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# $\int_0^{\infty} \frac{x^{k_1}}{\left(x^{n_1} + a_1 \right)^{m_1}} \cdot \frac{x^{k_2}}{\left(x^{n_2} + a_2 \right)^{m_2}}\:dx$

Recently I was able to find a result to a common definite integral: $$$$J(n_1, k_1, m_1) = \int_0^{\infty} \frac{x^{k_1}}{\left(x^{n_1} + a_1 \right)^{m_1}}\:dx = \frac{a^{\frac{k_1 + 1}{n_1} - m_1}}{n_1}\Gamma\left(m_1 - \frac{k_1 + 1}{n_1}\right)\Gamma\left(\frac{k_1 + 1}{n_1}\right)$$$$ The advantage with this integral is that if you can introduce a free parameter into your integral (ala Feynman's Trick) then under a given transformation using that parameter (i.e. derivatives, Laplace Transform, Fourier Transforms) that you can isolate out the parameter. This makes taking the inverse transform quite easy in most cases.

I'm hoping to expand this method to cater to more difficult integrals where multiple parameters are introduced. In doing so, I've come this very similar but much more complicated integral.

$$$$H\left(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2\right)= \int_0^{\infty} \frac{x^{k_1}}{\left(x^{n_1} + a_1 \right)^{m_1}} \cdot \frac{x^{k_2}}{\left(x^{n_2} + a_2 \right)^{m_2}}\:dx$$$$

Where $$a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2 \in \mathbb{R}^{+}$$

Unfortunately I'm stuck with this one and am interested to find out if anyone has encountered this form before and if so, if they know a method to find a solution expressed in terms of elementary or non-elementary functions.

Any starting points would be greatly appreciated.

Also, if the solution can be represented with certain restrictions on the parameters, please post up. The only absolute conditions are $$a_{1},a_{2},m_{1},m_{2} \gt 0$$. Keen on all solutions restricted or not.

Edit - Just thinking now that the one thing that can be done to simplify the integral is to let $$u = x^{n_1}$$ or $$u = x^{n_2}$$. Here I will use the later to render the integral as

\begin{align} &H\left(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2\right)= \int_0^{\infty} \frac{\left( u^{\frac{1}{n_2}}\right)^{k_1}}{\left(\left( u^{\frac{1}{n_2}}\right)^{n_1} + a_1 \right)^{m_1}} \cdot \frac{\left( u^{\frac{1}{n_2}}\right)^{k_2}}{\left(u + a_2 \right)^{m_2}}\frac{1}{n_2}u^{\frac{1 - n_2}{n_2}}\:du\\ \quad& = \frac{1}{n_2}\int_0^{\infty} \frac{u^{\frac{k_1 + k_2 + 1 - n_2}{n_2}}}{\left( u^{\frac{n_1}{n_2}} + a_1 \right)^{m_1}} \frac{1}{\left(u + a_2\right)^{m_1}}\:du = \frac{1}{n_2}\int_0^{\infty} \frac{u^{k_3}}{\left(u^{n_3} + a_1\right)^{m_1}} \frac{1}{\left(u + a_2\right)^{m_2}}\:du \end{align}

• Just as a joke : $a_2=0$ ! – Claude Leibovici Feb 3 at 9:56
• Haha oh, I wish I could set it that way :-) – user150203 Feb 3 at 10:14
• maybe preform $u\mapsto a_2 u$ then use a series for $\frac1{(u+1)^{m_2}}$ – clathratus Feb 5 at 1:40
• yeah break up the integral with $\int_0^\infty=\int_0^1 +\int_1^\infty$ then use $$(x+1)^\alpha=\sum_{n\geq0}{\alpha\choose n}x^n,\qquad x\in(-1,1)$$ and $$\left(\frac{x+1}x\right)^\alpha=\sum_{n\geq0}{\alpha\choose n}x^{-n},\qquad x\in\Bbb R\setminus [-1,1]$$ after you do the substitution I mentioned above. – clathratus Feb 5 at 1:45
• Thanks heaps for those pointers. I will apply tonight. – user150203 Feb 5 at 7:34

This is an integral of two Meijer G-functions which gives a Fox H-function. Let $$\nu_1 = 1/n_1$$, $$\,\nu_2 = 1/n_2$$, $$\,\sigma = k_1 + k_2 + 1$$. The Mellin transform is $$\mathcal M_{x \to p}[(x + a)^{-m}] = \frac {\Gamma(p) \Gamma(m - p)} {\Gamma(m)} a^{p -m}, \\ \mathcal M[(x^{n_1} + a_1)^{-m_1}] = \nu_1 \mathcal M[(x + a_1)^{-m_1}](\nu_1 p), \\ \mathcal M[x^{-\sigma} (x^{-n_2} + a_2)^{-m_2}] = \nu_2 \mathcal M[(x + a_2)^{-m_2}](\nu_2 (\sigma - p)).$$ The convolution is mapped to the product: $$\mathcal M[f * g] = \mathcal M {\left[\int_0^\infty f(x) \,g {\left( \frac \omega x \right)} \,\frac {dx} x \right]} = \mathcal M[f] \mathcal M[g].$$ The product is again a rational combination of linear gamma functions times $$z^p$$, the inverse transform of which is, by definition, the H-function. Evaluating it at $$1$$ ($$\omega = 1$$ above) gives $$\int_0^\infty \frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}} \frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx = \frac {\nu_1 \nu_2 a_1^{-m_1} a_2^{\nu_2 \sigma - m_2}} {\Gamma(m_1) \Gamma(m_2)} \times \\ \mathcal M^{-1}[ \Gamma(m_1 - \nu_1 p) \Gamma(\nu_2 \sigma - \nu_2 p) \Gamma(\nu_1 p) \Gamma(m_2 - \nu_2 \sigma + \nu_2 p) (a_1^{-\nu_1} a_2^{\nu_2})^{-p}](1) = \\ \frac {\nu_1 \nu_2 a_1^{-m_1} a_2^{\nu_2 \sigma - m_2}} {\Gamma(m_1) \Gamma(m_2)} H_{2, 2}^{2, 2} {\left( a_1^{-\nu_1} a_2^{\nu_2} \middle| {(1 - m_1, \nu_1), (1 - \nu_2 \sigma, \nu_2) \atop (0, \nu_1), (m_2 - \nu_2 \sigma, \nu_2)} \right)}.$$ The integral exists if $$\sigma < n_1 m_1 + n_2 m_2$$.

• Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail. – user150203 Feb 8 at 10:31
• – Harry Peter Oct 31 at 9:40
• @HarryPeter I know it's a minor thing, but I wish you'd make your links non-parsable, e.g. by doing http://google.com, which would show up as http://google.com. Otherwise there is a bunch of links to/from unrelated questions in the Linked section. If we add an extra exponential factor, this is now an integral of a product of three G-functions, formulas for which exist but require even more general functions than the H-function. If the integration range is $[1, \infty)$, this is also equivalent to multiplying by a third G-function. – Maxim Oct 31 at 13:15

To long for the comment.

Maybe it's too trite, but

In the case $$n_1=n_2$$ this integral can be rewrite through the hypergeometric function $$_2F_1$$. I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form $$I= \alpha\int_0^{\infty} \frac{u^{k_3}}{\left(u + 1\right)^{m_1}} \frac{1}{\left(u + c\right)^{m_2}}\:du$$ like it was done by DavidG.

After that one can do the change of variables $$y=u/(1+u)$$ $$I=\frac{\alpha}{c^{m_2}}\int_0^{1} \frac{y^{k_3}}{\left(1 - y\right)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}\:du=\frac{\alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)$$

I am not sure about the existance of the answer in the general case.

• Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it! – user150203 Feb 7 at 7:43

HINT

Let us start from the OP result \begin{align} &H_1\left(a_1, a_2, m_1, m_2, n_3, k_3\right) = \frac{1}{n_2}\int\limits_0^{\infty} \frac{u^{k_3}}{\left(u^{n_3} + a_1\right)^{m_1}} \frac{1}{\left(u + a_2\right)^{m_2}}\:du.\tag1 \end{align}

Taking in account, that identity

\begin{align} &\frac{x^n}{\left(x^n + a\right)^m} =\frac{x^n+a-a}{\left(u^n + a\right)^m} =\frac{1}{\left(x^n + a\right)^{m-1}} -\frac{a}{\left(x^n + a\right)^m}\tag2 \end{align}

allows to manage the degree in the numerator both decreasing and increasing, assume WLOG $$\mathbf{k_3

Also, due to the scaling substitution, let WLOG $$\mathbf{a_1=1}.$$

Under these conditions, enough to consider the integral \begin{align} &I(a,k,n,p,q) = \int\limits_0^{\infty} \frac{u^k}{\left(u^n + 1\right)^p} \frac{1}{\left(u + a\right)^{q}}\:du,\quad k1.\tag3 \end{align}

Is known the binomial decomposition

\begin{align} &(1-z)^{-d-1}=\sum\limits_{j=d}^\infty\binom{j}{d}z^{j-d} = \sum\limits_{j=d}^\infty\dfrac{\Gamma(j+1)}{\Gamma(d+1)\Gamma(j-d+1)}z^{j-d},\quad|z|<1.\tag4 \end{align}

If $$|u|<1$$ then from $$(4)$$ should Maclaurin series
\begin{align} &(1+u^n)^{-q} = \sum\limits_{j=q-1}^\infty(-1)^{j-q+1}\dfrac{\Gamma(j+1)}{\Gamma(q)\Gamma(j-q+2)}u^{n(j-q+1)}.\tag5 \end{align}

If $$|u|>1$$ then from $$(4)$$ can be obtained Laurent series \begin{align} &(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = \sum\limits_{j=q-1}^\infty(-1)^{j-q+1} \dfrac{\Gamma(j+1)}{\Gamma(q)\Gamma(j-q+2)}u^{-n(j-q+1)}.\tag6 \end{align}

At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function

$$\int_0^1\dfrac{x^k}{(x+a)^m}\:dx=\dfrac{a^{-m}}{k+1}\,{_2F_1\left(k+1,m,k+2,-\dfrac1a\right)}\tag7$$

$$\int_1^\infty\dfrac{x^k}{(x+a)^m}\:dx=\dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}\tag8$$

Splitting the issue integral to the intervals $$(0,1)$$ and $$(1,\infty)$$ and applying solutions $$(5)-(8),$$ one can present the result $$I(a,k,n,p,q)$$ as the sum of the pair of series, using gamma function and hypergeometric functions.

Thanks to the other members of the discussion for their interesting ideas.

• Thank you very much for your solution - very much appreciated. – user150203 Feb 8 at 11:09

This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $$\ldots$$

The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of $$\frac{1}{\beta^p (x)} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p - 1} e^{-u \beta (x)} \, du,$$ for suitable conditions on the function $$\beta (x)$$.

