Is analytic continuation well-defined as a summation method? I am not well versed in summation methods or complex analysis, so I will be presenting a detailed view of my question with examples to illustrate my point as well as a few guiding questions that got me to this point. The guiding questions are mostly wishful thinking on my part. I apologize in advance for the lack of focus on a specific question in the post, but I would prefer responses address the question given by the title if the answer is known, and comment on the line of guiding questions if the answer is unknown.

One specific way of assigning a value to a series that is divergent in the usual sense of partial sums is to see that it matches an expression for another series given in terms of some parameter or variable. As an example, consider the following series.
$$1-1+1-1+...$$
This series could be identified as the following series evaluated at $z=-1$.
$$\sum\limits_{k=0}^\infty {z^k}$$
This series is the well known geometric series, and is analytic on some open set of the complex plane, but -1 is not part of that set. As such, we analytically continue the function, which provides us with the function $f(z) = \frac{1}{1-z}$. This is useful because this function is well defined at the point $z=-1$ of interest, and we therefore assign the value $f(-1)=\frac{1}{2}$ to our original series.

What seemed important here were the following facts:
1.) We found a function (in this case a series) with series representation that matches our series of interest, term-by-term for the value of some input.
2.) The function we found could be analytically continued to a region of the complex plane with a well-defined value at our input of interest.

I am moving forward with the assumption that these are the only two requirements for the general technique that I refer to as "analytic continuation as a summation method". This led to the following line of questioning which inspired the title of this post.

Given a series $S$, is the existence of a function satisfying properties 1 and 2 guaranteed?
If so, is the function unique?
If not, is the value obtained for all such functions the same so that this method could be considered well-defined? If this is not true for all series, is there a known subset $U$ for which is is true?

Related but somewhat tangent questions that probably deserve their own post but which feel contingent on this post would include:
For each absolutely convergent series in $U$, does the value obtained by analytic continuation agree with the standard sum?
For two conditionally convergent series $S_1$ and $S_2$ in $U$ that are permutations of each other, does the analytic continuation method always assign the same value? I ask this question in particular because the standard sum would not satisfy this property in general due to the Riemann rearrangement theorem.

I've failed to find a counterexample that satisfies both properties, but there is an interesting series which has a function satisfying properties 1 and 2 but also has another that only satisfies property 1.
$$1+1+1+1+...$$
The above series could be attributed to a geometric series with ratio 1, but this in turn is still a point where the analytic continuation of the geometric series fails to give a finite value. However, if we use the series
$$\sum\limits_{n=1}^\infty \frac{1}{n^s}$$
we recognize that when $s=0$, we get term-by-term matching. So, we analytically continue this function to obtain the Riemann zeta Function, which gives $\zeta (0)=-\frac{1}{2}$. It would be interesting if there was a well-motivated way to obtain the value $-\frac{1}{2}$ from the geometric series continuation that worked for other such functions that satisfy property 1 but not property 2, but that is just some more wishful thinking beyond the already wishful thinking of my previous line of reasoning.

Here's the classical counterexample to your question. It was an important topic of discussion in the 18th century and the early 19th century, long before any infinite series had a firm foundation, let alone divergent series. It's adapted from Wikipedia:History of Grandi's series. Consider your motivating series:
$$1-1+1-1+1-1+\cdots$$
Define the functions $f$ and $g$ on the open unit disk $|z|<1$:
$$f(z)=1-z+z^2-z^3+z^4-z^5+\cdots=\frac{1}{1+z}$$
$$g(z)=1-z^2+z^3-z^5+z^6-z^8+\cdots=\frac{1+z}{1+z+z^2}$$
We don't have to analytically continue very far! Just take a limit at the boundary of the original domains of $f$ and $g$. Sadly, we get $\lim_{z\to1}f(z)=\frac12$ versus $\lim_{z\to1}g(z)=\frac23$.

To paraphrase Lagrange, the problem is that the series used to describe $g$ is not actually a power series. So this example shows why you'll need to restrict the kinds of series under consideration in order to get an interesting result.


Teaching myself differential topology and differential geometry I have a hazy notion of some stuff in differential geometry and a better, but still not quite rigorous understanding of basics of differential topology.

I have decided to fix this lacuna once for all. Unfortunately I cannot attend a course right now. I must teach myself all the stuff by reading books.

Towards this purpose I want to know what are the most important basic theorems in differential geometry and differential topology. For a start, for differential topology, I think I must read Stokes' theorem and de Rham theorem with complete proofs.

Differential geometry is a bit more difficult. What is a connection? Which notion should I use? I want to know about parallel transport and holonomy. What are the most important and basic theorems here? Are there concise books which can teach me the stuff faster than the voluminous Spivak books?

Also finally I want to read into some algebraic geometry and Hodge/Kähler stuff.

Suggestions about important theorems and concepts to learn, and book references, will be most helpful.

ADDITION: I have compiled what I think is a definitive collection of listmanias at Amazon for a best selection of books an references, mostly in increasing order of difficulty, in almost any branch of geometry and topology. In particular the books I recommend below for differential topology and differential geometry; I hope to fill in commentaries for each title as I have the time in the future.

If you want to have an overall knowledge Physics-flavored the best books are Nakahara's "Geometry, Topology and Physics" and above all: Frankel's "The Geometry of Physics" (great book, but sometimes his notation can bug you a lot compared to standards).

If you want to learn Differential Topology study these in this order: Milnor's "Topology from a Differentiable Viewpoint", Jänich/Bröcker's "Introduction to Differential Topology" and Madsen's "From Calculus to Cohomology". Although it is always nice to have a working knowledge of general point set topology which you can quickly learn from Jänich's "Topology" and more rigorously with Runde's "A Taste of Topology".

To start Algebraic Topology these two are of great help: Croom's "Basic Concepts of Algebraic Topology" and Sato/Hudson "Algebraic Topology an intuitive approach". Graduate level standard references are Hatcher's "Algebraic Topology" and Bredon's "Topology and Geometry", tom Dieck's "Algebraic Topology" along with Bott/Tu "Differential Forms in Algebraic Topology."

To really understand the classic and intuitive motivations for modern differential geometry you should master curves and surfaces from books like Toponogov's "Differential Geometry of Curves and Surfaces" and make the transition with Kühnel's "Differential Geometry - Curves, Surfaces, Manifolds". Other nice classic texts are Kreyszig "Differential Geometry" and Struik's "Lectures on Classical Differential Geometry".

For modern differential geometry I cannot stress enough to study carefully the books of Jeffrey M. Lee "Manifolds and Differential Geometry" and Liviu Nicolaescu's "Geometry of Manifolds". Both are deep, readable, thorough and cover a lot of topics with a very modern style and notation. In particular, Nicolaescu's is my favorite. For Riemannian Geometry I would recommend Jost's "Riemannian Geometry and Geometric Analysis" and Petersen's "Riemannian Geometry". A nice introduction for Symplectic Geometry is Cannas da Silva "Lectures on Symplectic Geometry" or Berndt's "An Introduction to Symplectic Geometry". If you need some Lie groups and algebras the book by Kirilov "An Introduction to Lie Groops and Lie Algebras" is nice; for applications to geometry the best is Helgason's "Differential Geometry - Lie Groups and Symmetric Spaces".

FOR TONS OF SOLVED PROBLEMS ON DIFFERENTIAL GEOMETRY the best book by far is the recent volume by Gadea/Muñoz - "Analysis and Algebra on Differentiable Manifolds: a workbook for students and teachers". From manifolds to riemannian geometry and bundles, along with amazing summary appendices for theory review and tables of useful formulas.

EDIT (ADDED): However, I would argue that one of the best introductions to manifolds is the old soviet book published by MIR, Mishchenko/Fomenko - "A Course of Differential Geometry and Topology". It develops everything up from $\mathbb{R}^n$, curves and surfaces to arrive at smooth manifolds and LOTS of examples (Lie groups, classification of surfaces, etc). It is also filled with LOTS of figures and classic drawings of every construction giving a very visual and geometric motivation. It even develops Riemannian geometry, de Rham cohomology and variational calculus on manifolds very easily and their explanations are very down to Earth. If you can get a copy of this title for a cheap price (the link above sends you to Amazon marketplace and there are cheap "like new" copies) I think it is worth it. Nevertheless, since its treatment is a bit dated, the kind of algebraic formulation is not used (forget about pullbacks and functors, like Tu or Lee mention), that is why an old fashion geometrical treatment may be very helpful to complement modern titles. In the end, we must not forget that the old masters were much more visual an intuitive than the modern abstract approaches to geometry.

NEW!: the book by Mishchenko/Fomenko, along with its companion of problems and solutions, has been recently typeset and reprinted by Cambridge Scientific Publishers!

If you are interested in learning Algebraic Geometry I recommend the books of my Amazon list. They are in recommended order to learn from the beginning by yourself. In particular, from that list, a quick path to understand basic Algebraic Geometry would be to read Bertrametti et al. "Lectures on Curves, Surfaces and Projective Varieties", Shafarevich's "Basic Algebraic Geometry" vol. 1, 2 and Perrin's "Algebraic Geometry an Introduction". But then you are entering the world of abstract algebra.

If you are interested in Complex Geometry (Kähler, Hodge...) I recommend Moroianu's "Lectures on Kähler Geometry", Ballmann's "Lectures on Kähler Manifolds" and Huybrechts' "Complex Geometry". To connect this with Analysis of Several Complex Variables I recommend trying Fritzsche/Grauert "From Holomorphic Functions to Complex Manifolds" and also Wells' "Differential Analysis on Complex Manifolds". Afterwards, to connect this with algebraic geometry, try, in this order, Miranda's "Algebraic Curves and Riemann Surfaces", Mumford's "Algebraic Geometry - Complex Projective Varieties", Voisin's "Hodge Theory and Complex Algebraic Geometry" vol. 1 and 2, and Griffiths/Harris "Principles of Algebraic Geometry".

You can see their table of contents at Amazon.
Hope this helps... good luck!

My take on it is like this:

Basics of Smooth manifolds

Loring Tu, Introduction to manifolds - elementary introduction,
Jeffrey Lee, Manifolds and Differential geometry, chapters 1-11 - cover the basics (tangent bundle, immersions/submersions, Lie group basics, vector bundles, differential forms, Frobenius theorem) at a relatively slow pace and very deep level.
Will Merry, Differential Geometry - beautifully written notes (with problems sheets!), where lectures 1-27 cover pretty much the same stuff as the above book of Jeffrey Lee

Basic notions of differential geometry

Jeffrey Lee, Manifolds and Differential geometry, chapters 12 and 13 - center around the notions of metric and connection.
Will Merry, Differential Geometry - lectures 28-53 also center around metrics and connections, but the notion of parallel transport is worked out much more thoroughly than in Jeffrey Lee's book.
Sundararaman Ramanan, Global calculus - a high-brow exposition of basic notions in differential geometry. A unifying topic is that of differential operators (done in a coordinate-free way!) and their symbols.

What I find most valuable about these books is that they try to avoid using indices and local coordinates for developing the theory as much as possible, and only use them for concrete computations with examples.

However, the above books only lay out the general notions and do not develop any deep theorems about the geometry of a manifold you may wish to study. At this point the tree of differential geometry branches out into various topics like Riemannian geometry, symplectic geometry, complex differential geometry, index theory, etc.

I will only mention one book here for the breadth of topics discussed

Arthur Besse, Einstein manifolds - reviews Riemannian geometry and tells about (more or less) the state of the art at 1980 of the differential geometry of Kähler and Einstein manifolds.

Differential topology

Amiya Mukherjee, Differential Topology - first five chapters overlap a bit with the above titles, but chapter 6-10 discuss differential topology proper - transversality, intersection, theory, jets, Morse theory, culminating in h-cobordism theorem.
Raoul Bott and Loring Tu,  Differential Forms in
Algebraic Topology - a famous classic; not a book on differential topology - as the title suggests, this is a treatment of algebraic topology of manifolds using analytic methods


Using induction to prove formula I am revising for my test from Discrete math. I have come to this problem.

I am to prove by using mathematical induction that

$6\times7^{n} - 2 \times 3^{n}$ is divisible by 4. for $n \ge 1$ ;

I created basic step :

$6\times7^{1} - 2\times3^{1} = 36 $

and induction step

$\forall n\ge 1, \exists K: 6\times7^{n} - 2\times3^{n} = 4K \Rightarrow  \forall n \ge 1, \exists l: 6\times7^{n+1} - 2\times3^{n+1} = 4l$

we can transform the formula into

$6\times7\times7^{n} - 2\times3\times3^{n}$

which is basicly

$42\times7^{n} -6\times3^{n}$

But what is the next step? I can i prove this fact?

