Mathematicianadda

Greatest Common Factor Word Problems

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HCF and LCM Worksheet

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"Six men and three women are standing in a supermarket queue. Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?"

I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.

$$ {3 \choose 1} \ \times \ {6 \choose 2} \ + \ {3 \choose 2} \ \times \ {6 \choose 1} \ + \ {3 \choose 3} $$

Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.

This approach is in fact correct. However, another approach came to my mind as well.

If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.

$$ {3 \choose 1} \ \times \ {8 \choose 2} $$

This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.

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Shaheer ziya

My question essentially has to do with the derivative of a Contour Integral's parameterized curve. $$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$ to be exact. Where $\partial \Omega(x)$ is a Jordan curve which is differentiable for any $x \in \mathbb{C}$, and $f(n, x): \mathbb{C}^2 \to \mathbb{C}$ integrable around the curve $\partial \Omega(x)$ in respect to $n$. Define $\gamma$ as the parameterized curve of $\partial \Omega$, and the terminology $f_x(n, x) = \frac{\partial f(n, x)}{\partial x}$ is used.

My work has essentially gotten down to these steps.

STEP 1: Turning the contour integral into the usual integral.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \frac{\partial}{\partial x} \int_{0}^{2\pi} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 2: Using the Liebniz rule.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \int_{0}^{2\pi} \frac{\partial}{\partial x} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 3: Taking the derivative.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) + \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) + \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 4: Separating the integrals.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta$$ $$+ \int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \mathrm{d}\theta$$ $$+ \int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 5: Simplifying the first integral into a contour integral.

$$\int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta = \oint_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n.$$

STEP 6: Plugging in the first integral to get the final answer.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$

$$= \int_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n+\int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \; \mathrm{d}\theta$$ $$+\int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

I am unsure of how to simplify this further or if this is even a decent approach. Does anybody have a good resource for this?

My goal is to write this derivative as multiple contour integrals, bar any $\gamma$-parameterized functions.

(Disclaimer: This same question has been posted by myself to MathOverflow)

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Jackson

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HCF and LCM Word Problems

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What's the range of function $a \sin(mx) + b \cos(nx)$ where $a,b,m,n \in R$?

Not hard to solve for the case where $m=n$. We can let $m=n=1$ WLOG

$a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin(x) + \frac{b}{\sqrt{a^2 + b^2}} \cos(x) \right)$

Then we can substitute $\sin\theta = \frac{b}{\sqrt{a^2 + b^2}}$

How do we handle the case when they are different?

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R.Yeh

I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds: $$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$ The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

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Matthew

How to Multiply Matrices

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Greatest Common Factor Worksheet

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A monohedral polyhedron is one whose faces are all congruent. Note that this is a weaker condition than being isohedral, or being face-transitive.

We have a classification of all convex isohedral polyhedra, consisting of 30 classes of assorted finite polyhedra and infinite families. See Wolfram Mathworld for a list.

In the process of writing this answer, I was trying to find instances of convex polyhedra which are monohedral but not isohedral, and struggled to find a classification of such shapes or even a list of known instances with an outline of which cases remain open.

The instances I know of beyond the isohedral polyhedra given above:

• When the faces are regular polygons, we can search among the Johnson solids to find three non-isohedral examples: the snub disphenoid, the triaugmented triangular prism, and the gyroelongated square bipyramid.

• The pseudo-deltoidal icositetrahedron, the dual of the pseudorhombicuboctahedron. In footnote 46 on page 185 of Advances in Discrete and Computational Geometry: Proceedings of the 1996 AMS-IMS-SIAM Joint Summer Research Conference, Discrete and Computational Geometry, it is remarked that no other convex non-isohedral monohedra are known with non-triangular faces (or at least, none were known in 1996). (Here is a Google Books link to the relevant section of the previous source.)

• However, in a result apparently unknown to the above source, Ed Pegg provides what seems to be a counterexample in this mathSE question; manually cutting out and folding the net, it does indeed seem to fold into a convex polyhedron.

• In the article The Bilinski Dodecahedron and Assorted Parallelohedra, Zonohedra, Monohedra, Isozonohedra, and Otherhedra (Grünbaum, B., 2010), the author includes the rhombic icosahedron and the Belinski dodecahedron, and cites Belinski as proving that these are the only convex monohedra with centrally symmetric faces (such polyhedra termed isozonohedra) not already listed among the isohedra.

• In David Eppstein's paper On Polyhedral Realization with Isosceles Triangles (link to arXiv abstract), three infinite families of convex monohedra not given above are listed, with isosceles triangle faces: one consisting of a heightened antiprism with pyramids glued to each large opposite face (called a "gyroelongated bipyramid"), a variant of the previous shape in which two halves are rotated about a skew hexagonal cross-section (called a "twisted gyroelongated bipyramid"), and a shape called the "biarc hull" which I can't describe very well but is related to an oloid. They cite the paper Some New Tilings of the Sphere with Congruent Triangles by Robert Dawson, most of which seem not to translate to monohedral polyhedra but which I haven't checked fully.

• The answers to this math.SE question, which provide monohedral polyhedra combinatorially equivalent to the icosahedron but with non-equilateral faces. (The scalene case is not always convex, but can be made so with small distortions to the angles of the triangles involved.) The book Advances in Discrete and Computational Geometry mentioned above goes into more detail about the possibility of such polyhedra.

Given the apparent lack of a classification of more restricted classes of polyhedra, I expect that no complete classification exists. However, I'd like to better understand the scope of which examples and impossibility results are known, as I haven't been able to find much in the way of a definitive source tackling this question. I'm hoping this question and its answers can at least serve as a better overview of known convex monohedra than the scattered state of information that seems to exist on the problem at present.