Let us use this parametrisation on your first integral $$J(a_1, n_1, m_1, k_1)$$. From Schwinger parametrisation we observe that $$\frac{1}{(x^{n_1} + a_1)^{m_1}} = \frac{1}{\Gamma (m_1)} \int_0^\infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} \, du,$$ (here $$\beta (x) = x^{n_1} + a_1$$ and $$p = m_1$$). Your integral for $$J$$ then becomes $$J(a_1, n_1, m_1, k_1) = \frac{1}{\Gamma (m_1)} \int_0^\infty u^{m_1 - 1} e^{-u a_1} \, du \int_0^\infty x^{k_1} e^{-u x^{n_1}} \, dx,$$ where a change in the order of integration has been made. Now if we let $$t = u x^{n_1}$$ we have \begin{align} J(a_1, n_1, m_1, k_1) &= \frac{1}{\Gamma (m_1)} \int_0^\infty u^{m_1 - 1} e^{-u a_1} \, du \frac{1}{n_1} u^{-\frac{(k_1 + 1)}{n_1}} \int_0^\infty t^{\frac{k_1 + 1}{n_1} - 1} e^{-t} \, dt\\ &= \frac{1}{n_1 \Gamma (m_1)} \Gamma \left (\frac{k_1 + 1}{n_1} \right ) \int_0^\infty u^{\frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} \, du. \end{align} Enforcing a substitution of $$u \mapsto u/a_1$$ leads to \begin{align} J(a_1, n_1, m_1, k_1) &= \frac{a_1^{\frac{k_1 +1}{n_1} - m_1}}{n_1 \Gamma (m_1)} \Gamma \left (\frac{k_1 + 1}{n_1} \right ) \int_0^\infty u^{\frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} \, du\\[2ex] &= \frac{a_1^{\frac{k_1 +1}{n_1} - m_1}}{n_1 \Gamma (m_1)} \Gamma \left (m_1 - \frac{k_1 + 1}{n_1} \right ) \Gamma \left (\frac{k_1 + 1}{n_1} \right ). \end{align}

Now attempting such an approach on the integral $$H$$. Here the following two Schwinger parametrisations are used: $$\frac{1}{(x^{n_1} + a_1)^{m_1}} = \frac{1}{\Gamma (m_1)} \int_0^\infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} \, dy,$$ and $$\frac{1}{(x^{n_2} + a_2)^{m_2}} = \frac{1}{\Gamma (m_2)} \int_0^\infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} \, dz.$$ So the integral for $$H$$, after a change in the order of integration has been made, becomes $$H = \frac{1}{\Gamma (m_1) \Gamma (m_2)} \int_0^\infty y^{m_1 - 1} e^{-y a_1} \, dy \int_0^\infty z^{m_2 - 1} e^{-z a_2} \, dz \int_0^\infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} \, dx.$$ The roadblock is what is to be done with that inner $$x$$-integral if $$n_1 \neq n_2$$?

• Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly – user150203 Feb 8 at 1:04
• What are the restrictions on such a function $\beta$ for that property to hold? – clathratus Feb 8 at 4:24
• @clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $\beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer. – omegadot Feb 8 at 7:46


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# Natural Language Processing (NLP)

Natural Language Processing (NLP) is a field of artificial intelligence that focuses on extracting and understanding important information from text. Some applications are as follows:

1. Long essay (>5,000 words) to read and understand we can easily extract what the major themes of the text (i.e., bag-of-words analysis).
2. A whole set of documents (i.e., corpus) that we know is about cooking, but we want to see what the major themes are in the corpus like does the corpus mention value Asian cooking or Western cooking more (i.e., term frequency-inverse document frequency).
3. Chatbot application where we need to understand what questions a person is asking, and then redirect him to the right resources to answer his question.
4. Need to translate a set of documents from English to French, or vice-versa.
5. Trying to analyze the general sentiment on a specific topic (i.e., immigration) by using a large set of documents written on that topic.

Specifically for finance, NLP can be used in the following applications * Extracting the buy/sell sentiment on a corpus of sell-side analysts reports on a specific stock. * Understanding economic sentiment from conference calls, online job postings, inflation chatter, e-invoicing, etc. * Extracting the amount of media attention to political events * Extracing a long-term view of the global economic sentiment from Warran Buffet's annual reports.

There are several popular NLP libraries as shown below and I found the nice summary below from a ActiveWizards:

Important packages to install using conda for Natural Language Processing are:

1. gensim
2. stop_words

In this blog series, we will go through the following packages where [x] is available and [ ] is in the pipeline:

1. [x] Regular expressions with re
2. [x] Tokenization with nltk
3. [x] Bag-of-Words (BoW) with nltk
4. [x] Bag-of-Words (BoW) with gensim
5. [x] Named-Entity-Recognition (NER) with nltk
6. [ ] Production grade nlp with spacy
7. [ ] Translation with polyglot
8. [x] Text classification with sklearn
9. [ ] Text scraping with BeautifulSoup:
10. [ ] [CNNs with t]

format: plain


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# Find a differentiable curve on the paraboloid $\,z= 2x^2 + y^2$ with minimum curvature

Let $S$ be the graph of the function $\,f(x,y) = 2x^2+ y^2$ in $\,\mathbb{R}^3$ (a paraboloid with vertex at the origin). It is clear that $S$ is a regular surface which can be parametrized by the function $\,\boldsymbol{x} : \mathbb{R}^2 \to S\,$ given by $\,\boldsymbol{x}\left(u,v\right) = \left(u, v, 2u^2 + v^2\right)$. Out of all the differentiable parametrized curves $\,\alpha : \left(- \varepsilon, \varepsilon\right) \to S\,$ such that $\,\alpha \left(0\right) = \left(0,0,0\right),\,$ find one with minimum curvature. Does there exist such a curve with maximum curvature?

This is a homework problem that talks about maximizing or minimizing all the possible values of the curvature of a smooth curve in the paraboloid that goes through the origin and not about the minimum or maximimum normal curvatures which is what confuses me. I guess that the question should still somehow be related to the normal curvature of a curve, but I'm kinda lost on how to attack the problem to be honest, including the question about the existence of a curve with maximum curvature. Any hints or ideas would be greatly appreciated. Thanks for all your answers and comments in advance.


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HW Helper
P: 3,225
 Quote by JasonRox In essence, we take for granted that when we write ordered pairs as vector, we are assuming that it's "foot" is (0,0) without even realizing it. (Well, I do, but most don't. )
Of course, if we think of a vector as a radius vector. But I don't understand what that has to do with ordered pairs. Ordered pairs of, let's say real numbers, have an binary operation of addition and scalar multiplication defined, and they satisfy the general properties of a vector space. So, it makes them vectors. Of course, not vectors in the 'popular' meaning.


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## NCERT Exemplar Problems Class 11 Mathematics Chapter 11 Conic Sections

Short Answer Type Questions
Q1. Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.
Sol:
Given that the circle of radius â€˜a’ touches both axis. So, its centre is (a, a).

Q2. Show that the point (x, y) given by  $x=\frac { 2at }{ 1+{ t }^{ 2 } }$ and  $y=\frac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }$ lies on a circle .

Q3. If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Sol: We have circle through the point A(0, 0), B(a, 0) and C(0, b).
Clearly triangle is right angled at vertex A.

So, centre of the circle is the mid point of hypotenuse BC which is (a/2, b/2)

Q4. Find the equation of the circle which touches x-axis and whose centre is (1,2).
Sol: Given that, circle with centre (1,2) touches x-axis.
Radius of the circle is, r = 2
So, the equation of the required circle is:
(x – l)2 + (y – 2)2 = 22
=>x2-2x + 1 + y2-4y + 4 = 4
=> x2 + y2 – 2x-4y + 1 = 0

Q5. If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
Sol: Given lines are 6x – 8y + 8 = 0 and 6x – 8y – 7 = 0.
These parallel lines are tangent to a circle.

Q6. Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.

Q7. If one end of a diameter of the circle x2 + y2 -4x -6y + 11 = 0 is (3,4), then find the coordinate of the other end of the diameter.
Sol: Given equation of the circle is:

Q8. Find the equation of the circle having (1, -2) as its centre and passing through 3x +y= 14, 2x + 5y = 18.
Sol: Given lines are 3x + y = 14 and 2x + 5y= 18.

Q9. If the line y= âˆš3 x + k touches the circle x2 + y2 = 16, then find the value of
Sol:
Given line is y = âˆš3 x + k and the circle is x2 + y2 = 16.

Q10. Find the equation of a circle concentric with the circle x2 +y2 – 6x + 12y + 15 = 0 and has double of its area.
Sol: Given equation of the circle is:

Q11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Q12. Given the ellipse with equation 9X2 + 25y2 = 225, find the eccentricity and foci.

Q13. If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.

Q14. Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and thecentre is (0, 0).

Q15. Find the distance between the directrices of the ellipse  $\frac { { x }^{ 2 } }{ 36 } +\quad \frac { { y }^{ 2 } }{ 20 } \quad =\quad 1$

Q16. Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.
Sol: Given parabola is y2 = 8x

Q17. Find the length of the line-segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle 6 to the x-axis.
Sol: Given equation of the parabola isy2 = 4ax.
Let the point on the parabola be P(x1,,y1).

Q18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Q19. If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.
Sol: Given that, liney = mx + 1 is tangent to the parabola y2 = 4x.

Q20. If the distance between the foci of a hyperbola is 16 and its eccentricity is âˆš2, then obtain the equation of the hyperbola.

Q21. Find the eccentricity of the hyperbola 9y2 – 4x2 =36
Sol: We have the hyperbola:9y2 – 4x2 = 36

Q22. Find the equation of the hyperbola with eccentricity 3/2  and foci at ( ±2, 0).

Long Answer Type Questions

Q23. If the lines 2x – 3y = 5 and 3x-4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
Sol: Given that lines 2x – 3y – 5 = 0 and 3x – 4y -1 = 0 are diameters of the circle. Solving these lines we get point of intersection as (1, -1), which is centre of the circle.

Q24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.
Sol:  Let the centre of the circle be C(h, k).
Given that the centre lies on the line y – 4x + 3 = 0.
k â€” 4h + 3 = 0 or k = 4h â€” 3
So, the centre is C(h, 4h – 3).

Q25. Find the equation of a circle whose centre is (3, -1) and which cuts off a chord of length 6 units on the line 2x â€” 5y+ 18 = 0.

Sol: Given centre of the circle 0(3, -1)
Chord of the circle is AB.

Q26. Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).

Q27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x -1.
Sol:
Given that circle passes through the point A(7, 3) and its radius is 3.

Q28. Find the equation of each of the following parabolas.
(i) Directrix, x = 0, focus at (6, 0)
(ii) Vertex at (0,4), focus at (0, 2)
(iii) Focus at (-1, -2), directrix x – 2y + 3 = 0
Sol: We know that the distance of any point on the parabola from its focus and its directrix is same.
(i) Given that, directrix, x = 0 and focus = (6, 0)
So, for any point P(x, y) on the parabola
Distance of P from directrix = Distance of P from focus =>   x2 = (x â€” 6)2 + y2
=>                 y2– 12x + 36 = 0
(ii) Given that, vertex = (0,4) and focus = (0, 2)
Now distance between the vertex and directrix is same as the distance between the vertex and focus.
Directrix is y – 6 = 0
For any point of P(x, y) on the parabola
Distance of P from directrix = Distance of P from focus

Q29. Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Sol: Let the coordinates of the variable point be (x, y).
Then according to the question,

Q30. Find the equation of the set of all points whose distance from (0,4) are 2/3 of their distance from the line y = 9.
Sol: Let the point be P(x, y).
According to the question

Q31. Show that the set of all points such that the difference of their distances from (4, 0)and (-4, 0) is always equal to 2 represent a hyperbola.

