Mathematicianadda

Areas of Regular Polygons Worksheet

from Math Blog http://bit.ly/2J6cNz8

Practice Questions on Similar Triangles

from Math Blog http://bit.ly/2PFAS10

Areas of Regular Polygons

from Math Blog http://bit.ly/2V6OCYp

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from math http://bit.ly/2UUlw9H

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from math http://bit.ly/2vu2uwL

Suppose I have smooth functions $f,g,y_0$ and $y_1$ from $\mathbb{R}$ to $\mathbb{R}$, such that $$y_1(x) = y_0(x) - \epsilon g(y_0(x))$$ Then I consider $$f(y_0(x)) = f(y_1(x) + \epsilon g(y_0(x)))$$ Is there a closed form expression for the Taylor series in the small parameter $\epsilon$ in terms of derivatives of $f$ and $g$ and only the function $y_1$?

The first few terms are

$$f(y_0) = f(y_1) + \epsilon f'(y_1) g(y_0) + \frac{1}{2}\epsilon^2 f''(y_1)g^2(y_0) +..$$ Where we interpret $f(y_0)$ as $f|_{y_0(x)}$ and treat $x$ fixed. Then we can again replace the $y_0$ in $g(y_0)$ with $$g(y_0) = g(y_1)+ \epsilon g'(y_1)g(y_0) +...$$ giving $$= f(y_1) + \epsilon f'(y_1) [g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]$$ $$+ \frac{1}{2}\epsilon^2 f''(y_1)[g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]^2 +...$$ Continuing to replace the $y_0$ with $g(y_0)$ like this and grouping terms gives $$f(y_0) = f(y_1)+ \epsilon [f'g](y_1) + \epsilon^2[f'g'g + \frac{1}{2}f''g^2](y_1) + \epsilon^3[f'g'^2g + \frac{1}{2}f'g''g + \frac{1}{2}f''g'g + \frac{1}{6}f'''g^3](y_1) + O(\epsilon^4)$$ But is there some way to write this as a more compact sum like $$f(y_0) \sim f(y_1) + g(y_1)\sum_{n=1}^\infty\sum_{m=0}^n \epsilon^n \alpha(n,m)f^{(n)}g^{(n-m)}(y_1)$$ I am having trouble identifying the pattern. I know there will be some product involved as well.

Edit:

Thinking about it some more it may suffice to just set $f=id$ and consider $$y_0 = y_1 + \epsilon g(y_0)$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_0))$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_0)))$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_1 + ...)))$$ and somehow use the chain rule $$[f_1\circ f_2 \circ .... \circ f_n]' = \prod_{i=1}^n(f'_{i}\circ f_{i+1}\circ ...\circ f_n)$$

from Hot Weekly Questions - Mathematics Stack Exchange

Upon reviewing the basic theorem that the number of elements in a fixed finite set is unique, I tried to determine what part of this proposition is in need of proof. It seems axiomatic. Nonetheless, BBFSK have a very long-winded, and seemingly convoluted discussion of this, and related ideas.

When I attempted to produce my own argument in support of the above proposition, the part which I am not able to state purely in the terminology of mappings (bijection, injection, etc) is that an injection of a finite set into itself is a mapping onto itself. The proof BBFSK give uses induction. After thinking about it for a while, that was the only approach I could come up with.

Is there a rigorous proof of the proposition that every injective mapping of a finite set into itself is a mapping of the set onto itself which does not involve induction?

from Hot Weekly Questions - Mathematics Stack Exchange

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from math http://bit.ly/2GWqGi1

In propositional logic, one could not correctly say that : $(A \& B) = (B \& A)$.

The reason is syntactic: the first conjunct of $(A \& B)$ is $A$, while the first conjunct of $(B \& A)$ is $B$. So the two formulas are not identical, they are not the same formula.

The only thing one can say is that the two formulas are equivalent.

My question is: does this syntactic argument also hold for fractions?

Remark. I write this question after having watched a video by Herbert Gross where he expresses his reluctancy to call two fractions like $1/5$ and $2/10$" equal". According to Gross, they would be better called "equivalent" inasmuch as they " name the same number"

Remark. I do not ask whether the equivalence relation between fractions is the same as the logical equivalence relation " formula X is true in exactly the same interpretations as Y ". My question is not :

does " 1/5 = 2/10 mean 1/5 <=> 2/10" ?

I simply ask whether the equal sign between fractions should be read as some sort of arithmetical equivalence ( not a logical one of course).

from Hot Weekly Questions - Mathematics Stack Exchange

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from math http://bit.ly/2vvjeny

Can we divide $\mathbb{R}^2$ into two connected parts such that each part is not simply-connected?

My attempt

Put $A= \{ (0,0) \} $ and $B$ is the punctured plane.

Since that $S^1$ is a deformation retract of the punctured plane, $B$ is not simply-connected. Thus we can find a division of $\mathbb{R}^2$ such that one part is simply-connected but the other is not.

But how to deal with the problem above which requires that each part is not simply-connected?

It seems to be related to contractible and holes. But I don't know how to convert these ideas into precise mathematical language.

Any hints? Thanks in advance!

Added:

As pointed out in the comment, the counterexample exists.

