Show mean of a sequence of sets is zero but mean of countably infinite set is infinity if increasing Let $\Omega$ be a countably infinite set, and let $F$ consist of all subsets of $\Omega$. Define $\mu(A) = 0$ if $A$ is finite and $\mu(A) = \infty$ if $A$ is infinite. Show that $\Omega$ is the limit of an increasing sequence of sets $A_n$ with $\mu(A_n) = 0$ for all $n$ but $\mu(\Omega) = \infty$.

Does anyone know which theorem might help me start to show this? I was already able to show that μ is finitely additive but not countable additive using finite subadditivity. Any help would be greatly appreciated. I haven't had a proof course in 10 years and am trying to study for the GRE.

$\Omega$ is infinit thus $\mu(\Omega) = \infty$

Now $\Omega$ is countable, thus let $f:\Bbb N \rightarrow \Omega$ be a bijection.

Let $A_n = \{f(k), k\in [0,n]\}$, what follows is that for any $n$, $A_n$ is finite thus $\mu(A_n) = 0$ but $\Omega = \lim_{n\to+\infty}A_n$

You have to show that for any countably infinite set $\Omega$, there exists some sequence of sets

$A_1, A_2, ... \in \mathscr F := 2^{\Omega}$ (I believe this is equivalent to: $A_1, A_2, ... \subseteq \Omega$)

s.t.

$$A_n \subseteq A_{n+1}$$
$$\bigcup_{n=1}^{\infty} A_n = \Omega$$

Relevant Theorem 1 Continuity of measure:

If we have a measure space $(\Omega, \mathscr F, \mu)$ and $A, A_1, A_2, ... \in \mathscr F$ s.t.

$$A_n \subseteq A_{n+1}$$
$$\bigcup_{n=1}^{\infty} A_n = A$$

then

$$\mu(A) = \lim_{n \to \infty} \mu(A_n)$$

Relevant Theorem 2

If we have a measurable space $(\Omega, \mathscr F)$ and a set function $\mu: \mathscr F \to [0,\infty]$ s.t.

$\mu(\emptyset) = 0$
If $A, A_1, A_2, ... \in \mathscr F$ s.t.

2.1. $$A_n \subseteq A_{n+1}$$
2.2. $$\bigcup_{n=1}^{\infty} A_n = A$$

then

$$\mu(A) = \lim_{n \to \infty} \mu(A_n)$$

then $\mu$ is a measure on $(\Omega, \mathscr F)$

It follows that $\mu$ defined as

$$\mu(\text{finite}) = 0$$

$$\mu(\text{infinite}) = \infty$$

is not a measure on $(\Omega, \mathscr F)$ for any countably infinite set $\Omega$ with $\mathscr F = 2^{\Omega}$

If $\Omega = \mathbb N$, then consider $$A_n:= \{1,2,\cdots,n\}$$

As for any other countably infinite set $\Omega = A$, we want to choose $A_n$ as follows:

$$A_n = \{\text{first n elements of A, assuming A has some...}\}$$

$$ \{\text{...order by which one element precedes another}\}$$

With $\Omega = A = \mathbb N$, we had the elements in ascending order. But what if we had $A = \mathbb Z$? What if we had some countably infinite set of names for instance such as $A = \{Elsa, Anna, ElsaA, AnnaA, ElsaB, AnnaB, \cdots\}$ ?

Solution 1:

A countably infinite set can be enumerated: $A=\{a_n\}_{n=1}^{\infty}$

So let $$A_n := \{a_1, ..., a_n\}$$

Solution 2:

Note that by definition (or a property, depending on the textbook) of a countably infinite set, there exists a bijection $f: A \to \mathbb N$.

This bijection is what allows us to give order to the elements of $A$:

$$A_n := \{f(1), ..., f(n)\}$$

Observe that $f(n)$ from Solution 2 equals $a_n$ from Solution 1.


Evaluate $\int\frac{1}{1+x^6} \,dx$ I came across following problem

  Evaluate $$\int\frac{1}{1+x^6} \,dx$$

When I asked my teacher for hint he said first evaluate

  $$\int\frac{1}{1+x^4} \,dx$$

I've tried to factorize $1+x^6$ as

$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing

$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$

However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps

I've also tried to reverse engineer the solution given by Wolfram Alpha

And I need to have terms similar to
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?

Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?

Here's a nice "trick" my former professor taught me

$$ \int\frac{dx}{1+x^6} = \frac{1}{2} \int \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} dx \\
= \frac{1}{2}\int \frac{dx}{1+x^2} + \frac{1}{2} \int \frac{x^2}{1+x^6} dx + \frac{1}{2} \int \frac{1-x^2}{1-x^2+x^4} dx \\
= \frac{1}{2}\int \frac{dx}{1+x^2} + \frac{1}{2} \int \frac{x^2}{1+x^6} dx - \frac{1}{2} \int \frac{1-\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}} dx  $$

The first integral is simply the arctangent of $x$. The second can be solved by substituting $u = x^3$. The third can be solved by substituting $t = x + \frac{1}{x}$

As suggested, you can transform your denominator $1+x^6$ into $(1+x^2)(1-x^2+x^4)$. You can then use partial fractions to split the following into something that can be integrated:

$\Large\int \frac{1}{(1+x^2)(1-x^2+x^4)}dx$

Hope that helps. It looks like a very nasty and messy integration to do. Best wishes.


Newton's Method to find point on function closest to origin Using Newton's Method, find the coordinates accurate to 6 decimal points of the point on the function y=e^x that is closest to the origin.

Hint

We want the $x$ that yields the minimum value of $D(x)$, the distance between $(0,0)$ and $(x, e^x)$. This is the value for which $D'(x)=0$ and $D''(x)>0$.

Here we would approximate $D'(x) = 0$ using Newton's Method as given by: $$ x_{n + 1} = x_n - \frac{D'(x_n)}{D''(x_n)} $$

Now we just need an expression for $D(x)$ and a first approximation, $x_1$.

Newton's Method will approximate the solution of $ f(x) = 0 $.
Draw a diagram, consider your function, and apply the formula: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ until you get your answer


Conjugate in Cauchy-Schwarz Inequality for Complex Numbers Simple question: do we really need the conjugate in the inequality?

$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 $$

The Cauchy-Schwarz inequality says that $|\langle a,b \rangle| \leq \| a \| \|b\|$.  In $\mathbb C^n$, the inner product is
\begin{equation}
\langle a, b \rangle = \sum_{j=1}^n a_j \bar{b_j}.
\end{equation}

That's why the Cauchy-Schwarz inequality in $\mathbb C^n$ has conjugates in it.

While it is true that you could omit the conjugates in your inequality and still have a true statement, that would only take us further away from the nice statement that $|\langle a,b \rangle| \leq \| a \| \|b\|$.

Simple answer, yes. Look back to the definition of the inner product (for complex numbers) and note the complex conjugate symmetry (as opposed to pure symmetry)


Prove that for all real numbers x and y there is a real number $z$ such that $x + z = y - z$ This one must be pretty basic, but...

  Prove that for all real numbers $x$ and $y$ there is a real number $z$
  such that $x + z = y - z$

I am quite confused about how you need to prove this.