True/False Type Questions

Q33. The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.
Sol: False
Given equation of the circle is x2 + y2 + 6x + 2y = 0
Centre = (-3, -1)
Clearly, it does not lie on the line x + 3y = 0 as -3 + 3(-l) = -6.
So, this line is not diameter of the circle.

Q34. The shortest distance from the point (2, -7) to the circle x +y2 – 1 4jc – lOy- 151 = 0 is equal to 5.
Sol: False

Q35. If the line lx + my = 1 is a tangent to the circle x2 + y2 = a2, then the point (1, m) lies on a circle.

Q36. The point (1,2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.
Sol: False

Q37. The line lx+ my + n = 0 will touch the parabola^2 = 4 ax if In = am2.
Sol: True

Fill in the Blanks Type Questions

Q41. The equation of the circle having centre at (3, -4) and touching the line 5x + 12y- 12 = 0 is ______.

Q42. The equation of the circle circumscribing the triangle whose sides are the  lines y = x + 2, 3y = 4x, 2y = 3x is   _______.
Given equation of line are:

Q43. An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are _____ .

Q44. The equation of the ellipse having foci (0,1), (0, -1) and minor axis of length 1 is ___ .

Q45. The equation of the parabola having focus at (-1, -2) and the directrix x – 2y + 3 = 0 is______ .
Sol:
Given that, focus at S(-l, -2) and directrix is x – 2y + 3 = 0

Q46. The equation of the hyperbola with vertices at (0, ±6) and eccentricity 5/3 ________ and its foci are _____    .

Objective Type Questions

Q47. The area of the circle centred at (1,2) and passing through (4, 6) is

Q48. Equation of a circle which passes through (3, 6) and touches the axes is

Q49. Equation of the circle with centre on the j-axis and passing through the origin and the point (2, 3) is

Q50. The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

Q51. If the focus of a parabola is (0, -3) and its directrix is y = 3, then its equation is

Q52. If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is

Q53. If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0, then its equation is
(a) y2 = 8(x + 3)
(b) x2 = 8(y + 3)
(c) y2 = -8(x + 3)
(d) y2 = 8(x + 5)

Q54. The equation of the ellipse whose focus is (1, -1), the directrix the line x-y-3 = 0 and eccentricity 1/2 is

Q55. The length of the latus rectum of the ellipse 3x2 +y2 = 12 is
(a) 4
(b) 3
(c) 8
(d) 4/âˆš3


text:
# Batting Average Calculator (Cricket)

LAST UPDATE: April 8th, 2018

## What is “Batting Average” in Cricket?

Batting Average is a cricket statistic used put a player’s number of runs in comparison with their number of times they have been out.
It is intended to be a more accurate representation of a player’s abilities than only the total number of hits or the total number of outs alone, as it compares their number of hits with their relative time at the plate.

## Equation

In Cricket, batting average calculates as follows:
$\text{Batting Average} = \frac{\text{Runs Scored}}{\text{Times Out}}$

## Example

If a batter has scored at bat 522 times and been out 27 times in that time, then:
522 divided by 27 is 19.33
Therefore, the player’s batting average is 19.33


text:
# MCQ Questions for Class 8 Maths Chapter 8 Comparing Quantities with Answers

Free PDF Download of CBSE Maths Multiple Choice Questions for Class8 with Answers Chapter 8 Comparing Quantities. Maths MCQs for Class 8 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 8 Maths Comparing Quantities MCQs Pdf with Answers to know their preparation level.

## Comparing Quantities Class 8 Maths MCQs Pdf

1. The ratio of speed of cycle 12 km per hour to the speed of scooter 36 km per hour is
a) 1:2
b) 1:3
c) 1:4
d) None of the above

Explanation: Speed of cycle/Speed of scooter = 12/36 = ⅓

2. The ratio of 10m to 10 km is:
a) 1/10
b) 1/100
c) 1/1000
d) 1000

Explanation: 10m/10km = 10m/10000m = 1/1000

3. The percentage of 3:4 is:
a) 75%
b) 50%
c) 25%
d) 100%

Explanation: 3:4 = ¾

(3×100)/(4×100) = ¾ x 100% = 0.75 x 100% = 75%

4. The percentage of 2:5
a) 20%
b) 50%
c) 60%
d) 40%

Explanation: 2:5 = ⅖ = ⅖ x 100% = 0.4 x 100% = 40%

5. If 50% of students are good at science out of 20 students. Then the number of students good at science is:
a) 10
b) 15
c) 5
d) 11

Explanation: 50% of students out of 20 students = 50% of 20

= (50/100) x 20

= ½ x 20

= 10

6. The price of a motorcycle was Rs. 34,000 last year. It has increased by 20% this year. The price of motorcycle now is:
a) Rs. 36,000
b) Rs. 38,800
c) Rs. 40,800
d) Rs. 32,000

Explanation: 20% of Rs.34,000 = 20/100 x 34,000 = Rs.6800

New price = Rs. 34,000+Rs.6800

= Rs. 40,800

7. An item marked at Rs. 840 is sold for Rs. 714. The discount % is:
a) 10%
b) 15%
c) 20%
d) 25%

Explanation: Discount = Marked Price – Sale Price

= 840 – 714

= Rs. 126

Discount % = (126/840) x 100% = 15%

8. A person got a 10% increase in his salary. If his salary was Rs. 50000, then the new salary is:
a) Rs. 55000
b) Rs. 60000
c) Rs. 45000
d) Rs. 65000

Explanation: Previous salary = Rs. 50000

10% of Rs.50000 = (10/100) x 50000 = Rs. 5000

New salary = Rs. 50000 + Rs. 5000

= Rs. 55000/-

9. The cost of the article was Rs. 15500 and Rs. 500 was spent on its repairing. If it is sold for a profit of 15%. The selling price of the article is:
a) Rs.16400
b) Rs.17400
c) Rs.18400
d) Rs.19400

Explanation: Total cost = 15500+500 = 16000

Profit % = (Profit/Cost price) x 100

Profit = (Profit% x Cost price)/100

P = (15×16000)/100 = 2400

Selling Price = Profit + cost price = 2400 + 16000 = Rs.18400

10. Waheeda bought an air cooler for Rs. 3300 including a tax of 10%. The price of the air cooler before VAT was added is:
a) Rs. 2000
b) Rs. 3000
c) Rs. 2500
d) Rs. 2800


text:
# What is the best training method for 15-30k records with 5-12 features to classify data into 2 groups?

I don't know if this is a suitable question here. I have data from 2 groups, neutral and positive.

Here how the data looks like

A   02.840  00.960  00.950  00.995  00.995  00.602  1
B   04.730  01.000  01.000  00.999  01.000  01.000  1
C   NULL    00.970  00.005  00.078  00.987  00.000  0
D   NULL    01.000  01.000  00.010  00.675  00.451  1
E   00.530  01.000  00.022  00.231  00.443  00.000  0

The first column is the record name/ID and the last column is its class/group. The six columns in the middle are the features which some of them might be NULL.

With the knowledge that I studied years ago, I'm gonna try ANN and SVM first. But I'm not sure if I am on the right track. So I wonder if anyone here can guide me.

-


text:
# Shape Calculus for Multi-Domain Formulation of Elliptic Problems and its Application to Electrical Machines

## DI Armin Fohler

April 24, 2018, 3:30 p.m. S2 416-1

This work is driven by requirements from applications in the framework of rotating electrical machines. For the simulation of electrical machines we have to solve a non-linear PDE in each rotation step of the motor. This is done via a Newton scheme. To reduce the computational costs of such a scheme one can construct a good initial guess in each rotation step. From an abstract point of view we are dealing with a parametrized family of PDEs with the rotation angle $\varphi$ as our
parameter. The construction of the initial guess comes down to computing the derivative of the solution with respect to
$\varphi$ - the so called Euler predictor. For this we have to extend the ideas of shape calculus presented in the book of
Sokolowski and Zolesio 'Introduction to shape optimization: shape sensitivity analysis' to the multidomain case with
different transformations acting on the individual parts.


text:
# Source code for sympy.ntheory.modular

from __future__ import print_function, division

from sympy.core.numbers import igcdex, igcd
from sympy.core.mul import prod
from sympy.core.compatibility import as_int, reduce
from sympy.ntheory.primetest import isprime
from sympy.polys.domains import ZZ
from sympy.polys.galoistools import gf_crt, gf_crt1, gf_crt2

