x}{2\,\text k\Omega} + 2.5\,\text{mA}\qquad$ (same as the traditional method)
What have we learned about the Thévenin form? The Thévenin form plots as a line in $i$-$v$ space. The tilt of the line is controlled by $\text R_\text T$. The line crosses the voltage axis at $\text V_\text T$. You can position the line anywhere you want in $i$-$v$ space by your choice of component values.
The open-circuit/short-circuit technique is pretty handy. We will use it again.
## Norton form
Now let’s look at the Norton form,
Work out a symbolic expression for port current $i$ in terms of $v,\text R_{\text N},\text I_{\text N}$.
$i = f(v,\text R_{\text N},\text I_{\text N}) =$ _____________
Then plot your function with these values, $\text I_{\text N} = 2\,\text{mA}$ and $\text R_{\text N} = 500\,\Omega$.
Now that you’ve had a chance to plot the function, I will derive the Norton $i$-$v$ expression by traditional circuit analysis, and then with EE cleverness like we did above.
#### Traditional circuit analysis
Write Kirchhoff’s Current Law for the top node; add up currents flowing into the node,
$+\text I_{\text N} - \dfrac{v}{\text R_{\text N}} - i = 0$
Solve for $i$ in terms of $v,\text R_{\text N}$, and $\text I_{\text N}$,
$i = -\dfrac{v}{\text R_{\text N}} + \text I_{\text N}$
This the equation of a line, which shouldn’t be a surprise. The $i$-$v$ graphs for a current source and a resistor are lines, so it makes sense to get a line when we add lines. The y-intercept is $\text I_{\text N}$ and the slope is $-1/\text R_{\text N}$.
Notice the resemblance of this equation to the one we derived for the Thévenin form.
Hmm, pretty similar. Hold that thought.
Here’s the plot with the given values,
$i = -\dfrac{v}{500\,\Omega} + 2\,\text{mA}$
#### More EE cleverness
Let’s identify two points with the same open-circuit/short-circuit thing we did above. Please give this a try before you peek,
Find two points where the $i$-$v$ line of this circuit crosses the $v$ axis and $i$ axis.
Put open and short circuits across the port and measure with your mental multimeter.
$(v_{oc}, 0) = ($ ________, $0)$
$(0, i_{sc}) = (0$, ________ $)$
The line crosses the $v$ axis when $i = 0$. To force $i = 0$ leave the port unconnected (open circuit) so no current can flow in or out of the port,
With the port open, all of $\text I_{\text N}$ is has to flow through $\text R_{\text N}$. The open-circuit voltage is,
$v_{oc} = \text I_{\text N}\,\text R_{\text N}$
$v_{oc} = 2\,\text{mA} \cdot 500 \,\Omega = 1\,\text V$
The $i$-$v$ line crosses the $v$ axis at $1\,\text V$. That’s our first point.
The line crosses the $i$ axis when $v = 0$. To force $v = 0$ connect a wire to short across the port.
With the wire shorting out the port there is $0$ voltage across the resistor. All of $\text I_{\text N}$ flows through the short and none through the resistor. What is the short-circuit current? By simple inspection,
$i_{sc} = \text I_{\text N} = 2\,\text{mA}$
The $i$-$v$ line crosses the $i$ axis at $2\,\text{mA}$. This is our second point.
Use these two points to construct an equation of the line. It should match what we did with the KCL analysis just above. Here is the plot,
What have we learned about the Norton form? The Norton form behaves just like the Thévenin form. It plots as a line in $i$-$v$ space. You can position the line anywhere you want by your choice of component values.
## Source transformation challenge 1
Thévenin and Norton forms forms both generate tilted lines on the $i$-$v$ plot.
1. Make the two circuits produce the same line.
Here are equations for the Thévenin and Norton forms,
$i = -\dfrac{v}{\text R_\text T} + \dfrac{\text V_\text T}{\text R_\text T}$
$i = -\dfrac{v}{\text R_{\text N}} + \text I_{\text N}$
Find relationships between the key parameters, $\text R_\text T$, $\text R_\text N$, $\text V_\text T$, and $\text I_\text N$ to make the two equations the same.
$\text R_\text N =$ ________
$\text I_\text N \,=$ ________
$\text V_\text T =$ ________
Two lines are the same if they have the same slope and the same y-intercept. Look at the two equations and match the things that need to match.
The slopes match if $\text R_\text N = \text R_\text T$. The resistors are the same.
The y-intercepts match if $\text I_\text N = \text V_\text T / \text R_\text T$.
This is the same as saying $\text V_\text T = \text I_\text N\,\text R_\text N$.
If you are given one form you instantly change it into the other with these relationships. Both circuits will have identical $i$-$v$ characteristics.
2. Use the component values from the Thévenin example above to create an equivalent Norton circuit.
$\text V_{\text T} = 5\,\text V$, $\text R_{\text T} = 2\,\text k\Omega$
$\text I_{\text N} =$ ________
$\text R_{\text N} =$ ________
$\text R_\text N = \text R_\text T = 2000\,\Omega$
$\text I_{\text N} = \dfrac{\text V_{\text T}}{\text R_{\text T}} = \dfrac{5}{2000} = 2.5\,\text{mA}$
3. Use the component values from the Norton example above to create an equivalent Thévenin circuit.
$\text I_{\text N} = 2\,\text{mA}$, $\text R_{\text N} = 500\,\Omega$
$\text V_{\text T} =$ ________
$\text R_{\text T} =$ ________
$\text R_\text T = \text R_\text N = 500\,\Omega$
$\text V_{\text T} = \text I_{\text N} \, \text R_{\text N} = 0.002 \cdot 500 = 1\,\text V$
Notice how the conversion process resembles Ohm’s Law.
If you “look into” these circuits from the port you can’t tell them apart. (You “look” with a voltmeter or ammeter.) They have identical behavior for any $v$ or any $i$. This means they are equivalent and therefore interchangeable. We will take advantage of this in the next article.
## Source transformation challenge 2
Use this simulation model to help you with this design challenge (open the link in another tab). Double-click on a component to change its value. At the appropriate step, click DC in the top menu bar to find the voltage and current.