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RavenclawPrefect

Let $p$ be an odd prime, and let $n_1,\dots,n_{p-2}, m$ be even integers such that $n_1 < n_2 < \dots < n_{p-2}$ and \begin{equation} 2m > \sum_{i=1}^{p-2} n_i^2. \end{equation} Consider the polynomial \begin{equation} g(x)=(x^2 + m)(x - n_1) \dots (x - n_{p-2}). \end{equation} From Rolle's Theorem, we know that for each $i=1,2,\dots,p-3$, there exists $x_i \in (n_i,n_{i+1})$ such that $g'(x_i)=0$. So $g'(x)$ has at least $p-3$ distinct real zeroes. My question is: can $g'(x)$ have more than $p-3$ distinct real zeroes?

I do not know the answer, but for sure the constraints on the parameters are relevant here. For example the polynomial $g(x)=(x^2+1)(x-4)(x-2)(x+2)$ has derivative $g'(x)=5x^4-16x^3-9x^2+24x-4=(x-1)(5x^3-11x^2-20x+4)$ which has four distinct real roots, as you can check on WolframAlpha.

NOTE This strange polynomial arises in the construction given by R. Brauer of a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $p$ whose Galois group over $\mathbb{Q}$ is isomorphic to the symmetric group $\mathcal{S}_p$: see Jacobson, Basic Algebra I, $\S 4.10$. The question I asked is clearly irrelevant for the construction, but has intrigued me, since I could not answer it in the negative nor I could find some counterexample.

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Maurizio Barbato

Suppose $f$ is a polynomial of degree $n$, and $f\geq0$ for all $x$, prove that $f+f'+f''+f'''+\cdots +f^n\geq0$. (so just functions of the form $ax^n+bx^{n-1}+.....c$. We are not going to include irrational functions, exponential, logs, etc.)

So for this problem, I looked at a few functions and graphed them to see why this statement would be true. The first thing I found out is that the polynomial's highest degree must be even, otherwise there would be some $x$ where $f(x)$ is negative. Now this implies the odd numbered derivatives will have an odd number for the term with the highest degree.

Having figured that out, I was wondering whether the sum of every successive would change the function so that $f+f'$ has negative values, and $f+f'+f''$ would have positive values and so on, and this does turn out to be true for the ones I've tried. If this is true in general, that would mean that $f+f'$ has some negative values, $f+f'+f''$ has not negative values, and $f+f'+f''+f'''+....f^{n-1}$ has negative values until we get to $f+f'+f''+f'''+....f^{n-1}+f^n$ which has no negative values.

Ok but having figured this out, I'm not sure as to how to proceed with the proofs because the above assumptions I've made are not lemmas or proofs that I was given.

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William

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Adding and Subtracting Matrices Worksheet

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Let $a,b$ be two positive integers, if $a^n-1$ and $b^n-1$ have the same set of prime factors for each $n\in\Bbb{Z}^+$, then $a=b$.

It seems to be not hard, but I have no idea. Any hint is welcome.

Edit: As the comment points out, it is probably not easy. I thought it is easy because it is kind of intuitively correct. I tried to study the Zsigomondy primes as $n$ increases, but it is not successful.

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MathEric

This is a super weird question coming from someone who is just starting out in mathematics.

If $n$ is a positive integer, one can notice that as $n$ grows, $\mathbb{Z}/n\mathbb{Z}$ starts to look more and more like $\mathbb{N}$. If I were a small integer $k$, the entire group would begin to look more and more indistinguishable from $\mathbb{N}$ as $n$ grows - my additive inverse is getting further and further away. Is this idea something that is studied or formalized anywhere?

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hahahahahahahababahaha

I'm trying to do the following exercise for my Real Analysis class:

Let $p$ be a given natural number. Give an example of a sequence $\left(x_{n}\right)$ that is not a Cauchy sequence, but that satisfies $\lim \left|x_{n+p}-x_{n}\right|=0$

However, I am in constant doubt in regard to the "let" word in math texts. Can I choose, say $p = 1$, or when one says "let" I am supposed to stick with $p \in \mathbb{N}$ and nothing more? What kind of "control" do I have over $p$?

For instance, when one says "Given $\epsilon \gt 0$", I usually see things like: Let $\epsilon = \frac{\epsilon}{2}$ so one can finish a certain argument.

Can someone help me? I need to solve this doubt once and for all, it bothers me very often.

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Lucas

Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, a >0 converges or not.

I used d'Alembert's criterion and I found $\lim_{n\to \infty} \frac{x_n}{x_{n+1}}=1$

Moving to Raabe–Duhamel's test, I found $\lim_{n\to \infty} n(\frac{x_n}{x_{n+1}} -1)$ = $\lim_{n\to \infty} n(\frac{1}{\sqrt[n+1]{a}} - 1) $

What can I do from there? Or is there a better way to find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, a >0 converges or not?

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vyavar ignut

Oblique shapes are extraordinarily common, but many people don’t know what they are. Oblique shapes are shapes made of oblique angles. Oblique angles are:

• Acute angles: angles that are 0-90 degrees
• Obtuse angles: angles that are 90-180 degrees

The sides that form the angles of an oblique shape are never perpendicular. In simple terms, these sides will never meet.

Oblique Shapes

When someone talks about oblique shapes, they refer to a shape, either plane or space, that has either an acute or obtuse (an oblique) angle.

You can figure out if two planes are oblique or not by looking at the angle they create when they intersect. When two plane figures intersect, squares, for example, and create an oblique angle, the planes are called oblique planes.

Oblique Triangles

The most straightforward example of an oblique plane figure to visualize is an oblique triangle. An oblique triangle is any triangle that is not a right triangle. This is because it has an oblique angle and does not contain an angle that is 90 degrees. You can instead call this triangle an acute or obtuse triangle.

An acute triangle has three internal angles of less than 90 degrees. Obtuse triangles have one internal angle greater than 90 degrees; the other two are less than 90. Regardless of the type of triangle, it still follows that all three internal angles will equal 180 degrees.

Finding the Area of Oblique Figures

Finding the area of oblique figures is not that different from finding the area of right figures.