[docs]def symmetric_residue(a, m): """Return the residual mod m such that it is within half of the modulus. >>> from sympy.ntheory.modular import symmetric_residue >>> symmetric_residue(1, 6) 1 >>> symmetric_residue(4, 6) -2 """ if a <= m // 2: return a return a - m
[docs]def crt(m, v, symmetric=False, check=True): r"""Chinese Remainder Theorem. The moduli in m are assumed to be pairwise coprime. The output is then an integer f, such that f = v_i mod m_i for each pair out of v and m. If symmetric is False a positive integer will be returned, else \|f\| will be less than or equal to the LCM of the moduli, and thus f may be negative. If the moduli are not co-prime the correct result will be returned if/when the test of the result is found to be incorrect. This result will be None if there is no solution. The keyword check can be set to False if it is known that the moduli are coprime. As an example consider a set of residues U = [49, 76, 65] and a set of moduli M = [99, 97, 95]. Then we have:: >>> from sympy.ntheory.modular import crt, solve_congruence >>> crt([99, 97, 95], [49, 76, 65]) (639985, 912285) This is the correct result because:: >>> [639985 % m for m in [99, 97, 95]] [49, 76, 65] If the moduli are not co-prime, you may receive an incorrect result if you use check=False: >>> crt([12, 6, 17], [3, 4, 2], check=False) (954, 1224) >>> [954 % m for m in [12, 6, 17]] [6, 0, 2] >>> crt([12, 6, 17], [3, 4, 2]) is None True >>> crt([3, 6], [2, 5]) (5, 6) Note: the order of gf_crt's arguments is reversed relative to crt, and that solve_congruence takes residue, modulus pairs. Programmer's note: rather than checking that all pairs of moduli share no GCD (an O(n**2) test) and rather than factoring all moduli and seeing that there is no factor in common, a check that the result gives the indicated residuals is performed -- an O(n) operation. See Also ======== solve_congruence sympy.polys.galoistools.gf_crt : low level crt routine used by this routine """ if check: m = list(map(as_int, m)) v = list(map(as_int, v)) result = gf_crt(v, m, ZZ) mm = prod(m) if check: if not all(v % m == result % m for v, m in zip(v, m)): result = solve_congruence(*list(zip(v, m)), check=False, symmetric=symmetric) if result is None: return result result, mm = result if symmetric: return symmetric_residue(result, mm), mm return result, mm
[docs]def crt1(m): """First part of Chinese Remainder Theorem, for multiple application. Examples ======== >>> from sympy.ntheory.modular import crt1 >>> crt1([18, 42, 6]) (4536, [252, 108, 756], [0, 2, 0]) """ return gf_crt1(m, ZZ)
[docs]def crt2(m, v, mm, e, s, symmetric=False): """Second part of Chinese Remainder Theorem, for multiple application. Examples ======== >>> from sympy.ntheory.modular import crt1, crt2 >>> mm, e, s = crt1([18, 42, 6]) >>> crt2([18, 42, 6], [0, 0, 0], mm, e, s) (0, 4536) """ result = gf_crt2(v, m, mm, e, s, ZZ) if symmetric: return symmetric_residue(result, mm), mm return result, mm
[docs]def solve_congruence(*remainder_modulus_pairs, **hint): """Compute the integer n that has the residual ai when it is divided by mi where the ai and mi are given as pairs to this function: ((a1, m1), (a2, m2), ...). If there is no solution, return None. Otherwise return n and its modulus. The mi values need not be co-prime. If it is known that the moduli are not co-prime then the hint check can be set to False (default=True) and the check for a quicker solution via crt() (valid when the moduli are co-prime) will be skipped. If the hint symmetric is True (default is False), the value of n will be within 1/2 of the modulus, possibly negative. Examples ======== >>> from sympy.ntheory.modular import solve_congruence What number is 2 mod 3, 3 mod 5 and 2 mod 7? >>> solve_congruence((2, 3), (3, 5), (2, 7)) (23, 105) >>> [23 % m for m in [3, 5, 7]] [2, 3, 2] If you prefer to work with all remainder in one list and all moduli in another, send the arguments like this: >>> solve_congruence(*zip((2, 3, 2), (3, 5, 7))) (23, 105) The moduli need not be co-prime; in this case there may or may not be a solution: >>> solve_congruence((2, 3), (4, 6)) is None True >>> solve_congruence((2, 3), (5, 6)) (5, 6) The symmetric flag will make the result be within 1/2 of the modulus: >>> solve_congruence((2, 3), (5, 6), symmetric=True) (-1, 6) See Also ======== crt : high level routine implementing the Chinese Remainder Theorem """ def combine(c1, c2): """Return the tuple (a, m) which satisfies the requirement that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2. References ========== - http://en.wikipedia.org/wiki/Method_of_successive_substitution """ a1, m1 = c1 a2, m2 = c2 a, b, c = m1, a2 - a1, m2 g = reduce(igcd, [a, b, c]) a, b, c = [i//g for i in [a, b, c]] if a != 1: inv_a, _, g = igcdex(a, c) if g != 1: return None b *= inv_a a, m = a1 + m1*b, m1*c return a, m rm = remainder_modulus_pairs symmetric = hint.get('symmetric', False) if hint.get('check', True): rm = [(as_int(r), as_int(m)) for r, m in rm] # ignore redundant pairs but raise an error otherwise; also # make sure that a unique set of bases is sent to gf_crt if # they are all prime. # # The routine will work out less-trivial violations and # return None, e.g. for the pairs (1,3) and (14,42) there # is no answer because 14 mod 42 (having a gcd of 14) implies # (14/2) mod (42/2), (14/7) mod (42/7) and (14/14) mod (42/14) # which, being 0 mod 3, is inconsistent with 1 mod 3. But to # preprocess the input beyond checking of another pair with 42 # or 3 as the modulus (for this example) is not necessary. uniq = {} for r, m in rm: r %= m if m in uniq: if r != uniq[m]: return None continue uniq[m] = r rm = [(r, m) for m, r in uniq.items()] del uniq # if the moduli are co-prime, the crt will be significantly faster; # checking all pairs for being co-prime gets to be slow but a prime # test is a good trade-off if all(isprime(m) for r, m in rm): r, m = list(zip(*rm)) return crt(m, r, symmetric=symmetric, check=False) rv = (0, 1) for rmi in rm: rv = combine(rv, rmi) if rv is None: break n, m = rv n = n % m else: if symmetric: return symmetric_residue(n, m), m return n, m


text:
# Chapter – 2: Elasticity of Demand – II (Numericals)

### ₹10 to ₹5 and its coefficient of price elasticity of demand is 2. How much quantity will be demanded at the changed price where the original quantity demanded is 40?

Ans. (a)

Price Total Expenditure Demand
₹ 4
₹ 200
₹ 50
₹ 3
₹ 300
₹ 100

Percentage method of price Elasticity

$${e_{d\,}} = \frac{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{demand}}}}{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{price}}}}$$

$$= \frac{\Delta Q}{\Delta P} \times \frac{P}{Q} = \frac{50}{1} \times \frac{4}{50} = 4$$

$$\begin{array}{l}\left( b \right)\,\,\,{e_{d\,}} = \frac{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{demand}}}}{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{price}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{12}}{5}\end{array}$$

= 2.4, i.e. more than unity or more elastic

$$\left( c \right)\,\,\,{e_{d\,}} = \frac{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{demand}}}}{{{\rm{Percentage}}\,{\rm{change}}\,{\rm{in}}\,{\rm{price}}}}$$

$$= \frac{\frac{\Delta Q}{Q}\times 100}{10} = \frac{\frac{4}{16}\times 100}{10}=2.5$$

(d) $${e_d} = \frac{{\Delta Q}}{{\Delta P}} \times \frac{P}{Q}$$

$$= \frac{x}{5} \times \frac{10}{40}= 2$$

$$= \frac{x}{20}=2$$ or $$x=40$$

$$\therefore$$ Quantity demanded will be: $$Q+\Delta Q=40=80$$


text:
## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 2 - Section 2.1 - The Additon Property of Equality - Exercise Set - Page 103: 83

#### Answer

A solution of an equation is a number that, when substituted for a variable, makes the equation a true statement.

#### Work Step by Step

That is what is meant by the solution of an equation.

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.


text:
Shape of long sequences in C(ω_1) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:23:50Z http://mathoverflow.net/feeds/question/21028 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21028/shape-of-long-sequences-in-c-1 Shape of long sequences in C(ω_1) David R. MacIver 2010-04-11T19:21:00Z 2010-04-13T19:13:33Z
Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!

This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega_1$, which I know relatively little about.

Let $V$ be a normed space. For $A \subseteq V$, define $r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$. Define a bad sequence in $V$ to be a sequence ${ v_\alpha : \alpha < \omega_1 }$ with the properties that:

$\forall \beta, r(\{ v_\alpha : \alpha < \beta \}) \leq 1$

$\inf_\beta r(\{ v_\alpha : \alpha \geq \beta \}) > 1$

An example of a space with a bad sequence is $c_0(\omega_1)$ (the set of all bounded real-valued sequences of length $\omega_1$ such that $\{ \alpha : |x_\alpha| > 0 \}$ is countable). The sequence $2 * 1_{\{\alpha\}}$ is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around $1_{[0, \alpha]}$, which is in $c_0(\omega_1)$ because $\alpha < \omega_1$.

I have two (three depending on how you count it) major examples of spaces which have no bad sequences:

Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.

Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:

Any reflexive space: If $U_\alpha$ forms a chain, the sets $F_\alpha = \bigcap_{v \in U_\alpha} \overline{B}(v, r + \epsilon)$ form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in $\overline{B}(c, r + \epsilon)$

any space with the property that $\textrm{diam}(A) = 2 r(A)$ (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.

So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega_1)$ contain a bad sequence?

I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.

For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.
 http://mathoverflow.net/questions/21028/shape-of-long-sequences-in-c-1/21055#21055 Answer by Joel David Hamkins for Shape of long sequences in C(ω_1) Joel David Hamkins 2010-04-12T01:20:32Z 2010-04-12T12:42:45Z
I claim that there are no bad sequences in C(ω1).

Suppose to the contrary that xα is bad. For any countable ordinal β, there is rβ in C(ω1) such that the distance between rβ and xα for α < β is at most 1. For any countable ordinal β and any positive rational number ε, there is a smaller ordinal γ < β such that all rβ(α) are within ε of rβ(β) for α in (γ,β]. For fixed ε, this is a regressive function on the countable ordinals. Thus, by Fodor's Lemma, there is a stationary and hence unbounded set of ordinals on which the function has constant value, which we may call γε. Since there are only countably many ε, we may find a countable ordinal γ above all γε. This ordinal has the property that for all ordinals β above γ, we have rβ(α) = rβ(β) for all α in the interval (γ,β], since the values are within every ε of each other. That is, every rβ function is constant from the same fixed γ up to β.

Let Cβ be the closed interval of values s such that the constant sequence s of length β lies within 1 of all xη(α) for all η ≤ β and all γ < α ≤ β. These are nested and not empty, since rβ(β) is in Cβ. By compactness, there is a value s in all Cβ. Thus, the number s is within xη(α) for all η and all α above γ.

Thus, we may form the desired sequence r by finding a center that works for the sequences up to stage γ, using the separability idea in your question, augmented with the constant value s at the stages above γ up to ω1. That is, we solve the problem separately on the first γ many coordinates, and then append the constant s sequence up to ω1. This sequence is continuous, and it lies within 1 of every xη, as desired.
 http://mathoverflow.net/questions/21028/shape-of-long-sequences-in-c-1/21082#21082 Answer by Sergei Ivanov for Shape of long sequences in C(ω_1) Sergei Ivanov 2010-04-12T08:48:42Z 2010-04-12T12:23:17Z
I can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point [Edit: $\omega_1$ has this property but not compact], then there are no bad sequences. Moreover the following holds:

If $F\subset C(X)$ is a family such that every countable subfamily has radius $\le 1$, then $r(F)\le 1$.

Define a function $S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$ (the "essential supremum" of $F$) as follows: $S(x)$ is the maximum $t$ such that for every neighborhood $U$ of $x$ one has $\sup\{f(y):f\in F,y\in U\}\ge t$. Observe that $S$ is upper semi-continuous: for every $t\in\widehat{\mathbb R}$, the set $\{x\in X:S(x)\ge t\}$ is closed.

Define the essential infimum $I_F$ similarly, this function is lower semi-continuous.

For every $x\in X$ there is a countable family $G\subset F$ such that $S_G(x)=S_F(x)$ and $I_G(x)=I_F(x)$. Indeed, using countable base at $x$, one can realize $S(x)$ by a sequence $x_i:i\in\mathbb N$ converging to $x$ and functions $f_i\in F$ such that $f_i(x_i)\to S(x)$.

It follows that $S(x)\le I(x)+2$ for all $x\in X$. Indeed, take $G$ as above, it is contained in a $(1+\epsilon)$-ball centered at some $f\in C(X)$, then $S_G(x)\le f(x)+1+\epsilon$ and $I_G(x)\ge f(x)-1-\epsilon$.

Fix $\epsilon>0$ and let us prove that $F$ is contained in a $(1+\epsilon)$-ball. For $x\in X$, define $C_x=\frac12(S(x)+I(x))$. Note that $S(x)\le C_x+1$ and $I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood $U_x$ of $x$ such that $S(y)<C_x+1+\epsilon$ and $I(y)>C_x-1-\epsilon$ for all $y\in U_x$. Choose a finite subcovering $V_i=U_{x_i}:i\le N$.