My attempt was:
$$\tag1 x + z = y - z$$
$$\tag2 2z = y -x $$
$$\tag3 z = \frac{y-x}{2}$$
Denominator doesn't equal to zero, hence $z$ is defined for all $x$ and $y$.

Is there anything else I need to show?

If this question is from an analysis book, you might have to explain it in addition to solving it. Strictly speaking, it is a mathematical statement. The first part of the statement, "for all $x$ and $y$," implies that values of $x$ and $y$ have already been stipulated. You have no control over them. The second part of the statement, "there is a real number $z$," implies that you only have to find one real number $z$ that satisfies $x + z = y - z$. Solving this equation for $z$ shows that you can always find this value of $z$.


Proof that integers (except $1$ and $-1$) don't have an integer multiplicative inverse This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:

How can it be proven that, among the integers, only $1$ and $-1$ have an integer multiplicative inverse? Or is that statement taken as an axiom?

Well, let's assume $ab=1$ when $a,b\in\mathbb{Z}$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|\geq |a|$, and in the same way $1\geq |b|$. So both $a$ and $b$ must belong to the set $\{-1,1\}$.

As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z.  Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
 a.  a^(-1) =1.
 Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.


order of operations with many level exponents I was wondering, what is the order of operations when it comes to multi level exponents. Couldn't find anything in google. Something like:

$$n^{n-1^{n-2^{\cdots^1}}}$$

In this case, if n equals 4, would it be correct to assume that 4^(3^(2^1)) is the correct order? And thus the answer is 262144?

Yes to both questions.

You should include parentheses around the differences, such as
$$n^{(n-1)^{(n-2)^{\cdots^1}}}$$

No.

4^3^2 = 4^6 ; It does not equal 4^9. Multiple exponents evaluated from left to right.  (4^3)^2. The power is 2*3=6. Because. First you get 4*4*4, and square that. It's alway true that you can multiply the exponents if there arent parenthesis grouping them in some way.  4^3^2^5 = 4^30


Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$ I found in this link the reduction formula
$$
\int_0^{\pi/2} \sin^m x \cos^n x \ dx =
\begin{cases}
\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2} (\frac{\pi}{2}) & m, n \text{ even}\\[5pt]
\frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)} & \text{otherwise}
\end{cases}
$$

How can I prove it?

Let $I_{m,n}=\int_0^{\pi/2} \sin^m x \cos^n x \ dx$, integrating by parts we find that
$$
I_{m,n}=\frac{n-1}{m+1} I_{m+2,n-2} \qquad (1) \\
I_{m,n}=\frac{m-1}{n+1} I_{m-2,n+2} \qquad (2)
$$

Using (1) when $n$ is odd,
$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 2}{(m+n-2) \cdots (m+1)}I_{m+n-1,1}=\frac{(n-1)(n-3) \cdots 2}{(m+n)(m+n-2) \cdots (m+1)} \quad (3)
$$
Interchaging $m$ and $n$ in (3) we find $I_{m,n}$ when $m$ is odd.

Then
$$
I_{m,n} = \frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)}
$$
when either $m$ or $n$ are odd.

If $n$ and $m$ are even

$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 1}{(m+n-1) \cdots (m+1)}I_{m+n,0}
$$

On the other hand, let $J_k=\int_0^{\pi/2} \sin^k x \ dx$. Again,integrating by parts we find:
$$
J_k=\frac{k-1}{k}J_{k-2}
$$

In particular when $k$ is even
$$
J_k=\frac{(k-1)(k-3) \cdots 1}{k(k-2) \cdots 2}\left(\frac{\pi}{2}\right)
$$

Then

$$
\begin{align}
I_{m,n}&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n-1)(m+n-3) \cdots 1} \left[\frac{(m+n-1)(m+n-3) \cdots 1}{(m+n)(m+n-2) \cdots 2} \left(\frac{\pi}{2}\right)\right]\\
&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2}\left(\frac{\pi}{2}\right)
\end{align}
$$

Hint-

break the expression into $\sin^{(m-1)} x \; \{ \cos ^n x  \sin x\}$.
Let 2nd function which is in curly bracket and 1st function which is out side. Now integrate using integration by part.


Describing maps $\mathbb{R}^n \to \mathbb{R}^m$ up to homotopy I'm trying to describe continuous maps $f : \mathbb{R}^n \to \mathbb{R}^m$, where $f^{-1}(0) = \{0\}$, up to homotopy.

Without the condition that $f^{-1}(0) = \{0\}$, it seems that all maps are homotopic to each other; since all the homotopy groups of $\mathbb{R}^n$ and $\mathbb{R}^m$ are trivial, and this is evidence that maps between the two spaces are trivial up to homotopy too (though I can't seem to come up with a proof of this fact?)

With the condition that $f^{-1}(0) = \{0\}$, I'm not sure how I would go about approaching this problem. The main thing that seems to cloud my vision is that I can only really imagine the cases where $n,m \leq 3$; and even then I'm not sure that my intuition holds.

This looks like it may be equivalent to computing $\pi_{n-1}(S^{m-1})$.
Each $f$ restricts to a map from $\Bbb  R^n-\{0\}$ to $\Bbb  R^m-\{0\}$ to
and these spaces are homotopy equivalent to $S^{n-1}$ and $S^{m-1}$.
Conversely each $g:S^{n-1}\to S^{m-1}$ extends radially to an $f$ of your form.

I haven't checked the details on whether these correspondences are preserved by homotopy etc.

Hint: For $f, g: R^n \to R^k$, consider, for each $P \in R^n$, the line segment from $f(P)$ to $g(P)$ as a parameterized arc (with parameter $s$, say, running from $0$ to $1$). Suppose that $f(0) = g(0) = 0$. What will this line segment look like?


Variance and Conditional Distribution Suppose that the conditional distribution of $Y$ given $X=x$ is Poisson with mean $E(Y|x)=x$, $Y|x\sim POI(x)$, and that $X\sim EXP(1)$

Find $V[Y]$

We know that

$$V[Y]=E_x[Var(Y|X)] + Var_x[E(Y|X)]$$

we also know that

$$V(Y|x)=E(Y^2|x)-[E(Y|x)]^2$$

I know we can find $V[Y]$ with those 2 formulas, but I am getting confused with so many expectances and variances around each other. Can anyone explain me how to solve this?

If the conditional distribution of $Y$ given $X$ is Poisson with parameter $X$ then the conditional expectation is also $X$. That is, if $X$ is of exponential distribution with parameter one then

$$E[Y]=\int_0^{\infty}E[Y\mid X=x\,]e^{-x}\ dx=\int_0^{\infty}xe^{-x}\ dx=E[X]=1.$$

Also,

$$E[Y^2]=\int_0^{\infty}E[Y^2\mid X=x\,]e^{-x}\ dx=\int_0^{\infty}(x^2+x)e^{-x}\ dx$$
because the second moment of the Poisson distribution is the square of the parameter plus the parameter.

Now,
$$\int_0^{\infty}(x^2+x)e^{-x}\ dx=\int_0^{\infty}x^2e^{-x}\ dx+\int_0^{\infty}xe^{-x}\ dx=E[X^2]+E[X]=2+1=3$$
because the second moment of the exponential distribution of parameter $1$ is $2$.