Triangles: base x height / 2 = area

Oblique Space Figures

An oblique space figure, like a rectangular prism, has faces that do not align with the bases. This makes the prism slanted with its base still flush. This is the case for any oblique prism, including a cylinder. The Leaning Tower of Pisa is a fun way to think about oblique space figures. If the figure leans like the tower, it’s oblique.

Note: an oblique space figure is not tipped so that the bottom is balancing on one of its corners. This is simply a tipped right space figure.

Finding the Volume of Oblique Space Figures

Like with plane figures, there really isn’t much of a difference in figuring out the volume of an oblique space figure.

Cylinder: base x height = volume

Pyramid: base x height (of apex when oblique) / ⅓ = volume

Essentially, all formula will remain the same. You must consider any measurement changes that could have happened when the right space figure turned into an oblique prism.

From Parallelogram to Rhombus

There are some shapes out there that we may not use quite often, but they are quite important to know in the realm of oblique planes. One of those figures is a rhomboid. A rhomboid is a parallelogram with oblique, internal angles. When a rhomboid has sides of equal length, it is a Rhombus.

But is a parallelogram oblique? Not always.

If you want to define a parallelogram, it is a quadrilateral with opposite sides equal in length and parallel. Technically, a rectangle can be a parallelogram. When the rectangle is not right, it becomes this oblique shape.

Oblique Pictorials

There is also a type of drawing called an oblique pictorial. You’ve likely drawn the most common oblique pictorial, a cube.

To create an oblique pictorial, you take a plane figure and make it 3D. You do this by extending parallel, angled lines from the figure and connecting the lines to mirror the shape. Usually, these angles are at a 35, 45, or 60-degree angle.

Can you guess why we refer to these drawings as oblique drawings? Look again at the standard angles used to make the plane figure 3D -- they are all less than 90 degrees. Therefore, we draw the extended lines at an oblique angle.

There are three types of oblique pictorial drawings, cabinet, cavalier, and general.

• Cavalier: represents the full depth
• Cabinet: represents half the depth
• General: concerned more with the idea of the pictorial rather than accuracy of depth

Outside of trying to impress your friends in middle school math class with a general oblique pictorial, they have very real-world applications. Eighteenth-century military artists in France used the cavalier pictorial method to design forts. The cabinet projection comes from its application in drawings used in the furniture business.

So when you look at building structures and even your living room furniture, look for the oblique angle that turned its face into a 3D figure.

In Conclusion

Think of what you can call an oblique shape without having to use a protractor to measure the angles -- any imperfect tortilla chip, for one. Identifying oblique shapes is simple once you know what you’re looking for: an acute or obtuse angle! Remember, if it’s not right or a prism is leaning, it’s oblique.

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Kathleen Knowles

I want to find a formula to calculate the number of dispersion lines I can draw from any one corner of a square dot matrix of size $n \times n$.

By dispersion lines, I mean the lines that connect a particular point to all other points in a dot diagram. Here's a picture of what I mean :

(That's an approximate diagram of what I mean; sorry for the crude diagram)
Here, the horizontal blue line on the top row (when considered a straight line) connects many dots to a single point (thus all those dots on the line are collinear), so when we count the number of dispersion lines, we have to count lines connecting many collinear points as one and we mustn't take the subunits into consideration.

So when we consider a square dot matrix, where the dots are arranged as a square, we can draw $3$ apparent dispersion lines, plus some more. What I would like to find is a formula to calculate the number of dispersion lines in square dot grids (or dot matrices).
What I have at hand is this one formula that I tried to formulate today:$$n^2 - (n + 2(n - 1)) + 3$$ The problem is that this formula isn't working for all values of $n$ (I mean, the number of dots in a column/row of the square dot matrix), and also I haven't been able to spend sufficient time to find the number of dispersion lines for $n > 5$. If asked for, I'll attach the numbers below.

Any help in formulating an accurate formula is appreciated. If such a formula exists, please do tell me.

Thanks in advance.

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Spectre

Anna Neufeld ’20 and fellow grad students in (bio-)statistics at the University of Washington are holding a Zoom graduate school panel (mostly for Stats and Math) on Monday, November 9, 2020 at 4:00 PM (PST). Follow this link for more information and registration.

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Bernhard Klingenberg

A sentence is called existential if it's of the form $\exists x_1 \cdots \exists x_n \varphi(x_1, \cdots, x_n)$, where $\varphi$ is quantifier-free formula.

I'm trying to prove a lemma left as an exercise in my lecture notes that says

Let $C$ be an axiomatizable class. Then the following conditions are equivalent: (i) $C$ is $\exists$-axiomatizable; (ii) If $A \in C$ and $A \leq B$ then $B \in C$.

One direction is simple - if $C$ is $\exists$-axiomatizable then it follows easily by the fact that if $A \models \varphi(a_1, \cdots , a_n)$ for an existential formula $\varphi(v_1, \cdots , v_n)$, then $B \models \varphi(a_1, \cdots , a_n)$.

I'm really not sure how to go about proving the other direction.

I've proved the equivalent version for universal axiomatisation - i.e. that $C$ is $\forall$-axiomatizable iff $B \in C$ and $A \leq B$ then $A \leq C.$ To do that, I showed that Th$(C) \cup $Th$_∃(A)$ is finitely satisfiable (where $A\models$ Th$_\forall$(C) - i.e. $A$ is a model of the universal sentences in the theory of $C$) where Th$_∃(A)$ are the existential sentences of the theory of $A$. It followed from that, and some other results I have, that Th$(C)\cup$Diag$(A)$ was satisfiable, from which it followed there is a model $B$ of Th$(C)$ such that $A \leq B$, which meant by the assumption that $A\in C$ and hence (since $A\models$ Th$_\forall$(C)) that $C$ was universally axiomatisable.

I'm really unsure how I'd go about taking a similar approach for the existential case - I haven't proved any similar results like the one that took me from Th$(C)\cup$Diag$(A)$ being satisfiable to there being a $B$ like I have here, and even if I did I'm not sure how I'd apply a similar kind of result - since in this case I have to start with a smaller model $A$ instead of a bigger one.