First, show that this is true for $n=1$:

$6\cdot7^{1}-2\cdot3^{1}=4\cdot9$

Second, assume that this is true for $n$:

$6\cdot7^{n}-2\cdot3^{n}=4\cdot{k}$

Third, prove that this is true for $n+1$:

$6\cdot7^{n+1}-2\cdot3^{n+1}=$

$6\cdot7\cdot7^{n}-2\cdot3\cdot3^{n}=$

$7\cdot6\cdot7^{n}-3\cdot2\cdot3^{n}=$

$7\cdot6\cdot7^{n}-(7-4)\cdot2\cdot3^{n}=$

$7\cdot6\cdot7^{n}-7\cdot2\cdot3^{n}+4\cdot2\cdot3^{n}=$

$7\cdot(\color\red{6\cdot7^{n}-2\cdot3^{n}})+4\cdot2\cdot3^{n}=$

$7\cdot\color\red{4\cdot{k}}+4\cdot2\cdot3^{n}=$

$4\cdot(7\cdot{k}+2\cdot3^{n})$

Please note that the assumption is used only in the part marked red.


The simplest way to prove that any left-invariant vector field on a Lie group is complete It's all in the question: I look for the most intuitive proof that the integral curves of any left-invaraint vector field on a Lie group can be extended for all values of "time". I realize  that the argument is always based on the existence of group multiplication; what I look for is the most straightforward proof available. Thanks in advance!

A vector field is complete if and only if its flow $\Phi_t$ exists for all $t \in \Bbb{R}$. So we can show that a left-invariant vector field $X$ on a Lie group $G$ is complete by computing its flow.
The flow $\Phi_t$ of $X$ satisfies the differential equation $\frac{ \rm{d}}{\rm{d}t} \Phi_t(g) |_{t=0} = X_g$ for all $g \in G$. Applying the flow to the identity element, we have $\Phi_t(1) = e^{tX}$, the Lie-theoretic exponential map. Moreover since $X$ is left-invariant, $\Phi_t$ is left-invariant $\Phi_t(g) = g\Phi_t(1) = ge^{tX}.$ So we see that $\Phi_t: G \to G$ is equal to right-multiplication by $e^{tX}$.

I do think this statement is true if $G$ is not connected. If $G$ is connected and compact, then you can prove this via Hopf-Rinow theorem using the exponential map. See Terence Tao's article on this in his blog.

But one still needs to prove the integral curves of a left invariant vector field is the one produced by the exponential map. While this seems standard in the wikipedia definition, it is not entirely clear to me why this is true unless one verify the exponential map of the Lie group coincide with the exponential map of the Riemannian geometry. But then it takes quite a bit of work to prove this "trivial" fact.


Complex integration along a curve I have to calculate this integral:
$$
\int_C e^z\,dz
$$
where $C$ is the circle $|z - jπ/2| = π/2$ from the point $z = 0$ to the point $z = jπ$.

I know how to calculate these with circles which central point is at $(0,0)$ but I don't know how it works with this.. So I will be very glad if I got any tips or help.
Thanks!

Use the complex version of the fundamental theorem of calculus: $F(z) = e^z$ is an anti-derivative of your $f$, so
$$
\int_C e^z\,dz = F(i\pi) - F(0) = -2.
$$

$e^z$ is an entire function so by Cauchy integral theorem, your integral is 0.


what is the Fourier cosine transform of $e^{-ax}$ What is the Fourier cosine transform of $e^{-ax}$

I got

$$
\int_{0}^{\infty}\cos(kx)e^{-ax}dx = \frac{e^{-ax}(k\sin(kx) -\cos(kx))} {a^{2}+k^{2}}\Bigr|_{0}^{\infty}
$$

But how do you continue from here?

The expression you obtained is incorrect. It is missing a factor of $a$ in front of the cosine term.

Assume $a > 0$. (Otherwise the integral does not converge.)
\begin{align}
\int_{0}^{\infty}\cos(k x)e^{-ax}\,dx &= Re \int_{0}^{\infty}e^{i k x}e^{-ax}\,dx\\
&= Re \int_{0}^{\infty}e^{(i k - a)x}\,dx\\
&= Re\; \frac{e^{(i k - a)x}}{i k - a}\; \Bigr|_{0}^{\infty}
\end{align}
Since $a>0$, $\lim_{x\rightarrow\infty}e^{(i k - a)x}$ is zero, and we are left with:
$$
\int_{0}^{\infty}\cos(k x)e^{-ax}\,dx \;=\; Re\; \frac{-1}{i k - a} \;=\;
\frac{a}{a^2 + k^2}
$$

As weird as it may sound for a question which is evidently purely-mathematical, one way to catch an error such as the missing $a$ above is by dimensional analysis. Imagine that the units of $x$ are length. Then for consistency, $a$ and $k$ must have units of inverse length. That means you can't  add or subtract two terms like $k \sin(k x)$ and $\cos(kx)$, since they have different units. (One has units of inverse length, the other is dimensionless.) If you arrive at such an expression, then there must have been a mistake somewhere.

Observe that if $a>0$ then
$$
\lim_{x\to+\infty}e^{-ax}(k\sin(kx) -\cos(kx))=0.
$$


Topological spaces vs. metric spaces Are there any "realistic" examples of topological spaces that are not metric spaces. You are free to invent your own definition of "realistic". But, at a minimum, a realistic example is one that occurs naturally as part of some other problem, not something that is fabricated to provide a counterexample or a topology homework exercise. One good example is the topological vector space of all functions $f:\mathbb{R} \to \mathbb{R}$ under pointwise convergence. Any others?

If such examples are difficult to find, maybe there are results in the opposite direction: are there results that say something like "every topology that satisfies conditions X, Y, Z can be generated from a metric"? One example: a compact Hausdorff space is metrizable if and only if it is second-countable. Any other theorems like that?

Edit
I found this page, which does a pretty good job of answering my second question. So, that just leaves the first one.

Ordinal spaces occur naturally, and any uncountable ordinal space is not metrizable.

You can also talk about Moore spaces, Stone-Cech compactification of $\Bbb N$ (which is compact but has is too big to be metrizable), there are Zariski topologies which are often non-metrizable, and there are plenty of Cantor cubes which are naturally occurring in set theory.

The projection of the complex plane onto the real axis gives a simple, naturally occurring, example of a topological space that is not a metric space.  Define d(z1, z2) = |Re(z1)-Re(z2)|.  This is not a metric on the complex plane because d(z1, z2) = 0 does not imply z1=z2.  But the open sets defined the obvious way does define a topology on C.  Any non-trivial projection onto a metric space can give a similar example.


Integration with trigonomitry Hi I was trying to find the area between the following curves (below) however I am unsure how to continue from the trigonometry which gets presented:

Curves:

$$y = 2\sin(x)\\y = \cos\left(\frac{x}{2}\right)$$

I have found that the curves meet at $x=\pi$ and $x=2 \arcsin(\frac14)$. I am supposed to find the area being $\frac94$ however this is what I got, I don't know how to simplify:
$$A = 2\cos \left( 2\arcsin\left(\frac14\right)\right) + 2\sin \left( 2\arcsin\left(\frac14\right)\right)$$
I tried letting $z = \arcsin(\frac14)$ meaning I get:
$$A = 2\cos(2z) + \frac12$$
which then can simplify to:
$$A = 2(1-\sin^2(z))+\frac12$$
but then I have no idea. Thanks and good luck!

Are you requested to find the red-coloured area?

Then it's simply $$A=\int_{2\arcsin\left(\frac14\right)} ^{\pi} 2\sin(x)-\int_{2\arcsin\left(\frac14\right)} ^{\pi} \cos\left(\frac{x}{2}\right)\\=\left[-2\cos(\pi)+2\cos\left(2\arcsin\left(\frac14\right)\right)\right]-\left[2\sin(\frac{\pi}{2})-2\sin\left(\frac{2\arcsin\left(\frac14\right)}{2}\right)\right]\\=2-2+2\sin\left(\frac{2\arcsin\left(\frac14\right)}{2}\right)+2\cos\left(2\arcsin\left(\frac14\right)\right)\\=\frac12+\frac74\\=\frac94=2.25$$

$$\sin\left(\arcsin\left(\frac14\right)\right) = \frac14$$ This is all you need to do from where you have reached.


solution for the following second order linear ODE Could anybody help me with the solution of:

$$a(1-x)^2x^2y''+(1-x)xy'+(x+b)y=0.$$
Consider $a>0$ and $b<0$.

Similary to $ax^2y''+bxy'+cy=0$ for which we look for solutions on the form $x^k$ and similary to $a(1-x)^2y''+b(1-x)y'+cy=0$ for which we look for solutions on the form $(1-x)^h$, this draw us to try a particular solution on the form $x^k(1-x)^h$

Let x= p/(1+p) and make the ansatz y(x)= 1/(1+p) u(p).


Evaluating $\int_0^{\infty}\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3 dx$ I’m looking for a closed form of this integral $$I_3=\int_0^{\infty}\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3 dx$$ I’ve managed to evaluate $$I_1=\int_0^{\infty} \frac{\ln(1+x^2)}{1+x^2}dx=\pi \ln 2$$ with the change of variable $x =\tan \theta $, $dx=(1+(\tan \theta)^2) d\theta $ giving $$I_1=-2\int_0^{\frac{\pi}{2}} \ln(\cos \theta ) d\theta$$ which is classic. But I’m stuck with the power $(\ln (\cos \theta))^3$ in $I_3$. Thanks for your help.

We have

  $$\int_0^{+\infty}\!\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3\!{\rm d}x=\frac{9\pi}{4}  \zeta(3)+\frac{3 \pi}{2} \ln^3 2-\frac{21\pi}{8}\ln^2 2-\frac{33\pi}{16} \ln 2-\frac{7 \pi ^3}{32}-\frac{57 \pi }{64}.$$

Proof. Observe that
$$
\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3=-\partial_s^3 \left.\left(\frac{1}{(1+x^2)^s}\right)\right|_{s=3} \tag1
$$
Then you may write
$$\begin{align}
\int_0^{+\infty}\frac{1}{(1+x^2)^s}{\rm d}x
&=\frac12\int_0^1u^s\left(\frac1u-1\right)^{-1/2}\frac{{\rm d}u}{u^2},\quad \color{purple}{u=\frac{1}{1+x^2},\, x=\left(\frac1u-1\right)^{1/2}}\\\\
&=\frac12\int_0^1u^{s-3/2}\left(1-u\right)^{-1/2}{\rm d}u\\\\
&=\frac12\frac{\Gamma\left(s-\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(s)}\\\\
&=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(s-\frac12\right)}{\Gamma(s)} \qquad \qquad (2)
\end{align}
$$ where, in the last step, we have used the classic Euler beta function integral representation.

Now, performing $\displaystyle -\partial_s^3 $ to $(2)$ and using $(1)$, you easily get a closed form expression for your integral with the help of Wolfram|Alpha for the computation of
$$
-\partial_s^3 \left.\left(\frac{\Gamma\left(s-\frac12\right)}{\Gamma(s)}\right)\right|_{s=3}
$$

Try this
$$
I_3 =\lim_{\beta\rightarrow 0}\frac{\partial^3}{\partial \beta^3}\int \left(1+x^2\right)^{\beta-3}dx
$$


Example of a $\kappa$-long sequence of disjoint club subsets of regular cardinal $\kappa$ I'm self-studying set theory and got stuck on this exercise:

  Let $\kappa$ be a regular cardinal. Give an example of a sequence $\langle C_\alpha\mid\alpha<\kappa\rangle$ such that $C_\alpha$ is club in $\kappa$ for every $\alpha<\kappa$, but
  $$\bigcap\{C_\alpha\mid\alpha<\kappa\}=\emptyset.$$

To make some progress at all, I've tried specific cases. For $\kappa=\omega$, I define $f:\omega\to\omega$ as taking $n$ to the $n$'th prime, and then define

$$C_n:=\{f(n)k\mid k<\omega\},$$

which is clearly unbounded ($|C_n|=\aleph_0$) and closed (every $n\cap C_n$ will be finite, so $\sup(n\cap C_n)=\max(n\cap C_n)\in C_n$). Furthermore due to the $f(n)$'s being primes, we have

$$\bigcap\{C_n\mid n<\omega\}=\emptyset.$$

I'm trying to do the case with $\kappa=2^{\aleph_0}$ now, where I'm trying to exploit that $|\mathbb{R}|=2^{\aleph_0}$ and $|[0,1]|=2^{\aleph_0}$, so if we set $C_n:=[n,n+1]$ then the $C_n$'s are clubs (unbounded due to cardinality and closed due to closed intervals preserve limits) and disjoint. But I can only construct countably many such $C_n$'s, so it doesn't satisfy the example of needing $2^{\aleph_0}$ many disjoint clubs.

If you guys would be so kind as to hint me in the right direction, whether it be about the $2^{\aleph_0}$ case, the general case, or about how I might generalize my methods from my special cases to the general, I would appreciate it a great deal.

Thanks.