The variance of $Y$ is then $$E[Y^2]-E^2[Y]=3-1=2.$$

The second moments of the distributions above can be calculated based calculated based on the corresponding MGFs. The MGFs can be found in Wikipedia. If you have the MGF then take the second derivative and substitute $0$.

We have been given that $Y|x$ is a Poisson Variate with parameter $x$. We know that, in Poisson distribution the mean and variance are equal to its parameter. Hence, the following will hold:

$E[Y|x]=V[Y|x]=x$.

Now, to obtain the variance of $Y$,

$V(Y)=E(V[Y|x])+V(E[Y|x]) = E(x)+V(x)$

Now, $X$ itself is a exponential Variate with parameter $1$. So, it's Expectation and variance both will be $1$. Hence, the required value of Variance of $Y$ after plugging in the appropriate Values will be:

$V(Y)=E(x)+V(x)=1+1=2$

Note: The mean and variance of $X~Exp(\beta)$ is:

$E(X)=\frac{1}{\beta}$,
$V(X)=\frac{1}{\beta^2}$


What are the differences between class, set, family, and collection? In school, I have always seen sets. I was watching a video the other day about functors, and they started talking about a set being a collection, but not vice-versa. I also heard people talking about classes. What is their relation? Some background would be nice.

It has to do with something called Russell's paradox, but I don't know what that is.

I think that the difference between a family and a set is that the former is a function and the latter is a set. Is this right?

The idea behind a "collection" is simply a notion of a bunch of mathematical objects which are collected into one big pile. Think of it as a big bin full of trash, diamonds and empty bottles of beer, it doesn't have to make sense what is in this collection, it's just a collection.
One of the problem to explain these things to people who are not mathematicians (or trying to "outsmart a set theorist", as I ran into several of those) is that the notion of a collection is not fully formal unless you already know what sets and class are, and even then it's not exactly what we mean.
Let me start over now. Doing mathematics we often have an idea of an object that we wish to represent formally, this is a notion. We then write axioms to describe this notion and try to see if these axioms are self-contradictory. If they are not (or if we couldn't prove that they are) we begin working with them and they become a definition. Mathematicians are guided by the notion but they work with the definition. Rarely the notion and the definition coincide, and you have a mathematical object which is exactly what our [the mathematicians] intuition tells us it should be.
In this case, a collection is a notion of something that we can talk about, like a mystery bag. We might know that all the things inside this mystery bag are apples, but we don't know which kind; we might know they are all Granny Smith, but we cannot guarantee that none of them is rotten. A collection is just like that. We can either know something about its elements or we don't, but we know that it has some.
Mathematician began by describing these collections and calling them sets, they did that in a relatively naive way, and they described the axioms in a rather naive way. To the non-mathematician (and to most of the non-set theorists) everything is still a set, and we can always assume that there is a set theorist that assured that for what we need this is true. Indeed, if we just wanted to discuss the real numbers, there is no worry at all we can assume everything we work with is a set.
This naive belief can be expressed as every collection is a set. It turned out that some collections cannot be sets, this was expressed via several paradoxes, Cantor's paradox; Russell's paradox; and other paradoxes. The exact meaning is that if we use that particular axiomatic description of "what is a set" then we can derive from it contradiction, which is to say that these axioms are inconsistent.
After this happened several people began working on ways to eliminate this problem. One method in common was to limit the way we can generate collections which are sets. This means that you can no longer derive such contradiction within the theory, namely you cannot prove that such collection even exists - or rather you can prove it doesn't.
The common set theory nowadays called ZFC (named after Zermelo and Fraenkel, the C denotes the axiom of choice) is relatively close to the naive way from which set theory emerges, and it still allows us to define collections which are not sets, though, for example "the collection of all sets". These collections are called classes, or rather proper classes.
What is definable? This is a whole story altogether, but essentially it means that we can describe it with a single formula (perhaps with parameters) of one free variable. "$x$ is taller than 1.68m" is an example to such formula, and it defines the class of all people taller than said height.
So in ZFC we may define a collection which is not a set, like the collection of all the singletons, or the collection of all sets. These are not sets because they are too big, in some sense, to be sets, but they are classes, proper classes. We can talk about collections which are not definable but that requires a lot more background in logic and set theory to get into.

To sum up

Classes are collections which can be defined, sets are particular classes which are relatively small and there are classes which are not sets. Collections is a notion which is expressed via both these mathematical objects, but need not be well-defined otherwise.
Of course when we say defined we mean in the context of a theory, e.g. ZFC. In this sense, sets are things which "really exist" whereas classes are collections we can talk about despite their possible nonexistence.

One last thing remains, families. Well, as you remarked families are functions. But functions are sets, so families are sets. We can make a slight adjustment to this, and we can, in fact, talk about class functions, and an index which is not a set but a proper class. We, therefore, can talk about families which are classes.
Generally, speaking, if so, a family is a correspondence from one collection into another which uses one collection as indices for elements from another collection.

To read more

What is the difference between a class and a set?

Why is "the set of all sets" a paradox, in layman's terms?

In predicate logic, we might have a statement like "for all values of $x$, $\phi(x)$ is a true statement", where $\phi(x)$ might be a statement such as $x=x$.  What are the "values of $x$" that we are talking about?  Well, those values of $x$ are "sets".

Set theory allows us to generate classes of sets: we can say, for example, "the collection of all sets $x$ that satisfy a condition $\phi(x)$".  In set builder notation, this is written $\{x:\phi(x)\}$.  The question might arise: can we assume this new class of sets is a set itself?

Intuitively, it may seem reasonable to consider $\{x:\phi(x)\}$ to be a value of $x$.  However, Russell's paradox shows us that such an assumption leads to a contradiction.

The paradox asks us to consider the collection of all sets $x$ that are not members of itself. That is $\{x:x\not \in x\}$; we will call this set $R$.  Is the resulting class a member of itself?  Well, if we accept $\{x:x \not \in x\}$ to be a possible value of $x$, then we arrive at the contradiction that this set, $$R \in R \iff R \not \in R$$

So, sets are the members of the universe of discourse: that is, they are all the "values of $x$" that we refer to when we create a statement using quantifiers like "for all $x$" or "there exists an $x$".

Classes on the other hand, are collections of sets.  Some classes are sets, such as $\{x\}$ (in most set theories), but others are not, such as the class $R$ described earlier.  All sets are classes, but not all classes are sets.


Draw a Square Without a Compass, Only a Straightedge -- Part Deux So, I previously asked the question Draw a Square Without a Compass, Only a Straightedge. From the comments and answers, it appears that that question is not solvable.
Given that the question I originally saw was on an actual exam (entrance exam for Cambridge undergraduate from the 90s, or maybe 80s), this got me thinking: odds are, I've misremembered the question!

I think I have remembered it correctly now, and so pose the following question.

  Is it possible to, given a square drawn on a plane, using only an unmarked straightedge, construct another square with twice the area? If so, how is this done?

From an arbitrary point A on the top half of the vertical side of the square, construct the sequence of points to finish with a square twice the area of the original square. That is, the diagonal of a smaller unit square being $\sqrt2$ and forming the side of the larger square with double the area.

I assume that, although the straightedge starts unmarked, we can mark lengths on it. Thus, if the side length is 1, we can mark any length in $\mathbb  N[\sqrt 2]$.