On each neighborhood $V_i$ we have a constant function $f_i:=C_{x_i}$ which works as a center within this neighborhood. It suffices to construct a function $g\in C(X)$ such that for every $x\in X$, $g(x)$ is between the minimum and maximum of these partially defined constant functions at $x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined $g=g_{n-1}$ that works on $\bigcup_{i<n} V_i$. By Urysohn's lemma there is $\phi:X\to[0,1]$ such that $\phi=0$ on $X\setminus V_n$ and $\phi=1$ on $X\setminus\bigcup_{i\ne n} V_i$. Then $g_n:=\phi f_n+(1-\phi)g_{n-1}$ works on $\bigcup_{i\le n}V_i$.
 http://mathoverflow.net/questions/21028/shape-of-long-sequences-in-c-1/21084#21084 Answer by David R. MacIver for Shape of long sequences in C(ω_1) David R. MacIver 2010-04-12T09:17:50Z 2010-04-13T19:13:33Z
Edit: This proof is wrong, but preserving for posterity.

Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).

Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.

Let $A \subseteq C(K)$ be non-empty. Define

$g(x) = \sup_{f \in A} f(x)$

$h(x) = \inf_{f \in A} f(x)$

$g$ is upper semicontinuous: $g(x) > a$ iff there exists $f \in A$ such that $f(x) > a$. Similarly $h$ is lower semicontinuous.

Further, $g(x) - h(x) \leq \textrm{diam}(A)$.

Therefore $g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$

But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with

$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$

But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.

But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.


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# Lebesgue measure

In measure theory, the Lebesgue measure, named after French mathematician Henri Lebesgue, is the standard way of assigning a measure to subsets of n-dimensional Euclidean space. For n = 1, 2, or 3, it coincides with the standard measure of length, area, or volume. In general, it is also called n-dimensional volume, n-volume, or simply volume.[1] It is used throughout real analysis, in particular to define Lebesgue integration. Sets that can be assigned a Lebesgue measure are called Lebesgue measurable; the measure of the Lebesgue measurable set A is denoted by λ(A).

Henri Lebesgue described this measure in the year 1901, followed the next year by his description of the Lebesgue integral. Both were published as part of his dissertation in 1902.[2]

The Lebesgue measure is often denoted dx, but this should not be confused with the distinct notion of a volume form.

## Definition

Given a subset ${\displaystyle E\subseteq \mathbb {R} }$, with the length of an (open, closed, semi-open) interval ${\displaystyle I=[a,b]}$ given by ${\displaystyle \ell (I)=b-a}$, the Lebesgue outer measure ${\displaystyle \lambda ^{*}(E)}$ is defined as

${\displaystyle \lambda ^{*}(E)=\operatorname {inf} \left\{\sum _{k=1}^{\infty }\ell (I_{k}):{(I_{k})_{k\in \mathbb {N} }}{\text{ is a sequence of open intervals with }}E\subseteq \bigcup _{k=1}^{\infty }I_{k}\right\}}$.

The Lebesgue measure is defined on the Lebesgue σ-algebra, which is the collection of all the sets E which satisfy the condition that, for every ${\displaystyle A\subseteq \mathbb {R} }$,

${\displaystyle \lambda ^{*}(A)=\lambda ^{*}(A\cap E)+\lambda ^{*}(A\cap E^{c})}$

For any set in the Lebesgue σ-algebra, its Lebesgue measure is given by its Lebesgue outer measure ${\displaystyle \lambda (E)=\lambda ^{*}(E)}$.

Sets that aren't included in the Lebesgue σ-algebra are not Lebesgue measurable. Such sets do exist, i.e., Lebesgue σ-algebra is strictly contained in the power set of ${\displaystyle \mathbb {R} }$.

### Intuition

The first part of the definition states that the subset ${\displaystyle E}$ of the real numbers is reduced to its outer measure by coverage by sets of intervals. Each of these sets of intervals ${\displaystyle I}$ covers ${\displaystyle E}$ in the sense that when the intervals are combined together by union, they contain ${\displaystyle E}$. The total length of any covering interval set can easily overestimate the measure of ${\displaystyle E}$, because ${\displaystyle E}$ is a subset of the union of the intervals, and so the intervals may include points which are not in ${\displaystyle E}$. The Lebesgue outer measure emerges as the greatest lower bound (infimum) of the lengths from among all possible such sets. Intuitively, it is the total length of those interval sets which fit ${\displaystyle E}$ most tightly and do not overlap.

That characterizes the Lebesgue outer measure. Whether this outer measure translates to the Lebesgue measure proper depends on an additional condition. This condition is tested by taking subsets of the real numbers ${\displaystyle A}$ using ${\displaystyle E}$ as an instrument to split ${\displaystyle A}$ into two partitions: the part of ${\displaystyle A}$ which intersects with ${\displaystyle E}$ and the remaining part of ${\displaystyle A}$ which is not in ${\displaystyle E}$: the set difference of ${\displaystyle A}$ and ${\displaystyle E}$. These partitions of ${\displaystyle A}$ are subject to the outer measure. If for all possible such subsets ${\displaystyle A}$ of the real numbers, the partitions of ${\displaystyle A}$ cut apart by ${\displaystyle E}$ have outer measures which add up to the outer measure of ${\displaystyle A}$, then the outer Lebesgue measure of ${\displaystyle E}$ gives its Lebesgue measure. Intuitively, this condition means that the set ${\displaystyle E}$ must not have some curious properties which causes a discrepancy in the measure of another set when ${\displaystyle E}$ is used as a "mask" to "clip" that set, hinting at the existence of sets for which the Lebesgue outer measure does not give the Lebesgue measure. (Such sets are, in fact, not Lebesgue-measurable.)

## Properties

Translation invariance: The Lebesgue measure of ${\displaystyle A}$ and ${\displaystyle A+t}$ are the same.

The Lebesgue measure on Rn has the following properties:

1. If A is a cartesian product of intervals I1 × I2 × ... × In, then A is Lebesgue measurable and ${\displaystyle \lambda (A)=|I_{1}|\cdot |I_{2}|\cdots |I_{n}|.}$ Here, |I| denotes the length of the interval I.
2. If A is a disjoint union of countably many disjoint Lebesgue measurable sets, then A is itself Lebesgue measurable and λ(A) is equal to the sum (or infinite series) of the measures of the involved measurable sets.
3. If A is Lebesgue measurable, then so is its complement.
4. λ(A) ≥ 0 for every Lebesgue measurable set A.
5. If A and B are Lebesgue measurable and A is a subset of B, then λ(A) ≤ λ(B). (A consequence of 2, 3 and 4.)
6. Countable unions and intersections of Lebesgue measurable sets are Lebesgue measurable. (Not a consequence of 2 and 3, because a family of sets that is closed under complements and disjoint countable unions need not be closed under countable unions: ${\displaystyle \{\emptyset ,\{1,2,3,4\},\{1,2\},\{3,4\},\{1,3\},\{2,4\}\}}$.)
7. If A is an open or closed subset of Rn (or even Borel set, see metric space), then A is Lebesgue measurable.
8. If A is a Lebesgue measurable set, then it is "approximately open" and "approximately closed" in the sense of Lebesgue measure (see the regularity theorem for Lebesgue measure).
9. A Lebesgue measurable set can be "squeezed" between a containing open set and a contained closed set. This property has been used as an alternative definition of Lebesgue measurability. More precisely, ${\displaystyle E\subset \mathbb {R} }$ is Lebesgue measurable if and only if for every ${\displaystyle \epsilon >0}$ there exist an open set ${\displaystyle G}$ and a closed set ${\displaystyle F}$ such that ${\displaystyle F\subset E\subset G}$ and ${\displaystyle \lambda (G\setminus F)<\epsilon }$.[6]
10. A Lebesgue-measurable set can be "squeezed" between a containing Gδ set and a contained Fσ set. I.e, if A is Lebesgue measurable then there exist a Gδ set G and an Fσ set F such that GAF and λ(G\A)=λ(A\F)=0.
11. Lebesgue measure is both locally finite and inner regular, and so it is a Radon measure.
12. Lebesgue measure is strictly positive on non-empty open sets, and so its support is the whole of Rn.
13. If A is a Lebesgue measurable set with λ(A) = 0 (a null set), then every subset of A is also a null set. A fortiori, every subset of A is measurable.
14. If A is Lebesgue measurable and x is an element of Rn, then the translation of A by x, defined by A + x = {a + x : aA}, is also Lebesgue measurable and has the same measure as A.
15. If A is Lebesgue measurable and ${\displaystyle \delta >0}$, then the dilation of ${\displaystyle A}$ by ${\displaystyle \delta }$ defined by ${\displaystyle \delta A=\{\delta x:x\in A\}}$ is also Lebesgue measurable and has measure ${\displaystyle \delta ^{n}\lambda \,(A).}$
16. More generally, if T is a linear transformation and A is a measurable subset of Rn, then T(A) is also Lebesgue measurable and has the measure ${\displaystyle |\det(T)|\,\lambda \,(A)}$.

All the above may be succinctly summarized as follows:

The Lebesgue measurable sets form a σ-algebra containing all products of intervals, and λ is the unique complete translation-invariant measure on that σ-algebra with ${\displaystyle \lambda ([0,1]\times [0,1]\times \cdots \times [0,1])=1.}$

The Lebesgue measure also has the property of being σ-finite.

## Null sets

Main article: Null set

A subset of Rn is a null set if, for every ε > 0, it can be covered with countably many products of n intervals whose total volume is at most ε. All countable sets are null sets.

If a subset of Rn has Hausdorff dimension less than n then it is a null set with respect to n-dimensional Lebesgue measure. Here Hausdorff dimension is relative to the Euclidean metric on Rn (or any metric Lipschitz[disambiguation needed] equivalent to it). On the other hand, a set may have topological dimension less than n and have positive n-dimensional Lebesgue measure. An example of this is the Smith–Volterra–Cantor set which has topological dimension 0 yet has positive 1-dimensional Lebesgue measure.

In order to show that a given set A is Lebesgue measurable, one usually tries to find a "nicer" set B which differs from A only by a null set (in the sense that the symmetric difference (AB) ${\displaystyle \cup }$(BA) is a null set) and then show that B can be generated using countable unions and intersections from open or closed sets.

## Construction of the Lebesgue measure

The modern construction of the Lebesgue measure is an application of Carathéodory's extension theorem. It proceeds as follows.

Fix nN. A box in Rn is a set of the form

${\displaystyle B=\prod _{i=1}^{n}[a_{i},b_{i}]\,,}$

where biai, and the product symbol here represents a Cartesian product. The volume of this box is defined to be

${\displaystyle \operatorname {vol} (B)=\prod _{i=1}^{n}(b_{i}-a_{i})\,.}$

For any subset A of Rn, we can define its outer measure λ*(A) by:

${\displaystyle \lambda ^{*}(A)=\inf \left\{\sum _{B\in {\mathcal {C}}}\operatorname {vol} (B):{\mathcal {C}}{\text{ is a countable collection of boxes whose union covers }}A\right\}.}$

We then define the set A to be Lebesgue measurable if for every subset S of Rn,

${\displaystyle \lambda ^{*}(S)=\lambda ^{*}(S\cap A)+\lambda ^{*}(S\setminus A)\,.}$

These Lebesgue measurable sets form a σ-algebra, and the Lebesgue measure is defined by λ(A) = λ*(A) for any Lebesgue measurable set A.

The existence of sets that are not Lebesgue measurable is a consequence of a certain set-theoretical axiom, the axiom of choice, which is independent from many of the conventional systems of axioms for set theory. The Vitali theorem, which follows from the axiom, states that there exist subsets of R that are not Lebesgue measurable. Assuming the axiom of choice, non-measurable sets with many surprising properties have been demonstrated, such as those of the Banach–Tarski paradox.