Now I want to ask another question

Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?

from Hot Weekly Questions - Mathematics Stack Exchange

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from math http://bit.ly/2UPKUxt

This spring, I was asked by the Santa Ana Unified School District (SAUSD) School Board president, Valerie Amezcua, to be “Principal for a Day.” This is an experience where local business leaders and other community members are invited to visit a school, tour classrooms, meet students and teachers, and learn about the great things being done in education.

I arrived at Jackson Elementary with a smile on the morning of March 22. It was already a special day because I was visiting one of the very same schools I’d attended as a child! I was greeted by the office team, Ms. Maria Benoun and Ms. Graciela Rodriguez, and immediately made to feel important. I had reserved parking, a main seat in the principal’s office, and was assigned to greet school families.

Principal Cervantes reporting for duty!

A Chance Encounter

When I met the principal I’d be shadowing for the day, I was shocked! There I stood, face-to-face with a familiar Mr. Norris Perez. You see, my first job after graduating high school was as a paraprofessional within SAUSD, and I was tasked to support a new teacher: Mr. Perez. Twenty years later, I was spending my day with him in his first year as principal at Jackson Elementary. We were both a little surprised!

 

I still can’t believe that I got to shadow Principal Perez for the day! What are the odds?

One of the first things we did was participate in a twenty-minute staff meeting. The faculty does this on a daily basis to prep for the day ahead. I was impressed by how productive it was and how staff made sure to stick to twenty minutes. Immediately after, I attended their Friday school-wide assembly.

I was honored to be able to give a brief speech. When I said who my boss was, all the kids went wild! They immediately asked, “Is JiJi a boy or a girl?” Students started coming up to me, proudly sharing their ST Math® progress and asking for my support if they were stuck on a challenging puzzle.

“Is JiJi really your boss?” a little girl asked.

“Well, yes,” I responded.

“So does JiJi only talk to adults in real life?” she asked. “Because JiJi doesn’t talk in the games and JiJi needs to talk to you in order to tell you what to do at work.” Kids are so smart and observant!

An Emphasis on Academic Success

I then toured the classrooms, from transitional kindergarten through fifth grade, and met with teachers and students. During my tour, I quickly noticed a strong presence of colleges represented by shirts, banners and signs in classrooms. It turns out, each class represented a college campus. I also learned that students refer to themselves as scholars. It was apparent that the Jackson Elementary community strongly values academics.

Look at these scholars’ University of Washington pride!
I wonder if this will motivate many of them to apply there?

Jackson Elementary has been an ST Math school since 2008 through the Orange County Math Initiative and the JiJi culture there is strong. One of the classes I walked into had just finished working on dioramas and a few students had decided to create penguin habitats. I also heard stories about scholars spending their lunch or after-school time playing ST Math because they really wanted to get to 100% progress. In every class I visited, there were at least three students who had already achieved this goal!

Looks like JiJi has inspired all kinds of penguin-themed projects at this school.

A Well-Rounded Student Body

My tour then included a visit to the school’s $100,000 fitness center that they received as the grand prize winner of the 2010 Governor's Fitness Challenge, and to their mini-pitch soccer field, courtesy of LA Galaxy. Both of these gifts are part of a state-wide initiative for healthy eating and living. The school utilizes exercise as a reward for students, and it’s having an especially positive impact on those with attendance issues. The exciting fitness opportunities encourage scholars to come to school more often and stay engaged.   

This mini-pitch soccer field blew me away! My son would love it.

The highlight of my day was meeting Ms. Juarez’s class. As soon as I entered her classroom, I felt an uplifting energy. The room didn’t look like a traditional classroom. Ms. Juarez had flexible seating including bean bag chairs, balance balls, and carpet space. At the time of my visit, she was teaching students how to become fundraisers for a meaningful cause. This was the second year they were collecting change for the Leukemia & Lymphoma Society. Last year, they collected almost $3,000 in change and they were expecting to surpass that goal this year. Isn’t that impressive?   

An Outstanding Day

My morning as principal flew by. At noon, I headed to a luncheon where I joined the rest of the schools that participated in this amazing day. McFadden Middle School hosted a nice Italian lunch for everyone and we had great entertainment from the local high schools.  

Seeing friends in the community at the luncheon was the perfect way to end a fantastic day.

As the Community Partnerships Director at MIND Research Institute, I’m grateful for the opportunities our leadership provides to get involved with our partner schools. It’s just one of the things that’s kept me at this organization for over 17 years. MIND offers paid volunteer time (5% of working hours) to all colleagues through our Open MINDs program. Youth and women’s empowerment are causes close to my heart, and I was glad to be able to use Open MINDs hours to spend the day with students and their families. There are amazing things happening at our local schools!

After my visit, Principal Perez sent a wonderful message to my colleagues and I at MIND:

"What a wonderful experience it was having Mrs. Cervantes from MIND Research Institute serve as Principal for a Day at Jackson Elementary School. She connected with students right away, asking questions about their projects and how they use ST Math to learn math. We were thrilled to share about Jackson's Soccer Academy and ask her advice about how to grow the program over the next few years. She was thrilled with the strong focus on creating a pathway for students to earn college scholarships while leveraging the love our scholars have for soccer. We are grateful for MIND's many years of support here in SAUSD and are looking forward to many more years of success, innovation, and dynamic team work."