Suppose we have square ABCD, with A =(0,0) B = (1,0), C = (1,1), D = (0,1). By the line AC and taking the point $\frac {3\sqrt 2}{2}$ from the origin, we can construct the point E = (1.5,1.5). By taking the ray DB and taking the point  $\frac {3\sqrt 2}{2}$  along it, we can construct F = (1.5,-.5). The intersection of the diagonals of the square gives us G = (.5,.5). We can extend the sides of the square to give us the $x$ and $y$ axes, and then taking the distance from F to the $x$-axis gives us .5 (note that we can create the line segment between F and the $x$-axis by first taking the line segment EF). We can then take H = (1,.5), and draw the ray GH. Taking the point that is distance 2 away from G gives us I = (2.5,.5). GFIE is then a square with area 2.


Holomorphic function with constant absolute value Show that if a holomorphic function has a constant absolute value, it must be a constant.

Suppose $f(z)=u(z)+iv(z)$ is holomorphic (where $z=x+iy$ is complex), and that $(u(z))^2+(v(z))^2=C$ for some constant $C$.

We have $\dfrac{\partial((u(z))^2+(v(z))^2)}{\partial x}= 0$. Applying the chain rule, we find that $$u(z)\cdot\dfrac{\partial u(z)}{\partial x} + v(z)\cdot\dfrac{\partial v(z)}{\partial x} = 0$$

Similarly, $$u(z)\cdot\dfrac{\partial u(z)}{\partial y} + v(z)\cdot\dfrac{\partial v(z)}{\partial y} = 0$$

Using the Cauchy-Riemann differential equations satisfied by any holomorphic function:
$$u(z)\cdot\left(-\dfrac{\partial v(z)}{\partial x}\right) + v(z)\cdot\dfrac{\partial u(z)}{\partial x} = 0$$

We can write as a matrix form:

$$\begin{pmatrix} u(z) & v(z) \\ v(z) & -u(z)\end{pmatrix}\begin{pmatrix}\frac{\partial{u(z)}}{\partial{x}}\\\frac{\partial{v(z)}}{\partial{x}}\end{pmatrix} = 0$$

For values of $z$ such that the matrix $A=\begin{pmatrix} u(z) & v(z) \\ v(z) & -u(z)\end{pmatrix}$ is invertible, we get $\dfrac{\partial{u(z)}}{\partial{x}}=\dfrac{\partial{v(z)}}{\partial{x}}=0$. But what can we do for values of $z$ such that the matrix $A$ is not invertible?

If $C = 0$, then $u$ and $v$ are necessarily constant $0$, so we assume that $C \neq 0$. The determinant of $A$ is

$$u(z)(-u(z)) - v(z) v(z) = - (u(z)^2 + v(z)^2) = - C \neq 0$$

for any $z$. So you've got a an invertible matrix and a vector in its null space; so what must that vector be?

This is a well-known theorem in Complex Analysis:

Liouville's Theorem: If a holomorphic function of one variable is bounded and entire, the function is necessarily constant.

Since your function has constant absolute value, it is bounded. By Liouville's Theorem it is therefore constant.


Proving the series sin(n+1)-sin(n+2) converges or diverges As the title states, I'm trying to determine whether the series $$\sum_{n=1}^\infty \sin(n+1)-\sin(n+2)$$ converges or diverges. My intuition is saying diverging since the sines are oscillating. How will I go about 'formally' proving this?

I searched online for this and I got an 'explanation' saying $s_n=\sin(2)-\sin(n+2)$. Where did this even come from, and how does this prove anything?

First: your sum diverges. To see this, write it out and note that it is a telescoping series- all but two of the terms cancel. Thats how you get your $s_n$. Then note that the value of the nth partial sum is actually a linear combination of $\sin(n)$, whose limit does not exist, therefore the limit of $s_n$ does not exist.

In order that $\sum_{n\geq 1}a_n$ is convergent, $\lim_{n\to +\infty}a_n = 0$ is needed.
On the other hand
$$\sin(n+1)-\sin(n+2) = -\cos\left(n+\frac{3}{2}\right)\cdot 2\sin\frac{1}{2} $$
and $\left\{\cos\left(n+\frac{3}{2}\right)\right\}_{n\geq 1}$ is dense in $(-1,1)$, hence $\limsup_{n\to +\infty}a_n = 2\sin\frac{1}{2}>0$.

Proof of density:

$\pi $ is irrational
$\{e^{in}\}_{n\geq 1}$ and $\{e^{in+3i/2}\}_{n\geq 1}$ are dense in the unit circle (equidistributed, actually)
The projection $z\to\text{Re}(z)$ is a continuous map, hence it preserves density
$\left\{\cos\left(n+\frac{3}{2}\right)\right\}_{n\geq 1}$ is dense in $(-1,1)$.

We may also avoid to go through the irrationality of $\pi$ and the previous density argument. If we assume that $\sin(n+1)-\sin(n+2)$ has an absolute value smaller than $\varepsilon>0$ for any $n\geq N_0$ we get a contradiction, since the function
$$ f(x)=\left|\sin(x+1)-\sin(x+2)\right|+\left|\sin(x+2)-\sin(x+3)\right|+\left|\sin(x+3)-\sin(x+4)\right| $$
is $\geq\frac{3}{2}$ for any $x\in\mathbb{R}$.

Summarizing, the sequence given by
$$ S_N = \sum_{n=1}^{N}\left[\sin(n+1)-\sin(n+2)\right]\stackrel{\text{telescopic sum}}{=}\sin(2)-\sin(N+2) $$
is not convergent.


What is the right terminology / methods for this standard signal processing task? Say we have a signal s = [1,2,2,1,3,3,2,2,1,1] or similar (only 1,2,3 can occur). I want to look for a particular pattern, e.g. [1,3,3,2], which can occur anywhere (0 or many times) in the signal. I would like to perform the computation as a convolution with some filter f.

I've convinced myself that f = [100,1000,1000,10] will do the job, i.e. sending the pattern p = [1,3,3,2] to f = 10.^fliplr(p), in matlab syntax. That is, the convolution s*f will take the value 1*10 + 3*1000 + 3*1000 + 2*100 = 6210 if and only if we have the pattern p in the right place, at least as long as the pattern we are looking for is short enough.

For a long enough pattern, e.g. [1,1,1,1,1,1,1,1,1,1,2] (ten 1s followed by a 2),  the corresponding filter f = [100,10,..,10] would give a "false positive" for the pattern [2,2,2,2,2,2,2,2,2,2,1], since the convolution would give the values 10*10 + 2*100 = 300 for the first, and 2*10*10 + 1*100 = 300 for the second. (That I needed 10 1s to "trick" the filter is obviously related to the fact that my filter "encodings" of the 1s and 2s differ by a factor 10. Changing the encoding to e.g. {1,2,3} ->  {10,1000,100000} should let me handle correspondingly longer patterns).

I have a strong feeling that this is a very standard signal processing task. What is the standard terminology and methods concerning this problem? (I.e. what key words should I google.)

The thing you are looking for is the matched filter, which is essentially the (optimal) linear approach to look for a known pattern in a signal (which may be impacted by additive zero mean noise).