First of all, when talking about clubs we generally require that $\kappa$ has uncountable cofinality. There is a problem with clubs for cardinals whose cofinality is $\omega$, because one unbounded sequence without limit points is not really a club (but still closed and unbounded). It's true this question is true for the countable case, but its solution doesn't necessarily generalize. It's better to start with the uncountable case, and deduce from it the countable case (and seeing how you solved the countable case, you only need to solve the uncountable case anyway).

Secondly, you can't use $\Bbb R$. It's not a well-order. The idea is to use the order topology of the cardinal, not of an ordered set of the same cardinality. If you note, the real numbers is an ordered set whose cofinality is $\omega$, as the natural numbers is a cofinal sequence in that order. Furthermore, $[0,1]$ is not unbounded in $\Bbb R$ as it is very much bounded by $0$ and by $1$.

Thirdly, $2^{\aleph_0}$ can be singular. It is consistent, for example that $2^{\aleph_0}=\aleph_{\omega_1}$.

Fourthly, clubs cannot be disjoint. The whole idea is that clubs represent "almost everything", and when you intersection two "almost everything", you should get "almost everything" again. The exercise asks for you to show that it is possible to have a sequence of $\kappa$ clubs, that the intersection of all of them is empty.

Finally, to hint you for the solution, think about end-segments of $\kappa$. (with its order as an ordinal!)

For $\alpha < \kappa$ let $C_\alpha =  \kappa - \alpha$. And you're done.


Calculating $\pi$ using $\pi=\lim_{n\to\infty}n\sin\frac{180^\circ}{n}$ How can the fact that
$$\pi=\displaystyle\lim_{n\to\infty}n\sin\frac{180^\circ}{n}$$
be useful when calculating $\pi$? I mean, $180^\circ =\pi$, isn't that a circular reasoning?

I got the idea that Lagrange interpolation might help, since every argument is divided by the corresponding power of $\pi$, therefore avoiding the circularity when we choose to interpolate
$$0\to 0,\, \frac{\pi}{2}\to 1,\, \pi\to 0,\, \frac{3\pi}{2}\to -1,\, 2\pi\to 0.$$
This interpolation yields
$$\sin x\approx \dfrac{8x^3}{3\pi ^3}-\dfrac{8x^2}{\pi ^2}+\dfrac{16x}{3\pi}.$$
But this is problematic since it's a polynomial and its behavior at $\infty$ is very different from $\sin$ at $\infty$, so that can't be used. Using
$$\sin x=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n\, x^{2n+1}}{(2n+1)!}$$
or
$$\sin x=x\displaystyle\prod_{n=1}^\infty \left(1-\dfrac{x^2}{n^2\pi^2}\right)$$
doesn't help, since not every $x$ is divided by corresponding power of $\pi$; using such series or products to calculate $\pi$ would be circular. So, how can the formula in the question be used to calculate $\pi$?

One way of using this fact is to stick to a certain subset of $\mathbb N$, namely $n = 2^k$, and evaluate $\sin\frac{\pi}{2^{k+1}}$ in terms of $\sin\frac{\pi}{2^k}$. This essentially is what Viete did to arrive to his formula.

Notice that you don't need to know $\pi$ to compute $\sin \frac{\pi}{4}$.

Another way is to interpret it in purely geometrical way: divide the half circle into $n$ congruent arcs, construct a corresponding sine, and replicate it $n$ times to get a geometric approximation of $\pi$.

I'm not prepared to touch on the bulk of your post, but with regard to the circularity of reasoning you seem primarily concerned about, I would beg to differ. Everything else you posted on seems to be tied to examples for which your concern holds consequences, rather than as separate questions in their own right.

Degrees as a unit of measure for angles predate the radian. The trigonometric functions are defined on angles, and not on any particular unit of measure, thus they can be measured in any type of unit we wish.

Trigonometric functions have been computed for centuries in degrees.  No, in fact thinking of trigonometric functions in terms of radians rather than more abstractly in terms of angles, in my view, IS where the circularity of reasoning is taking place. This is because it puts the radian before the function, rather than as the result of a need for a special unit in the treatment of trigonometric functions in calculus. Perhaps this perspective is too respecting of history to hold in mathematics, but I like to take math in the chronological order it was discovered; it does tend to resolve a lot of the circularity.

There is no circularity in defining $\sin(x)$ when $x$ is in degrees.  The only time there is a conflict is when $x$ also exists outside of the argument of the function - such as $x\sin(x)$, clearly the unit changes the solution. Or when by way of some sort of transformation (such as differentiation) a constant is brought out of the argument - if $x$ is in radians then $\frac{d}{dx}\sin(x)=\cos(x)$ but if $x$ is in grads then $\frac{d}{dx}\sin(x) = \frac{\pi}{200}\cos(x)$.  Also of note are the polynomial representations of trigonometric functions, approximates or otherwise, but this is because they are built on differentiation. The list goes on, I'm sure. This isnt exhaustive.  Im only suggesting that if an expression is somehow derived from a trigonometric function, especially using calculus, then care needs to be taken.

I could prove that $\frac{d}{dx}\sin(x) = \frac{\pi}{180}\cos(x)$ if I wanted to use only degrees.  The math is perfectly valid and perfectly rigorous, but if you were just blindly going through the motions you might not see why. If you thought only in terms of radians you'd probably call it wrong. But that is what the identity would look like if we never created the radian. Thus many of those expressions you wrote would contain constants involving ratios of pi and 180, and it would just get messy. That's a part of the reason for why the radian was inventned and standardized as a unit of measure for angle: it eliminates those constants. And yes, it was invented, not discovered. We chose it specifically because it plays well with calculus, but the units of angle measure are completely arbitrary.

Im sure this isnt news to you. I just thought it was worth pointing out the natural evolution of the math to resolve the apparent contradiction.  The contradiction only exists to you because, if I may speculate, you view the function not only as a function requiring a radian argument, but perhaps even as a function only computable with a Taylor (or similar) series? This is, of course, problematic. Not only does it imply circularity but it fails to appreciate history and, frankly, limits your own ability to abstract angle.


Convince Me:The sum of two independent random variables (X, Y) is normal iff X and Y are normally distributed. The sum of two independent random variables (X, Y) is normal  iff X and Y are normally distributed.

I came across this fact recently, and it totally baffles me.  I am not convinced it is true.  Is there a proof for this somewhere?

UPDATE:

The direction of starting with two independent random variables that are normally distributed ($X$, $Y$) to proofing that $X+Y$ is also normally distributed is not confusing.  I am more confused on the other direction of the proof. It seems like we could have two random variables that are independent and  NOT normally distributed were the sum could be normally distributed.  Suppose we have a random variable $Y$ that is standard normal, and $U$, $V$ which are iid with finite mean and variance from a non-normal distribution.  If we have another variable $W = Y -U$, and then take  $W + V$ is it really the case that $W + V$ would not be normally distributed?  I have done a few arbitrary examples that concur with the statement. I am curious what the proof would look like so I know it is true always.

UPDATE #2

I think I found a counter example.  Suppose that $X\sim I(2)$ that is, $X$ is a degenerate random variable, and $Y \sim N(0,1)$.  Then the MGF of $Y$ is $exp(t^2/2)$ and the MGF of $X$ is $exp(2t)$, hence the MGF of $X+Y$ is $exp(2t + t^2/2)$, which implies that $X+Y \sim N(2, 1)$

Suppose $X+Y$ is normal and $X,Y$ are independent.

$X+Y \sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$ since we know its mean and variance.

Thus the MGF of $X+Y$ is $e^{(\mu_X+\mu_Y)t+\frac{1}{2}t^2(\sigma_X^2 + \sigma_Y^2)}$.

By independence this is equal to $E[e^{tX}]E[e^{tY}]$.
I can now deduce $E[e^{tX}]$ is proportional to $e^{\mu_X t+\frac{1}{2}t^2 \sigma_X^2 }$. Hence it is a normal random variate. The same can be said of $Y$.

The other direction is quite easy (see comments).


How can I solve this limit using L'Hospital's rule somewhere? I have to solve this limit $\lim _{x\to 0+}\left(\left(\left(1+x\right)^x-1\right)^x\right)$ as $x$ approaches $0$ from the positive numbers. I have tried to change it to e to the limit of the exponent and apply L'Hospital's on it but to no avail. Can someone help me?

We begin by writing for $x>0$

$$\begin{align}
\left((1+x)^x-1\right)^x&=e^{x\log\left((1+x)^x-1\right)}\\\\
&=e^{x\log\left(e^{x\log(1+x)}-1\right)}
\end{align}$$

Next, using $\log(1+x)=x+O(x^2)$ and $e^{x\log(1+x)}-1=x^2+O(x^3)$, we see that

$$\left((1+x)^x-1\right)^x=e^{2x\log(x)}\underbrace{e^{x\log(1+O(x))}}_{\to 1 \,\text{as} \,x\to 0}$$

Finally, since $\lim_{x\to 0^+}x\log(x)\overbrace{=}^{LHR}=-\lim_{x\to 0^+} x=0$

we conclude that

$$\left((1+x)^x-1\right)^x=1$$


Shows that $EFD$ is also equilateral. In the following figure, it is known that ABC is equilateral and that $AE = BD = CF$, shows that $EFD$ is also equilateral.

Attemp:
I got a little lost in proving these triangles to be similar, but if you can do that, it's simple. Call the angles EAB, DBC and FCA x and the angles DBA, FCB and EAC y. (We know they are the same because we assume the triangles to be similar.) We know that x + y + x + y + x + y = 180 °, so x + y = 60 °. DEF is the outer angle of AEB, so DEF = x + y = 60 °. Similarly the other angles are 60 °, thus being DEF equilateral.

I also think this looks like Morley's triangle. But I don't know if DBC is 1/3 ABC:

Let us embed the construction in the complex plane by taking $A=0, B=1, C=\omega=\frac{1+i\sqrt{3}}{2}$.
Given $E=z$, the positions of $D$ and $F$ are fixed via
$$ D = B+|z|\frac{E-B}{|z-1|} = 1+\frac{|z|}{|z-1|}(z-1) $$
$$ F = C+|z|\frac{D-C}{|D-C|} = \omega+\frac{|z|}{|1+|z|(z-1)-\omega|}(1+|z|(z-1)-\omega)$$
and $F$ is a real multiple of $z$, since $F\in AE$. This implies that
$$\omega+\frac{|z|}{|1+|z|(z-1)-\omega|}(1-|z|-\omega)$$
is a real multiple of $z$ too. After a few manipulations, this leads to the fact that $z$ and $\omega(1-z)$ are real multiples of each other, i.e. to the fact that  $\widehat{AEB}=120^\circ$. This gives that $AEB,BDC$ and $CFA$ are similar, hence congruent and such that
$$\widehat{FED}=\widehat{EDF}=\widehat{DFE} $$
QED.

EDIT!:

Assume that each side is rotated by $\delta$ and when lines are drawn in the way shown there is an angle $ (60-\delta) $  at $A$ and the angle of reduced amount $\delta$ at $C$ as shown.

In the triangle $CAF$  external angle equals sum of two opposite angles.

$$\angle DFE= \delta + 60-\delta = 60\, deg $$ and the same holds for other two vertices $E,D$ by rotational symmetry, thus creating a smaller equilateral triangle with scaled down ratio $\dfrac{CF}{CA}$ that can be found by Sine Law.

$$ \dfrac{CF}{CA}=\dfrac{\sin (60-\delta)}{\sin 60}= \cos\delta- \sin \delta/\sqrt 3 $$

$$ \dfrac{AF}{CA}=\dfrac{\sin \delta}{\sin 60}$$

scale ratio

$$ \rho= \dfrac{CF-AF}{CA}= f(\delta) $$

which is a single variable monotone function of $\delta.$

It is given that a constant proportion is valid for

$$ AE:BD:CF = AB: BC : CA $$

or

$$\dfrac{AE}{AB}=\dfrac{BD}{BC}=\dfrac{CF}{CA} = constant = \rho , say$$

When this proposition is shown by converse logic it can be concluded that each line must rotate by a constant amount,say $\delta$, establishing scaling/similarity.


Prove: matrix A is diagonalizable iff exp(A) is diagonalizble I need to prove: matrix A is diagonalizable iff $\exp(A)$ is diagonalizble. exp means exponent function.

I know to prove that if $A$ is diagonalizable so $\exp(A)$ is diagonalizable, but have a problem with the other side.

can I write $P^{-1}.\exp(A).P=D$ (since $\exp(A)$ is diagonalizable  ) and operate log on both sides of the equation? if yes I'm done, or any other hints?

Since you already know that "$A$ is diagonalisable" implies that "$\exp(A)$ is diagonalisable", we show that if $A$ is not diagonalisable, then neither is $\exp(A)$.

  $A$ is not diagonalisable $\Rightarrow$ $\exp(A)$ is not diagonalisable.