In 1970, Robert M. Solovay showed that the existence of sets that are not Lebesgue measurable is not provable within the framework of Zermelo–Fraenkel set theory in the absence of the axiom of choice (see Solovay's model).[7]

## Relation to other measures

The Borel measure agrees with the Lebesgue measure on those sets for which it is defined; however, there are many more Lebesgue-measurable sets than there are Borel measurable sets. The Borel measure is translation-invariant, but not complete.

The Haar measure can be defined on any locally compact group and is a generalization of the Lebesgue measure (Rn with addition is a locally compact group).

The Hausdorff measure is a generalization of the Lebesgue measure that is useful for measuring the subsets of Rn of lower dimensions than n, like submanifolds, for example, surfaces or curves in R³ and fractal sets. The Hausdorff measure is not to be confused with the notion of Hausdorff dimension.

It can be shown that there is no infinite-dimensional analogue of Lebesgue measure.


text:
What is wrong with the following equation?

Choose one of the following and enter the letter below:
(a) There is nothing wrong with this equation.
(b) This equation does not make sense at $x=2$.
(c) None of the above.


text:
# How do you write a slope-intercept equation for a line perpendicular to 3x + y - 2 = 0 with point (2,1)?

Aug 14, 2017

$y - 1 = \left(\frac{1}{3}\right) \left(x - 2\right)$
$y = \left(\frac{1}{3}\right) x + \frac{1}{3}$


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Volume 67 , issue 2 ( 2015 ) back
 An iterative approximation of fixed points of strictly pseudocontractive mappings in Banach spaces 79$-$91 Yekini Shehu

Abstract

We prove strong convergence of an iterative scheme for approximation of fixed point of $\lambda$-strict pseudocontractive mapping in a uniformly smooth real Banach space (which is not necessarily uniformly convex). We apply our result to approximation of common fixed point of a finite family of strictly pseudocontractive mappings. Our result extends the results of Li and Yao [M. Li, Y. Yao, Strong convergence of an iterative algorithm for $\lambda$-strictly pseudocontractive mappings in Hilbert spaces, An. St. Univ. Ovidius Constanta 18 (2010), 219-228] and complements other new interesting results in the literature.

Keywords: Strong convergence; strictly pseudocontractive mappings; uniformly smooth Banach spaces; uniformly convex Banach spaces.

MSC: 47H06, 47H09, 47J05, 47J25

 Generalized derivations as a generalization of Jordan homomorphisms acting on Lie ideals 92$-$101 Basudeb Dhara, Shervin Sahebi and Venus Rahmani

Abstract

Let $R$ be a prime ring with extended centroid $C$, $L$ a non-central Lie ideal of $R$ and $n\geq 1$ a fixed integer. If $R$ admits the generalized derivations $H$ and $G$ such that $H(u^2)^n=G(u)^{2n}$ for all $uın L$, then one of the following holds: {(1)} $H(x)=ax$ and $G(x)=bx$ for all $xın R$, with $a,bın C$ and $a^n=b^{2n}$; {(2)} char$(R)\neq 2$, $R$ satisfies $s_4$, $H(x)=ax+[p,x]$ and $G(x)=bx$ for all $xın R$, with $bın C$ and $a^n=b^{2n}$; {(3)} char$(R)=2$ and $R$ satisfies $s_4$. As an application we also obtain some range inclusion results of continuous generalized derivations on Banach algebras.

Keywords: Prime ring; generalized derivation; extended centroid; Utumi quotient ring; Banach algebra.

MSC: 16W25, 16N60, 16R50, 16D60

 The rainbow domination subdivision numbers of graph 102$-$114 N. Dehgardi, S. M. Sheikholeslami and L. Volkmann

Abstract

{$2$-rainbow dominating function} (2RDF) of a graph $G$ is a function $f$ from the vertex set $V(G)$ to the set of all subsets of the set $\{1,2\}$ such that for any vertex $vın V(G)$ with $f(v)=\emptyset$ the condition $\cup_{uın N(v)}f(u)=\{1,2\}$ is fulfilled. The {weight} of a 2RDF $f$ is the value $\omega(f)=\Sigma_{vın V}|f (v)|$. The {$2$-rainbow domination number} of a graph $G$, denoted by $\gamma_{r2}(G)$, is the minimum weight of a 2RDF of G. The {$2$-rainbow domination subdivision number} $\tx{\rm sd}_{\gamma_{r2}}(G)$ is the minimum number of edges that must be subdivided (each edge in $G$ can be subdivided at most once) in order to increase the $2$-rainbow domination number. In this paper, we initiate the study of $2$-rainbow domination subdivision number in graphs.

Keywords: domination number; $2$-rainbow domination number; $2$-rainbow domination subdivision number.

MSC: 05C69

 Common fixed point theorems for expansive mappings satisfying an implicit relation 115$-$122 Abdelkrim Aliouche and Ahcene Djoudi

Abstract

We prove common fixed point theorems in metric spaces for expansive mappings satisfying an implicit relation without non-decreasing assumption and surjectivity using the concept of weak compatibility which generalize some theorems appearing in the recent literature.

Keywords: Weakly compatible mappings; common fixed point; metric space.

MSC: 47H10, 54H25

 Faber polynomial coefficient estimates for analytic bi-Bazilevič functions 123$-$129 Jay M. Jahangiri and Samaneh G. Hamidi

Abstract

A function is said to be bi-univalent in the open unit disk $\Bbb{D}$ if both the function and its inverse are univalent in $\Bbb{D}$. By the same token, a function is said to be bi-Bazilevič in $\Bbb{D}$ if both the function and its inverse are Bazilevič there. The behavior of these types of functions are unpredictable and not much is known about their coefficients. In this paper we use the Faber polynomial expansions to find upper bounds for the coefficients of classes of bi-Bazilevič functions. The coefficients bounds presented in this paper are better than those so far appeared in the literature. The technique used in this paper is also new and we hope that this will trigger further interest in applying our approach to other related problems.

Keywords: Faber polynomials; bi-Bazilevič functions; univalent functions.

MSC: 30C45, 30C50

 Proximity structures and ideals 130$-$142 A. Kandil, S. A. El-Sheikh, M. M. Yakout and Shawqi A. Hazza

Abstract

In this paper, we present a new approach to proximity structures based on the recognition of many of the entities important in the theory of ideals. So, we give a characterization of the basic proximity using ideals. Also, we introduce the concept of $g$-proximities and we show that for different choice of $g$'' one can obtain many of the known types of generalized proximities. Also, characterizations of some types of these proximities -- ($g_0,h_0$) -- are obtained.

Keywords: ideals; basic proximity; proximity space; g-proximities; nearness; topological space.

MSC: 54E05

 Generalized relative lower order of entire functions 143$-$154 Sanjib Kumar Datta, Tanmay Biswas and Chinmay Biswas

Abstract

The basic properties of the generalized relative lower order of entire functions are discussed in this paper. In fact, we improve here some results of Datta, Biswas and Biswas [Casp. J. Appl. Math. Ecol. Econ., {\bf1}, 2 (2013), 3--18].

Keywords: Entire function; generalized relative lower order; Property~(A).

MSC: 30D20, 30D30, 30D35


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# A brief look at how some groups we’ve supported are responding to COVID-19

Organizations supporting and delivering public health and poverty alleviation programs have been impacted by the COVID-19 pandemic in many ways. Here, we provide a brief look at how some of the groups we’ve supported are responding to the pandemic.

We share an example of a charity that continues to implement its health program, with modifications for social distancing and safety; charities that have paused their programs, but continue supporting their staff; and a charity that is allocating funding to pandemic-response efforts.

This is neither a comprehensive update nor a static one. The situation is evolving rapidly. We will share a more complete update on what the pandemic means for our top charities and their funding needs closer to the end of the year. We continue to recommend that donors give to “Grants to recommended charities at GiveWell’s discretion,” which we allocate among our recommended charities where we see the greatest need.

What follows is a snapshot of what’s happening now.

## Examples of how charities are responding

A charity that is continuing its program, but with modifications: Malaria Consortium’s seasonal malaria chemoprevention program

Seasonal malaria chemoprevention (SMC) is the delivery of anti-malarial medication to children under age five during the time of year when malaria transmission is highest.[1] The World Health Organization (WHO) recommends that programs to prevent malaria, including SMC, continue during the pandemic.[2]

Malaria Consortium is a GiveWell top charity that supports SMC campaigns (primarily door-to-door) in the Sahel region of Africa.[3] We expect Malaria Consortium’s next SMC distributions to take place during the upcoming July-to-October rainy season.[4]

GiveWell-directed funding to Malaria Consortium’s SMC program is flexible. We don’t restrict its use to specific activities. That means Malaria Consortium can use GiveWell-directed funds to quickly adapt its distribution model to prevent the spread of COVID-19.

Malaria Consortium will aim to provide personal protective equipment, including masks and gloves, to its community SMC distributors. It also plans to provide alcohol-based hand sanitizers, disinfecting wipes, soap, bio-waste bags, and additional T-shirts and hijabs.[5] Malaria Consortium has instructed community distributors to maintain two meters of distance in their interactions and to share information on how to prevent COVID-19 while visiting each household.[6]

Image from Malaria Consortium’s visual aid for community distributors. SPAQ (sulfadoxine-pyrimethamine and amodiaquine) is medication to prevent malaria.

Overall, Malaria Consortium expects slightly higher costs for its campaigns as a result of these changes.[7] We expect these additional costs to amount to a modest increase in the cost per child treated. In our projections of Malaria Consortium’s 2020 spending, we use an assumption of 20% higher costs due to program modifications.[8] This reduces our estimate of the cost-effectiveness of past donations to Malaria Consortium that are now being used for the 2020 SMC campaigns. Even with that adjustment, we still consider these past donations to be highly cost-effective, given our assessment of Malaria Consortium’s cost-effectiveness overall.

We expect that donations made to Malaria Consortium today will support its SMC campaigns in 2022, at which point we hope that these additional measures are no longer necessary. Our estimate of the cost-effectiveness of donations made to Malaria Consortium today thus remains high at $2,000 to$3,000 per death averted.[9]

Charities that are pausing programs and retaining staff: organizations that treat parasitic worm infections (deworming)

Added December 2020: While deworming programs were paused at the time this post was published (July 16, 2020), the World Health Organization updated its guidance shortly thereafter to advise considering the costs and benefits to decide whether to do neglected tropical disease mass drug administrations (including deworming) during COVID-19 (source, p. 1; archive). As of December 2020, we understand that all of our deworming top charities have resumed either undertaking deworming distributions or planning for upcoming distributions.

Our top charities list includes four organizations that support deworming programs: Evidence Action’s Deworm the World Initiative, SCI Foundation, Sightsavers, and the END Fund. The WHO has recommended pausing deworming programs to mitigate the spread of COVID-19.[10] Our top charities are not distributing deworming treatments at this time.

As with Malaria Consortium, the donations we’ve directed to the deworming organizations we recommend have been allocated flexibly, so they may support operations as well as delivery of deworming treatments. We understand that our top charities plan to retain most staff using this and other funding. Maintaining stability in uncertain times benefits staff. It will also enable a quick and smooth transition back to providing deworming treatments once campaigns restart, as hiring and training new staff can be difficult and time-consuming.