The impulse response (i.e., the sequence you are convolving with) of the matched filter is just the pattern reversed, i.e. $f = [2, 3, 3, 1]$ will do the job. If the input signal is normalized in RMS amplitude, the matthed filter will give maximum  output if and only if the pattern occurs.


Is trigonometry part of synthetic approach to geometry? I used to think Trigonometry is an analytic way to solve math problems but now I'm reading a book with only synthetic methods to solve geometry problems and it contains trigonometry too.
I can't understand the difference between analytic  and synthetic approaches to geometry.

"I can't understand the difference between analytic and synthetic approaches to geometry."

I think a good way to understand the difference between the two is that in analytic geometry, mathematicians assign a a real number to each pair of points in whatever space they are working in (representing the distance between the given points).  Similarly for angle measure. When one assigns real numbers to represent things like angle and distance one has entered the realm of analytic geometry.

So, while precise definitions change from century to century (and even author to author in the same century), in synthetic geometry, we might claim that two angles are congruent, meaning one could be superimposed on the other with no mention of a numerical quantity.  In analytic geometry we might claim that two angles are equal meaning that the numbers used to measure those angle are the same.

Again, the precise meanings of particular words can vary depending on the setting, but the use of real numbers to describe geometric entities is the hallmark of the analytic approach.


How do I find the angle a vector makes to the $+x$ axis? You are given two vectors $\vec a = -3.00\hat i + 7.00\hat j$ and $\vec b= 4.00\hat i + 2.00\hat j$. Let the counterclockwise angles be positive.

What angle $\theta (\vec a)$ where $0^\circ \le \theta (\vec a) < 360^\circ $, does $\vec a$ make with the $+x$-axis?

I drew a right triangle with a $\vec ax$ component of $-3$ and an $\vec ay$ component of $7$. Do I just use trig to find the angle off the $x$-axis?

A general rule for finding the counterclockwise angle from the positive $x$ axis to the direction of a two-dimensional vector is explained in
How to convert components into an angle directly (for vectors)?
The basic idea is that you first take the arc tangent of the $y$-component
divided by the $x$-component:
$$
\theta_1 = \arctan\frac{a_y}{a_x}.
$$
One complication is that this procedure gives the same resulting $\theta_1$
for the vector $\langle a_x,a_y \rangle$ and the vector $\langle -a_x,-a_y \rangle,$ which are two vectors in directions $180$ degrees apart.
So $\theta_1$ may be in the direction you want, or in the direction
$180$ degrees opposite.
In fact $\theta_1$ will be in the correct direction whenever $a_x > 0,$
since that is how the arc tangent function is designed to work.
(It always produces angles in the range $-\frac\pi2$ to $\frac\pi2$ radians,
that is, $-90$ to $90$ degrees,
which correspond to vectors with positive $x$-components.)
The cases where $\theta_1$ is wrong are precisely the cases where
$a_x < 0,$ so if $a_x < 0$ (as it is in your particular question)
you have to add $180$ degrees to $\theta_1.$
A second complication (which does not actually come up in your particular
question) occurs because the procedures given above produce a result in the range $-90$ degrees to $270$ degrees, but people usually want an answer in the range $-180$ to $180$ or $0$ to $360.$ The solution to this, of course, is to add or subtract $360$ degrees as needed to get an answer in the desired angle range.

Yeah you just use trig. Tan is the easiest. I saw your question about whether it is arctan(7/3) or arctan(7/-3) it is the 7/-3. Now this will give you a negative angle. This is outside your range so add 360 to it to get the answer that is in your range.


How to prove that $\sin(\sqrt{x})$ is not periodic? How to prove that $\sin(\sqrt{x})$ is not periodic?
THe definition of a periodic function is $f(x+P)=f(x)$.

So I assume that $\sin(\sqrt{x+P})=\sin(\sqrt{x})$. This is equivalent to $\sin(\sqrt{x+P})-\sin(\sqrt{x})=0$. This implies $2cos(\frac{\sqrt{x+P}+\sqrt{x}}{2})\sin(\frac{\sqrt{x+P}-\sqrt{x}}{2})$. What should I do next?

If $\sin {\sqrt x}$ is period with period $P$ then so is $\cos {\sqrt x}$ because $\cos {\sqrt x} = \sqrt {1 - \sin^2 \sqrt x}=\sqrt {1 - \sin^2 \sqrt {x+P}}=\cos (\sqrt{x+P})$.  Likewis so is the derivative of $\sin {\sqrt x}$ because $ \lim \frac {\sin (\sqrt {x + h}) -\sin(\sqrt{x})}h=  \lim \frac {\sin (\sqrt {x +P + h}) -\sin(\sqrt{x+P})}h$.

But the derivative of $\sin (\sqrt{x})$ is $\frac{cos(\sqrt{x})}{2\sqrt{x}}$ and if $\cos (\sqrt{x + P}) = \cos (\sqrt{x})$ then $\frac{cos(\sqrt{x+P})}{2\sqrt{x+P}}=\frac{cos(\sqrt{x})}{2\sqrt{x+P}} \ne \frac{cos(\sqrt{x})}{2\sqrt{x}}$

This is a contradiction.

$$\sin(\sqrt{x+P})=\sin(\sqrt{x})$$

$$\implies \sqrt{x+P}=\sqrt{x}+2k\pi  \text { or }\sqrt {x+P}=2k\pi+ \pi - \sqrt{x} $$

Upon squaring we get  $$ x+P = x+4 k^2\pi ^2 +4k\pi \sqrt x $$

or $$x+P =((2k+1) \pi) ^2 + x-2(2k+1)\pi \sqrt x$$

Note that neither of the above holds for a constant $P$ and every $x$


exotic smooth structures I have difficulty imagining how two manifolds can be homeomorphic but not diffeomorphic, it seems for a homeomorphism $f:M\to N$ and charts $(U,\phi)$ on $M$ and $(V,\psi)$ on $N$, one can always approximate $\phi^{-1} \circ f\circ \psi:U\to V$ by a smooth function and do this in such a way that although one purturbs $\phi$ and $\psi$,but eventually it is possible to reconcile the changes on overlap of any two charts, so that this gives a diffeomorphism as well. Of course this is not correct, but I am not sure how I should think about exotic smooth structures to make it apparent how two manifolds potentially can have different smooth structures.

Here there is a huge difference between low-dimensional manifolds ($1$, $2$, $3$) and high-dimensional manifolds ($\geq 4$). This difference is illustrated by the following theorem.

  Theorem: Let $M$ and $N$ be smooth manifolds of like-dimension $1$, $2$, or $3$. Then $M$ and $N$ are homeomorphic if and only if they are diffeomorphic as smooth manifolds.

The important detail here is that $M$ and $N$ are smooth manifolds, in higher dimensions this result does not hold.

For a simple, visual example of two homeomorphic but non-diffeomorphic manifolds, consider $\Bbb S^1$, the standard unit circle and $C^1$, the square. Clearly these are homeomorphic, but does it seem sensible to claim that they are diffeomorphic? After all, the square has those pointy vertices...

Why doesn't the theorem apply to $\Bbb S^1$ and $C^1$? Because $C^1$ is not a smooth manifold, it is simply a topological manifold.

I'm afraid I will be little help in visualizing exotic structures, since they only exist in dimension $4$ and up!