1. Design your own Thévenin form. Pick any values for $\text V_\text T$ and $\text R_\text T$.
2. Design the Norton equivalent to your Thévenin circuit,
3. Determine the open-circuit voltage and short-circuit current of both forms,
$v_{oc} =$ _________ $\quad i_{sc} =$ _________
4. Write the $i$-$v$ equation and plot it on an $i$-$v$ graph. It’s the same equation for both forms,
$i =$ __________________
(Adjust the scales so your particular graph look nice.)
5. Connect a load resistor $\text R_\text L$ to both forms. Use the same resistance for both load resistors.
6. Compute the voltage and current for both load resistors,
$v_\text{RLTh} =$ _________ $\quad i_\text{RLTh} =$ _________
$v_\text{RLN} =$ _________ $\quad i_\text{RLN} =$ _________
7. Plot this point on your $i$-$v$ graph. It’s not the same point as $v_{oc}$ or $i_{sc}$, but it should fall somewhere on the $i$-$v$ line.
If you used the simulation model notice the simulator didn’t tell you how to do the problem. You had to figure out the Norton equivalent on your own. The simulator did provide some help because you could confirm if your answer was right. Simulators are great tools, but they don’t think for you.
## Summary
Thévenin’s circuit is a voltage source in series with a resistor.
Norton’s circuit is a current source in parallel with a resistor.
Thévenin and Norton forms have identical $i$-$v$ behavior if you set,
$\text R_{\text{Thévenin}} = \text R_{\text{Norton}}$
$\text V_{\text{Thévenin}} = \text I_{\text{Norton}}\, \text R$
When circuits produce the same $i$-$v$ curve from the viewpoint of a selected port, we say they are equivalent (from the perspective of the port).
To derive the equation of the $i$-$v$ line for a complicated circuit we found two points on the line. We left the port open and measured the open circuit voltage, $v_{oc}$. Then we placed a short across the port and measured the short circuit current, $i_{sc}$. From the equation we derive the Thévenin and Norton component values.
Caution: Shorting out real electronic equipment to find $i_{sc}$ is a recipe for smoke. Be super careful before you do this.
In the next article we apply source transformation to a problem.
What about other circuits with one source and one resistor?
After so much talk about Thévenin and Norton forms, it’s an obvious question to ask about the other two possibilities. Consider,
• Voltage source in parallel with a resistor
• Current source in series with a resistor
Give this a little thought for a second or two. What do the resistors do?
The resistors don’t do anything in either circuit.
If you put a resistor in parallel with a voltage source it has no effect on the voltage, and it doesn’t influence $i$. All you do is pull some extra current out of the ideal voltage source, a current we can’t observe from the port. Who cares if there’s a resistor in parallel with the ideal voltage source?
The same goes for the resistor in series with a current source. The source pushes its current through the resistor no matter what the resistor value is. The resistor just forces the ideal current source to create some extra voltage to drive the required current. We can’t observe the voltage across the current source from the port.
These two variations don’t make sense. | 2019-08-18T03:47:03 | {
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http://stanford.edu/class/ee364a/quizzes/problems.html | $\newcommand{\ones}{\mathbf 1}$
# Convex problems
Consider the optimization problem $\begin{array}{ll} \mbox{minimize} & x_1^2 \\ \mbox{subject to} & x_1 \leq -1, \quad x_1^2 + x_2^2 \leq 2, \end{array}$ with variable $x=(x_1,x_2)$.
The point $(-1,1)$ is a solution.
1. True.
Correct!
2. False.
Incorrect.
The optimal value is
1. $0$.
Incorrect.
2. $1$.
Correct!
3. $-1$.
Incorrect.
The problem is convex.
1. True.
Correct!
2. False.
Incorrect.
If an optimization problem is feasible, its optimal value $p^\star$ satisfies $p^\star > -\infty$.
1. True.
Incorrect.
2. False.
Correct! The correct conclusion is $p^\star < \infty$.
If the optimal value $p^\star$ of an optimization problem satisfies $p^\star < \infty$, then the problem is feasible.
1. True.
Correct!
2. False.
Incorrect.
Geometric programming
$f(x,y) = x/y + y/x$ is a posynomial function ($x$ and $y$ are positive variables).
1. True.
Correct!
2. False.
Incorrect.
The squareroot of a monomial function is a monomial function.
1. True.
Correct!
2. False.
Incorrect.
Suppose $f$, $g$, and $h$ are posynomial functions of a positive variable $z$. The constraint $f(z) + g(z) \leq h(z)$ can be handled by Geometric Programming.
1. True.
Incorrect.
2. False.
Correct!
Multi-objective optimization
Suppose $\tilde x$ uniquely minimizes $\max\{f(x),g(x)\}$. Then $\tilde x$ is Pareto optimal for the bi-criterion optimization problem $\begin{array}{ll} \mbox{minimize} & (f(x),g(x)). \end{array}$
1. True.
Correct!
2. False.
Incorrect. | 2016-02-10T17:53:56 | {
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http://math.stackexchange.com/questions/116514/is-there-a-name-for-this-bipartite-graph-problem | # Is there a name for this bipartite graph problem?
Is there a name for the following problem?
Given a bipartite graph $G = (U,V,E)$:
What is a minimum subset $U'$ of $U$ that covers all of $V$?
(i.e. every vertex of $V$ is connected to at least on vertex of $U'$)
For example, in the graph above, the set $[u_2, u_3]$ (the vertices 2 and 3 of $U$) is a solution.
-
I am not sure if it has got a name in graph theory terms, but you can say it is equivalent to the Set Cover problem.
Given a family of subsets $S = \{S_1, S_2, \dots, S_m\}$ of $\{1,2,\dots, n\}$, you find the smallest subset of $S$ such that the union of those sets is $\{1,2,\dots, n\}$.
You can consider $U=S$, where you represent each vertex by the set of its neighbours and $V=\{1,2,\dots, n\}$.