Any advice or suggestions you could offer would be much appreciated.

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yeet

Greatest Common Factor - Concept - Methods - Examples

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Source: https://xkcd.com/2303/ A brief explanation can be found at https://ift.tt/2HMWys6.

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John Quintanilla

A conversation about mathematics inspired by a lottery machine. Presented by Katie Steckles and Peter Rowlett.

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Katie Steckles and Peter Rowlett

I have read around about bicategories, lax functor, lax natural transformation and modifications. I know that we have a 1category of Bicategories and lax functors. I know why we do not have a bicategory or 2category of bicategories, lax functor, and lax natural transformations, and I know that using ICONS instead of lax natural trasformations solves this problem. (or oplax, it doesn't matter)

What I cannot see is why we do not get a 3categories or tricategories of bicategories, Lax functors, ICONS and modifications? What fails? Where can I find a reference about it?

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Lolman

Least Common Multiple Worksheet

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I have this limit: $$ \lim_{x \to 0} \frac{x(e^x+1)-2(e^x-1)}{x^3} $$ By using L'Hospital rule and Maclaurin expansion I both found that limit equals to $1/6$.

I want to simulate this to my student by Excel but I failed. This is my computation in Excel.

I did not know why Excel doesn't work in this situation?

Please help me. Thanks alot.

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Nguyen van tien

Least Common Multiple - Concept - Methods - Examples

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From what I know, Lie wanted to generalize properties of differentials, but (ab)' = a'b + ab', so why exactly did they construct an algebra that follows the jacobian identity and anti-skew symmetry? If they were working with one-parameter groups then why would generalizing cross products or jacobians help?

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newmathquestions

Adding and Subtracting Matrices

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Let $a_1=1$ and $$a_{n+1} = \sqrt{a_1+a_2+...+a_n}, \;\forall \; n\in \mathbb{N}^*.$$ Show $$\lim_{n\to\infty}\frac{a_n}{n} = \frac{1}{2}.$$

I found that at $n=2$, $\frac{a_n}{n}=\frac{1}{2}$ and decreases until $n=5$, where it starts increasing. I used induction to prove that the sequence monotonically increases after $n=5$ but I am having trouble proving its limit is $\frac{1}{2}$. Any insight is appreciated, thanks!

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yaymath

Venn Diagram Method for HCF and LCM - Concept - Example

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How to Find HCF and LCM Easily - Concept - Examples

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Finding LCM and HCF of Polynomials

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My attempt:

$z^2 = 3(x^2 + 2y^2)$

so $3|z^2$ and thus $3|z$

letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$

Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.

Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even

If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.

But I don't know how to proceed from here. Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.

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A_miracle

Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$ where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

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Mike D

What does the research suggest about racial equity issues in mathematics education?

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The pandemic has postponed or canceled a lot of things this year, but luckily learning isn’t one of them. Check out these picks for new Wolfram Language books that will help you explore new software, calculus, engineering and more from the comfort of home.

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With this book, you’ll learn how to:

• Quickly create notebooks that combine calculations, graphics, interactive examples and notes.
• Enter free-form input and get solutions for a variety of calculations (e.g. arithmetic, algebra, calculus, linear algebra).
• Access step-by-step solutions, suggestions for next steps and related computations.
• Create 2D, 3D and interactive graphics with controls to dynamically change the parameters.
• Use previous results in future calculations, assign variables and define functions.
• Create dynamic slide show presentations with interactive elements that can be changed on the fly.

Analysis with Mathematica: Volume 1: Single Variable Calculus

In this De Gruyter textbook, Galina Filipuk and Andrzej Kozłowski explore how Mathematica is able to do much more than just numerics. This text shows how to tackle real mathematical problems from basic analysis. Readers will learn how Mathematica represents domains, qualifiers and limits to implement actual proofs—a requirement to unlock the huge potential of Mathematica for a variety of applications. While this volume focuses on single variable calculus, readers should also keep an eye out for Volume 2: Multi-variable Calculus, which is scheduled to be published in 2021.

Symmetry in Optics and Vision Studies: A Data-Analytic Approach

In this book, authors Marlos A. G. Viana and Vasudevan Lakshminarayanan present an introduction to the foundations, interpretations and data-analytic applications of symmetry studies with an emphasis on applications in optical sciences. Symmetry studies connect group theoretic and statistical methods for data summary and inference. This book is written for an audience familiar with calculus and linear algebra as well as introductory statistics. The book reviews finite group theory in the introductory chapters. Computational tools used in the text are also available for download as Mathematica notebooks, offering readers an interactive experience.

Advanced Engineering Mathematics with Mathematica

Advanced Engineering Mathematics with Mathematica presents advanced analytical solution methods that are used to solve boundary value problems in engineering and integrates these methods with Mathematica procedures. In this book, author Edward B. Magrab emphasizes the Sturm–Liouville system and the generation and application of orthogonal functions, which are used by the separation of variables method to solve partial differential equations. It introduces the relevant aspects of complex variables, matrices and determinants, Fourier series and transforms, solution techniques for ordinary differential equations, the Laplace transform and procedures to make ordinary and partial differential equations used in engineering non-dimensional. Magrab presents numerous and widely varied solved boundary value problems to show the diverse applications of the material.

Computer Modeling and Simulation of Dynamic Systems Using Wolfram SystemModeler

Geared toward students and professionals in the field of dynamic system modeling, this book briefly discusses the main provisions of the theory of modeling. It also describes in detail the methodology for constructing computer models of dynamic systems using the Wolfram visual modeling environment, System Modeler, and provides illustrative examples of solving problems of mechanics and hydraulics. This book serves as a supplement to university courses in modeling and simulation of dynamic systems, but it’s also a great resource for anyone looking for examples and explanations of System Modeler.