We’re unclear how long the pause will last and thus how disruptive it will be. The hope is that it will be short-lived.

Deworming is done on an annual basis in many locations, or at most twice per year.[11] Many rounds of deworming are scheduled for later in the year and so have not yet been missed. It’s possible that in many of these cases, deworming will take place as scheduled.[12]

If the pause persists beyond the end of the year, it will impact our estimated cost-effectiveness of deworming. We estimate that the maintenance of staff costs in the absence of program implementation during a 12-month interruption of deworming would lead to a roughly 10-25% decrease in the cost-effectiveness for our deworming top charities over three years (our typical deworming funding period).[13] This estimate relies upon a number of simplifying assumptions and judgment calls; more information is in the following footnote.[14] We believe that deworming would remain a cost-effective intervention overall.

A charity that is allocating funds to respond to COVID-19: Abdul Latif Jameel Poverty Action Lab’s Innovation in Government Initiative

In December 2018, we recommended a grant of $1 million to the Innovation in Government Initiative (IGI), a grantmaking entity within the Abdul Latif Jameel Poverty Action Lab (J-PAL) focused on scaling evidence-based policies in collaboration with low- and middle-income country governments.[15] This grant was part of a new area of our work to explore approaches to improve government policies.[16] Our grant supported IGI’s general operations and two requests for proposals (RFPs). The RFPs sought proposals primarily for technical assistance to governments to scale up evidence-based policies.[17] In March 2020, IGI reached out to us about allocating approximately$150,000 of the remaining funding from the GiveWell-directed grant to COVID-19 response. Specifically, IGI was interested in funding off-cycle, rapid-response grants to mitigate the impact of COVID-19 in low- and middle-income countries by supporting technical assistance and analysis for governments to inform COVID-19 policy responses in areas like health and social protection.[18] This goal is in line with the original grant purpose, and we approved. A list of COVID-19 projects IGI has funded so far can be found here.

This update highlights the importance of allocating funding flexibly so that grantees can respond to unforeseen events. But it doesn’t address all of the work our top charities and other groups we support are doing in response to the pandemic. GiveWell standout charity Development Media International is supporting public health broadcasts to promote behaviors to mitigate the spread of the pandemic. We’ve provided grant funding to expand this work. GiveDirectly has launched COVID-19 cash transfer programs in the United States, Kenya, and Uganda, and it plans to expand its COVID-19 response into five more countries.[19] [July 22 update: GiveDirectly informed us that it is now also running COVID-19 response programs in Liberia, Malawi, and Rwanda.]

Other charities we support have adjusted their operations in large and small ways. We plan to share a more detailed look at the pandemic’s impact on charities we support later in the year.

## References

Sources and footnotes for this post may be found here.

• ofhsoupkitchen on July 22, 2020 at 9:52 am said:

These are unprecedented times, and we are all learning as we go along. Here’s what I learned from this blog: that there is indeed a way for nonprofits to offer support and to seek it at the same time. I’ve since come to see a few other organizations set a similar tone. One statewide mental health organization I know compiled and shared a list of COVID-19 resources for the benefit of its traditional clientele, certainly, but also for donors and the community at large. A performing-arts center wrote to me and its other members, asking after our welfare, expressing sadness at not being able to share artistic experiences together, recounting past memorable performances, and only then gently asking “those of you who can” to consider a gift to help the organization through this fraught time.

• NICO CHAVEZ on November 8, 2020 at 8:29 pm said:

Buenas noches, consulta como puedo ver las actividades que han realizado, el tema es, me ha llegado una invitación para trabajar con ustedes, quiero estar seguro, gracias.

• Erin Wolff (GiveWell) on November 10, 2020 at 12:41 pm said:

Hola Nico,

Si recibió un correo electrónico sobre una donación de 2 millones de dólares, esto es un intento de fraude. La persona que le contacto no tiene ninguna afiliación con GiveWell. Le sugerimos que por favor ignore este mensaje y que no comparta ningún tipo de información personal.


text:
Lemma in an email of Matthew Emerton dated June 15, 2016

Lemma 92.9.11. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces and $\mathcal{Y}$ is a stack in groupoids, then $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ is a stack in groupoids.

Proof. The property of a morphism being representable by algebraic spaces is preserved under base-change (Lemma 92.9.8), and so, passing to the base-change $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ over $\mathcal{Y}$, we may reduce to the case of a morphism of categories fibred in groupoids $\mathcal{X} \to \mathcal{Y}$ which is representable by algebraic spaces, and whose target is a stack in groupoids; our goal is then to prove that $\mathcal{X}$ is also a stack in groupoids. This follows from Stacks, Lemma 8.6.11 whose assumptions are satisfied as a result of Lemma 92.9.2. $\square$

There are also:

• 2 comment(s) on Section 92.9: Morphisms representable by algebraic spaces

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Chapter 64

A 65-year-old man with a history of coronary artery disease, high cholesterol, type 2 diabetes, and hypertension presents with a question about a dietary supplement. He is in good health, exercises regularly, and eats a low-fat, low-salt diet. His most recent laboratory values show that his low-density lipoprotein (LDL) cholesterol is still slightly above goal at 120 mg/dL (goal < 100 mg/dL) and his hemoglobin A1C is well controlled at 6%. His blood pressure is also well controlled. His medications include simvastatin, metformin, benazepril, and aspirin. He also regularly takes a vitamin B complex supplement and coenzyme Q10. He asks you if taking a garlic supplement could help to bring his LDL cholesterol down to less than 100 mg/dL. What are two rationales for why he might be using a coenzyme Q10 supplement? Are there any supplements that could increase bleeding risk if taken with aspirin?

*The industry marketing these materials is replacing the terms "herbal medication" and "botanical medication" with the term "dietary supplement" in order to avoid legal liability and added governmental regulation. For the purposes of this chapter, they are identical.

The medical use of plants in their natural and unprocessed form undoubtedly began when the first intelligent animals noticed that certain food plants altered particular body functions. While there is a great deal of historical information about the use of plant-based supplements, there is also much unreliable information from poorly designed clinical studies that do not account for randomization errors, confounders, and—most importantly—a placebo effect that can contribute 30–50% of the observed response. Since the literature surrounding dietary supplements is evolving and much of it is not peer-reviewed, it is recommended that reputable evidence-based resources be used to help guide treatment decisions. An unbiased and regularly updated compendium of basic and clinical reports regarding botanicals is Pharmacists Letter/Prescribers Letter Natural Medicines Comprehensive Database (see references). Another evidence-based resource is Natural Standard, which includes an international, multi-disciplinary collaborative website, http://www.naturalstandard.com. Unfortunately, the evidence available to these objective and unbiased evaluators is rarely adequate to permit clear conclusions. As a result, all statements regarding positive benefits should be regarded as preliminary and even conclusions regarding safety should be considered tentative at this time.

For legal purposes, "dietary supplements" are distinguished from "prescription drugs" derived from plants (morphine, digitalis, atropine, etc) by virtue of being available without a prescription and, unlike "over-the-counter medications," are legally considered dietary supplements rather than drugs. This distinction eliminates the need for proof of efficacy and safety prior to marketing and also places the burden of proof on the Food and Drug Administration (FDA) to prove that a supplement is harmful before its use can be restricted or removed from the market. Although manufacturers are prohibited from marketing unsafe or ineffective products, the FDA has met significant challenges from the supplement industry largely due to the strong lobbying effort by supplement manufacturers and the variability in interpretation of the Dietary Supplement Health and ...

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# Nouns and Verbs

I’ve been quoted as saying that there is no such thing as “a homosexual”.  And, frankly, I’m quoting that from Dr. Joe Nicolosi and Richard Cohen.  But much ado was made about “Such were some of you” at the Restored Hope Network Conference this summer.  They’re quoting a verse which says “Do not be deceived, neither fornicators, nor idolaters, nor adulterers, nor effeminate, nor homosexuals, nor thieves, nor the covetous, nor drunkards, nor revilers, nor swindlers, will inherit the kingdom of God.  Such were some of you;” “but God.”  I’m going to assume Christians will know what the “but God” part stands for.  It would appear that “the Bible says there’s such a thing as a homosexual”.  But our modern English definition of “a homosexual” as a noun and the Greek language’s definition of “a homosexual” as a noun are different.  Let’s look at all of these words:

fornicator: a man who committed fornication

idolater: a man who idolized something other than God

adulterer: a man who had an extra-marital erotic sexual relations

effeminate: a man who submitted for prostitution

homosexual: a man who committed erotic activities with other men

thief: a man who stole something from another man

the covetous: a man who coveted a belonging of another man

drunkard: a man who drank to the point of intoxication

reviler: a man who insulted another man angrily

swindler: a man who cheated another man

See what they all have in common?  “a man” followed by a verb.  The noun part of the definitions are all “a man”.  Now, truthfully, I’ve tweaked the list.  Most of these can be “a human who”, but in the case of the “effeminate” it actually is specifically men, so I just adapted them all so it’s easier to see the distinction:  Is the verb part a noun?  Are our verbs what we are?  Or are our being part, the noun part, the ‘man’ part what we are? and the rest what we do?  I tend to think of it the second way.  Especially, for men with pre-oedipal disorder, their homosexuality is mainly caused by a deficiency of feeling what they are.  Teaching men who have been wounded in this way, that it’s what they are is counterproductive.


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ScaleSoftLayer_Scale_Network_Vlan

Overview

A scale network VLAN is a VLAN that scaled members will be placed on.

Methods

Create a network VLAN for a scale group. Once created, the VLAN will be used to scale with.
Delete this network VLAN reference. Note, this does not affect existing scaled members. Once deleted however, future scaled members will not be placed on this referenced VLAN.
Retrieve the network VLAN to scale with.
Retrieve a SoftLayer_Scale_Network_Vlan record.
Retrieve the group this network VLAN is for.