Thank you, Jackson Elementary, for providing me with this opportunity to give back and to learn about the strengths, challenges, and successes of your school. It was a privilege meeting your scholars and seeing that our future is in really great hands.

from MIND Research Institute Blog http://bit.ly/2WhUtaa

Check out this Advanced Knowledge Problem of the Week.
Be sure to let us know how you solved it in the comments below or on social media!
Solution below.



from The Center of Math Blog http://bit.ly/2UQ83Qf

Our department is working on some curricular projects involving both developmental algebra and pre-calculus.  This work has involved some discussion of what “rigor” means, and has increased the level of conversation about algebra in general.  I’ve posted before about pre-calculus College Algebra is Not Pre-Calculus, and Neither is Pre-calc and College Algebra is Still Not Pre-Calculus for example, so this post will not be a repeat of that content. This post will deal with algebraic identities.

So, our faculty offices are in an “open style”; you might call them cubicles.  The walls include white board space, and we have spaces for collaboration and other work.  Next to my office is a separate table, which one of my colleagues uses routinely for grading exams and projects.  Recently, he was grading pre-calculus exams … since he is heavily invested in calculus, he was especially concerned about errors students were making in their algebra.  Whether out of frustration or creative analysis, he wrote on the white board next to the table.  Here is the ‘blog post’ he made:

 

 

 

 

This picture is not very readable, but you can probably see the title “Teach algebraic identities”, followed by “Example:  Which of the following are true for all a, b ∈ ℜ.  In our conversation, my colleague suggested that some (perhaps all) of these identities should be part of a developmental algebra course.  The mathematician part of my brain said “of course!”, and we had a great conversation about the reasons some of the non-identities on the list are so resistant to correction and learning.

Here are images of each column in the post:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

When we use the word “identities” in early college mathematics, most of us expect the qualifier to be “trig” … not “algebraic”.  I think we focus way too much on trig identities in preparation for calculus and not enough on algebraic identities.  The two are, of course, connected to the extent that algebraic identities are sometimes used to prove or derive a trig identity.  We can not develop rigor in our students, including sound mathematical reasoning, without some attention to algebraic identities.

I think this work with algebraic identities begins in developmental algebra.  Within my own classes, I will frequently tell my students:

It is better for you to not do something you could … than to make the mistake of doing something ‘bad’ (erroneous reasoning).

Although I’ve not used the word identities when I say this, I could easily phrase it that way: “Avoid violating algebraic identities.”  Obviously, few students know specifically what I mean at the time I make these statements (though I try to push the conversation in class to uncover ‘bad’, and use that to help them understand what is meant).  The issue I need to deal with is “How formal should I make our work with algebraic identities?” in my class.

I hope you take a few minutes to look at the 10 ‘identities’ in those pictures.  You’ve seen them before — both the ones that are true, and the ones that students tend to use in spite of being false.  They are all forms of distributing one operation over another.  When my colleague and I were discussing this, my analysis was that these identities were related to the precedence of operations, and that students get in to trouble because they depend on “PEMDAS” instead of understanding precedence (see PEMDAS and other lies and More on the Evils of PEMDAS!   ).  In cognitive science research on mathematics, the these non-identities are labeled “universal linearity” where the basic distributive identity (linear) is generalized to the universe of situations with two operations of different precedence.

How do we balance the theory (such as identities) with the procedural (computation)?  We certainly don’t want any mathematics course to be exclusively one or the other.  I’m envisioning a two-dimensional space, where the horizontal axis if procedural and the vertical axis is theory.  All math courses should be in quadrant one (both values positive); my worry is that some course are in quadrant IV (negative on theory).  I don’t know how we would quantify the concepts on these axes, so imagine that the ordered pairs are in the form (p, t) where p has domain [-10, 10] and t has range [-10, 10].  Recognizing that we have limited resources in classes, we might even impose a constraint on the sum … say 15.

With that in mind, here are sample ordered pairs for this curricular space:

• Developmental algebra = (8, 3)           Some rigor, but more emphasis on procedure and computation
• Pre-calculus = (6, 8)                         More rigor, with almost equal balance … slightly higher on theory
• Calculus I to III = (5, 10)                   Stronger on rigor and theory, with less emphasis on computation

Here is my assessment of traditional mathematics courses:

• Developmental algebra … (9, -2)         Exclusively procedure and computation, negative impact on theory and rigor
• Pre-calculus … (10, 1)                           Procedure and computation, ‘theory’ seen as a way to weed out ‘unprepared’ students
• Calculus I to III … (10, 3)                      A bit more rigor, often implemented to weed out students who are not yet prepared to be engineers

Don’t misunderstand me … I don’t think we need to “halve” our procedural work in calculus; perhaps this scale is logarithmic … perhaps some other non-linear scale.  I don’t intend to suggest that the measures are “ratio” (in the terminology of statistics; see http://bit.ly/2UTkXwR https://www.questionpro.com/blog/ratio-scale/ ).  Consider the measurement scales to be ordinal in nature.

I think it is our use of the ‘theory dimension’ that hurts students; we tend to either not help students with theory or to use theory as a way to prevent students from passing mathematics.  The tragedy is that a higher emphasis on theory could enable a larger and more diverse set of students to succeed in mathematics, as ‘rigor’ allows other cognitive strengths to help a student succeed.  The procedural emphasis favors novice students who can remember sequences of steps and appropriate clues for when to use them … a theory emphasis favors students who can think conceptually and have verbal skills; this shift towards higher levels of rigor also serves our own interests in retaining more students in the STEM pipeline.