Applied Engineering Mathematics

This textbook is written for undergraduate engineering students (specifically those in years 2–4 of an engineering degree course) and it places a strong emphasis on visualization and the methods and tools needed across the whole of engineering. Author Brian Vick emphasizes the visual approach while avoiding excessive proofs and derivations. The visual images explain and teach the mathematical methods. The book’s website provides Mathematica notebooks with dynamic and interactive code to accompany the examples for the reader to explore with Mathematica or the free Wolfram Player, and it provides access for instructors to a solutions manual.

If you’re interested in finding more books that use the Wolfram Language, check out the full collection on the Wolfram Books site. If you’re working on a Mathematica or Wolfram Language book, contact us to find out more about our options for author support and to have your book featured in an upcoming blog post!

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 During these difficult pre-election coronavirus days I have been turning to poetry, and especially favoring -- for their brevity -- Haiku.  The January 2018 issue of the Journal of Humanistic Mathematics offers a folder "Math in Seventeen Syllables:  A Folder of Mathematical Haiku" -- with more than thirty poets sharing poetic insights using this ancient form.  Here is one, by Laura Kline, that spoke to me today:

          Peaceful living and
          Nicely balanced equations
          How we long for both

from Intersections -- Poetry with Mathematics
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noreply@blogger.com (JoAnne Growney)

In my latest Bloomberg post, I make the case that, in the best case scenario that Trump is gone in January, we have a massive amount of work to catch up on, especially with regard to combatting the power and malevolence of big tech.

If Biden Wins, Prepare for an Epic Policy Hangover

There’s so much to fix beyond what Trump has broken.

For more of my Bloomberg columns, go here.

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Cathy O'Neil, mathbabe

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Consider the following one-person game:

A player starts with score $0$ and writes the number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$) and writes two positive integers (call them $b$ and $c$) such that $b + c = a$. The player then adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$s on the whiteboard. Then the game is over, and the final score is calculated.

Example: At the first step, a player erases $20$ and writes $14$ and $6$, and gets a score of $14 × 6 = 84$. In the next step, she erases $14$, writes $9$ and $5$, and adds $9 × 5 = 45$ to her score. Her score is now $84 + 45 = 129$. In the next step, she may erase any of the remaining numbers on the whiteboard: $5$, $6$ or $9$. She continues until the game is over.

Alya and Bob play the game separately. Alya manages to get the highest possible final score. Bob, however, manages to get the lowest possible final score. What is the difference between Alya’s and Bob’s final scores?

I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.

from Hot Weekly Questions - Mathematics Stack Exchange
Lucius

I have read multiple posts on here and in other places where most people seem to recommend to learning linear algebra before abstract algebra. Is that because vector spaces are simpler to understand than groups and rings? I am having some challenges with understanding how certain aspects of vector spaces work, I was wondering if learning about rings and/or groups can help me better understand how vector spaces work?

from Hot Weekly Questions - Mathematics Stack Exchange
bernando_vialli

In the Principles of Mathematical Analysis, 3rd ed. by Rudin, an ordered field is defined as:

"An ordered field $\mathbb{F}$ is a field $F$ which is also an ordered set, such that for all $x, y, z \in \mathbb{F}$:

1. $x+y<x+z$ whenever $y<z$
2. $xy>0$ if $x>0, y>0$."

The idea of these additional axioms (apart from the field axioms for addition, multiplication and distributivity) is that it tells us what happens to the relation or ordering between two elements of the field when any of the two binary operations are applied.

I wanted to know why the second axiom is defined with respect to the additive identity. Why not any arbitrary member of the field? As in, why not define it as $xy>a$ if $x>a, y>a; a \in \mathbb{F}$?

In more abstract terms, if I create my own field with two binary operations, will ordering relative to one of the operations always be with respect to the identity element of the other binary operation? Is this a consequence of the field axioms?

Has it something to do with the following consequences:

1. $0x=0$ $\forall$ $x \in \mathbb{F}$ (multiplying by the additive identity is always $0$)
2. $(-x)y=-xy$ (multiplying two objects on opposite sides of the additive identity always yields an object 'lesser than' the additive identity)
3. $(-x)(-y)=xy$ (multiplying two objects on the same side of the additive identity always yields an object 'greater than' the additive identity)

---

For some context on the kind of explanation I'm seeking, I had a similar question about why the additive identity was preferentially excluded when defining the multiplicative identity and multiplicative inverse. As in, if I wanted to create my own field (an abelian group with two binary operations and a distributive law that binds the two operations), would I always have to exclude one of the identities? This is how I resolved it.

The exclusion is because of the way distributivity is defined to place multiplication over addition. Specifically, the field axiom for distributivity demands that $x(y+z)=xy+xz$, and not $x+yz=(x+y)(x+z)$.

The consequence of our definition is that $$0x + 0x = (0+0)x,$$ or $$0x + 0x = 0x,$$ meaning that $0x = 0$, $\forall$ $x \in \mathbb{F}$.

Thus, multiplication by the additive identity always returns the additive identity, so multiplication cannot be an invertible binary operation if $0$ is included. A similar resolution would have been necessary if we had defined the distributive law otherwise, where we would have excluded $1$ from our definition of the additive identity and additive inverse.

from Hot Weekly Questions - Mathematics Stack Exchange
PhutureFysicist

Let $A$ be a set and let $ f: A \to B $ be a surjective function. Prove that there exists a subset $ C \subseteq A $ and a function $ \phi: C \to B $ such that $ \phi $ is bijective.

I did it by giving particular examples, giving $ A = \{1,2,3 \} $ and $ B = \{1,2 \} $ and the function $f(x)=\begin{cases} 1& \text{ if } x=1 \\ 1& \text{ if } x=2 \\ 2& \text{ if } x=3 \end{cases}$ it is clearly a surjective function. Now if $ C = A \setminus \{1 \} $ what is desired remains. Guiding me from particular examples, I thought to consider the set $ C $ of all the images that are repeated because the function is not necessarily injective, as in the previous example removing 1, since it repeated with 2. But I don't know if the idea is well and I don't know how to express that set either. I tried as follows, $ C = A\setminus \{f (a) = f (b) \wedge a \neq b \mid \forall a, b \in A \} $ but I don't know if it's ok. Help.