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# BC_1753_EMONTAGU_GW

<Q A 1753? FO GW EMONTAGU>
<X ELIZABETH MONTAGU>
[}ELIZABETH MONTAGU TO GILBERT WEST. 1753? NOVEMBER 1. bMS ENG 1365 (157)}]
<P1>
Nov=ber= ye 1=st= [\ADDED 1753\] My most honor'd Cousin
Most truly hath the wise man said there is a time to rejoyce & a time to mourn, but we shortsighted mortals know not always the proper Seasons, thus was my excellent Cousin rejoycing in my recovery at the very time I was become ill again: but still his joy is not thrown away, & is a cordial of such virtue as to restablish my new returning health. Last saturday noon soon after my return from taking the air in the post chaise I was seized with a nervous tremor & great disorder, upon which I went to bed, in about two hours the agitation abated, & left no complaint but weakness & languor, & such a degree of fever as generally succeeds to these violent [\WORD INTO attacks\] . I have been very carefull of myself, have kept my person within the [\WORD INTO limits\] of [\my/] dressing room, my mind [\but\] have only Gently recreated in the fields of fancy. I order'd my apothecary to send from Newbury along with some Nervous medicines, Homers Odyssey, imagining my confinem=t=
<P2>
And solitude would be well [\inliven'd INTO enliven'd BY WHOM?\] by the adventures of the Man who, wandering from clime to clime, observant stray Their manners noted & their states survey'd. I was the more desirous to read them as two poets for whom I have great [\honor\] prefer this tunefull bard as a teacher of morals to the severer philosophers, for whose precepts the mind enervated by sickness is too weak; and indeed this has answerd as all must do which is undertaken by so good direction, and I have passd my hours in no unpleasing indolence, lolling in my great chair with the [\oddyssey INTO odyssey BY WHOM?\] in my hand. [\but INTO But\] unfortunate is the mind that is too delicate to bear the voice of [\wise DELETED\] [\wisedom/] [\widedom ORIGINALLY?\] unless attempted by Apollo's Lyre & must have an example of virtue & patience beautified with all the coloring of fancy & soften'd with the [\skelf\] touch of art [\WORDS INTO to make it\] a fit object for their tender sight. Ulysses refuses to listen to the Syrens tho they promise him instruction, so wary is wisedom, when the voice is too soothing to the ear, but he had a soul that could not [\be/] immersed in the waves of affliction, mine not able to buffet the stream, must borrow even the [\bladder\] of vanity to bear it up. I have given you this notable
<P3>
Account of my amusements to shew you [\first that/] I have not been not been very ill, secondly that there is no time nor occasion in which your friendship does not profit me, & thirdly & lastly, because I had nothing else to say & was not wise enough to take that hint to be silent. You are now a man of pleasure, visiting Ladies, feasting in princely bowers, & living (I was going to say being full of the odyssey) like Penelopes Suitors rioting at the expence of a widow'd dame, consuming the fruits of the earth, & soothing your cares with the music of her Ladyships harpsicord; but when you return to Wickham, to your usefull books & serious muse, I must grow more discreet, & not break in upon your sacred hours with letters of such intolerable length. I am very glad of the account you give of M=r= Pitt during his stay at Hagley, I hope the Bath waters will quite set him right either with or without the gout, as to the second part of your story it pleases me not so well, for I do not think we shall come to Town in november, certainly we shall not, unless there be something of more than ordinary business going forward in the house of commons, you will
<P4>
Say perhaps I may be ill again! why if that should happen I must take more nervous medicine & read the Iliad I think. Shaw or his great Sire [\OEsculapius\] could do me no service; I have had the same disorder so often in my life I know how to manage it as well as the learned faculty; warmth & quiet are the best remedies, the [\WORD INTO woods\] offer both. I have quite left off my evening airings, & shall no more like [\WORD UNCLEAR\] drive my chariot thro the realms of night till sultry days recommend the evening hour. I am very glad you have got safe to Stoke, I was not without anxiety for you while at such a distance from home at this season of the year. I hope you will enjoy good health this winter after all [\the INTO your?\] exercise you have had, which must certainly be very beneficial. as soom as Xmass is over I promise myself the happiness of seeing you, till then, I can have no [\greater\] pleasure than hearing you are well. Your (\amour\) with a certain person goes on well? What to Fulham again? Smiles from reason flow, & are of love the food. but really you love as if every month were may
<P5>
To quit your habitation [\at/] this time of the year to go to that damp cold place. I have been reading Ariosto, & I can find nothing [\in/] Orlando furioso, or (\non furioso\), that equals it. You are very good in thinking of poor M=r= Botham. how many would think themselves unhappy if they had not money to adorn their gardens & parks, to have collections or rare plants & fine aromatic shrubs. [\how INTO How BY WHOM?\] much more may he reasonably desire to have what is necessary for the education of his pretty Babes! to cultivate their [\UNCLEAR\], perdure their talents, & give them the noble virtues that are rarely found but in persons of liberal education & competent circumstances! when once the virtues have taken root they grow perhaps the faster fixd for being shaken by the storms of fortune, but while they are young & tender, the cold blast hinders their shoot, & kills the opening bud. My sweet little Harry is laying his careless head upon my lap at this instant, nothing sensible of your generous or my anxious care for his future wellbeing, happy season
<P6>
When neither hopes nor fears dance before the mental eye [\diverting\] its alteration from present objects. I do not expect M=r= Botham here before tuesday, I will certainly tell him all the kind & friendly things you say, he is a truly [\pious\] & virtuous man, & we [\hope\] of the righteous shall not fail but when or where they shall be answerd we know not, it ought to satisfy us that we know the time will come. M=r= Montagu desires his best comp=ts= to you & M=rs= West my most affectionate respects attend
her.
I am
My most esteem'd Cousins
very faithfull & affection=te= Friend
and Most Obliged H=ble= Serv=t=
Eliza: Montagu.


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# Math equation could help find missing Malaysian plane

Bayes’ Theorem helped researchers locate Air France Flight 447’s black box in 2011

(Video: How Bayesian Search found the USS Scorpion)

Days after a Malaysian airliner with 239 people aboard went missing en route to Beijing, searchers are still struggling to find any confirmed sign of the plane. Authorities have acknowledged that they didn’t even know what direction it was heading when it disappeared.

As frustrations mount over the failures of the latest technology in the hunt for Malaysia Airlines Flight MH370, some scientists say an 18th-century mathematical equation – used in a previous search for an Air France jetliner’s black box recorder – could help pinpoint the location of the Malaysian plane.

Indonesian Air Force officers examine a map of the Malacca Strait during a briefing following a search operation for the missing Malaysia Airlines Boeing 777, at Suwondo air base in North Sumatra, Indonesia, on Wednesday.AP2014

In 2009, Air France Flight 447 en route to Paris from Rio de Janeiro vanished over the Atlantic Ocean, triggering the most expensive and exhaustive search effort ever conducted for a plane. After two years, officials could only narrow the location of the plane’s black box down to an area the size of Switzerland.

But Flight 447’s black box was found in just five days after authorities contacted scientific consultants who applied a centuries-old equation called Bayes’ Theorem.

## What is Bayes’ Theorem

Mathematically, Bayes’ theorem gives the relationship between the probabilities of A and B, P(A) and P(B), and the conditional probabilities of A given B and B given A, P(A|B) and P(B|A). In its most common form, it is: (Wikipedia)

$\displaystyle\boxed{P(A|B)=\frac{P(B|A)P(A)}{P(B)}}$

## Proof of Bayes’ theorem (Theorem useful for finding MH370 plane)

The proof of Bayes’ theorem is actually relatively simple, the only requirement is to know the formula for conditional probability (Learnt in H1/H2 Maths): $\displaystyle \boxed{P(A|B)=\frac{P(A\cap B)}{P(B)}}$

From this, we have $\displaystyle \boxed{P(A\cap B)=P(A|B)P(B)}$

Similarly, $\displaystyle \boxed{P(B\cap A)=P(B|A)P(A)}$

But since $\displaystyle P(A\cap B)=P(B\cap A)$, we have $P(A|B)P(B)=P(B|A)P(A)$. Dividing throughout by $P(B)$ gives Bayes’ Formula: $\displaystyle\boxed{P(A|B)=\frac{P(B|A)P(A)}{P(B)}}$

Sincerely wishing that the MH370 plane will be found soon, and hopefully the passengers are still alive.

Also see: Bayesian search theory (Bayesian search theory is the application of Bayesian statistics to the search for lost objects. It has been used several times to find lost sea vessels, for example the USS Scorpion. It also played a key role in the recovery of the flight recorders in the Air France Flight 447 disaster of 2009.)


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# zbMATH — the first resource for mathematics

An LMI approach to robust $$H_\infty$$ filtering for uncertain systems with time-varying distributed delays. (English) Zbl 1201.93126
Summary: The problem of robust $$H_\infty$$ filtering for uncertain systems with time-varying distributed delays is considered. The uncertainties under discussion are time varying but norm bounded. Based on the Lyapunov stability theory, a sufficient condition for the existence of full order $$H_\infty$$ filters is proposed by a Linear Matrix Inequality (LMI) approach such that the filtering error system is asymptotically stable and satisfies a prescribed attenuation level of noise. A numerical example is given to demonstrate the availability of the proposed method.

##### MSC:
 93E11 Filtering in stochastic control theory 93B36 $$H^\infty$$-control 93D05 Lyapunov and other classical stabilities (Lagrange, Poisson, $$L^p, l^p$$, etc.) in control theory
##### Keywords:
uncertain; delay; robust $$H_\infty$$ filtering; LMI
LMI toolbox
Full Text:
##### References:
 [1] Chen, J.D., Delay-dependent robust H∞ control of uncertain neutral systems with state and input delays: LMI optimization approach, Chaos solitons fractals, 33, 595-606, (2007) · Zbl 1136.93018 [2] Fridman, E., New lyapunov – krasovskii functionals for stability of linear retarded and neutral type systems, Syst. control lett., 43, 309-319, (2001) · Zbl 0974.93028 [3] Hu, G.D., Some simple stability criteria of neutral delay-differential systems, Appl. math. comput., 80, 257-271, (1996) · Zbl 0878.34063 [4] Zhao, Z.; Wang, W.; Yang, B., Delay and its time-derivative dependent robust stability of neutral control system, Appl. math. comput., 187, 1326-1332, (2007) · Zbl 1114.93076 [5] Chen, W.H.; Zheng, W.X., Delay-dependent robust stabilization for uncertain neutral systems with distributed delays, Automatica, 43, 95-104, (2007) · Zbl 1140.93466 [6] Liu, X.G.; Wu, M.; Martin, R.; Tang, T.L., Delay-dependent stability analysis for uncertain neutral systems with time-varying delays, Math. comput. simulation, 75, 15-27, (2007) · Zbl 1128.34048 [7] Yue, D.; Han, Q.L., Robust H∞ filter design of uncertain descriptor systems with discrete and distributed delays, IEEE trans. signal process., 52, 3200-3212, (2004) · Zbl 1370.93111 [8] Xu, S.; Chen, T., An LMI approach to the H∞ filter design for uncertain systems with distributed delays, IEEE trans. circuits systems-II: express briefs, 51, 195-201, (2004) [9] Xu, S.; Lam, J.; Chen, T.; Zou, Y., A delay-dependent approach to robust H∞ filtering for uncertain distributed delay systems, IEEE trans. signal process., 53, 3764-3772, (2005) · Zbl 1370.93109 [10] Gahinet, P.; Nemirovski, A.; Laub, A.; Chialali, M., LMI control toolbox User’s guide, (1995), The Mathworks Massachusetts [11] Zhang, X.-M.; Han, Q.-L., Robust H∞ filtering for a class of uncertain linear systems with time-varying delay, Automatica, 44, 157-166, (2008) · Zbl 1138.93058 [12] Hale, J.; Verduyn Lunel, S.M., Introduction to functional differential equations, (1993), Springer New York · Zbl 0787.34002 [13] Xu, S.; Lam, J., Improved delay-dependent stability criteria for time-delay systems, IEEE trans. automat. control, 50, 384-387, (2005) · Zbl 1365.93376 [14] Xu, S.; Lam, J., On equivalence and efficiency of certain stability criteria for time-delay systems, IEEE trans. automat. control, 52, 95-101, (2007) · Zbl 1366.93451 [15] Fridman, E.; Shaked, U., A new H∞ filter design for linear time delay systems, IEEE trans. signal process., 49, 2839-2843, (2001) · Zbl 1369.93622
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.