-
In graph theory terms, the problem is called the "Hitting set problem". – Gabor Retvari Mar 5 '12 at 22:25
@GaborRetvari: Thanks. But isn't hitting set a non-graph theory term, just like set cover too? – Aryabhata Mar 5 '12 at 22:32
@Aryabhata, thanks for your answer. The equivalence with Set Cover is very interesting. – Eelvex Mar 6 '12 at 0:44
@GaborRetvari it seems the "Hitting set problem" is the closest I can get. Why don't you add this as an answer so I can accept it? – Eelvex Mar 6 '12 at 0:45
@Eelvex: I am not so sure :-) Set cover seems more appropriate. Read this: en.wikipedia.org/wiki/Set_cover_problem#Hitting_set_formulation. – Aryabhata Mar 6 '12 at 1:01 | 2013-12-21T10:15:16 | {
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http://mathhelpforum.com/algebra/103920-projectile-flight.html | 1. ## Projectile Flight
Okay so I have this problem and am not sure where to start. Can anyone help?:
An arrow shot verticaly into the air from ground level with a cross bow reaches a maximum height of 484 feet after 5.5 seconds of flight. Let the quadratic function d(t) represent the distance above ground in feet t seconds after the arrow is released.
(a)Find d(t) and state its domain.
(b)At what times (to two decimal places) will the arrow be 250 ft. above the ground?
Okay so I have this problem and am not sure where to start. Can anyone help?:
An arrow shot verticaly into the air from ground level with a cross bow reaches a maximum height of 484 feet after 5.5 seconds of flight. Let the quadratic function d(t) represent the distance above ground in feet t seconds after the arrow is released.
(a)Find d(t) and state its domain.
(b)At what times (to two decimal places) will the arrow be 250 ft. above the ground?
Use the kinematic equations. In this case initial speed is not known, final speed is known (it stops before falling), the height and the time are all known so the following equation is known and can be used to find u.
$s = ut + \frac{1}{2}at^2$
This is already a quadratic in t. For the domain think about time, do you get negative time for arrows?
For b set s to 250 and solve for t | 2017-04-24T10:54:56 | {
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https://www.shaalaa.com/question-bank-solutions/quadratic-equations-if-x-2-3-x-3-are-roots-equation-ax2-7x-b-0-find-values-aand-b_23087 | Share
Books Shortlist
# If X = 2/3 And X = −3 Are the Roots of the Equation Ax2 + 7x + B = 0, Find the Values Of Aand B. - CBSE Class 10 - Mathematics
#### Question
If x = 2/3 and x = −3 are the roots of the equation ax2 + 7x + b = 0, find the values of aand b.
#### Solution
We have been given that,
ax2 + 7x + b = 0, x = 2/3, x = -3
We have to find and b
Now, if x = 2/3 is a root of the equation, then it should satisfy the equation completely. Therefore we substitute x = 2/3 in the above equation. We get,
a(2/3)2 + 7(2/3) + b = 0
(4a + 42+9b)/9=0
a=(-9b-42)/4 ......... (1)
Also, if x = -3 is a root of the equation, then it should satisfy the equation completely. Therefore we substitute x = -3 in the above equation. We get,
a(-3)2 + 7(-3) + b = 0
9a - 21 + b = 0 ......... (2)
Now, we multiply equation (2) by 9 and then subtract equation (1) from it. So we have,
81a + 9b - 189 - 4a - 9b - 42 = 0
77a - 231 = 0
a = 231/77
a = 3
Now, put this value of ‘a’ in equation (2) in order to get the value of ‘b’. So,
9(3) + b - 21 = 0
27 + b - 21 = 0
b = 21 - 27
b = -6
Therefore, we have a = 3 and b = -6.
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [5]
Solution If X = 2/3 And X = −3 Are the Roots of the Equation Ax2 + 7x + B = 0, Find the Values Of Aand B. Concept: Quadratic Equations.
S | 2019-07-22T04:34:53 | {
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https://mathoverflow.net/questions/282596/characterisation-of-bell-shaped-functions | # Characterisation of bell-shaped functions
This is an open problem that I learned from Thomas Simon. I will completely understand if the question is judged as non-research level (and it is indeed not related to my research), but I believe a solution would result in some nice, publishable mathematics. The point of this post is to popularise the problem.
We need two definitions.
Definition 1: We say that a function $f$ is bell-shaped on $\mathbb{R}$ if it is infinitely smooth, converges to zero at $\pm \infty$ and the $n$-th derivative of $f$ has $n$ zeroes in $I$ (counting multiplicity).
Definition 2: We say that the derivative is interlacing for $f$ on $\mathbb{R}$ if $f$ is infinitely smooth and for every $n \geqslant 0$ the zeroes of $f^{(n)}$ and $f^{(n+1)}$ interlace (counting multiplicity).
When we say that the zeroes of $f$ and $g$ interlace, we mean that each connected component of $\{x \in I : f(x) \ne 0\}$ contains exactly one zero of $g$, except perhaps for unbounded components, which are allowed to contain no zero of $g$ — at least when all zeroes of $f$ and $g$ are simple. The definition in the general case is somewhat more involved, but hopefully clear.
Problem: Describe all bell-shaped functions. Describe all functions $f$ such that the derivative is interlacing for $f$.
Natural modifications are allowed. For example:
• $f$ can be assumed to be analytic or entire;
• the problem can be restricted to a finite or semi-infinite interval.
By description we mean essentially arbitrary condition that is easier to check than the definition. Ideally, one could hope for a result similar to Bernstein's characterisation of completely monotone functions as Laplace transforms, which can be re-phrased as follows: if $f^{(n)}$ has no zeroes on $\mathbb{R}$, then $f$ is the Laplace transform of either a measure on $[0, \infty)$ (a completely monotone function), or a measure on $(-\infty, 0]$ (a totally monotone function). However, the answer for bell-shaped functions is likely much more complicated.
Easy observations:
• It is clear that $f$ is bell-shaped if and only if it has no zero, it converges to zero at $\pm \infty$ and the derivative is interlacing for $f$.
• By Rolle's theorem, if $f$ has no zeroes and $f$ converges to zero at $\pm \infty$, then $f^{(n)}$ has at least $n$ zeroes.
• The function $\exp(-x^2)$ is bell-shaped: its $n$-th derivative is $P_n(x) \exp(-x^2)$ for a polynomial $P_n$ of degree $n$, so it has no more than $n$ zeroes.
• Similarly, the function $f(x) = (1 + x^2)^{-p}$ is bell-shaped for $p > 0$ (and the derivative is interlacing for $f$ also when $p \in (-\tfrac12, 0)$): the $n$-th derivative of $f$ is of the form $P_n(x) = (1 + x^2)^{-p-n}$ for a polynomial $P_n$ of degree $n$.