Compactness Invariant between normed spaces Let $X$ and $Y$ be finite dimensional normed spaces. Let $D:\X \rightarrow Y$ be an isometric isomorphism then if $X$ is compact the $Y$ is also compact.

I have started by choosing a sequence in $Y$ and then taking its inverse image but I am stuck.

Any hints are greatly appreciated!

Chival mentioned in the comments that a normed space cannot be bounded (since it is a vector space).
Rephrasing the question to overcome this issue, we get

  Let $X$ and $Y$ be finite dimensional normed spaces. Let $D:X \to Y$ be an isometric isomorphism. Prove that if $Z \subset X$ is compact then $D(Z)$ is also compact.

Then, by noticing $D$ is continuous, the conclusion follows, because as stated by Frank in another answer the following theorem holds:

  If $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is continuous, and if $X$ is compact, then $Y$ is compact.

Look up the following (more general) theorem: If $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is continuous, and if $X$ is compact, then $Y$ is compact.


What are the Ideals of $Z/5Z$. Now $Z/5Z$ is going to form a quotient ring .Will this ring have an ideal? My intuition is the ideal will be 0 (the only ideal)?.

To find the ideal structure of a ring $\mathbb Z/n\mathbb Z$ you can consider the ideal structure of $\mathbb Z$ which contains $n\mathbb Z$. for example

So the ideals of $\mathbb Z/5\mathbb Z$ are $\mathbb Z/5\mathbb Z$ and 0

The only ideals in $\Bbb Z/n\Bbb Z$ are those of the form $\Bbb Z/m\Bbb Z$, where $m|n$.

Since $5$ is prime, there are only $0$ and $\Bbb Z/5\Bbb Z$, which, incidentally, means it's a field.


Is being good at mathematic contests necessary to pursue a career in mathematics or physics? I am extremely poor in solving a problem quickly, which leaves me doing badly in olympiads and other contests. But, if I just don’t give up on a problem, no matter how much time it takes, I usually end up solving it. To have a future in mathematics or physics, is it necessary for me to excel at contests, because these days children excelling at contests have an advantage over those who don’t (not complete advantage, but advantage nonetheless)?

Mathematics takes place at different time-scales.  If you can solve a problem in $5$ minutes that others need an hour to solve, you can probably get a good job.  If you can solve a problem in a month that others might need a year to solve, you will probably do well as a graduate student.  But if you can solve a problem in $10$ years that nobody else can solve in a lifetime, you could be a great mathematician.

Solving "exercises" that just require formulaic application of pre-learned tricks (as is the case in most of these contests) quickly or not will likely have very little bearing on any future that you might have in mathematics.  I would suggest reading "A Mathematician's Lament" by Paul Lockhart (https://www.maa.org/external_archive/devlin/LockhartsLament.pdf) if you want to hear a little bit more (targeted at the educated layperson) about how the "mathematics" that you have been exposed to in school is very different from what mathematicians engage in.


The $n$-disk $D^n$ quotiented by its boundary $S^{n-1}$ gives $S^n$ Define $D^n = \{ x \in \mathbb{R}^n : |x| \leq 1 \}$. By identifying all the points of $S^{n-1}$ we get a topological space which is intuitively homeomorphic to $S^n$.
If $n = 2$, this can be visualised by pushing the centre of the disc $D^2$ down so you have a sack, then shrinking the boundry of the sack to a point which gives you a teardrop shaped object which is clearly homeomorphic to $S^2$.

I am new to algebraic topology. How do I prove that the quotient space is actually homeomorphic to $S^n$. I haven't been able to write down explicitly a continuous map between $D^n$ and $S^n$ which maps $S^{n-1}$ to a point on $S^n$, which at the moment is the only way I know how to begin showin that two spaces are homeomorphic.
Is more machinery needed? If so I am interested to hear what is needed. If not, please tell me how stupid I am and give me a hint!

None of the answer give any map explicitely so I'll write one here. Note that $S^{n−1}\times I→D^n,(u,t)\mapsto tu$ (this is scalar product) is a quotient map and thus one can define (after verifying some things) the continuous map $D^n\to S^n,tu\mapsto (\sin(tπ)u,\cos(tπ))$ which can be verified to work.

You know that $S^n$ is a one point compactification of $R^n$. That is $S^n = R^n \cup \{\infty\}$ with appropriate topology. Now define a map $f\colon D^n \to S^n$ as follows if $x$ is in interior of $D^n$ then $f(x) = \frac{x}{1 -\mod{x}}$ but if $x$ is in boundary then $f(x) = \infty$. Now it is easy to see that $f$ is continuas in the interior and as well as in boundary. Sequential criterion may help you to conclude the statement. Now $D^n$ is compact so the map is a closed map and therefore a quotient map. Hence it induces a homeomorphism from $D^n$ to $S^n$.


$f$ is T periodic and $f(x) + f'(x) \ge 0 \Rightarrow f(x) \ge 0$ Let $f: \Bbb R \to \Bbb R$ be a function such that $f'(x)$ exists and is continuous over $\Bbb R$. Moreover, let there be a $T > 0$ such that $f(x + T) = f(x)$ for all $x \in \Bbb R$ and let $f(x) + f'(x)\ge 0$ for all $x \in \Bbb R$.

  Show that $f(x) \ge 0$ for all $x \in \Bbb R$.

My attempt:
$f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)$.

Thus, it is enoguh to show that $0 \ge f'(x)$.

$\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$

I do not know how to proceed from here. I know that $f'$ also has a periodicity of T but I do not know how to use that here.

Am I on the right track? How can I use the periodicity of $f$ to solve the problem?

Let $g(x)=\mathrm e^xf(x)$, then $g'(x)=\mathrm e^x(f'(x)+f(x))\geqslant0$ hence: $$(1)\ \textit{The function $g$ is nondecreasing.}$$
Since $f$ is continuous and periodic, $f$ is bounded, say $|f(x)|\leqslant C$ for every $x$, hence $|g(x)|\leqslant C\mathrm e^x\to0$ when $x\to-\infty$, that is: $$(2)\ \textit{The function $g$ has limit $0$ at $-\infty$.}$$
Properties (1) and (2) of $g$ imply together that $g\geqslant0$ everywhere, hence $f\geqslant0$ everywhere.

Examples: Consider $$f(x)=c\,\mathrm e^{w\cos(ux+v)},$$ for every  $c\geqslant0$, $u\ne0$, $|uw|\leqslant1$ and $v$, then $f$ has period $2\pi/|u|$ and, for every $x$, $$f'(x)+f(x)=(1-wu\sin(ux+v))\,f(x)\geqslant0.$$
For example, if $c=w=u=1$, $v=0$, one gets the function $f$:

$\qquad\qquad\qquad$

...And the function $f'+f$:

$\qquad\qquad\qquad$

$ f \ne 0$

$f(x)=0 \Rightarrow f'(x) \ge 0$, hence $f$ can cross the X-axis at most once. periodicity means that it cannot cross  at all.

if $f$ is non-positive then $f' \ge 0$ so $f$ is monotone increasing. again this contradicts periodicity


What are the elements in this group? I think I am just getting thrown off by notation. I am not sure what the following means.
What is the group $G = (\mathbb{Z}/12\mathbb{Z})^2$? (What are the elements?)