Let $A$ be non-diagonalisable and let $A=XJX^{-1}$ be its Jordan form, where $J$ is a direct sum of elementary Jordan blocks: $J=\oplus_{i=1}^l J_i$. The function $\exp(A)$ can then be written in the form $\exp(A)=X\exp(J)X^{-1}$, where $\exp(J)$ is again a direct sum of exponentials of elementary Jordan blocks: $\exp(J)=\oplus_{i=1}^l \exp(J_i)$. Since $A$ is non-diagonalisable and so is $J$, $J$ contains at least one non-trivial Jordan block
$$
\tilde{J}=\lambda I + N\in\mathbb{C}^{k\times k}, \quad k\geq 2,
$$
where $\lambda\in\mathbb{C}$ is an eigenvalue of $A$ and $N$ is the nilpotent matrix
$$
N=\begin{bmatrix}
0 & 1 &        &        &   \\
  & 0 & 1      &        &   \\
  &   & \ddots & \ddots &   \\
  &   &        & 0      & 1 \\
  &   &        &        & 0
\end{bmatrix}.
$$
If we show that $\exp(\tilde{J})$ of such a non-trivial block is non-diagonalisable, we are done.
Looking again here how $\exp(J)$ looks like, we see that
$$
\exp(\tilde{J})=\exp(\lambda)\exp(N)
=\exp(\lambda)\sum_{i=0}^{k-1}\frac{1}{i!}N^i.
$$
Note that $1$ is the only eigenvalue of the upper triangular matrix $\exp(N)$.
So $\exp(N)$ is diagonalisable iff the kernel of the $k\times k$ matrix $\exp(N)-I$ is $\mathbb{C}^{k}$. This is, however, not the case as $\exp(N)-I$ is not a zero matrix. Therefore, $\exp(N)$ and consequently $\exp(\tilde{J})$ and consequently $\exp(A)$ is not diagonalisable.

Try the Taylor series expansion for the exponent. For small $t$, $e^{tA} \approx I + tA$


Prove that $\tan5 \theta = \frac {5\tan \theta -10 \tan ^3 \theta +\tan ^5 \theta} {1-10\tan ^2 \theta +5\tan ^4 \theta}$ As the title suggests, what is required to prove is that
$$\tan5 \theta = \frac {5\tan \theta -10 \tan ^3 \theta +\tan ^5 \theta} {1-10\tan ^2 \theta +5\tan ^4 \theta}$$

I was looking back through my old high school tests and came across this monster and have now -as I did then- no idea where to start with it. This was a test on a 'Complex numbers' unit so using complex numbers is required to prove it (which sadly, I'm quite out of practice on). I'd be really interested to see a solution to problem which has been on the back of my mind ever since I put my eyes on it for the first time. I'm assuming the first step should be to open $\tan 5\theta = \frac {\sin 5\theta} {\cos 5\theta}$ but I could be mistaken.

Hint:

$$\cos5x+i\sin5x=\left(\cos x+i\sin x\right)^{5}=$$$$\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x+i\left(5\cos^{4}x\sin x-10\cos^{2}x\sin^{3}x+\sin^{5}x\right)$$

(De Moivre)

$$\tan (3\theta + 2\theta)= (\tan 3\theta + \tan 2\theta)/(1-\tan 3\theta  \tan2\theta )$$

Put the value
Just expand the terms and u will get the answer


How to find multiples of numbers under a certain range I recently found a 'question' that requires me to find the sum of all multiples of 3 and 5 under 1000, I sadly cheated and found some code online to help build a code in python:

sum=0
for i in range(999):
        if not(i %3 == 0) or not (i % 5):
                sum=sum+i
print sum

However, even though it works, I feel guilty that I had to cheat to answer this question. I do not understand what the if not means, and why I cannot use the regular if statement. Can someone please clear this up for me? Thanks waco001

P.S. In other languages like C, this:

int problem1()
{
    int sum=0;

    for(int i=0;i<1000;i++)
    {
        if(i%3 == 0 || i%5 == 0)
        {
            sum+=i ;
        }
    }

    return sum ;
}

Works perfectly?!?

You don't need such a brute force calculation.  You can do this by hand.  First, some notation.  Let
$$
M_3 = 3\mathbb{Z} \cap [1, 999] = \{ \text{multiples of 3 between 1 and 999, inclusive} \}
$$
This definition works for any $k \in \mathbb{Z}$.  The quantity that you are looking for is
$$
S = \sum_{n \in M_3 \cup M_5} n
$$

Which numbers are in $M_3 \cup M_5$?  Multiples of $3$ or $5$, of course, but you must be careful.  Multiples of $15 = 3 \cdot 5$ are multiples of both $3$ and $5$.  In fact, $M_3 \cap M_5 = M_{15}$, so
$$
\begin{align}
S &= \sum_{n \in M_3} n + \sum_{n \in M_5} n - \sum_{n \in M_{15}} n \\
\end{align}
$$
I will show you how to calculate one of these sums.  (The other two are strictly analogous.)  Consider
$$
\sum_{n \in M_3} n = 3 + 6 + \cdots + 996 + 999
$$
How many summands are there?  $\left\lfloor \frac{999}{3} \right\rfloor = \left\lfloor 333 \right\rfloor = 333$.  Thus, by pairing terms à la Aryabhata (and famously young K.F. Gauss), you have
$$
\sum_{n \in M_3} n = \frac{333(3 + 999)}{2} = 166\,833.
$$

For a range $(n, m)$ both inclusive,number of multiples of a number $x$ can be
calculated as $x = (m-n+1)/(given \ number)$


Quick Area of Cross Section Question "The base of a solid is the region in the first quadrant between the graph of $y=2x$ and the $x$-axis for $0\leq1\leq2$. For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the $x$-axis and one radius in the $xy$-plane. What is the volume of the solid?"
I guess I'm a bit confused about both the wording and where I'd have to go in this question, I understand how to find the area of a cross-section when these cross-sections might be a square or a rectangle, but I'm a bit lost when it comes to circles as well. Any help would be appreciated!

THe method of using integraion to find find volume that you are using is

Pick an axis.
Slice the volume in slices perpendicular to that axis.
Integrate the volume of the slices with respect to the coordinate along that axis.

(Generally, the axis need not be a coordinate axis, but it is convenient when this happens.)
When the problem has

For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the x-axis and one radius in the $xy$-plane.

You are told that the slice is a quarter circle, so has one quarter the area of a circle, and that (1) all the radii for that quarter circular are perpendicular to the $x$-axis and (2) one radius is in the $xy$-plane.  This means that in a slice, one can take the distance perpendicular to the $x$-axis to the line $y = 2x$ as the radius of the circle, one quarter of which area is accumulated into the integral.  (In the $xy$-plane, a perpendicular to the $x$-axis is parallel to the $y$-axis.  A ray starting on the $x$ axis, in the first quadrant of the $xy$-plane, and proceeding perpendicular to the axis crosses the triangle base of the volume, reaching the end of the base at the line $y = 2x$.  (If this is not clear, sketch a graph of the base.)
Suppose $r$ is the radius of a circle.  The area of that circle is $\pi r^2$.  The area of one-quarter of that circle is $\frac{1}{4} \pi r^2$.  The distance from the $x$-axis to the line $y = 2x$ (the radius described by the ray in the previous paragraph) is $2x$, so the radius we are to take for the slice at coordinate $x$ is $r = 2x$.  Then the area contribution from the slice at coordinate $x$ is $\frac{1}{4} \pi (2x)^2$.  That is the integrand that we integrate from $x = 0$ to $x = 2$.  (Again, if it is not clear that this is the range of coordinate that includes all the slices, sketch the base.)
The following diagram of the volume is from a point in the $(+++)$-octant looking back throught the volume at the origin.

Example slices (for $x \in \{1/2, 1, 3/2, 2\}$ are shown as fairly clear quarter circles.  The base is the $xy$-plane, at the bottom of the diagram.  The dashed gray line is in the $xz$-plane and marks the ends of the radii of the quarter circles in that plane.
We can more clearly see the "slicing" viewing from the $(+-+)$-octant nearly parallel to the positive $y$-axis.

This angle reinforces that we are given the axis, we are given the area formula for the slices, and all we have to find is the relation between the $x$-coordinate and the area of the corresponding slice.

The hardest part of this problem is understanding the wording. Trying to describe a $3$-d shape using words is sometimes difficult, but what it is trying to describe is a pretty simple shape. The base is a right triangle with a length of two units and a height of $4$ units. To find the total volume we need to find how this shape looks in the $z$-direction. If you take a cross-section perpendicular to the $x$-axis you should see a quarter circle, with radius $2x$, because that is the distance from the $x$-axis to the given line. from this we can tell that one cross section has volume $(\pi/4) \cdot (2x)^2 \, dx$. This can be simplified to the $\pi \cdot x^2 \, dy$. Now we simply take the integral from $0$ to $2$ of this volume of a single cross section to add up all the cross sections from $x=0$ to $x=2$. This equation looks like the following:
$$ \pi \, \int_{0}^{2} x^2 \, dx = \frac{8 \, \pi}{3}.$$


Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I tried simplifying the equation but got stuck. Can someone help me simplify this inequality into one that's easier to solve the actual problem above? I don't want/need the answer to the actual problem, but just need help simplifying the inequality. (A hint would be fine.) Any help would be appreciated! :)

Let $f(x)=\frac{1}{x}-3+2\sqrt{x}$
Then its derivative is defined and positive real numbers and equals $f'(x)=\frac{x^2-\sqrt{x}}{x^2 \sqrt{x}}$
You can easily show that $f'<0$ if $x<1$ and $f' \ge 0$ if $x\ge 1$.
Then $f$ reaches a minimum at $1$, which is $0$.
By considering the limits of $f$ at $0$ and $+\infty$, you can conclude.


Proof that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$ It is asked to prove that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$, and suggested to use the following two inequalities:

$$a+b\leq a+ 2\sqrt{a}\sqrt{b}+b$$
$$\sqrt{b}\sqrt{b}\leq \sqrt{a}\sqrt{b}$$

The second inequality holds iff $a\geq b \geq 0$.

I've tried different possibilities, but couldn't figure out how to either take the limit sign out of the square root, or take the limit sign into the square root. Would appreciate some hints, but not an entire solution please.

You have to make sure that $\forall n\in\mathbb{N}^*$, $x_n\geq 0$. Otherwise the limit doesn't exist because the sequence $\{\sqrt{x_n}\}_{n=1}^{\infty}$ isn't even defined(well, in $\mathbb{R}$).
Suppose the statement above is true. Let $A=\lim\limits_{x\to\infty}x_n$ Then there are 2 cases:
(1)$A=0$. $\forall\epsilon>0$,$\exists N\in\mathbb{N}^*$ so that $\forall n\geq N$ there is $x_n<\epsilon^2$,that is, $\sqrt{x_n}<\epsilon$, so that $\lim\limits_{x\to\infty}\sqrt{x_n}=0=\sqrt{\lim\limits_{n\to\infty}x_n}$.
(2)Otherwise,$A>0$. $\forall\epsilon>0$,$\exists N\in\mathbb{N}^*$ so that $\forall n\geq N$ there is $|x_n-A|<\sqrt{A}\epsilon$.This means that $\epsilon>\frac{x_n}{\sqrt{A}}\geq|\frac{x_n-A}{\sqrt{x_n}+\sqrt{A}}|=|\sqrt{x_n}-\sqrt{A}|$, therefore $\lim\limits_{n\to\infty}\sqrt{x_n}=\sqrt{A}=\sqrt{\lim\limits_{n\to\infty}x_n}$.
This proof uses solely the definition of the limit of a sequence :)

It's wrong of course.

For $x_n\geq0$ it's true because $f$ is a continuous function, where $f(x)=\sqrt{x}$.


Sum of the digits of a perfect square Prove that the sum of the digits of a perfect square can't be 2, 3, 5
  , 6, or 8.

I'm completely stumped on this one, how would I go about proving it?

The sum of the digits of a number is invariant mod $9$. So this is the same as asking what are and what are not the squares mod $9$?

Let's consider: the squares mod $9$ are $1,4, 7$ (coming from $1,2,4$). The complement of these are $2,3,5,6,8$, which are exactly the parts from your question. These are the numbers which do not appear as squares mod $9$, and thus do not appear as sums of digits of squares.

If your square number had the sum of digits equal to 3 or 6, then it must be a multiple of 3. But a square that is a multiple of 3 must be a multiple of 9. Therefore the sum of its digits must be a multiple of 9. 3 or 6 are not multiples of 9.

Therefore 3 and 6 are impossible.


Get equation for a curve which intersects x at seemingly randomly distributed points? Is there any type of function that when graphed would show a curve which intersects the x axis multiple times, with each point being an arbitrary distance from the last?