• In the same vein, $f(x) = x^{-p} \exp(-x^{-1})$ is bell-shaped on $(0, \infty)$ for any $p > 0$: $f^{(n)}(x) = P_n(x) x^{-p - 2n} \exp(-x^{-1})$ for a polynomial $P_n$ of degree $n$.
• The derivative is interlacing for a polynomial if and only if it only has real roots.
• More generally, the derivative is interlacing for any locally uniform limit of polynomials with only real roots. This class includes $\sin x$, $\exp(x)$, $\exp(-x^2)$, some Bessel functions and many more; see Chapter 5 in Steven Fisk's book.
This topic appeared at least in probability literature, in an erroneous article by W. Gawronski, followed by two articles by T. Simon on stable laws and passage times (with W. Jedidi).
EDIT:
Having read through Widder and Hirschman's book and other references pointed out in Alexandre Eremenko's answer (which was really helpful!), I was able to extend Thomas Simon's work and prove that certain functions are bell-shaped. The preprint is available at arXiv; all comments welcome.
The class of functions that I was able to handle includes density functions of infinitely divisible distributions with Lévy measure of the form $\nu(x) dx$, where $x \nu(x)$ and $x \nu(-x)$ are completely monotone on $(0, \infty)$. Such functions are called extended generalised gamma convolutions (EGGC) by L. Bondesson, and it includes all stable distributions (thus correcting an error in W. Gawronski's article).
Of course this does not answer the question about characterisation of all bell-shaped functions: this remains an open problem.
However, I am not aware of any bell-shaped function outside this class. Is there any?
It allows one to show that certain simple functions are bell-shaped. For example, if I did not make a mistake, $f(x) = \dfrac{1}{(1 + x^2)(4 + x^2)}$ is an EGGC, so it is bell-shaped. Out of curiosity: is there any elementary way to prove that this particular function $f$ is bell-shaped?
• This is a nice question. The survey that I know on the topic is Polya, On the zeros of derivatives of a function and its analytic character, 1942, but it is a little bit out of date. – Alexandre Eremenko Oct 3 '17 at 20:49
• @AlexandreEremenko: I have to admit I am completely ignorant in this area, and so I was not aware of Pólya's article. Thanks for the reference: even though it is outdated, it is still a very good read! – Mateusz Kwaśnicki Oct 4 '17 at 21:35
• @AlexandreEremenko: I know that "thank you" comments are to be avoided here, so I will expand a bit. Problem solved (see my answer below), but this would not have been possible without your comments here. Thank you, this has been one of the most fascinating mathematical journeys in my life! – Mateusz Kwaśnicki Oct 18 '19 at 6:28
• Congratulations! If these comments were helpful, you may mention them in your paper and refer to MO. People do this. – Alexandre Eremenko Oct 18 '19 at 12:56
• @AlexandreEremenko: Of course I did that in the first paper! (The second one is of somewhat different character.) I also mentioned your name and MO in a number of talks I gave on the bell-shape. – Mateusz Kwaśnicki Oct 18 '19 at 13:02
This is a set of comments (the problem of completely describing the bell-shaped functions is probably too hard and its complete solution will probably be more than just a "publishable paper":-)).
1. In the definition of bell-shaped it is better to say that ALL derivatives tend to $0$ as $x\to\pm\infty$. This makes more sense, and most authors define it like this.
2. The large class of examples of bell-shaped functions is given by the so-called Polya frequency functions (Laplace transforms of reciprocals of Laguerre Polya functions), see Hirschman and Widder, Convolution Transform, Ch. IV, sect. 5.1. It seems reasonable to conjecture that all bell-shaped functions are of this kind.
3. Assumption that the bell-shaped function is entire is a restrictive one. There are bell-shaped functions which are not entire, $1/(1+x^2)$ for example. If the function is assumed to be entire, AND of the form $g(x)e^{-x^2}$, where $g$ is a real canonical product of genus $1$, then it has derivative-interlacing property only if $g$ has all zeros real. This follows from a deep result of Haseo Ki and Young-One Kim, Duke Math J. 104 (2000), no. 1, 45–73. As a corollary, we obtain that $e^{-z^2}$ is the only bell-shaped function of this form.
4. Hirschman (Proc. AMS, 1, (1950). 63–65.) proved the conjecture of Schoenberg that there are no bell shaped functions with compact support. But there are bell shaped functions which are zero on a ray.
EDIT. My conjecture in 2 is certainly wrong: according to the theorem of T. Simon (second reference in the question text), all positive stable densities are bell-shaped, but they are not Polya frequency functions.
• Thanks for these comments and references! I plan to write a follow-up with some more specific conjectures, and your comment no. 3 already answers the very first one. Regarding comment no. 1, if $f$ converges to zero at infinity and $f'$ is ultimately monotone, then $f'$ converges to zero too, so in fact all derivatives of a bell-shaped function automatically go to zero at $\pm\infty$. But you are right in that it is perhaps more natural to say this straight away. – Mateusz Kwaśnicki Oct 4 '17 at 21:41
• Regarding the edit, an even simpler counterexample is $(1+x^2)^{-s}$, which is not log-concave, and hence it is not a Pólya frequency function. – Mateusz Kwaśnicki Oct 14 '17 at 21:37
• @Mateusz Kwasnicki: Yes. All this shows that this class of bell-shaped functions is large, and that there is no plausible conjecture how to describe all of it. – Alexandre Eremenko Oct 14 '17 at 22:24
Together with Thomas Simon from Lille we just posted to arXiv a paper Characterisation of the class of bell-shaped functions, which answers this question.
To my great surprise, the class of bell-shaped functions studied in the paper mentioned in the edit, A new class of bell-shaped functions, exhausts all bell-shaped functions.
For the record, let me copy Corollary 1.5 from the new paper (with minor editing):
Theorem: The following conditions are equivalent:
(a) $$f$$ is a weakly bell-shaped function;
(b) $$f$$ is the convolution of a Pólya frequency function $$h$$ and a locally integrable function $$g$$ which converges to zero at $$\pm \infty$$, which is absolutely monotone on $$(-\infty, 0)$$, which is completely monotone on $$(0, \infty)$$, and which may contain an additional atom at zero;
(c) $$f$$ is a locally integrable function which converges to zero at $$\pm \infty$$, which is non-decreasing near $$-\infty$$ and non-increasing near $$\infty$$, and which satisfies $$\hat{f}(\xi) = \exp\biggl(-a \xi^2 - i b \xi + c + \int_{-\infty}^\infty \biggl( \frac{1}{i \xi + s} - \biggl(\frac{1}{s} - \frac{i \xi}{s^2} \biggr) \mathbb{1}_{\mathbb{R} \setminus (-1, 1)}(s) \biggr) \varphi(s) ds \biggr) ,$$ with $$a \geqslant 0$$, $$b \in \mathbb{R}$$, $$c \in \mathbb{R}$$, and $$\varphi : \mathbb{R} \to \mathbb{R}$$ a Borel function such that $$\varphi - k$$ changes its sign at most once for every integer $$k$$.