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Andrei Agamez

In Jacobson’s book BAII, he gave two exercises to show that there exists functor that does not preserve monic or epic.

Ex-1.Let M and N be monoids as categories with a single object. Show that in this identification,a functor is a homomorphism of M into N.

Ex-2.Use Ex-1 to construct a functor $F$ and a monic(epic) $f$ such that $F(f)$ is not monic(epic)

I have solved Ex-1. However, I don’t know how to construct concrete example to solve Ex-2.

I do hope someone can give me some hints. Thank you very much!

from Hot Weekly Questions - Mathematics Stack Exchange
zik2019

My problem is the following: "Suppose that $f\colon[-1,1]\to\mathbb{R}$ is continuous and differentiable on $(-1,1)$, $f(1) = 0$ and $f(x)>0$ for $x\in(-1,1)$. Show that there exists $c\in(-1,1)$ such that $mf(c)f'(-c) = nf'(c)f(-c)$, in which $m\in N$ and $n\in N$."

I can prove if when $M=N$: there exists $c\in(-1,1)$ for which $f(c)f'(-c) = f'(c)f(-c)$ using $g(x) = f(x)f(-x)$, which satisfies $g(1) = g(-1) = 0$ and Rolle's theorem, but I can't prove it when $m ≠ n$. Can someone please tell me how to prove it?

from Hot Weekly Questions - Mathematics Stack Exchange
notgoodatmath49

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Difference of Two Squares Worksheet

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Difference of Two Squares - Concept - Examples

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Remark: throughout this post we work in the smooth category, so that all manifolds, bundles, maps etc are assumed to be smooth.

An exercise asks me to show that there exists no principal $S^1$-bundle isomorphism between the Hopf bundle $S^1 \to S^3 \xrightarrow\pi S^2$ (where $S^1$ acts on $S^3$ by multiplication as usual), and the bundle with the same fiber structure but equipped with the "inverted" action $(w_1,w_2) \cdot z = (w_1z^{-1}, w_2z^{-1})$.

Aiming to reach a contradiction I suppose an $S^1$-bundle isomorphism between these two bundles exists. This is a diffeomorphism $\Phi: S^3 \to S^3$ satisfying: \begin{array}{rc} \pi \circ \Phi = \pi, & (*)\\[0.2cm] \Phi(w_1 z,w_2 z) = \Phi(w_1,w_2)z^{-1}. & (**) \end{array}

As far as I can see, the conditions $(*)$ and $(**)$ alone do not violate each other, since $\pi$ is invariant under multiplication by elements of $S^1$ (and hence their inverses) anyway. So I guess there should be issues with the diffeomorphy of $\Phi$. More particularly, I would guess smoothness/continuity of $\Phi$ (or its inverse) is an issue, since again I don't see why the equivariance condition $(**)$ would contradict the bijectivity of $\Phi$.

But I'm failing to see how to derive a contradiction here. Am I missing something simpler or am I on the right track?

from Hot Weekly Questions - Mathematics Stack Exchange
MSDG

On the upcoming test I will be given a problem of type:

Find all normal subgroups $H$ in $F_n$ such that $F_n/H \cong G$.

Here $n$ is a small integer, likely 2 or 3, and $G$ is an Abelian group given as product of some cyclic groups.

Question: is there a more-or-less routine algorithm for such problems?

I do know an approach. I count two parameters: the number of epimorphisms $\#(F_n \twoheadrightarrow G)$ and the number of automorphisms $\#Aut(G)$. The answer then is $\frac{\#(F_n\twoheadrightarrow G)}{\#Aut(G)}$. This solution is based on the following facts: 1) every normal subgroup is a kernel of some homomorphism; 2) $F_n/Ker\,\phi \cong G$ if $\phi$ is an epimorphism; 3) $Ker\,\phi=Ker\,\psi$ iff $\psi = \alpha\circ\phi$ for some $\alpha \in Aut(G)$.

The problem is, very often I'm not sure how do I count one or both of these parameters. I could go through all homomorphisms and automorphisms manually, but the numbers in problem are usually too large.

Here are some cases to show how big numbers might be on the actual test. No need to solve them!

1. $n=3$, $G=\mathbb Z_{13}^{3}$
2. $n=3$, $G=\mathbb Z_{10}^3\times \mathbb Z$
3. $n=3$, $G=\mathbb Z_{70}\times \mathbb Z_{30} \times \mathbb Z_{15}$
4. $n=2$, $G=\mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_5$
5. $n=3$, $G=\mathbb Z_{55} \times \mathbb Z_{7} \times \mathbb Z_{77}$
6. $n=3$, $G=\mathbb Z_{75} \times \mathbb Z_{375} \times \mathbb Z_{125} \times \mathbb Z_{333}$

from Hot Weekly Questions - Mathematics Stack Exchange
dnes

I would like to know if there is a nice proof of the fact that every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without use of the Lefschetz fixed point theorem.

from Hot Weekly Questions - Mathematics Stack Exchange
Math

Special Products of Binomials - Concept - Examples

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Special Products of Binomials Worksheet

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Find Degree of Sum and Difference of Two Polynomials

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Let $G$ be a group and $g \in G$. Define $\lambda_g: G \rightarrow G$ by $\lambda_g(x) = gx$ for all $x \in G$. Similarly, define $\tau_g: G \rightarrow G$ by $\tau_g(x) = xg$ for all $x \in G$. If $\theta: G \rightarrow G$ is a bijection such that $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, show that $\theta = \tau_h$ for some $h \in G$.

My proof is as follows:

Since $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, we have $$ g \theta(x) = \lambda_g(\theta(x)) = \theta (\lambda_g(x)) = \theta(gx) $$ for all $g, x \in G$. In particular, we have $$x^{-1} \theta(x) = \theta(x^{-1}x) = \theta(1)$$ or $\theta(x) = x\theta(1)$ for all $x \in G$. Hence $\theta = \tau_{\theta(1)}$ as required.