I mean, not like a trig function where each intersect is the same distance from the last. (2,0); (4,0); (6,0); (8,0). And not like a spiral where the distance gets bigger and bigger (or smaller) (2,0); (4,0); (8,0); (16,0);

But for example, some curve which intersects x-axis at (2,0); (6,0); (14,0); (15,0); (20,0); (122,0)...

Does that type function exist?

If so, is it possible to solve/get the equation, given only those intersect points?

I wouldn't need the exact equation of any particular curve. Just the equation of any curve that happens to intersect x-axis at whatever given arbitrary x values. Is that at least that possible to do?

If you specify an arbitrary finite set of points $\{a_1, a_2, a_3, \dots, a_n\}$,  the polynomial function
$$f(x)=(x-a_1)(x-a_2)(x-a_3) \cdots (x-a_n)$$
has the desired property.

you can find the curves using matrix method by taking the columns as the x polynomials and multiply with the constant matrix which finally equated to given y.
but on which does the degree of x which have to be taken. on what does that depend.


When is $C_c^\infty(\mathbb R^d \setminus \{ 0 \})$ dense in $C_c^\infty(\mathbb R^d)$? Let $d\in \mathbb N$. Under what constraints for $d$ and $p$  is $C_c^\infty(\mathbb R^d \setminus \{ 0 \})$ dense in $C_c^\infty(\mathbb R^d)$ with respect to the $L^p(\mathbb R^d)$-norm?

Let $b_n$ be a sequence of smooth functions $b_n: \mathbb{R} \to [0,1]$ such that $b_n(x) = 1$ for $|x|> { 2\over n}$ and $b_n(x) = 0$ for $|x|< {1 \over n}$.

If $f \in C_c^\infty(\mathbb R^d)$, then $b_n\cdot f \in C_c^\infty(\mathbb R^d \setminus \{0\})$ and for $p< \infty$ we have $\|f-b_n \cdot f\|_p \to 0$.

If $f \in C_c^\infty(\mathbb R^d)$ such that $f(0) = 1$, then we must have
$\|f-g\|_\infty \ge 1$ for any $g \in C_c^\infty(\mathbb R^d \setminus \{0\})$.

I'm not sure that your question makes sense, since the first set is not a subset of the second. I don't think you can embed it in any reasonable way, since it is not always possible to extend a continuous function from the first set to the other. I'd guess for $d=1$, something like $sin(1/x)$ smoothed out to some compact domain should be an example.

Edit: I guess setting $f(0)=0$ should work as an embedding, since any compact set in $\mathbb{R}^d$ should have empty intersection with some ball around $0$.


Evaluating Riemann-Stieltjes integral $\int_a^bf(x)g(x)d\alpha(x)$ I am looking to evaluate the following Riemann-Stieltjes integral:
$$\int_a^bf(x)g(x)d\alpha(x)$$
and I am unsure how to deal with the product on the inside. I have considered the following:
$$\int_a^bf(x)g(x)\,d\alpha(x)=\int_a^bf(x)\,dG(x)$$
Where $G(x):=\int_a^xg(x)\,d\alpha(x)$, but I don't think this really helps me out. I also tried doing integration by parts, but this left me with another integral and an $\alpha(x)$ not in a differential so I'm not sure if that helps either. Is there a good method of evaluating this?

Assuming $g\ge 0$, we can define a Borel measure on $[a,b]$ by setting $\mu(a,x]=G(x)$. For a simple $\phi=\sum_{i=1}^n a_i \chi_{A_i}$, the integral $\int \phi\text{d}G=\int\phi \text{d}\mu=\sum a_i\int_{A_i}g\text{d}\alpha=\int\sum a_i\chi_{A_i}g\text{d}\alpha=\int \phi g\text{d}\alpha$. For $f\in L^+$, find a sequence of simple $\phi_n \to f$ and apply the MCT to obtain $\int f\text{d}G=\int fg\text{d}\alpha$.

$\left| \int_a^bf(x)g(x)d\alpha(x) \right| \leq F.G.A'.(b-a)$
where $|f(x)|\leq F \ ; \  |g(x)| \leq G \ ; \  |\alpha'(x)| \leq A'$.


Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$

Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?

Hint:

Use differential binomial. Note that $$\sqrt{1+\frac{1}{4x}}=0.5x^{-0.5}(1+4x)^{0.5}$$


Why are symbols not written in words? We could have written = as "equals", + as "plus", $\exists$ as "thereExists" and so on. Supplemented with some brackets everything would be just as precise.

$$\exists x,y,z,n \in \mathbb{N}: n>2 \land x^n+y^n=z^n$$

could equally be written as:

ThereExists x,y,z,n from theNaturalNumbers suchThat
     n isGreaterThan 2 and x toThePower n plus y toThePower n equals z toThePower n

What is the reason that we write these words as symbols (almost like a Chinese word system?)

Is it for brevity? Clarity? Can our visual system process it better?

Because not only do we have to learn the symbols, in order to understand it we have to say the real meaning in our heads.

If algebra and logic had been invented in Japan or China, might the symbols actually have just been the words themselves?

It almost seems like for each symbol there should be an equivalent word-phrase that it corresponds to that is accepted.

"Integral From a x squared Plus b Plus c To Infinity of a Times LeftParenthesis x Plus LeftParenthesis x Plus c RighParenthesis Over LeftParenthesis x Plus One RightParenthesis Differential x IsGreater Integral From a x squared Plus b Minus c To Infinity of a Times LeftParenthesis x Plus LeftParenthesis x Plus c RighParenthesis Over LeftParenthesis x Minus One RightParenthesis Differential x Implies Integral From a x squared Minus b Plus c To Infinity of a Times LeftParenthesis x Plus LeftParenthesis x Plus c RighParenthesis Over LeftParenthesis x Plus One RightParenthesis Differential x IsGreater Integral From a x squared Plus b Minus c To Infinity of a Times LeftParenthesis x Plus LeftParenthesis x Plus c RighParenthesis Over LeftParenthesis x Minus One RightParenthesis Differential x" ?

Quiz:

Do you recognize this one ?

Summation On j From 1 Upto N Of y Index k Times the Product On k From 1 And k NotEqual to j Upto N Of x Minus x Index k Over x Index j Minus x Index k.

Yves Daoust's answer demonstrates how unreadable a formula with words instead of symbols tends to be, but doesn't really explain why it is so unreadable.

Certainly brevity is one thing: the more symbols there are in total, the more processing the brain needs to do to even lex the expression (to use the CS term), before we can even begin to parse it.

Reducing the length of the words intuitively reduces the total amount of information on the page – though actually this only works because if all parentheses etc. are written out as a word, there's lots and lots of duplication. Efficient coding avoids such duplication; the simplest way is to choose short indentifiers for everything that's used often. Well, mathematical notation drives this to an extreme by basically using only a single glyph for everything; that's not always viable, but in many formulas there are only few different variables and most of them are repeated quite a lot, so single-character symbols do in maths generally make more sense than whole words for everything.

But that's not all. Yves Daoust's example could be compressed with a coding

a: a
b: b
c: c
d: Derivative
f: From
g: IsGreater
i: Integral
l: LeftParenthesis
m: Times
n: Infinity
o: Of
p: Plus
q: Minus
r: RightParenthesis
s: squared
t: to
u: one
v: Over
x: x
y: Implies

to an extremely short form, that would nevertheless be almost completely unreadable:

ifaxspbpctnoatlxplxpcrvlxpurdxgifaxspbqctnoamlxplxpcrvlxqurdx
yifaxsqbpctnoatlxplxpcrvxpurdxgifaxspbqctnoamlxplxpcrolxqurdx

The real problem isn't solved:

  it is completely unusable because it is lacking a geometric layout

Well, I wouldn't phrase it quite this way. Geometric layout is a helpful visual aid for humans, but really the point is that mathematical language is for the most part not “linear stream” like, as a prose story. Rather, it is organised as an abstract syntax tree, and that's where the maths notation shines: it uses geometry to make that structure evident much clearer than with open/close parentheses that need to be found. Note that this can also be done with the compressed string above, with the technique that programmers have for the purpose: indentation and whitespace.

  i f axs p b p c
    t n
    o a m l
           x p lxpcr
           v
           lxpur
          r
    dx
 g
  i f axs p b q c
    t n
    o a m l
           x p lxpcr
           v
           lxqur
          r
    dx
y
  i f axs p b p c
    t n
    o a m l
           x p lxpcr
           v
           x p u
          r
    dx
 g
  i f axs p b q c
    t n
    o a m l
           x p lxpcr
           v
           lxqur
          r
    dx

This would now actually be quite readable, with some practise. The main problem is that it takes up vastly more space than the standard notation, because of all that whitespace. Maths notation avoids this by just making the “top level separators” a bit larger or otherwise visible, instead of making them stand out in the indentation.


Tangent Line to the curve that is Parallel to the line Find the equation of the tangent line to the curve $ y=1-x^2$ that is parallel to the line $2x-4y=8.$

I really don't know how to start this problem, can you help me please to solve this problem? Thank you.

Determine the slope of the given line: Since we want the tangent line to be parallel to the line $2x - 4y = 8$, we need for it to have the same slope. We can rewrite the equation in slope-intercept form to easily find slope: $4y= 2x- 8\iff y = \frac 12 x - 2$. Hence the desired slope of our line is $m = \frac 12$.
Then find the derivative of $y = 1-x^2$ and set it equal to the desired slope. $y' = -2x$. So put $-2x = \frac 12$
Solve for the value of $x$. $$-2x = \frac 12 \iff x = -\frac{1}{4}$$
That will give you the x-coordinate of a point on the desired line. Find the corresponding y-coordinate, and you'll have both coordinates of a point on the desired line. $$y = 1 - x^2 \implies y\left(-\frac 14\right) = 1 - \frac 1{16} = \frac {15}{16}$$
We know the slope of the desired line is equal to $\frac 12$, and we now have a point on the desired line, as well as on the given curve: i.e., the point of tangency $\left(-\frac 14, \frac {15}{16}\right)$.
Use the point-slope form to obtain the equation of the desired line. Given slope $m$ and a point on the desired line,$(x_0, y_0)$, the equation of the line is given by $$(y-y_0) = m(x-x_0)$$

This gives us: $$y - \frac {15}{16} = \frac 12\left(x + \frac 14\right) \iff 16y - 15= 8x + 2 \iff 8x - 16y= -17$$

We can graph the function, the given line, and the desired tangent line to see "what we've got". The "brownish-grey" line is the desired line:

The slope is not $0.5$ that is the slope for the line parallel to the line. The slope that should be used when solving its equation.
$y-y_1=m(x-x_1)$ the slope is 2 from $y= 2x$ which is the slope of the tangent line
.


What is the name of the area formed by two intersecting circles? When two circles intersect they form an area which is "ellipse-like" in shape.  What is the name of this shape?

The technical term is lens.${}{}{}{}{}{}{}$

It's called a Vesica Piscis, which is Latin for "fish bladder," a reference to the shape of a fish's swim bladders.

Furthermore, for art history buffs, in Italian it is called a Mandorla, which is also the name used for it in art. Mandorlas were used in medieval and renaissance art as a frame for Christ and other heavenly figures, thought of as the shape of the gate to the Kingdom of Heaven.


Interesting variant of Monty Hall I was wondering how we could think about the following variant of the classic Monty hall theorem,

Suppose now that the host actually does not remember what is behind any of the doors, once you choose one of the three doors, he will open at random one of the other two doors. If he reveals the prize the show ends, if he does not reveal the prize, should you still switch doors?

I was trying to formulate it but I am just not sure if I am making correct assumptions.

For example,

say we choose door $A$, then the probability that he opens door $B$ is equal to the probability he chooses door $C$ is equal to $0.5$.

The probability that the prize is behind door A, B and C is 1/3

the probability that Monty ends up opening any given door is also $1/3$

So should one just condition on something else now?

Looking forward to hearing any opinions on this,

Thanks

Suppose you choose door 1. And Monty reveals door 2, but Monty has no prior information.

Then the conditional probability that the prize is in fact behind door one, given that it is not behind door 2 is $50\%$

If Money has no information, then not enough information gained by the reveal, to justify switching.

There is a long discussion of this on The Straight Dope by Cecil Adams. I will add the link here:

http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3


Help in proof of theorem about Riemann-Liouville Fractional Calculus Theorem:

Let,
$$\left[D^{nv}+a_{1}D^{\left(n-1\right)v}+\dots+a_{n}D^{0}\right]\left(y\right)=0$$
be a fractional differential equation of order $\left(n,q\right)$,
where $v=\frac{n}{q}$, and let
$$P\left(x\right)=x^{n}+a_{1}x^{n-1}+\dots+a_{n}$$
be the corresponding indicial polynomial.