Furthermore, $$f$$ is strictly bell-shaped if and only if $$f$$ is smooth and weakly bell-shaped.
The function $$\varphi$$ necessarily satisfies an appropriate integrability condition at $$\pm\infty$$, and has some regularity at $$0$$; see Theorem 1.1 in the new paper for details.
In the theorem:
• a function $$f$$ is strictly bell-shaped if it is non-negative, it converges to zero at $$\pm \infty$$, and $$f^{(n)}$$ changes its sign $$n$$ times for $$n = 0, 1, 2, \ldots$$;
• a function $$f$$ is weakly bell-shaped if the convolution of $$f$$ with every Gaussian is strictly bell-shaped;
• a function $$g$$ is absolutely monotone on $$(-\infty, 0)$$ if $$g^{(n)}(x) \geqslant 0$$ for every $$x < 0$$;
• a function $$g$$ is completely monotone on $$(0, \infty)$$ if $$(-1)^n g^{(n)}(x) \geqslant 0$$ for every $$x > 0$$;
• a Pólya frequency function $$h$$ is a convolution of a (possibly degenerate) Gaussian and a finite or infinite number of exponentials, that is, functions of the form $$a e^{-1 - a x} \mathbb{1}_{(0, \infty)}(1 + a x)$$ (this is the density function of a centred exponential random variable with variance $$1 / |a|$$, with a negative sign if $$a < 0$$). | 2021-05-15T19:51:00 | {
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https://byjus.com/rs-aggarwal-solutions/rs-aggarwal-class-10-solutions-chapter-19-volume-and-surface-areas-of-solids-exercise-19-3/ | # RS Aggarwal Solutions Class 10 Ex 19C
QUESTION-1: A hemispherical bowl of internal radius 9cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6cm. Find the height of water in the cylindrical vessel.
Solution:
We have,
The radius of the hemispherical bowl, R = 9cm and
The internal base radius of the cylindrical vessel, r = 6cm
Let the height of the water in the cylindrical vessel be h.
As,
Volume of water in the cylindrical vessel = Volume of hemispherical bowl
$\Rightarrow \pi r^{2}h=\frac{2}{3}\pi R^{3}$
$\Rightarrow r^{2}h=\frac{2}{3} R^{3}$
=> 6 x 6 x h = $\frac{2}{3}$ x 9 x 9 x 9
=> h = $\frac{2}{3}\times \frac{9\times 9\times 9}{6\times 6}$
=> h = $\frac{27}{2}$
$Therefore, h=13.5cm$
So, the height of the water in the cylinder vessel is 13.5cm
QUESTION-2: A hemispherical tank, full of water, is emptied by a pipe at the rate of 25/7 litres per second. How much time will it take to empty half the tank if the diameter of the base of the tank is 3m?
Solution:
We have,
The radius of the hemispherical tank, r = 3/2m
Volume of the hemispherical tank = $\frac{2}{3}\pi r^{3}$
$=\frac{2}{3}\times \frac{22}{7}\times \frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}$
$=\frac{99}{14}m^{3}$
Now,
Volume of half tank = $=\frac{1}{2}\times \frac{99}{14}$
$=\frac{99}{14}m^{3}$
$=\frac{99}{14}kL$
$=\frac{99000}{28}L$
As, the rate of water emptied by the pipe = $=\frac{25}{7}$ L/s
So, the time taken to empty half the tank = $=\frac{\left ( \frac{99000}{28} \right )}{\left ( \frac{25}{7} \right )}$
$=\frac{99000}{25\times 4}$
= 990 s
= $=\frac{990}{60}$ min
= 16.5 min
= 16 min 30 seconds
So, the time taken to empty half the tank is 16 minutes and 30 seconds.
QUESTION-3: The rain water from a roof of 44m x 20m drains into a cylindrical tank having the diameter of base 4m and height 3.5m. If the tank is just full, find the rainfall in-cm.
Solution:
We have,
The length of the roof, l = 44m,
The width of the roof, b = 20m,
The height of the cylindrical tank, H = 3.5m and
The base radius of the cylindrical tank, R = 4/2 = 2m
Let the height of the rainfall be h.
Now,
Volume of rainfall = Volume of cylindrical tank
$\Rightarrow lbh=\pi R^{2}H$
$\Rightarrow 44\times 20\times h=\frac{22}{7}\times 2\times 2\times 3.5$
$\Rightarrow h=\frac{22}{7}\times \frac{2\times 2\times 3.5}{44\times 20}$
$\Rightarrow h=\frac{1}{20}m$
$\Rightarrow h=\frac{100}{20}cm$
$Therefore, h=5cm$
So, the height of the rainfall is 5cm.
QUESTION-4: The rain water from a 22m x 20m roof drains into a cylindrical vessel of diameter 2m and height 3.5m. If the rain water collected from the roof fills 4/5th of the cylindrical vessel then find the rainfall in centimeter.
Solution:
We have,
The length of the roof, l = 22m,
The width of the roof, b = 20m,
The base radius of the cylindrical vessel, R = 2/2 = 1m and
The height of the cylindrical vessel, H = 3.5m
Let the height of the rainfall be h.
Now,
Volume of rainfall = Volume of rain water collected in the cylindrical vessel
=> lbh = 4/5 x Volume of cylindrical vessel
=> 22 x 20 x h = $\frac{4}{5}\times \pi R^{2}H$
=> 440h = 4/5 x 22/7 x 1 x 1 x 3.5
=> h = 0.02m
$Therefore, h=2cm$
So, the height of the rainfall is 2cm.
QUESTION-5: A solid right circular cone of height 60cm and radius 30cm is dropped in a right circular cylinder full of water, of height 180cm and radius 60cm. Find the volume of water left in the cylinder, in cubic meters.