It looks pretty simple, but I don't seem to use the condition that "$\theta$ is a bijection".

Is my proof correct? Thank you.

from Hot Weekly Questions - Mathematics Stack Exchange
Enya Hsu

I have this series and I need to determine whether it is divergent or convergent, are my calculations correct and it is divergent or am I not seeing something

$$ \sum_{n=2}^\infty (-1)^n \left( \frac{n}{2n+1}\right)^2 $$ since it is an alternating series I find the limit of a_n $$ \lim_{n\to \infty}\left( \frac{n}{2n+1}\right)^2 = \frac{n^2}{(2n+1)^2} = \frac{n^2}{4n^2+4n+1} = \frac{1}{4} $$ and since it does not equal to 0 by the alternating divergence test it s divergent ??

from Hot Weekly Questions - Mathematics Stack Exchange
GregoryStory16

Let $(X,\mathscr{O}_X)$ be a locally ringed space. If it is a scheme, the natural notion of dimension is the dimension of the subjacent topological space (the size of the biggest chain of irreducible closed subsets). But if $X$ is a manifold, I think that the natural notion of dimension is perhaps the dimension of the Zariski tangent space.

Is there a "good" notion of dimension in locally ringed spaces? If so, how does this notion relates to the dimension of the subjacent topological space, the dimension of the Zariski tangent space and the Krull dimension of the stalks?

from Hot Weekly Questions - Mathematics Stack Exchange
Gabriel

Let $\{f_n\}_{n\in \mathbb N}$ be a sequence of uniformly convergent functions on the interval [0,1]. Find the limit of the following integral: $$\lim_{n\to \infty} \int_0^1nf_n(t)e^{-nt}\,dt$$

Here is my thought process so far. In isolation l'hospital's rule would indicate that $\lim_{n\to \infty} \frac{n}{e^{nt}}$ would tend to zero, and so I expect that this exponential term will be more dominating.

Normally Lebesgue's Dominated convergence theorem (or at the very least his monotone convergence theorem) would allow us to say that this uniformly convergent sequence of functions would converge to f...however I am guessing the surrounding terms sufficiently mess this up to the point where we no longer have convergence. My guess is the point of this question is to prove that this is the case.

We know that $\lim_{n\to \infty}\{f_n\}_{n\in \mathbb N}$ is measurable, but If I can show that $g_n= \frac{n}{e^{nt}}$ is not measurable for $t\in[0,1]$ (because it tends to zero), then can I say this limit does not exist? I'm pretty confused, as is abundantly clear hahahaha.

from Hot Weekly Questions - Mathematics Stack Exchange
Chair

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I am reading Jörg Brendle's Bogota note, and the author claimed that Sacks forcing with countable support product does not satisfy a product lemma. Especially, he mentioned that if $\mathbb{S}_I$, $|I|$-many countable support product of Sacks forcing $\mathbb{S}$, adds $|I|$ Sacks reals $\langle s_i\mid i\in I\rangle$ and $i\neq j$, then $s_j$ is not $\mathbb{S}$-generic over $V[s_i]$.

Let me take $I=2$ for simplicity. One subtle point of the claim is that $\mathbb{S}$ need not be absolute between $V$ and $V[s_0]$. In fact, the product lemma for general forcing shows that $s_1$ would be $\mathbb{S}^V$-generic over $V[s_0]$. Hence I think the author intends to claim that $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$.

However, $\mathbb{S}^V\neq\mathbb{S}^{V[s_0]}$ does not automatically mean $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$. My question is:

How to prove $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$?

I am new at Sacks forcing, so I have no idea how to start. I would appreciate your help!

from Hot Weekly Questions - Mathematics Stack Exchange
Hanul Jeon

I would like to prove the following inequalities on Laguerre polynomials evaluated at point 2: $$ (-1)^n \text{Laguerre}_n(2) \leq 1 $$ This seems to hold numerically. I tried to use the recurrence relation between Laguerre polynomials but I was not successful. Any ideas? I also tried expression via Bessel function $J_0$ and then contour integral but it was not successful.

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Plussoyeur

$$\sum_{n=1}^\infty\bigl(3\log(n^2+1)-2\log(n^3+1)\bigr)$$

I tried the limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$ and then applied L'Hopital's Rule. What I ended up was a limit which was a real number and as the series of $\frac1{n^2}$ converges, my original series converges as well. However, I was wondering if there is another way of solving the problem, one that involves fewer computations.

from Hot Weekly Questions - Mathematics Stack Exchange
abigmistake

Powers of Monomials Worksheet

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Let $p$ be an odd prime, and let $n_1,\dots,n_{p-2}, m$ be even integers such that $n_1 < n_2 < \dots < n_{p-2}$ and \begin{equation} 2m > \sum_{i=1}^{p-2} n_i^2. \end{equation} Consider the polynomial \begin{equation} g(x)=(x^2 + m)(x - n_1) \dots (x - n_{p-2}). \end{equation} From Rolle's Theorem, we know that for each $i=1,2,\dots,p-3$, there exists $x_i \in (n_i,n_{i+1})$ such that $g'(x_i)=0$. So $g'(x)$ has at least $p-3$ distinct real zeroes. My question is: can $g'(x)$ have more than $p-3$ distinct real zeroes?

I do not know the answer, but for sure the constraints on the parameters are relevant here. For example the polynomial $g(x)=(x^2+1)(x-4)(x-2)(x+2)$ has derivative $g'(x)=5x^4-16x^3-9x^2+24x-4=(x-1)(5x^3-11x^2-20x+4)$ which has four distinct real roots, as you can check on WolframAlpha.

NOTE This strange polynomial arises in the construction given by R. Brauer of a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $p$ whose Galois group over $\mathbb{Q}$ is isomorphic to the symmetric group $\mathcal{S}_p$: see Jacobson, Basic Algebra I, $\S 4.10$. The question I asked is clearly irrelevant for the construction, but has intrigued me, since I could not answer it in the negative nor I could find some counterexample.