 

 

from Developmental Mathematics Revival! http://bit.ly/2ZKVrOk

from Maths & Physics News

from Intersections -- Poetry with Mathematics

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from math http://bit.ly/2GRruo4

Let $(X,\mathscr T)$ be a metrizable space such that every metric that generates $\mathscr T$ is bounded. Prove that $X$ is compact.

My attempt:- We know that $(X,\mathscr T)$ is metrizable. So there is a metric on $x$ such that collection of all open sets with respect to the metric is the $\mathscr T$. Let $\{d_\alpha\}_{\alpha \in \Lambda}$ be the collection of metric that generates $\mathscr T$. We know that $\forall \alpha \in \Lambda, (X,d_\alpha)$ is bounded.

Suppose on contrary $X$ is not compact. So $(X,d_\alpha)$ is not sequantially compact. So, there is a $\{x_n\}$ be a sequence in $X$ such that none of its subsequence converges. How do I make contradiction with our assumption?

from Hot Weekly Questions - Mathematics Stack Exchange

Statistics and Probability Practice Problems for Grade 10

from Math Blog http://bit.ly/2GJGu6q

Angle Measures in Polygons Worksheet

from Math Blog http://bit.ly/2V28iNj

Probability Word Problems for Grade 10

from Math Blog http://bit.ly/2ZN4mP1

Angle Measures in Polygons

from Math Blog http://bit.ly/2DGIpHY

Today my professor said something interesting, replace $x$ in $f(x)$ with matrix \begin{bmatrix}x&1\\0&x\end{bmatrix}

Then $$f(\begin{bmatrix}x&1\\0&x\end{bmatrix}) = f(x)\begin{bmatrix}1&0\\0&1\end{bmatrix} + f^{'}(x)\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$

and asked us to find the higher order terms ($f^{''}(x), f^{'''}(x) ....$) and extend it to multi variable functions.

I didn't understand how he came up with this.

After some googling, this seems similar to matrix exponential from lie groups.

from Hot Weekly Questions - Mathematics Stack Exchange

Statistics Problems with Solutions for Grade 10

from Math Blog http://bit.ly/2XUJIuI

Finding Missing Frequency When Mean is Given

from Math Blog http://bit.ly/2GRfnr8

Problem: In the following 30 days you will get 46 homework sets out of which you will do at least one every day and - of course - all during the 30 days. Show that there must be a period of consecutive days during which you will do exactly 10 homework sets!

Solution: Let $f_n$ denote the number of homeworks from day $1$ to day $n$, where $n\le 30$. So, let us consider from $f_1$ up to $f_{11}$. There are ten possibilities for the remainder when each is divided by $10$. By the pigeonhole principle, there must exist two that have the same remainder, call these $f_i$ and $f_j$, for some $i,j\in [1,11]$. Therefore $$f_i - f_j \equiv 0 \pmod{10}.$$ But also $f_i - f_j \not = 20$. Hence $f_i - f_j = 10$.

I think this is on the right track. However, I have not convinced myself that $f_i - f_j \not = 20$

from Hot Weekly Questions - Mathematics Stack Exchange

Laplace transform of functions multiplied by variables. Here I’ll talk about the Laplace transform of functions multiplied by variables. Have a look!! If you are looking for more in Laplace transform of functions, do check in: First shift theorem in Laplace transform Laplace transform of functions divided by a variable Cover up rule in inverse Laplace...

The post Laplace transform of functions multiplied by variables appeared first on Engineering math blog.

from Engineering math blog http://bit.ly/2ZCwu7s

Word Problems on Mensuration for Grade 10 Worksheet

from Math Blog http://bit.ly/2URsYCw

Solving Word Problems Mensuration for Grade 10

from Math Blog http://bit.ly/2vuFo9e

Solving Literal Equations Worksheet

from Math Blog http://bit.ly/2GUYNa5

Solving Literal Equations

from Math Blog http://bit.ly/2Lm8Klc

Consider the series: $$\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$$ show this converges to:

(a) $\frac{1}{(1-z)^2}$ for $|z|<1$

(b) $\frac{1}{z(1-z)^2}$ for $|z|>1$

Finally, show that this convergence is uniform for $|z| \leq c <1$ in (a) and for $|z| \geq c >1$ in (b).

This is my first course in complex analysis and I'm struggling a bit with seeing why this is true. Question (a) suggests recognising the product of two geometric series $$\frac{1}{1-z} \cdot \frac{1}{1-z}= \sum z^ k \cdot \sum z^ k $$ But I do not see how to rewrite the expression. I have no idea how one would derive the expression in question $(b)$ any hints on that would be greatly appreciated. Once I know how to derive these expressions I will maybe also know a way to apply the Weierstrass M-test for uniform convergence. Since the uniform limit must equal the pointwise limit all we need to show that the series converges uniformly on some domain of definition. Can someone provide some guidance on how to tackle this question, drop some small hints so I can maybe proceed?

from Hot Weekly Questions - Mathematics Stack Exchange

I spent the last few days trying to solve this exercise with little success, so I really hope someone here might be able to assist:

Denote Moore plane by $M$, the $x$-axis by $R$ and the upper half-plane by $H$ ($M = R \cup H$). Let $A, B \subseteq M$ be closed and disjoint subsets of $M$. Suppose $|A \cap R| < \infty$. Prove that $A$ and $B$ can be separated by disjoint open neighborhoods.