Solution:
We have,
Height of cone, h = 60cm,
The base radius of cone, r = 30cm,
The height of cylinder, H = 180cm and
The base radius of the cylinder, R = 60cm
Now,
Volume of water left in the cylinder = Volume of cylinder – Volume of cone
$=\pi R^{2}H-\frac{1}{3}\pi r^{2}h$
= 22/7 x 60 x 60 x 180 – 1/3 x 22/7 x 30 x 30 x 60
= 22/7 x 30 x 30 x 60 (2 x 2 x 3 – 1/3)
= 22/7 x 54000 (12 – 1/3)
= 22/7 x 54000 x 35/3
= 1980000 $cm^{3}$
= 1980000/1000000 $m^{3}$
= 1.98 $m^{3}$
So, the volume of water left in the cylinder is 1.98 $m^{3}$
QUESTION-6:Â Water is flowing at the rate of 6 km/hr through a pipe of diameter 14cm into a rectangular tank which is 60m long and 22m wide. Determine the time in which the level of water in the tank will rise by 7cm.
Solution:
We have,
Speed of the water flowing through the pipe, H = 6 km/hr = 600000 cm/3600 s = 500/3 cm/s,
Radius of pipe, R = 14/2 = 7 cm
Length of the rectangular tank, l = 60 m = 6000 cm,
Breadth of the rectangular tank, b = 22 m = 2200 cm and
Rise in the level of water in the tank, h = 7 cm
Now,
Volume of the water in the rectangular tank = lbh
= 6000 x 2200 x 7 = $92400000\:cm^{3}$
Also,
Volume of the water flowing through the pipe in 1 s = $\pi R^{2}H$
= 22/7 x 7 x 500/3 = 77000/3 $cm{3}$
So,
The time taken = Volume of the water in the rectangular tank / Volume of the water flowing through the pipe in 1 sec.
= $\frac{92400000}{\left ( \frac{77000}{3} \right )}cm^{3}$
= 92400 x 3/77 = 3600s = 1 hr
So, the time in which the level of water in the tank will rise by 7cm is 1 hour.
QUESTION -7: Water in a canal, 6m wide and 1.5m deep, is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes if 8 cm of standing water is needed irrigation?
Solution:
We have,
Width of the canal, b = 6 m,
Depth of the canal, h = 1.5 m,
Height of the standing water needed for irrigation, H = 8 cm = 0.08 m,
Speed of the flowing water, l = 4 km/hr = 4000/60 = 200/3 m/min
Now,
Volume of water flowing out from canal in 1 min = lbh
= 200/3 x 6 x 1.5
= $600\:m^{3}$
=> Volume of water out from canal in 10 min = 600 x 10 = $6000\:m^{3}$
So, the area of irrigation = Volume of water flowing out from canal in 10 min / Height of the standing water needed for irrigation
= 6000/0.08
= $75000\:m^{2}$
= 7.5 hectare (As, 1 hectare = $10000\:m^{2}$
Hence, it will irrigate 7.5 hectare of area in 10 minutes.
QUESTION -8: A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/hr, in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of $Rs0.07\:m^{3}$
Solution:
We have,
The radius of the cylindrical tank, R = 12/2 = 6m = 600cm,
The depth of the tank, H = 2.5m = 250cm,
The radius of the cylindrical pipe, r = 25/2 = 12.5cm,
Speed of the flowing water, h = 3.6 km/hr = 360000m/3600s = 100cm/s
Now,
Volume of water flowing out from the pipe in a hour = $\pi r^{2}h$
= 22/7 x 12.5 x 12.5 x 100 $cm^{3}$
Also,
Volume of the tank = $\pi R^{2}H$
= 22/7 x 600 x 600 x 250 $cm^{3}$
So, the time taken to fill the tank = Volume of the tank / Volume of water flowing out from the pipe in a hour
$=\frac{\frac{22}{7}\times 600\times 600\times 250}{\frac{22}{7}\times 12.5\times 12.5\times 100}$ = 5760 s = 5760/3600 = 1.6h
Also, the cost of water = 0.07 x 22/7 x 6 x 6 x 2.5 = Rs. 19.80
QUESTION-9: Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 liters per minute. Find the rate of flow of water in the pipe in km/hr.
Solution:
We have,
Radius of cylindrical pipe, r = 7/2 cm and
The rate of flow of water = 192.5 L/min
= 192.5L / 1min
= 192.5 x 1000 $cm^{3}$ / 1min (As, 1L = 1000$cm^{3}$)
= 192500 $cm^{3}$ / min
=> The volume of water flowing out from the cylindrical pipe in 1 min = 192500 $cm^{3}$
Now, the rate of flow of water in the pipe
= The volume of water flowing out from the cylindrical pipe in 1 min / $\pi r^{2}$
$=\frac{192500}{\left ( \frac{22}{7}\times \frac{7}{2} \times \frac{7}{2}\right )}$
= 192500 x 2 / 77 = 500 cm/min = 5000 x 60 / 1 x 100000 km/hr = 3 km/hr
So, the rate of flow of water in the pipe is 3 km/hr.
QUESTION -10: 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which is completely immersed in water. Find the rise in the level of water in the vessel.
Solution:
We have,
The radius of spherical marble, r = 1.4/2 = 0.7 cm and
The radius of the cylindrical vessel, R = 7/2 cm = 3.5 cm
Let the rise in the level of water in the vessel be H.
Now,
Volume of water raised in the cylindrical vessel = Volume of 150 spherical marbles
$\Rightarrow \pi R^{2}H =150\times \frac{4}{3}\pi r^{3}$
$\Rightarrow R^{2}H =200r^{3}$
=> 3.5 x 3.5 x H = 200 x 0.7 x 0.7 x 0.7
=> H = 200 x 0.7 x 0.7 x 0.7 / 3.5 x 3.5
Therefore, H = 5.6 cm
So, the rise in the level of water in the vessel is 5.6 cm.
QUESTION -11: Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.
Solution:
Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble = $\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \times \left ( 0.7 \right )^{3}cm^{3}$
The water rises as a cylindrical column.