Let
$$y_{1}\left(t\right)=\mathscr{L}^{-1}\left\{ P^{-1}\left(s^{v}\right)\right\} $$
Then if $N$ is the smallest integer with the property $N\geq n\cdot v$,
then
$$y_{1}\left(t\right), y_{2}\left(t\right)=D^{1}y_{1}\left(t\right),...,y_{N}\left(t\right)=D^{N-1}y_{1}\left(t\right)$$
are $N$ independent solutions.

This theorem is from the book "An Introduction to Fractional Calculus and Fractional Differential Equations" by Miller and Ross. I don't manage to understand the proof provided in the book.

Proof:

We take the Laplace transform of the equation:

$\mathscr{L}\left\{ P\left(D^{v}y\left(t\right)\right)\right\} =0$,

and also
$$\mathscr{L}\left\{ P\left(D^{v}y\left(t\right)\right)\right\} =P\left(s^{v}\right)Y\left(s\right)-\sum\limits_{r=0}^{N-1}B_{r}\left(y\right)s^{r}$$
where $B_{r}\left(y\right)$ is a linear combination of terms of the
form:

$$D^{kv-\left(r+1\right)}y\left(0\right)$$
$k=rq+1,\dots,n$ $r=0,1,\dots,N-1$

In particular $B_{0}\left(y\right)=P\left(D^{v}\right)D^{-1}y\left(0\right)-a_{n}D^{-1}y\left(0\right)$.

We have,

$$Y\left(s\right)=\frac{\sum\limits_{r=0}^{N-1}B_{r}\left(y\right)s^{r}}{P\left(s^{v}\right)}$$
and
$$y\left(t\right)=\mathscr{L}^{-1}\left\{ Y\left(s\right)\right\} $$
is a solution of the equation.

Now let,

$y_{1}\left(t\right)=\mathscr{L}^{-1}\left\{ P^{-1}\left(s^{v}\right)\right\} $,
then

$\mathscr{L}\left\{ P\left(D^{v}\right)y_{1}\left(t\right)\right\} =P\left(s^{v}\right)-\sum\limits_{r=0}^{N-1}B_{r}\left(y_{1}\right)s^{r}$.

The initial value theorem for the Laplace transform says that,
$\underset{s\rightarrow+\infty}{\lim}s^{v+1}\mathscr{L}\left\{ f\left(t\right)\right\} =D^{v}f\left(0\right)$
for all $v\in\mathbb{R}$.
Thus, $B_{0}\left(y_{1}\right)=1$, $B_{r}\left(y_{1}\right)=0$ for $r>1$

Is this last deduction in italics where I'm stuck. It is not clear to me.

As you might know, we have Laplace's transform of various sense of fractional derivatives. So it is not clear to me as to whether you really refer to Riemann Liouville, Caputo or something like that, please just clarify a little and may be I can help.

By definition, $(D^vf)(x)=\dfrac{1}{\Gamma(1-v)}\dfrac{d}{dx}\displaystyle\int^x_0\dfrac{f(t)dt}{(x-t)^{v-[\Re(v)]}}$ ......$(1)$
and
$\mathscr{L}(D^k\phi(t))(s)=s^k(\mathscr{L}\phi)(s)-\displaystyle\sum_{j=0}^{k-1}s^{k-j-1}(D^j\phi)(0)$ ......$(2)$,   $k\in{\mathbb{N}}$.
Then with these two, a more general relation is drived:
$\mathscr{L}(D^vf(t))(s)=s^v(\mathscr{L}f)(s)-\displaystyle\sum_{r=0}^{N-1}s^{N-r-1}(D^v(I^{N-v}f)(0)$ ......$(3)$, where $(\Re(s)>q_0)$. Study these three equations by passing $\lim_{s\rightarrow{\infty}}$ throughout equation $(3)$ and continue with your work.

But remember, this is only true on the half axis and I omitted the $+0$ subscript sign that indicates Riemann Liouville on the half axis.


Step-by-Step Solution for $x^{1/x}=2^{1/2}$ I came across the equation $$x^{1/x}=2^{1/2}$$ where $x\in\mathbb R$. One can immediately see that $x=2$ is a solution, but it is easy to miss that $x=4$ satisfies the equation as well. Verfiying that $2,4$ are solutions is not hard, but how would one go about formally solving this equation, i.e. how could one solve for $x$? I am asking because if the equation was  say $x^{1/x}=2^{1/3}$ then the two solutions would not be obvious.

Basically, how to fill in the dots:
$$x^{1/x}=2^{1/2} \iff \ldots\iff x=2 \text{ or } x=4$$

Notice

$$x ^{1/x} = 2^{1/2} \Rightarrow x = 2^{x/2} \Rightarrow x^{2} = 2^x$$

then you may look want take a look here.

I suppose it is necessary to consider what happens when $x < 0.$ There is a question of interpretation of $x^{\frac{1}{x}}$ for $x \leq 0,$ ( for example, $(-1)^{r}$ could reasonable interpreted as a non-real complex number if $r$ is an irrational positive real number, if we consider $-1$ to be $e^{i \pi}$), so I assume the function only to be defined for $x >0$, where it is defined to be $e^{\frac{\log x}{x}}.$ As far as this question is concerned, it seems that the case $x >0$ is the most relevant case anyway.

Note that $x >0$ is a solution of $x^{\frac{1}{x}} = 2^{\frac{1}{2}}$ if and only if $\frac{\log x}{x} = \frac{\log 2}{2}.$ The derivative of $\frac{\log x}{x}$ is $\frac{ 1 - \log x}{x^{2}}$ which is positive when $x <e$ and negative when $x > e.$ Hence $\frac{\log x}{x}$ increases when $x <e$, then decreases when $x >e$, taking a maximum value when $x = e.$ The values of $\frac{ \log x}{x}$ are positive on $(1, \infty)$  and the function is differentiable on that interval.Now if $\frac{log{x}}{x} = \frac{\log{y}}{y}$ for $1 < x < y,$ then the derivative vanishes somewhere on $(x,y)$ by the Mean Value Theorem. Since the derivative vanishes only at $e$, we see that any value of $\frac{\log x}{x}$ can be attained at most twice. Since $\frac{\log{4}}{4} = \frac{2\log 2}{2 \times 2} = \frac{\log 2}{2},$ we see that $2$ and $4$ are the only values of $x$ for which $x^{\frac{1}{x}} = 2^{\frac{1}{2}}.$


poisson random variable distribution using probability Let $X$ be the number of emails that a company receives in a day. Assume that $X$ is a Poisson random variable with parameter $\lambda$. The company classifies each email as spam or not spam. The probability that a single email is spam is $p$. Let $Y$ be the number of spam emails that the company receives in a day. Assume that for any two emails, whether or nor they are spam are events.
  (a) Find $E[Y]$.
  (b) Show that $Y$ is also a Poisson random variable and find its parameter (in terms of $\lambda$ and $p$). Hint: Use the partition theorem: $$P(Y=k)=\sum\limits_{n=k}^{\infty}{P(Y=k|X=n)P(X=n)}.$$

I have tried it out, but I am finding it difficult to post them in proper format.
Please help me with the solution.

Firstly we will prove $(b)$, then $(a)$ can be deduced easily. We will show that $Y \sim Poisson(\lambda p)$ using the fact that the Moment-generating function of $Y$ will be the Poisson one.

We have $X \sim Poisson(\lambda)$ and $\displaystyle M_X(t)=e^{\lambda\left(e^t - 1\right)}.$
Now

$$\begin{align}
M_Y(t)&=E[e^{it}]\\
&=\sum\limits_{i=0}^{\infty}{e^{it}P\left( Y=i \right)}\\
&=\sum\limits_{i=0}^{\infty}{e^{it}\left( \sum\limits_{j=i}^{\infty}{P(  Y=i|X=j)P\left(X=j\right)}\right)}
\end{align}$$

but $\displaystyle P(Y=i|X=j)=\binom{j}{i}p^iq^{j-i}=\frac{j!}{(j-i)!i!}p^iq^{j-i}$, i.e. binomial, because the events asociated with probability $p$ are independet and $\displaystyle P(X=j)=\frac{e^{-\lambda} \lambda^j}{j!}$. So

$$\begin{align}
M_Y(t)
&=\sum\limits_{i=0}^{\infty}{e^{it}\left( \sum\limits_{j=i}^{\infty}{\frac{j!}{(j-i)!i!}p^iq^{j-i}\frac{e^{-\lambda} \lambda^j}{j!}}\right)} \\
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{e^{it}\frac{1}{i!}\frac{p^i}{q^i}\sum\limits_{j=i}^{\infty}{\frac{q^{j}\lambda^j}{(j-i)!}}} \\
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q}\right)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^{j}}{(j-i)!}}} \\
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q} \right)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^i(q\lambda)^{j}}{(q\lambda)^i(j-i)!}}} \\
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q} \right)^i(q\lambda)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^{j-i}}{(j-i)!}}} \\
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i\sum\limits_{j=0}^{\infty}{\frac{(q\lambda)^{j}}{j!}}} \\
\end{align}$$

but we know $\displaystyle \sum\limits_{k=0}^{\infty}{\frac{x^{k}}{k!}}=e^x,$ hence

$$\begin{align}
M_Y(t)
&=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i e^{q\lambda}}
=e^{-\lambda}e^{q\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i }
=e^{-\lambda(1-q)}e^{\lambda e^{t}p}=e^{\lambda p\left( e^t -1\right)},
\end{align}$$

therefore $Y \sim Poisson(\lambda p)$.

$\large p \gtrsim 0\,,\quad X \gg 1$.
\begin{align}
{X \choose Y}\,p^{Y}\,\left(1 - p\right)^{X - Y}
&=
{X\left(X - 1\right)\ldots\left(X - Y + 1\right) \over Y!}\,
p^{Y}\exp\left(\vphantom{\LARGE A}\left[\,X - Y\,\right]\,\ln\left(1 - p\right)\right)
\end{align}
Since
$$
\left\lbrace%
\begin{array}{rcl}
X\left(X - 1\right)\ldots\left(X - Y + 1\right)
& = &
X^{Y}\left(1 - {1 \over X}\right)\ldots\left(1 - {Y - 1 \over X}\right)
\approx
X^{Y}
\\
\exp\left(\vphantom{\LARGE A}\left[X - Y\right]\,\ln\left(1 - p\right)\right)
& \approx &
\exp\left(\vphantom{\LARGE A}X\,\left[-p\right]\right)
= {\rm e}^{-Xp}
\end{array}\right.
$$

we get
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
{X \choose Y}\,p^{Y}\,\left(1 - p\right)^{X - Y}
\color{#000000}{\ \approx\ }
{X^{Y} \over Y!}\,p^{Y}{\rm e}^{-Xp}
\color{#000000}{\ =\ }
{\left(Xp\right)^{Y}{\rm e}^{-Xp} \over Y!}\quad}
\\ \\ \hline
\end{array}
$$


Gradient of a function Gradient of a function is $\langle f_x(x,y),f_y(x,y) \rangle$. But I don't understand this gradient vector shows what. When I find gradient of some function, that vectors represents what? Thank you.

The gradient is a fancy word for derivative, or the rate of change of a function. It’s a vector (a direction to move) that

Points in the direction of greatest increase of a function.
Is zero at a local maximum or local minimum (because there is no single direction of increase).

The term “gradient” is typically used for functions with several inputs and a single output (a scalar field).

We know the definition of the gradient: a derivative for each variable of a function.

Suppose,

If we want to find the direction to move to increase our function the fastest, we plug in our current coordinates (such as 3,4,5) into the equation and get:

So, this new vector (1, 8, 75) would be the direction we’d move in to increase the value of our function.

The other use of gradient is to find the max/min of multivariable functions.

For a multivariable function $f(x,y)$, that vector is called $\textbf{grad}f$, the gradient of $f$, sometimes denoted $\nabla f$.

It's a vector that points in the direction of greatest increase of the function.


Sets with same Cardinality, but no Explicit Bijection? Are there any good examples of sets where we know that they have the same cardinality, but have not found any explicit bijection between them?

The real numbers have the same cardinality as the elements of a basis of the real numbers regarded as a vector space over the rationals. The existence of such a basis needs some appropriate form of the axiom of choice, so the basis cannot be given explicitly, let alone any bijection.

The set of transcendental numbers, set of irrational numbers and the set of real numbers are all uncountable and have the same cardinality. It is not clear (at least to me) how to construct a bijections among them.

The set of rational numbers  and the set of algebraic numbers are both infinite countable sets. Here also it is not clear how to construct a bijection.


Ways to define a curve I'm trying to give shapes in my physics engine roundness/ curvature.

I am aware of various methods for mathematically defining curvature such as bezier-curves, ellipses, etc; but I'm not sure which methods are most appropriate for use in a physics engine. I must consider the speed, flexibility, and difficulty of constructing/joining the curves when choosing my approach. The ideal is a system where a user could easily construct and collide a large amount of curved, aesthetic polygon bodies together, yet I could still maintain a satisfactory framerate. However it's inevitable that ease-of-use and other attributes will suffer.