I am aware of the fact that $M$ is not $T_4$, and in the previous exercises I proved the following facts:

• $R$ as a subspace of $M$ has the discrete topology.
• $H$ as a subspace of $M$ is homeomorphic to $H$ as a subspace of $\mathbb{R}^2$ (with the standard topology).
• $H$ is open in $M$.
• The closure in $M$ of each element of the topology's basis is the same as its closure in the Euclidean plane.
• $M$ is a regular Hausdorff space ($T_3$).

I also showed that it suffices to prove the claim for the case where $A \cap R = \varnothing$ and $B = R$ (although other approaches might also work), but I'm having trouble showing that there exists an open neighborhood of $A$ whose closure does not intersect with $R$.

Your help will be much appreciated!

from Hot Weekly Questions - Mathematics Stack Exchange

Word Problems on Mensuration for Grade 10

from Math Blog http://bit.ly/2ZMlKDQ

Solid Mensuration Practice Problems

from Math Blog http://bit.ly/2XVwEFu

Check out this fantastic essay by Maria Roman ’22, which was featured American Mathematical Society’s blog!  It’s titled “Trip to the National Museum of Mathematics”.

from Blog – Mathematics & Statistics

Worksheet on Locus

from Math Blog http://bit.ly/2vus2Ke

Locus

from Math Blog http://bit.ly/2JbuVrB

If $f$ is continuous on $[a,b]$ and $g$ equals $f$ except at $x=c \in (a,b)$ where $g$ is defined arbitrarily. Then prove that $g$ is integrable on $[a,b]$ and $\int_{a}^{b} f = \int_{a}^{b} g$.

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Here's my attempt:

I will be using this to prove it:

Let $f$ be bounded on $[a,b]$ and continuous everywhere except at one point $c$ in $[a,b]$. Then $f$ is integrable on $[a,b]$.

Clearly, if $f$ is continuous on $[a,b]$, then there is some $M\in\mathbb{R}^{+}$ such that $|f(x)| \le M$. Hence, it must be that $|g(x)| \le \max \{ M , g(c) \} $. Hence, $g(x)$ is bounded above and is continuous on $[a,b]$ except at $x=c$. Thus, by the theorem above, it is integrable.

Now, we need to show that $\int_{a}^{b} f =\int_{a}^{b} g$. If we assume that $f(c) < g(c)$ (the other case may be handled similarly), then $f(x) \le g(x)$ for all $x \in [a,b]$ and so, $\int_{a}^{b} f \le \int_{a}^{b} g$.

Let $\varepsilon >0$ be given. Then there exists a partition $P$ of $[a,b]$ such that $\int_{a}^{b} g - \varepsilon < L(P,g)$. Now we claim that for any partition $P$ of $[a,b]$, we have $L(P,g)=L(P,f)$ (we prove this claim later). Hence it follows that $\int_{a}^{b} g - \varepsilon < L(P,g) = L(P,f) \le \int_{a}^{b} f$. Since, $\varepsilon$ was arbitrary, we have $\int_{a}^{b} g \le \int_{a}^{b} f$. Hence, $\int_{a}^{b} f = \int_{a}^{b} g$.

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Proof of my claim: Let $P$ be any partition of $[a,b]$. We need to show that $L(P,g) = L(P,f)$. First, we note that there must be some $i$ such that $c\in[x_{i-1} , x_{i}]$. Also, $f(x) \le g(x)$ on $[x_{i-1} , x_{i}]$, hence, $\inf_{[x_{i-1} , x_{i}]} f(x) \le \inf_{[x_{i-1} , x_{i}]} g(x)$. Also, since $f$ is continuous on $[x_{i-1} , x_{i}]$, there must be some $y\in[x_{i-1} , x_{i}]$ such that $f(y) =\inf_{[x_{i-1} , x_{i}]} f(x)$ by the Extreme Value Theorem. If $y \ne c$ then it is clear that $f(y) = g(y) =\inf_{[x_{i-1} , x_{i}]} g(x)$ otherwise if $y=c$, then for any $\varepsilon >0$, $\inf_{[x_{i-1} , x_{i}]} g(x) \le g(z)=f(z) < f(c)+\varepsilon$ for some $z \in [x_{i-1}, x_{i}]\setminus \{c\}$ ( for otherwise it would contradict our assumption that $f(c) < g(c)$). Since, $\varepsilon$ was arbitrary,$\inf_{[x_{i-1} , x_{i}]} g(x) \le f(c)$. Thus, $\inf_{[x_{i-1} , x_{i}]} f(x) = \inf_{[x_{i-1} , x_{i}]} g(x)$. Since, this is true for the rest of the subintervals as well, we must have that $L(P,g) = L(P,f)$.

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Is my proof correct? I'm a bit unsure about the last part. Any alternative proofs?

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EDIT: Found a easier way to complete the second part . Thanks to the hint provided by Kavi Rama Murthy.

Since, we've established that $g$ is integrable, consider the function $h:[a,b] \to \mathbb{R}$ given by $h(x)=f(x)-g(x)$. Then $h$ is integrable. Also, note that we're assuming here that $f(c) < g(c)$.