Volume of cylindrical column filled with water = $\pi r^{2}h=\pi \times \left ( \frac{7}{2} \right )^{2}\times 5.6cm^{3}$
Total number of marbles
= Volume of cylindrical water column / Volume of marble
$=\frac{\pi \times \left ( \frac{7}{2} \right )^{2}\times 5.6}{\frac{4}{3}\pi \times \left ( 0.7 \right )^{3}}$
$=\frac{7\times 7\times 5.6\times 3}{2\times 2\times 4\times 0.7\times 0.7\times 0.7}$ = 150
QUESTION -12: In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here?
Solution:
We have,
Radius of well, R = 10 / 5 = 5 m,
Depth of the well, H = 14 m and
Width of the embankment = 5 m,
Also, the outer radius of the embankment, r = R + 5 = 5 + 5 = 10 m
And, the inner radius of the embankment = R = 5 m
Let the height of the embankment be h.
Now,
Volume of the embankment = Volume of the earth taken out
=> Volume of the embankment = Volume of the well
$\Rightarrow \left ( \pi r^{2}-\pi R^{2} \right )h=\pi R^{2}H$
$\Rightarrow \pi \left ( r^{2}-R^{2} \right )h=\pi R^{2}H$
$\Rightarrow \left ( r^{2}-R^{2} \right )h=R^{2}H$
$\Rightarrow \left ( 10^{2}-5^{2} \right )h=5\times 5\times 14$
=> (100 – 25) h = 25 x 14
=> 75h = 25 x 14
=> h = 25 x 14 / 75
$Therefore, h=\frac{14}{3}m$
So, the height of the embankment is 14/3 m.
#### Practise This Question
Select the salts out of the following elements:
1. HCl 2. NaCl 3. Cu(OH)2 4. CuSO4 5.CH3COOH
6. CH3COONa 7. Na2SO4 8. H2SO4 9.Ca(OH)2 10. CaCl2 | 2019-02-21T09:23:07 | {
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http://davidlowryduda.com/2013/06/ | # Monthly Archives: June 2013
## A proof from the first sheet (SummerNT)
In class today, we were asked to explain what was wrong with the following proof:
Claim: As $x$ increases, the function
$displaystyle f(x)=frac{100x^2+x^2sin(1/x)+50000}{100x^2}$
approaches (gets arbitrarily close to) 1.
Proof: Look at values of $f(x)$ as $x$ gets larger and larger.
$f(5) approx 21.002$
$f(10)approx 6.0010$
$f(25)approx 1.8004$
$f(50)approx 1.2002$
$f(100) approx 1.0501$
$f(500) approx 1.0020$
These values are clearly getting closer to 1. QED
Of course, this is incorrect. Choosing a couple of numbers and thinking there might be a pattern does not constitute a proof.
But on a related note, these sorts of questions (where you observe a pattern and seek to prove it) can sometimes lead to strongly suspected conjectures, which may or may not be true. Here’s an interesting one (with a good picture over at SpikedMath):
Draw $2$ points on the circumference of a circle, and connect them with a line. How many regions is the circle divided into? (two). Draw another point, and connect it to the previous points with a line. How many regions are there now? Draw another point, connecting to the previous points with lines. How many regions now? Do this once more. Do you see the pattern? You might even begin to formulate a belief as to why it’s true.
But then draw one more point and its lines, and carefully count the number of regions formed in the circle. How many circles now? (It doesn’t fit the obvious pattern).
So we know that the presented proof is incorrect. But lets say we want to know if the statement is true. How can we prove it? Further, we want to prove it without calculus – we are interested in an elementary proof. How should we proceed?
Firstly, we should say something about radians. Recall that at an angle $theta$ (in radians) on the unit circle, the arc-length subtended by the angle $theta$ is exactly $theta$ (in fact, this is the defining attribute of radians). And the value $sin theta$ is exactly the height, or rather the $y$ value, of the part of the unit circle at angle $theta$. It’s annoying to phrase, so we look for clarification at the hastily drawn math below:
Note in particular that the arc length is longer than the value of $sin theta$, so that $sin theta < theta$. (This relies critically on the fact that the angle is positive). Further, we see that this is always true for small, positive $theta$. So it will be true that for large, positive $x$, we’ll have $sin frac{1}{x} < frac{1}{x}$. For those of you who know a bit more calculus, you might know that in fact, $sin(frac{1}{x}) = frac{1}{x} – frac{1}{x^33!} + O(frac{1}{t^5})$, which is a more precise statement.
What do we do with this? Well, I say that this allow us to finish the proof.
$dfrac{100x^2 + x^2 sin(1/x) + 50000}{100x^2} leq dfrac{100x^2 + x + 50000}{100x^2} = 1 + dfrac{1}{100x} + dfrac{50000}{100x^2}$
, and it is clear that the last two terms go to zero as $x$ increases. $spadesuit$
Finally, I’d like to remind you about the class webpage at the left – I’ll see you tomorrow in class. | 2018-01-16T11:51:00 | {
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"openwebmath_perplexity": 185.9844695643189,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
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https://math.stackexchange.com/questions/1308627/about-the-definition-of-sobolev-spaces | # About the definition of Sobolev Spaces
I'm studying Sobolev Space and I have a question about the definition:
Def.: The Sobolev Space $W^{k,p}(U)$ consists of all locally summable functions $u:U\to \mathbb{R}$ such that for each multiindex $\alpha$ with $|\alpha|\geq k,$ $D^\alpha u$ exist in the weak sense and belongs to $L^p(U).$
Observation: If $k=0$ and $p=2$, then $W^{0,2}(U)=L^2(U)$.
We have that $u\in W^{k,p}(U)$ since $u\in L^1_{loc}(U),$ so every $u\in L^2(U)$ belong to $L^1_{loc}(U)?$
How can I show that? Thanks.