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Maurizio Barbato

Worksheet on Multiplying Polynomials

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Multiplying Polynomials Worksheet

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Multiplying Polynomials - Concept - Examples

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My question essentially has to do with the derivative of a Contour Integral's parameterized curve. $$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$ to be exact. Where $\partial \Omega(x)$ is a Jordan curve which is differentiable for any $x \in \mathbb{C}$, and $f(n, x)$ is any function integrable around $\partial \Omega(x)$ in respect to $n$.

My work has essentially gotten down to:

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \int_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n$$ $$+\int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \; \mathrm{d}\theta$$ $$+\int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

Where $\gamma$ is the parameterized curve of $\partial \Omega$, and $f_x(n, x) = \frac{\partial f(n, x)}{\partial x}$. I am unsure of how to simplify this further or if this is even a decent approach. Does anybody have a good resource for this?

My goal is to write this derivative as multiple contour integrals, without using $\gamma$ within them.

from Hot Weekly Questions - Mathematics Stack Exchange
Jackson

This year as schools and communities explore ways to celebrate Halloween beyond trick-or-treating, ghost stories, joke telling and spooky math problems might feature in.  Here’s one to get started!

Q: Are any Halloween monsters good at Math?

A: Yes, if you Count Dracula!

from CUNYMath Blog
Ashleigh Thompson

Multiplying and Dividing Monomials - Concept - Examples

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Multiplying and Dividing Monomials Worksheet

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How can I find the function $f(t)$, $t\geq0$ continuous and everywhere differentiable that minimises the following integral

$$L = \int_0^\infty f(t)^2 + f'(t)^2dt$$

such that $f(0)=x$.

Intuitively I am asking what is the function $f$ that starts at $x$ and wants to approach $0$ the fastest but keeping the derivative smaller. My intuition tells me that somehow the function $f$ should be a decreasing exponential. But how to prove it? how does one tackle this problem?

Here is what I tried to do: I tried to think of a similar discrete problem: find the sequence $h_n$ that minimises

$$L = \sum_{i=1}^\infty h_i^2 + (h_{i-1}-h_i)^2$$ such that $h_0=x$

Some observations: If $x=0$ then trivially $h_i=0$ $\forall i$

On the other hand, if $h_i=0$ $\forall i$, then $L = x^2$. (here the function would suddenly drop to $0$)

If instead we had gradually decreased it with one step: $h_1=\frac{x}{2}$ and $0$ otherwise, then $L=\frac{x^2}{4}+\frac{x^2}{4}+\frac{x^2}{4}=\frac{3x^2}{4}$ which is lower than the previous example, showing that it is better to decrease it gradually...

Then my intuition tells me that if $h_i = \alpha h_{i - 1}$ for $\alpha<1$, then $h_i=x\alpha^i$ and $h_{i-1}-h_i = x\alpha^{i-1}(1-\alpha)$.

For this case, this means that $L = \sum_{i=1}^\infty (x\alpha^i)^2 + (x\alpha^{i-1}(1-\alpha))^2=x^2\sum_{i=1}^\infty (\alpha^2)^i+x^2(1-\alpha)^2\sum_{i=1}^\infty (\alpha^2)^{i-1}$

$$L=x^2\frac{\alpha^2}{1-\alpha^2}+x^2(1-\alpha)^2\frac{1}{1-\alpha^2}=\frac{x^2}{1-\alpha^2}(2\alpha^2-2\alpha + 1)$$

Now if we try to find $\alpha$ that minimises this expression we can take its derivative

$\frac{dL}{d\alpha}=x^2\frac{(4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)}{(1-\alpha^2)^2}=0 \Leftrightarrow (4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)=0 \Leftrightarrow 4\alpha-4\alpha^3-2+2\alpha^2+4\alpha^3-4\alpha^2+2\alpha=0\Leftrightarrow -2\alpha^2+6\alpha-2=0$

$\alpha=1+\frac{1-\sqrt{5}}{2}$ which seems to be related to the golden ratio

from Hot Weekly Questions - Mathematics Stack Exchange
gota

Consider a bijection $f: \mathbb{N} \to \mathbb{N}$:

We define the (possible infinite) number of "crosses" or "intersections" by the following:

For every pair $i,j$ with $i<j, $ define:

$$ \delta_{ij}=\begin{cases} 0&\ \text{if} \ f(j) > f(i)\\[8pt] 1&\ \text{if} \ f(j) < f(i)\\[8pt] \end{cases} $$

Then the number of (possibly infinite) intersections is $\ s(f) = \displaystyle \sum_{i=1}^\infty \left( \sum_{j=i+1}^\infty\delta_{ij} \right)$.

(s depends only on the bijection $f$).

Also, define a bijection eventually equal to the identity to be one in which $ \exists N$ such that $f(n) = n \quad \forall n \geq N$.

My conjecture is that a bijection $f: \mathbb{N} \to \mathbb{N}$ is eventually equal to the identity $ \iff s$ is finite, but I don't know how to prove this formally.

Note that my conjecture would be false with bijections $f: \mathbb{Z} \to \mathbb{Z}$, for example:

The pigeonhole principle came to mind, but I'm not sure if it's useful or necessary for a proof.

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Adam Rubinson

Multiplying a Binomial by a Trinomial - Concept - Examples

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I find it confusing that all of the following statements are true :

1. The computable real numbers are countable. $-\hspace{-3pt}-$ Alan Turing, "On Computable Numbers, with an Application to the Entscheidungsproblem"
2. In constructive analysis, the real numbers are uncountable. $-\hspace{-3pt}-$ Everett Bishop, Foundations of Constructive Analysis
3. "every mathematical statement [in constructive analysis] ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results" $-\hspace{-3pt}-$ Ibid.

Perhaps I am misunderstanding something.

I suppose I really have two questions. In constructive analysis :

1. Why isn't every real number computable?
2. How is it possible to construct an uncountable set?

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simple jack