Let $P$ be any partition of $[a,b]$. Then $U(P,h)=0$ because $M_k(h)=0$ for any subintervals. Hence, it must be that $\int_{a}^{b} h = \overline{\int_{a}^{b}} h = \inf \{ U(P,h) \} =0$. Thus, again by linearity, we have $\int_{a}^{b} f = \int_{a}^{b}g$.

from Hot Weekly Questions - Mathematics Stack Exchange

[True/False]The polynomial $x^4+7x^3−13x^2+11x$ has exactly one real root.

I want to solve it without drawing the graph. Here is my idea. Note that $f(1)=1+7-13+11=6>0$ and $f(-1)=1-7-13-11=-30<0$

So we have at least one real root. Now since degree is $4$ we have $4$ roots but rest three can not be complex as they occur in pairs, so we must have another real root.

So the statement is False

Am I right?

Thanks for reading and all the help.

from Hot Weekly Questions - Mathematics Stack Exchange

I'm working on the proof of Theorem 4.26 (pg. 215) in Rotman's "Advanced Modern Algebra". The theorem is stated as follows:

Let $f(x)\in k[x]$, where $k$ is a field, and let $E$ be a splitting field of $f(x)$ over $k$. If $f(x)$ is solvable by radicals, then its Galois group $\operatorname{Gal}(E/k)$ is a solvable group.

Here $E$ is the splitting field of $f(x)$ over $k$. The proof relied on Theorem 4.20 (pg.213)

Let $k$ be a field and let $f(x)\in k[x]$ be solvable by radicals, so there is a radical extension $k=K_0\subseteq K_1\subseteq \dots \subseteq K_t$ with $K_t$ containing a splitting field $E$ of $f(x)$. If each $K_i/K_{i-1}$ is a pure extension of prime type $p_i$, where $p_i\neq \operatorname{char}(k)$, and if $k$ contains all $p_ith$ roots of unity, then the Galois group $\operatorname{Gal}(E/k)$ is a quotient of a solvable group.

(Hence it is a solvable group.)

In the proof, he constructed an extension $E^*$ of $E$ and an extension $k^*$ of $k$ so that $k^*$ contains appropriate roots of unity, thus $\operatorname{Gal}(E^*/k^*)$ is solvable by Theorem 4.20. The problem is, I don't know why he can get rid of the extra hypothesis $p_i\neq \operatorname{char}(k)$ for all $i$. I read the chapter twice, but still no clue.

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The first issue of the twice-yearly newsletter from the International Mathematical Union Committee for Women in Mathematics has been published. It contains an interview with Marie-Francoise Roy, news, upcoming events and a book announcement (World Women in Mathematics 2018).

Download a copy of the CWM Newsletter here.

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Bin Cheng undertook a month long research visit to America this spring (3 to 28 March) travelling from Maryland on the east coast to California on the west coast. He gave a talk at the University of Maryland (6 March), the University of Texas at Dallas (15 March), and Arizona State University (22 March).  The talk title in all three cases was “Analysis of nonlinear dynamics with three time scales“. The picture left shows Bin in Sequoia National Park in California.

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Equations of Circles Worksheet

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Equations of Circles

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     Thanks to poet/mathematician Scott Williams who alerted me to this work by "a good poet and friend" Stephen Lewandowski.  Steve speaks of his poem this way:  "SPELL" exists because I fear the misuse of algorithms to standardize people . . ."

A SPELL AGAINST AL-GORITHMS     by Stephen Lewandowski

Named for a man, Abu Ja-far Muhammed ibn Musa,
and the Persian city Khwarizma where he lived
in the year 800, pursuing calculations
arithmetical and al-gebraical.

Begins admirably as
“how to solve a class of problems” and
proceeds through disambiguation to specification by
massaging a mass of data.
If the data are people, then
the massage is called a “census.”  
Read more »

from Intersections -- Poetry with Mathematics

Suppose that $$f(x)=1+\sum_{n=1}^\infty a_n \frac{x^n}{n!}\ \ \forall \ x\in \mathbb{R}$$ where $\sup_{x>0}\left|e^xf(x)\right| < \infty$ and $\sup_{n\in\mathbb{N}} |a_n|< \infty$.

Prove that $a_n = (-1)^n$ , $\forall n\in \mathbb{N}$

It seems amazing to me. What we need to prove is $f(x)=e^{-x}$. It seems insufficient to prove this strong conclusion, but actually it is true and all the "counterexamples" I found were wrong.

My attempt

Put $g(x)=e^x f(x)$. $$\left|g^{(n)}(0)\right|=\left|\sum_{k=0}^n \binom{n}{k} f^{(k)}(0)\right|\le 2^n\sup_{n\in\mathbb{N}} |a_n| $$ Put $h(x)=g(\frac{x}{2})$. Thus $$h^{(n)}(0)=\frac{1}{2^n}g(0) \le \sup_{n\in\mathbb{N}} |a_n|$$ which implies that $$h(x)=1+\sum_{n=1}^\infty b_n \frac{x^n}{n!}\ \ \forall \ x\in \mathbb{R}$$ where $$|b_n|\le \sup_{n\in\mathbb{N}} |a_n| \ \ \forall \ n\in\mathbb{N} \,\,\,\,\,\& \,\,\,\,\, \sup_{x>0}\left|h(x)\right| < \infty $$

And hence if $b_k<0$, then there exists $l>k$ such that $b_l>0$. I want to yield a contradiction by supposing this, but I failed.