• Every $u\in L^2(U)$ is in $L^1_{loc}(U)$, which has nothing to do with Sobolev space. – user99914 Jun 2 '15 at 1:35
• It's a question about the definition, just it. – Irddo Jun 2 '15 at 1:37
• So are you asking why an $L^2$ function is in $L^1_{loc}(U)$? – user99914 Jun 2 '15 at 1:48
• Yes, is it. Cause, in the book $W^{0,2}(U)$ is identified with $L^2(U)$, and I don't see why $(\int_U |f(x)|^2dx)^{1/2}$ finite implie $\int_V |f(x)|dx$, for every $V\subset\subset U$. Sorry if I was not clear. – Irddo Jun 2 '15 at 1:54
Indeed this is just Holder's inequality: Pick $V \subset U$ with $|V|<\infty$, then
$$||u||_{L^1(V)}=\int_V |u| dx\le \sqrt{\int_V 1^2 dx}\sqrt{\int_V |u|^2 dx} = \sqrt{|V|}\ ||u||_{L^2(U)}$$
then $u \in L^1_{loc}(U)$. | 2020-02-25T01:38:39 | {
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https://abakbot.com/algebra-en/polynomial-and-matrix-as-an-argument | The original polynomial f(x) (its coefficients) The argument is a square matrix with elements
Polynomial Variable x= Calculation result
We will consider in this material one of the time-consuming tasks in higher mathematics, which sounds like this: Find what the polynomial is given
$f(x)=a_0x^{n}+a_1x^{n-1}+a_2x^{n-2}+.....+a_{n-1}x+a_n$
if the argument is a square matrix, then there is $x=\begin {pmatrix}x_{00}&x_{10}&...&x_{i0}\\x_{01}&x_{11}&...&x_{i1}\\ ... & ...&...&... \\ x_0j} & x_{1j}&...&x_{ij}\end{pmatrix}$
And if the calculation principle itself is clear, especially if you have perfectly understood how to multiply matrices, then direct calculation, for me personally, is considered routine, which should be avoided if possible.
I would like to say right away where this calculator will come in handy. For teachers, teachers, for textbook creators, for those who need to create original tasks on this topic.
It is also useful for students or postgraduates who write essays, term papers, diplomas.
For everyone else, this is an easy way to check the error in a given example, to solve, without long intermediate calculations, the task.
When the calculator was written, it turned out that the sites that were dedicated to this topic contained errors in intermediate calculations and as a result were incorrect.
This calculator, I hope, is free from errors and you will be able to safely solve any examples.
Like the vast majority of calculators on this site, the values of both the coefficients of the polynomial and the elements of the matrix can be complex values.
Such a thing at the end of 2017, you will not find anywhere else, except of course for special created mathematical programs.
Shall we proceed to the examples?
Find the value of the polynomial $f (x)=2x^2-3x+4$ from matrix $x=\begin {pmatrix}-1&2&0\\2&1&-3\\{0}&-1&2\end{pmatrix}$
Polynomial $f (x) = (2)* x^{2}+(-3)*x+(4)$ Variable x= $x = \begin{pmatrix}-1 & 2 & 0 \\ 2 & 1 & -3 \\ 0 & -1 & 2 \\ \end{pmatrix}$ Calculation result $f(x) = \begin{pmatrix}17 & -6 & -12 \\ -6 & 17 & -9 \\ -4 & -3 & 12 \\ \end{pmatrix}$
Another example
What is the polynomial $f (x)=ix^5+(2-i)x^2-11x$ if $x=\begin {pmatrix}2 &2-i&3\\-7&0&i\\1+i&2&0\end{pmatrix}$
Polynomial $f (x) = (i)*x^{5}+(2-i)*x^{2}+(-11)*x$ Variable x= $x = \begin{pmatrix}2 & 2-i & 3 \\ -7 & 0 & i \\ 1+i & 2 & 0 \\ \end{pmatrix}$ Calculation result $f(x) = \begin{pmatrix}2774-2058i & -1092-741i & -66-1293i \\ -1336+2039i & 1937+1391i & 995+2236i \\ 1300+279i & 389+117i & 543+401i \\ \end{pmatrix}$
Find the value of the polynomial $f(x)=x^4-x-1$ from the complex matrix
$x=\begin{pmatrix}-1 & 0 & 0 & -1 & 0 & 0 \\ 0 & -1 & 1 & 0 & -1 & 1 \\ -1 & 0 & i & i & i & -1 \\ -1 & -1 & i & -1 & -1 & i \\ 0 & 1 & -1 & 0 & -1 & i \\ 1 & 0 & 0 & 0 & i & 0 \\ \end{pmatrix}$
Polynomial $f (x) = x^{4}+(-1)*x+(-1)$ Variable x= $x = \begin{pmatrix}-1 & 0 & 0 & -1 & 0 & 0 \\ 0 & -1 & 1 & 0 & -1 & 1 \\ -1 & 0 & i & i & i & -1 \\ -1 & -1 & i & -1 & -1 & i \\ 0 & 1 & -1 & 0 & -1 & i \\ 1 & 0 & 0 & 0 & i & 0 \\ \end{pmatrix}$ Calculation result $f(x) = \begin{pmatrix}7+2i & 3 & -2-2i & 6+i & 1+2i & -3-7i \\ 4+7i & -3-5i & 6+i & 5 & 3-2i & 11 \\ 3-5i & 3+i & -7+i & -2-2i & -4+9i & -4+i \\ 7+6i & 5-i & -1 & 5+4i & 0+6i & 1-2i \\ -2-5i & 3+4i & -8-5i & -2-i & -9-9i & -8+i \\ -5-i & -2+i & -2+3i & -5 & -8+4i & 0+2i \\ \end{pmatrix}$
Successful calculations !
Copyright © 2023 AbakBot-online calculators. All Right Reserved. Author by Dmitry Varlamov | 2023-01-30T10:54:50 | {
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"lm_q1_score": 0.9817357184418847,
"lm_q2_score": 0.6654105653819836,
"lm_q1q2_score": 0.6532573194641024
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https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-6-applications-of-the-derivative-chapter-review-review-exercises-page-351/11 | ## Calculus with Applications (10th Edition)
$f'(x)=-3x^{2}+12x$ $f'(x)=0 \rightarrow -3x^{2}+12x=0 \rightarrow x=4, x=0$ $f(4)=33$ $f(0)=1$ $f(-1)=8$ $f(6)=1$ The absolute maximum of f(x) on [-1,6] is 33 at x=4. The absolute minimum of f(x) on [1,6] is 1 at x=0 and x=6. | 2019-12-07T14:32:38 | {
"domain": "gradesaver.com",
"url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-6-applications-of-the-derivative-chapter-review-review-exercises-page-351/11",
"openwebmath_score": 0.5807904005050659,
"openwebmath_perplexity": 537.0146533277505,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357184418847,
"lm_q2_score": 0.6654105653819836,
"lm_q1q2_score": 0.6532573194641024
} |
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