I know the elements of $(\mathbb{Z}/12\mathbb{Z})$ are just $\{\bar0, \bar1, \bar2, \bar3, ... ,\bar{11}\}$. Would the elements of $G = (\mathbb{Z}/12\mathbb{Z})^2$ simply be $\{\bar0^2, \bar1^2, \bar2^2, \bar3^2, ... ,\bar{11}^2\}$?...but then those numbers mod 12?

$G=H^2$ means the direct product $G=H\times H$, so the elements are $(h,k)$ with $h,k\in H$. Now take $H=\mathbb{Z}/12$. If $H$ has $m$ elements, then $H\times H$ has $m^2$ elements.

Yes you're right. It means the direct product. But keep the modulo 12 in mind while computing.


Set containing set containing set containing... I was wondering if anyone knew whether or not in ZFC (or any other set theory) if the object,
$$\Bigg\{ \Big\{ \big\{ \{ \cdots \} \big\} \Big\} \Bigg\}$$
can exist.  That is, it is the set containing the set containing the set containing...ad infinitum.  Is this even a set?  Can it make sense to talk about such a thing?  I'm simply curious.

The axiom of foundation (or regularity) explicitly bans your set. This axiom is independent of the other axioms of ZFC and, depending on your point of view is either just a technical convenience or a monster-barring axiom. Proponents of non-well-founded set theory like these monsters.

To see why your set gives rise to a violation of the axiom of foundation, note that (as explained on the wikipedia page), the axiom of foundation implies that that the membership relation is well-founded: i.e., there is no infinite sequence of sets $A_1, A_2, \ldots$ such that $A_1 \ni A_2 \ni A_3 \ldots$, but your set gives rise to such a sequence.

It depends very much on how you handle the informal "..." and the words ad-infinitum.  The short answer is that the notation is not clear enough to answer whether it is a set or a class.

If you define your construct to be "a thing which has the property of having 1 element, which is itself," you have constructed a class, not a set, because a set cannot have those properties in most set theories.

On the other hand, if you construct it using a sequence S which starts with $S_0=\emptyset$ and is recursively constructed according to $S_{n+1}=\{S_n\}$, and define the thing you want to be $S_\omega$, then what you are constructing is not all that far from the standard set theory construction of natural numbers ($S_{n+1}=S_n \cup \{S_n\}$)


is this language context-free? a tricky one Is this language context-free?

$$L = \{a^nb^nc^{2n} \mid n \ge 0\}$$

It's tricky in my opinion because I know that $a^nb^nc^n$ is not context-free, but can I determine from this that $L$ is not context-free, too?

Thanks in advance

EDIT:
I can use these closures: all the closures of regular languages, the closure of intersection of regular language with a context-free language, and closure under homorphishms

It is not context-free. You can show this using the pumping lemma for context-free languages; the proof is very similar to the one for the language $\{a^nb^nc^n:n>0\}$, which is given in the linked article.

Added: Since you’re restricted to closure properties, perhaps the easiest argument is to use closure under inverse homomorphisms, using the homomorphism $h$ such that $h(a)=a$, $h(b)=b$, and $h(c)=cc$. If $L$ were context-free, $\{a^nb^nc^n:n\ge 0\}$ would also be context-free.

$$L=\{a^nb^nc^2n/n\ge1\}$$
is not context free but it is context sensitive language.
here we can write like below  and think it is cfl
$a^n b^n c^n c^n$
no of $a$'s plus no of $b$'s is equal or not to no of $c$'s of twice of $a$'s and b's
it should be possible bcz we can maintain a count between no of $b$'s and $c$'s then $a$'s and $c$'s but it is not correct .
reason is we are maintaining count between no of $a$'s plus no of $b$'s but not no of $a$'s equal to no of $b$'s hence it is cfl.


How to deduce a result assuming prime number theorem I am unable to derive this result assuming prime number theorem. Can someone please tell how to do it.

Edit -> Here is a proof from stackexchange >but I couldn't think how last line is true. Can someone please  tell how it's true

  If $d_n = \text{lcm} \{1,2,3,... n\}$ then prove that the prime number theorem implies $$\lim _{n\to \infty} \frac {\log(d_n)} {n} =1$$

Can someone please tell how to derive it.

with

$lcm(1,2,3,...,x)= \prod\limits_{p\leq x} p^{max(\nu_p(1),\nu_p(2),\nu_p(3),...,\nu_p(x))}$

Since $$max(\nu_p(1),\nu_p(2),\nu_p(3),...,\nu_p(x))$$ is the maximum exponent of $p$ you'll find in numbers $\leq x$, it is $\lfloor \log_px \rfloor$

$$lcm(1,2,3,...,x)=\prod\limits_{p\leq x} p^{\lfloor \log_px \rfloor}$$
$$\log lcm(1,2,3,...,x)=\sum\limits_{p\leq x} {\lfloor \log_px \rfloor}\log p$$
and with

$\psi(x)=\sum\limits_{p\leq x}\lfloor \log_px \rfloor \log p $ (see EDIT)

$$\log lcm(1,2,3,...,x)=\log(d_x)=\psi(x)=\log e^{\psi(x)}$$
$$lcm(1,2,3,...,x)=e^{\psi(x)}$$

and from $\log(d_x)=\psi(x)$ you have your result

EDIT (not clear enought in the link):

$\psi(x)=\sum\limits_{p^k\leq x}\log p$

Since the maximum exponent of $p^k\leq x$ is $\lfloor \log_px \rfloor$ you have $\lfloor \log_px \rfloor$ powers of $p$ in that range: $\{p^1, p^2, p^3, ..., p^{\lfloor \log_px \rfloor}\}$, so $\log p$ is counted accordingly for each $p$:

$\psi(x)=\sum\limits_{p^k\leq x}\log p=\sum\limits_{p\leq x}\lfloor \log_px \rfloor \log p $

Well, I don't know why you do not mention books, if this troubles you. Here is an excerpt from Ingham. On average, if a mathematics book is old enough you can find a "preview" online, and this may include pages discussing your interest.


How to prove linearity? Given a third-order differential equation, of the form $y''' + f(t,y,y',y'') = 0.$ that admits solution:
$$Y(t) = y(t) + C_1 f_1(t) + C_2 f_2(t) + C_3 f_3(t)$$
where $y(t)$ is a particular solution, and  $f_1(t), f_2(t), f_3(t)$ are linearly independent functions, then
How to prove that the differential equation is linear?,
that is I want to prove that $$f(t,y,y',y'') = A(t) y'' + B(t) y' + C(t) y + D(t)$$

Write

$$\begin{cases}Y(t)- y(t) = C_1 f_1(t) + C_2 f_2(t) + C_3 f_3(t)\\
Y'(t)- y'(t) = C_1 f_1'(t) + C_2 f_2'(t) + C_3 f_3'(t)\\
Y''(t)- y''(t) = C_1 f_1''(t) + C_2 f_2''(t) + C_3 f_3''(t)\\
Y'''(t)- y'''(t) = C_1 f_1'''(t) + C_2 f_2'''(t) + C_3 f_3'''(t).\end{cases}$$

Now you can express that this system has nontrivial solutions for $C$ by the condition

$$\Delta=0$$ which can be developed as

$$(Y(t)- y(t))M_0(t)+(Y'(t)- y'(t))M_1(t)+(Y''(t)- y''(t))M_2(t)+(Y'''(t)- y'''(t))M_2(t)=0$$ (where $M_k$ are minors built on the RHS), a linear ODE.