NOTE: my physics engine is continuous, which means I must pre-calculate the times at which polygons will collide. Keep in mind that the curve's route my not be linear.

For each suggestion please give the positives/negatives!

What techniques are there for mathematically defining a 2D curve?
Advantages/Disadvantages (speed, flexibility, ease of construction/use)?
Is the technique feasible for an engine where predictability of collisions is crucial?

TL;DR
If you want to specify a curve for computation in a computer system, use actual points rather than equations. It will be better anytime. And for analysis, use piecewise linear (polygonal) models.

NOW THE LONGER PART

Stick with piecewise linear as far as possible.
Why? Absolutely simple to construct, good speed, large flexibility. Accuracy can be easily improved using a higher resolution.

Your ease of use seems to be a problem. The solution to that will be to make an interface so that user has to specify only the points of the boundary of the body with a sufficient accuracy. He can use whatever methods he wants to do so. Once that is done, you can take those points and design whatever curves you want from it. As in this case, you can see, piecewise linear is quiet simpler than other curves and also, it imposes least restriction on the user.

There are many other advantages of piecewise linear. You can use mesh libraries to construct a matrix based description of your system, and then use it predict crashes. Once the crash has occured, if your bodies are not getting deformed, great, you can just reuse the descriptions but with a little change in the equations.

If you can get your functions in the form of:

f(t)=... and

g(t) =...

and you want the intersection, all you have to do is use Desmos Graphs

You could also do something like f(t)-g(t) = 0 and the graph of the result will show the intersection as well.

After the graph is displayed, you can place the mouse on the graph to show the point of interest. It helps when you color the graphs.


Functional Equation $f(f(x))=\frac{x}{2}+3$ Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ of class $C^1$ where $f\circ f(x)=\dfrac{x}{2}+3$.

I need help with this problem.

Let $f^{\circ n}$ be the shorthand for composing $f$ with itself for $n$ times.

Let $g(x) = f(x+6)-6$. Since $f \in C^1$, so does $g$. Notice

$$g\left(\frac{x}{2}\right) = f\left(\frac{x+6}{2}+3\right)-6 = f(f^{\circ 2}(x+6)))-6
= f^{\circ 2}(f(x+6)) - 6\\ = \frac{f(x+6)}{2} + 3 - 6 = \frac{f(x+6)-6}{2} = \frac{g(x)}{2}$$

When $x = 0$, this implies $g(0) = 0$.

When $x \ne 0$, divide both sides by $\frac{x}{2}$ gives us
$\displaystyle\;\frac{g(x)}{x} = \frac{g(x/2)}{x/2}$.
Repeat applying this, we find
$\displaystyle\;\frac{g(x)}{x} = \frac{g(x/2^k)}{x/2^k}$ for any integer $k > 0$.
As a result,
$$\frac{g(x)}{x} = \lim_{k\to\infty}\frac{g(x/2^k)}{x/2^k}
\stackrel{\color{blue}{\because\text{ RHS exists}}}{=} \lim_{h\to 0} \frac{g(h)}{h} = \lim_{h\to 0}\frac{g(h)-g(0)}{h} \stackrel{\color{blue}{g \in C^1}}{=} g'(0)
$$

This means for some constant $a$,

$$g(x) = ax \quad\implies\quad f(x) = a(x-6)+6
\quad\implies\quad f(f(x)) = a^2(x-6) + 6
$$
Compare rightmost expression with the functional equation $f(f(x)) = \frac{x}{2} + 3$,
we find
$$a = \pm \frac{1}{\sqrt{2}}
\quad\implies\quad f(x) = \pm\frac{1}{\sqrt{2}}(x-6) + 6$$

This means the original functional equation has two and only two solutions.

Let $f(x)=ax+b$. Then, $f(f(x))=a^2x+ab+b$. We can set up a system of equations:

$$a^2=\frac{1}{2},$$

$$(a+1)b=3.$$

The first equation gives $a=\pm\frac{1}{\sqrt{2}}$, while the second results in $b=\frac{3\sqrt{2}}{\sqrt{2}\pm1}$ or $b=6\pm3\sqrt{2}$.


Elementary Papers at ArXiv Inspired by this question, at MO i am asking this question.

Can anyone list some elementary articles at ArXiv which can be understood by High-School/Undergrad Students. I am asking this because, i would like to see some interesting papers and articles to learn something new. If the paper is very much advanced then its of no use to me since i haven't learnt enough Math. I have been searching around for past two hours and i couldn't find a single paper which i could comprehend. Article which i would like to see:-

Interesting proofs of some Elementary number Theory results.
Interesting identity involving Infinite series.
Interesting articles in Basic Abstract Algebra ( concerning Groups and Rings.)
Your favorite article. ( Make sure its elementary enough!)

The best example of what type of articles i am looking for can be found in the answer given by Bill here:

Characterizing continuous functions based on the graph of the function

This is an interesting article. I really enjoyed reading it since i could understand it.

A similar paper to Yuval's answer, which is also quite elementary, is Doyle (and Conway)'s Division by Three. I don't think it requires any advanced knowledge, and is rather interesting.

Sometime in the near future I plan on posting a paper in the General Mathematics section that may be what you're looking for, but for the time being you can find the paper here.


Finding the area between 2 curves using Green's Theorem Find the area bounded by $y=x^2$ and $y=x$ using Green's Theorem.

I know that I have to use the relationship $\int_c Pdx + Qdy = \int\int_D1dA$. But I don't know what my boundaries for the integral would be since it consists of two curves.

I know that if it were a circle, I would simply parametrize the circle equation, which I understand since the curve consists of only one equation. However, since there are two equations, I'm not too sure how to approach this.

Thanks

$$Area=\frac{1}{2}\oint_C xdy-ydx=\frac{1}{2}\oint_{C_1} xdy-ydx+\frac{1}{2}\oint_{C_2} xdy-ydx$$ where $C$ is the curve consisting of the curve $y=x^2$ from $0$ to $1$ and the line $y=x$ from $1$ to $0$, say $C_1$ and $C_2$ respectively.then $$\frac{1}{2}\oint_C xdy-ydx=\frac{1}{2}\int_0^1 x(2xdx)-x^2dx+\frac{1}{2}\int_{1}^0 xdx-xdx=\frac{1}{2}\int_0^1x^2dx=\frac{1}{6}$$

Given $y=x^2$ and $y=x$, we can have a line integral $1/2\int(xdy-ydx)$.
By Green's theorem, $\int_C(Mdx+Ndy) =\int_R[
∂N/∂x-∂M/∂y]dxdy$.
Now, $M=-y/2$ and $N=x/2$.

$∂N/∂x=1/2$ and
$∂M/∂y=-1/2$
so,$[
∂N/∂x-∂M/∂y]=1/2-(-1/2)=1$.

Thus, the area $$A=\int_C(Mdx+Ndy) =\int_R[
∂N/∂x-∂M/∂y]dxdy=\int_D dxdy,$$
where $D$ is defined as the region where $y$ goes from $y=x^2$ to $y=x$ and $x$ from 0 to 1. Then,the integral equals $1/6$.


Show that $\sum \pm X_n<\infty$ a.s. implies $\sum X_n^2<\infty$ a.s. Suppose $\{X_n\}$ are arbitrary random variables such that $\sum \pm X_n$ converges almost surely for all choices $\pm1$. Show that $\sum X_n^2$ converges almost surely.

Denote $\{B_n\}$ Bernoulli random variables, the statement above is to say $\sum B_n X_n$ converges a.s. implies $\sum X_n^2$ converges a.s.

Can anybody give a solution to this problem?

This problem comes from Kai Lai Chung's A Course in Probability Theory, Third Edition, pp.129

Since $\sum\limits_nB_nX_n<\infty$ $a.s.$, Thus $\exists K$ $s.t.$ $|B_nX_n|\le K$ $a.s.$

Notice that $E(B_nX_n)=EB_n\cdot EX_n=0$.

By Theorem 12.2 in Probability with Martingales (Williams)(Refer to  The convergence of $\sum \pm a_n$ with random signs) we obtain $\sum\limits_nVar(B_nX_n)<\infty$. (This may have some problems without the independence of $\{X_n\}$ as mentioned by the comments)

Thus $\sum\limits_nEX_n^2=\sum\limits_nE(B_nX_n)^2<\infty$.

Using Fubini's Theorem we have $E(\sum\limits_nX_n^2)<\infty$, therefore $\sum\limits_nX_n^2<\infty$ $a.s.$


The order of permutation groups and alternating groups The question was:

  True or False: $\forall{n}\in{\mathbb{N}}$ the group $S_n$ and $A_n$ have different sizes.

My answer is False. That is since  both $A_1 =(\text{id})$ and $S_1 =(\text{id})$.

Can any one confirm my answer please? Thank you very much.

Of course $S_n$ in general is much larger than $A_n$. $A_n$  is the group of permutations with even parity only, while $S_n$ is the group of all permutations of $\{1,\cdots,n\}$. Not all permutations are of even parity, e.g., consider
$$(1,2)\circ(1,3)\circ(1,5)$$
the product of three transpositions that give rise to a cycle. In fact any cycle of odd length does not belong to $A_n$.


Constructing a martingale from an SDE Let $dX_t=u(t,w)dt+dB_t$, where $dB_t$ is a Brownian Motion, $u$ is bounded and measurable with respect to the filtration $F_t$ and $u$ be an ito process on $(\Omega, \{\mathcal{F}\},\mathcal{P})$. Find a martingale $M_t$ s.t. $M_0=1$ and $Y_t=X_tM_t$ is an $\mathcal{F}_t$-martingale.

It is easily derived that $X_t=X_0+\int^t_0 u(s,w)ds+B_t$.

Try
$$
M=\exp\left(-\int_0^\cdot u_s\,\mathrm dB_s-\frac12\int_0^\cdot u_s^2\,\mathrm ds\right).
$$
Then $M$ is an exponential martingale. Furthermore, $\mathrm dX=\mathrm dB+u\mathrm dt$ and $\mathrm dM=-uM\mathrm dB$ hence $\mathrm d\langle M,X\rangle=-uM\mathrm dt$ and $Y=MX$ is such that
$$
\mathrm dY=X\mathrm dM+M\mathrm dX+\mathrm d\langle M,X\rangle=M(1-uX)\mathrm dB.
$$
In particular, $Y$ is a martingale.

Let $dM_t=\sigma (t,w)M_t dB_t$ then let $Y_t=X_t M_t$. we arrive at $dY_t=M_t(u(t,w)+\sigma(t,w))dt + (... )dB_t$, then we see that for $dY_t$ to be a martingale, we need to have  zero drift. Hence $\sigma (t,w)=-u(t,w)$
thus $dM_t=-u (t,w)M_t dB_t$

So $M_t=e^{-\int_0^t u(s,w) dW_s}$

not too sure if this is the actual solution to the SDE.


Solve $x\sqrt{x\sqrt{x\sqrt{x\dots}}} = 4$ Today I faced a strange equation and I didn't manage to find a solution to it:
$$x\sqrt{x\sqrt{x\sqrt{x\dots}}} = 4$$
Maybe someone will help me to find a way to solve it. By the way, this equation is from high school course.

The power of given variabe is in the term of sum of G.P . The indices of x is such that 1+1/2+1/4+......
Because we know that the infinite sum of G.P:
Sn = a/(1-r) where r=common ratio


Proof that the Kelly Criterion is optimal? Let's say you are betting on the outcome of some random variable. You can invest any amount $X$, and each outcome multiples that amount $X$ by a certain amount $ \ge 0$. You must choose $X$ less than how much capital you have. A Kelly Bet is one in which you bet an amount that maximizes the logarithm of your capital. The Kelly Bet will allow your capital to grow larger than any other betting scheme in the long term (i.e., as the number of bets approaches infinity).

What is the proof that previous statement?

The statement is false.

For example consider the bet where you double your money with probability $2/3$ and lose it otherwise. Now consider the strategy "bet \$1 more than the Kelly bet on the first turn, and then use the Kelly bet after that". If this wins on the first turn it will forever have more money than the Kelly strategy, and if it loses on the first turn it will forever have less money than the Kelly strategy. So no matter how large the number of bets becomes, it has a $2/3$ chance of having a larger amount than the Kelly strategy.

Say in total I have 1 dollar to start with and I bet $X$ amount of my current net worth each time. Say if I win a bet (with probability $p$), I will get $X b$ back (net odd of $b$). Then after the bet with probability $p$, I got back

$1+Xb$

and with probability $1-b$, I will have

$1-X$

left.