Any hints or other new ideas? Thanks in advance!

(I heard that this problem can be solved by complex analysis. This is the reason why I attach the complex-analysis tag.)

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Show that number of triples $(a,b,c)$ with $a,b,c\in [1,n]$ such that $ab=c$ is given by

$$\bigl|\bigl\{(a,b,c)\in [1,n]^3:ab=c\bigr\}\bigr|=2\sum_{i=1}^{\left\lfloor\sqrt{n}\right\rfloor}\Big(\left\lfloor\frac ni\right\rfloor-i\Big)+\left\lfloor\sqrt{n}\right\rfloor$$

Someone told me that this can be solved using hyperbolas. I could not really understand how that would help. Please help!

It could have been easier if the interval were not a requirement, but that is the part that is confusing me the most.

from Hot Weekly Questions - Mathematics Stack Exchange

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to […]

from Mean Green Math

Trigonometry Problems with Solutions

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Trigonometry Practice Questions for Grade 10

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Coordinate Geometry Practice Questions for Grade 10

from Math Blog http://bit.ly/2V2waAd

I have encountered myself with the following exercise:

Let $\langle A, <_R\rangle$ and $\langle B, <_S\rangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'\subseteq A$ and $B'\subseteq B$ such that:

$$\langle A,<_R\rangle\cong\langle B',<_S\cap(B'\times B')\rangle\qquad\&\qquad\langle B,<_S\rangle\cong\langle A',<_R\cap(A'\times A')\rangle$$

Is it necessarily true that $\langle A,<_R\rangle\cong\langle B,<_S\rangle$?

The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $\langle B',<_S\cap(B'\times B')\rangle$ is not isomorphic to $\langle B,<_S\rangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $\langle B,<_S\rangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.

However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?

I have tried with many examples between subsets of $\mathbb{R}$ and other subsets of $\mathbb{R}$, but it seems that they are all isomorphic to each other.

I ran out of ideas. Are there any counterexamples to this statement? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?

Thanks in advance for your time.

P.S.: I have thought, for instance, that a closed interval of $\mathbb{R}$ shouldn't be isomorphic to the whole $\mathbb{R}$. How can we prove this assertion, if true?

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Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$

I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar. Anyway I thought it's quite a trivial integral so I'm gonna solve it quickly, but I am having some hard time to finish it. I went on with Feynman's trick:

$$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{dx}{a+x+ax^2}$$ $$=\frac1a\int_0^\infty \frac{dx}{\left(x+\frac{1}{2a}\right)^2+1-\frac{1}{4a^2}}=\frac{1}{a}\frac{1}{\sqrt{1-\frac{1}{4a^2}}}\arctan\left(\frac{x+\frac{1}{2a}}{\sqrt{1-\frac{1}{4a^2}}}\right)\bigg|_0^\infty$$$$=\frac{\pi}{\sqrt{4a^2-1}}-\frac{2}{\sqrt{4a^2-1}}\arctan\left(\frac{1}{\sqrt{4a^2-1}}\right)=\frac{2\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}$$ We can prove easily via the substitution $x\to \frac{1}{x}$ that $I(0)=0$ so we have that: $$I=I(1)-I(0)=2\int_0^1 \frac{\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}da$$ Now I thought about two substitutions: $$ \overset{a=\frac12\cosh x}=\int_{\operatorname{arccosh}(0)}^{\operatorname{arccosh}(2)} \arctan(\sinh x)dx$$ $$\overset{a=\frac12\sec x}=\int_{\operatorname{arcsec}(0)}^{\frac{\pi}{3}}\frac{x}{\cos x}dx$$ But in both cases the lower bound is annoying and I think I am missing something here (maybe obvious). So I would love to get some help in order to finish this.

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Edit: We can apply once again Feynman's trick. First consider: $$I(t)=\int_0^1 \frac{2\arctan(t\sqrt{4a^2-1})}{\sqrt{4a^2-1}}da\Rightarrow I'(t)=2\int_0^1 \frac{1}{1+t^2(4a^2-1)}da$$ $$=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2at}{\sqrt{1-t^2}}\right)\bigg|_0^1=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)$$ So once again we have $I(0)=0$, so $I=I(1)-I(0)$. $$\Rightarrow I=\int_0^1\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)dt\overset{t=\sin x}=\int_0^\frac{\pi}{2}\frac{\arctan(2\tan x)}{\sin x}dx$$ At this point Mathematica can evaluate the integral to be: $$I=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$ I didn't try the last integral yet, but I am thinking of Feynman again $\ddot \smile$.

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Edit 2: Found that I already was on it some time ago, and actually posted it here, which means I have solved it before using Feynman's trick, but right now I can't remember how I did it.

So given the circumstances I am positive that it can be solved starting with my approach, but if you have any other ways then feel free to share it.

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I am trying to find a closed form of the following integral $$ \int _0^{\infty }\int _0^x\int _0^y\int _0^z \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2}-\frac{d w^2}{2} \right) \,\mathrm{d}w\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x $$ where $a,b,c,d>0$ are some constants.