Now, consider betting for $N$ times, then there will be about $Np$ win and $N(1-p)$ loss. So, the final amount is about

$(1+Xb)^{Np} (1-X)^{N(1-p)}=((1+Xb)^{p} (1-X)^{1-p})^N.$

To maximize this gain, we just need to maximize $(1+Xb)^{p} (1-X)^{1-p}$ with respect to $X$. Or we can maximize $f(X)=\ln [(1+Xb)^{p} (1-X)^{1-p}]$ instead. Setting $\frac{d f}{d X}=0$, we have $\frac{pb}{1+Xb}-\frac{1-p}{1-X}=0\Rightarrow X= \frac{bp-(1-p)}{b}.$


What is $E[E[Y|X]|X]$? What is $E[E[Y|X]|X]$ ?
Can I use the law of total expectation and say that it's equal to $E[Y|X]$ ?
If it's correct, could someone explain why?

Let $\mathscr F$ be the $\sigma$-algebra generated by $X.$ Then
$E[Y|X]=E[Y|\mathscr F]$ is an $\mathscr F$-measurable random variable and we know that if $Z$ is a $\mathscr F$-measurable random variable, we have $E[Z|\mathscr F]=Z.$

The law of total expectation would $E[Y]=E[E[Y|X]]$. In this expression $E[Y|X]$ is a random variable that is a function of $X.$ $E[Y|X]$ changes with $X,$ assuming different realizations depending on different outcomes of $X=x,$ and by the law of total expectation, its expected value, across the support of $X$ is just $E[Y].$ Conditioning on $X$ the expression $E[Y|X]|X$ is redundant $E[Y|X]|X=E[Y|X],$ and in this sense it amounts to applying the same law of total expectation $E[E[Y|X]]=E[E[Y|X]|X]=E[Y].$ $\;$


Proof of Ascoli's theorem In the Section 45 of Munkres' Topology, after the classical form of the Ascoli's Theorem have been proven, the author gave an exercise to show the proof is still valid if $\Bbb{R}^n$ is replaced by any metric space in which all closed bounded subspaces are compact.
The classical form of the Ascoli's theorem is as follows:

Theorem. Let $X$ be a compact space and let $(\Bbb{R}^n,d)$ denote the $n$-dimensional Euclidean space with the Euclidean metric. Suppose the space $C(X,\Bbb{R}^n)$ is equipped with the uniform topology. Then a subset $\mathcal{F}\subseteq C(X,\Bbb{R}^n)$ has compact closure if and only if $\mathcal{F}$ is equicontinuous and pointwise bounded under $d$.

The sketch of the proof is as follows:

Sketch of Proof. ($\Longrightarrow$). Suppose that the closure ${\rm Cl}(\mathcal{F})$ is compact. Then it is clearly totally bounded and equicontinuous. In particular, it is bounded so we have $\rho(f,g)\leq M$ for every $f,g\in{\rm Cl}(\mathcal{F})$ where $\rho$ is the sup metric. Then it follows that ${\rm Cl}(\mathcal{F})$ is also pointwise bounded. Since $\mathcal{F}\subseteq{\rm Cl}(\mathcal{F})$, the collection $\mathcal{F}$ is also equicontinuous and pointwise bounded.

($\Longleftarrow$). Assume that $\mathcal{F}$ is equicontinuous and pointwise bounded under $d$. Then it suffices to show that ${\rm Cl}(\mathcal{F})$ is complete and totally bounded. The completeness of ${\rm Cl}(\mathcal{F})$ follows from the completeness of $C(X,\Bbb{R}^n)$. As for the total boundedness, we can easily show that ${\rm Cl}(\mathcal{F})$ is equicontinuous and pointwise bounded as well. Then we can find a closed ball $Y$ large enough centered at origin containing all $g(X)$ where $g\in{\rm Cl}(\mathcal{F})$. The total boundedness of ${\rm Cl}(\mathcal{F})$ follows from the following lemma:

Lemma. Let $X$ and $Y$ be two compact spaces where $Y$ is metrizable by $d$. If a collection $\mathcal{F}\subseteq C(X,Y)$ is equicontinuous under $d$, then $\mathcal{F}$ is totally bounded under both the uniform and sup metrics corresponding to $d$.

As far as I can see, the proof used the property of $\Bbb{R}^n$ twice.
On the one hand, it used the completeness of $\Bbb{R}^n$ when proving the completeness of ${\rm Cl}(\mathcal{F})$, because the completeness of $\Bbb{R}^n$ implies the completeness of $C(X,\Bbb{R}^n)$. Since ${\rm Cl}(\mathcal{F})$ is a closed subspace of $C(X,\Bbb{R}^n)$, it is also complete.
On the other hand, it used the Heine-Borel property of $\Bbb{R}^n$; that is, every closed and bounded subspace of $\Bbb{R}^n$ is compact, to show that the union of $g(X)$ where $g\in{\rm Cl}(\mathcal{F})$ is contained in some compact space of $\Bbb{R}^n$ in order to apply the lemma.
If we replace $\Bbb{R}^n$ by an arbitrary metric space $Z$ satisfying Heine-Borel property, apparently the second part will still go, but I doubt whether the first part is valid. If $Z$ is not complete, can we still deduce that ${\rm Cl}(\mathcal{F})$ is complete?

There are several extensions of this result, depending on the application. Here is a version that works in many settings:
Theorem: Let $(X,\tau)$ be a compact topological space and let $(S,d)$ be a complete metric space. $\mathcal{F}\subset\mathcal{C}(X,S)$ is relatively compact iff
$\mathcal{F}$ is equicontinuous and $\{f(x):f\in\mathcal{F}\}$ is relatively compact in $S$ for each $x\in X$. (metric in $\mathcal{C}(X,S)$ is defined as $\rho(f,g)=\sup_{x\in X}d(f(x),g(x))$.
Here is a paper that has a beautiful proof of the result as quoted above.
Section 2 (preliminaries) of the paper contains very nice ideas that you may find useful

If $Z$ is not complete, you can find a counter example to the property using constant functions : Take some bounded subset $C$ of $Z$. Consider the constant functions with values in $C$. This defines an equicontinuous family of pointwise bounded maps. But its closure is not complete.


How to Evaluate $\int\frac1{x \ln x+ 7 \ln x} \,\mathrm dx$ I have tried many methods but do not know how to integrate this:

$$ \int \frac{1}{x\ln x + 7\ln x} dx $$

with respect to x.

I would factor the denominator into (ln x)(x+7). Then, you can split the fraction into (1/(x+7))*(1/ln x). From here, it is fairly easy to integrate by parts.


Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$. Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$

I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.

Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t} $

Thus the equation becomes,

$ t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$

If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\implies 1-x/2=0\implies x=2$

If $t=5-2\sqrt 6 \implies (\sqrt{5+2\sqrt{6}})^{x}=5-2\sqrt 6=\frac{1}{5+2\sqrt 6}\implies (5+2\sqrt 6)^{1+x/2}=1\implies x=-2$

Thus, $x=2,-2$ are the two solutions.

Verification:

Putting $x=2$ in the original equation gives L.H.S=$5+2\sqrt 6+5-2\sqrt 6=10=$R.H.S

Similarly, putting $x=-2$ also gives L.H.S=$5-2\sqrt 6+5+2\sqrt 6=10=$R.H.S


Prove Inequality $p^n+q^n+r^n \geq p^m+q^m+r^m$ when $pqr=1, n > m \geq 0$ Prove Inequality $$p^n+q^n+r^n \geq p^m+q^m+r^m$$ when $pqr=1, n > m \geq 0, p,q,r >0$

Help me to prove this thm... I think it use AM-GM, But It work not useful to me...

Hint : With $\alpha = \frac13\left(1-\frac{m}n\right)$ weighted AM-GM gives $(1-2\alpha) p^n +\alpha q^n + \alpha r^n \geqslant p^{n(1-2\alpha)}\cdot q^{n\alpha}\cdot r^{n \alpha} = p^m \cdot(pqr)^{n\alpha}$

$RHS=(p^m+q^m+r^m)(pqr)^\frac{n-m}{3}=\dfrac{1}{2}\sum_{sym} p^\frac{n+2m}{3}q^\frac{n-m}{3}r^\frac{n-m}{3} \\ LHS=\dfrac{1}{2}\sum_{sym} p^nq^0r^0 \\ \because (n,0,0)\succ(\frac{n+2m}{3},\frac{n-m}{3},\frac{n-m}{3}) \\ \therefore LHS \ge RHS$

By Muirhead's Inequality.


Reference request: Clean proof of Fermat's last theorem for $n=3$. I have seen a proof for FLT, $n=3$ using factorisation in the ring of Eisenstein integers, but it's quite long and convoluted; I am wondering if there is a more 'advanced' proof which avoids infinite descent/messy repeated calculations.

What's your favorite proof of Fermat's last theorem for $n=3$, and where can it be found written down?

Recently I have found two simple proofs of FLT for n=3.
They use the ring $\mathbb{Z}[\sqrt[3]{2}]$.
If $x^3+y^3+z^3=0$ then

(1) $x^6-4 (yz)^3=a^2$,

where $a=y^3-z^3$.

The idea is to deduce from (1) that the algebraic number $x^2-\sqrt[3]{4} yz$ is a square, more precisely:

(2) $x^2-\sqrt[3]{4} yz=u \beta^2$,

where $\beta \in \mathbb{Z}[\sqrt[3]{2}]$, and $u$ is a unit.

Once this is established, then it is not difficult to get a contradiction via infinite descent.
It is similar to Euler's proof, but another number field is used.

The (2) follows from the unique factorization in the Euclidean ring $\mathbb{Z}[\sqrt[3]{2}]$

My second proof deduces (2) without using the unique factorization.

The idea is simple: find a number $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$, such that $\alpha^2$ is divided by the $x^2-\sqrt[3]{4} yz$ and $\alpha^2/(x^2-\sqrt[3]{4} yz)$ has a small norm.
It turns out that the last number may have small norm indeed: less than 9.
Then the norm of it can be either 1 or 4, and (2) easily follows.
The method can be used to prove FLT for other concrete $n$ as well.

It is easy to get such $\alpha$ because $\alpha$ belongs to $I$, where $I$ is the ideal of norm $a$, generated by $a$ and $x^2-\sqrt[3]{4} yz$.
We know from the theory of algebraic numbers, that there is an $\alpha \in I$,
such that $N(\alpha)<m N(I)$, where $m$ is easily calculated constant, that turns to be less than 3.
Then $N(\alpha^2/(x^2-\sqrt[3]{4} yz))<(3 a)^2/a^2=9$.

This proof is shorter and easier than that of Euler and Kummer.

Now I'll explain how to get contradiction from (2).
We may assume, that $x$ is odd, otherwize take $y$ to play the role of $x$.

Units in the ring $\mathbb{Z}[\sqrt[3]{2}]$ have the form: $u=\pm(\sqrt[3]{2}-1)^k$, where $k$ is integer.
Therefore, from (2) follows, that either

(3) $x^2−\sqrt[3]{4} yz=\beta_1^2$

or

(4) $(\sqrt[3]{2}-1)(x^2−\sqrt[3]{4} yz)=\beta_2^2$,

where $\beta_1$ and $\beta_2$ belong to the ring $\mathbb{Z}[\sqrt[3]{2}]$.

However (4) is impossible, because the coefficient at $\sqrt[3]{2}$ turns out to be odd in the left part and even in the right part of (4).
Therefore (3) takes place, that is:

(5) $x^2−\sqrt[3]{4} yz=(a_0+a_1 \sqrt[3]{2}+a_2 (\sqrt[3]{2})^2)^2$, where $a_0$, $a_1$ and $a_2$ are integers.

From (5) we get:

(6) $a_0^2+4 a_1 a_2=x^2$

(7) $2 a_2^2+2 a_0 a_1=0$

(8) $a_1^2+2 a_0 a_2=-yz$

It follows, that all three numbers $a_0$, $a_1$ and $a_2$ are not zero.
We may assume, that $a_2>0$, otherwise change sign of all three numbers.

It follows from (7) that $a_0=\pm m b_0^2$, $a_1=\mp m b_1^2$, $a_2=m b_0 b_1$,
where $b_0$ and $b_1$ are relatively prime positive integers and $m$ is a positive integer.

Now it follows from (6) that:

(9) $m^2 b_0 (b_0^3 \mp 4 b_1^3)=x^2$

From (9) it follows that $b_0$ is a square, that is $b_0=x_1^2$, and $(b_0^3 \mp 4 b_1^3)$ is a square, that is:

(10) $x_1^6 \mp 4 b_1^3$ is a square

It can be shown, that $x_1^6 \mp 4 b_1^3 \neq 1$ (a separate lemma).
Then from (9) it follows, that $b_0<x^2$, that is $x_1<x$ and we've got the infinite descent.

Proof based on method of infinite descent: Method of infinite descent and proof of Fermat's last theorem for $n=3$. Canadian Journal on Computing in Mathematics, Natural Sciences, Engineering and Medicine (CMNSEM), Vol.1, No.6, Septermber 2010, pp. 181-186.