My idea is to change variable to to polar system by letting $$ x=\frac{r \cos (\alpha ) \cos (\beta ) \cos (\theta )}{\sqrt{a}}, \quad y=\frac{r \cos (\alpha ) \cos (\beta ) \sin (\theta )}{\sqrt{a}} $$ $$ z=\frac{r \sin (\alpha ) \cos (\beta )}{\sqrt{c}}, \quad w=\frac{r \sin (\beta )}{\sqrt{d}} $$

This reduces the original integral into $$ \int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta $$ But then I get stuck here.

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PS: I am interested in this because I found that $$ \int _0^{\infty }\int _0^x \exp \left(-\frac{a x^2}{2}-\frac{b y^2}{2}\right)\,\mathrm{d}y\,\mathrm{d}x = \frac{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)}{\sqrt{a b}} $$ and $$ \int _0^{\infty }\int _0^x\int _0^y \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2} \right) \,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x = \frac{\sqrt{\pi/2 } }{\sqrt{a b c}} \left(\tan ^{-1}\left(\sqrt{\frac{c}{b}}\right)-\tan ^{-1}\left(\sqrt{\frac{a c}{b (a+b+c)}}\right)\right) $$ So I am trying to generalize this. Maybe this is already known?

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Stone duality, one of many dualities between certain lattices and certain topological spaces, asserts that there is a contravariant categorical equivalence between the category $\text{Bool}$ of boolean algebras and the category $\text{Stone}$ of stone spaces. For those who are not familiar with this, here is a brief statement of what this says:

Definition: A boolean algebra is a partially ordered set $A$ such that

1) $A$ has finite meets and finite joins, including the empty join.

2) Meets distribute over joins in $A$

3) For each $a \in A$, there is $b$ such that $a \vee b$ is the greatest element, and $a \wedge b$ is the least element.

(1) says that $A$ is a bounded lattice, (1) and (2) say that $A$ is a bounded distributive lattice, and (3) says that $A$ is complemented.

Definition: A stone space $X$ is a space occuring as the cofiltered limit $\text{limit}\ X_i$ of discrete spaces in the category of topological spaces.

One functor $\text{Spec} : \text{Bool} \rightarrow \text{Stone}$ sends a boolean algebra $A$ to $\text{Bool}(A, \{ 0, 1 \})$ of maps of boolean algebras from $A$ to the boolean algebra $\{ 0, 1\}$. The other sends a stone space $X$ to $\text{Stone}(X, \{ 0, 1\})$, the set of maps of stone spaces from $X$ to $\{ 0, 1\}$.

I am interested in how this might work for $\sigma$-algebras. A $\sigma$-algebra is a subalgebra of the boolean algebra $P(X)$ (power set of a set) closed under countable meets (and therefore countable joins). By the categorical equivalence, $\sigma$-algebras $A$ on $X$ correspond to certain quotient objects of $\hat{X}$, the profinite completion of $X$ (inverse limit over all quotients onto a finite set).

Question: Let $X$ be a set. Can we characterize the quotient stone spaces of $\hat{X}$ corresponding to $\sigma$-algebras in $P(X)$?

Note: it was mentioned in the comments that $\sigma$-algebra can also refer to an ambient boolean algebra which has countable meets (and therefore countable joins). Here I mean to specifically fix an embedding into $P(X)$, and for joins and meets in the $\sigma$-algebr to match joins and meets in $P(X)$.

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If x was a positive 4 digit number and you divide it by the sum of its digits to get the smallest value possible, what is the value of x? For example (1234 = 1234/10)

I got 1099 as my answer however I don't know if this is right or how to prove it.

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Solve the following... $309|(20^n-13^n-7^n)$ in $\mathbb{Z}^+$. I invested lotof time to it and finally went to WolframAlpha for help by typing... Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!

from Hot Weekly Questions - Mathematics Stack Exchange

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Can you tell from its opening weekend how much the new movie, Avengers: Endgame, might earn in theaters?  This activity can be used anytime and is especially timely when there is a new blockbuster movie released.  We've left blank spaces in our chart for students to add their favorites or for future movies. The data for this activity can be found at: boxofficemojo.com.  Just click on any movie and look for this info: 
Students compare "Opening Weekend" amounts to "Total Lifetime Gross (Domestic)" data to try to  predict how much money the new Avengers movie (which is expected to earn almost $300,000,000 in its opening weekend in the U.S.) will make at theaters. This task is accessible for any student who can plot points on a grid. It could be used for students as young as 4th graders.  Older students can create a line of best fit and equation to model the data and make predictions.  By plotting and analyzing the data students decide if they can, in fact, predict the future.

The Activity: Avengers-Endgame.pdf

CCSS: 8.SP.1, 8.SP.2, 8.SP.3, HSS.ID.B.6

For members we have an editable Word docx, an Excel data sheet with charts, and solutions

Avengers-Endgame.docx      Avengers-Endgame.xlsx    Avengers-Endgame-solution.pdf

from Yummy Math

Consequences of the Prime Number Theorem tell us the probability of $n$ being prime is $1/\ln(n)$. This also means that the number of expected primes between $n$ and $n+\ln(n)$ is close to $1$, but not always. What is the probability there is no prime between $n$ and $n+\ln(n)$?

from Hot Weekly Questions - Mathematics Stack Exchange

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Solving Word Problems Trigonometry

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