Assume you have two solutions, prove that their sum is a solution, proceed with the other requirements of linearity.


How do I differentiate $f^3(x^2)$? I am learning about the chain rule and I'm not sure about kind of having to do it twice?
So say I have the function $f^{3}(g(x))$, how would I differentiate this to find $f'(x)$ using the chain rule?
Ok so what I have tried to do is say:
$$\frac{d}{dx}(f^3(g(x)) = 3f^2(g(x))f'(g(x))g'(x).$$
However, something tells me this is incorrect.

HINT
Let's differentiate $f^2(x^2) = f(f(x^2))$. You get
$$
\frac{d}{dx} f(f(x^2))
 = f'(f(x^2)) \frac{d}{dx} f(x^2)
 = f'(f(x^2))f'(x^2)\frac{d}{dx} x^2
 = f'(f(x^2))f'(x^2) \cdot 2x
$$
Another possible interpretation is that you mean $f^2(x^2) = (f(x^2))^2$ (but this is unlikely since we usually write that as $f(x^2)^2$, not as $f^2(x^2)$. If this is what you mean though, you would get
$$
\frac{d}{dx} (f(x^2))^2
 = 2f(x^2)\frac{d}{dx}f(x^2)
 = 2f(x^2) f'(x^2) \frac{d}{dx} x^2
 = 4x f(x^2) f'(x^2)
$$


The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$ The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$. Find $f(n)$.

I started plugging small values in and I got:
$$f(1)=1$$
$$f(2)=5$$
$$f(3)=11$$
I don't see any pattern so far, and I don't know another way to solve this question.  Solutions are greatly appreciated!

Hint: Let $g(x)=f(x)-x^2$. Then $g(x)+g(y) = g(x+y)-1$ for all $x,y$, so in particular $g(x+1) = g(x)+1$.


How to minimize $(x-a)^2+(y-b)^2$ subject to $ \sqrt{a}+\sqrt{b}=\sqrt{2}$? Let $x,y$ be positive real numbers satisfying $xy \le \frac{1}{4}$.

I am trying to find
$$ \min_{\sqrt{a}+\sqrt{b}=\sqrt{2}}(x-a)^2+(y-b)^2$$

as a function of $x$ and $y$.

Is there any way to do that analytically?

Edit:

Is there a way to use a computer for this? I tried running Mathematica, but it never finishes the computation...

Lagrange’s multipliers method gives

$$
(a-x)=\frac{\lambda}{\sqrt a}\\ (b-y)=\frac{\lambda}{\sqrt b}.
$$

In particular,
$$ (a-x)=\sqrt{\frac{b}{a}}\cdot(y-b).$$

I tried various ways to proceed but got stuck.

With $a:=2(\frac12+t)^2,b:=2(\frac12-t)^2$ we minimize

$$(x-2(\tfrac12+t)^2)^2+(y-2(\tfrac12-t)^2)^2.$$

By differentiating on $t$ we get (after simplification)

$$8t^3+2(3-x-y)t+y-x=0.$$

This is a general (depressed) cubic equation, and there is no hope for a nice solution.

Note that $\sqrt a+\sqrt b=1$ describes an arc of parabola (the axis is the first bissector), and you are looking for the shortest distance between the point $(x,y)$ and this arc.

So we want to minimize $\left(\frac1{ a}+\frac1{ b}\right)\lambda^2$.
If $ab\le \frac14$, we can of course take $\lambda=0$, i.e., $x=a$, $y=b$ and achieve the minimum value $0$.
So assume $ab>\frac14$.
From
$$ \frac14\ge xy=\left(a-\frac\lambda{\sqrt a}\right)\left(b-\frac\lambda{\sqrt b}\right)=ab-\left(\sqrt{\frac ab}+\sqrt{\frac ba}\right)\lambda+\frac{\lambda^2}{\sqrt{ab}},$$
we read off that we shold take the smaller of the two (necessarily positive) roots of the quadratic
$$ X^2-\left(a+b\right)X+\left(ab-\frac14\right)\sqrt {ab},$$
i.e.,
$$\lambda=\frac{a+b-\sqrt{a^2+b^2+2ab-4ab\sqrt{ab}+\sqrt{ab}}}2=\frac{a+b-\sqrt{(a-b)^2+(2\sqrt {ab}-\sqrt[4]{ab})^2}}2 $$


Differential equations in function Equations (1) : $xy'+(1-x)y=1$  let $z=xy+1$

determine and solve the differential equation (2) whose general solution is the function $z$
.

-determine the general solution of (1)

If $z=xy+1$ then
\begin{eqnarray}
\frac{dz}{dx} &=& \frac{d}{dx}\left(xy+1\right) \\
              &=& y+x\frac{dy}{dx}
\end{eqnarray}
Now:
\begin{eqnarray}
x\frac{dy}{dx}+y-xy       &=& 1 \\
\implies x\frac{dy}{dx}+y &=& 1+xy\\
\implies \frac{dz}{dx}   &=& z
\end{eqnarray}
Now you solve this equation for $z$ and then change back the variables to $x$ and $y$ at the end.

Compute $z'=y+xy'=y+1-(1-x)y=z$

So the general solution of the above equation (call it eqn (2))is

$$z(x)=z(0)e^x=e^x$$

because $z(0)=1$. Solving then for $y$:

$$y(x)=\frac{z-1}{x}=\frac{e^x-1}{x}$$


Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.

I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?

Historically, when Leibniz conceived of the notation, $\frac{dy}{dx}$ was supposed to be a quotient: it was the quotient of the "infinitesimal change in $y$ produced by the change in $x$" divided by the "infinitesimal change in $x$".

However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given any positive real number $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an "infinitesimal" $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibniz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying "Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points." But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says "However...", though.)

So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit:
$$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$
And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient.

However, Leibniz's notation is very suggestive and very useful; even though derivatives are not really quotients, in many ways they behave as if they were quotients. So we have the Chain Rule:
$$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$
which looks very natural if you think of the derivatives as "fractions". You have the Inverse Function theorem, which tells you that
$$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$
which is again almost "obvious" if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an actual quotient, it now represents a single limit.  In fact, Leibniz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for centuries in mathematics and science because, due to the fight between Newton's and Leibniz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibniz notation and stuck to Newton's... and got stuck in the mud in large part because of it.

(Differentials are part of this same issue: originally, $dy$ and $dx$ really did mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did.)

So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it like a fraction, it isn't really a fraction (it just plays one on television).

However... There is a way of getting around the logical difficulties with infinitesimals; this is called nonstandard analysis. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of real numbers that includes infinitesimals and a bunch of other things. If you do that, then you can, if you are careful, define derivatives exactly like Leibniz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ as if it were a fraction are justified because, in that setting, it is a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems.

If I give my answer from an eye of physicist ,then you may think of following-
For a particle moving along $x$-axis with variable velocity, We define instantaneous velocity $v$ of an object as rate of change of x-coordinate of particle at that instant and since we define "rate of change", therefore it must be equal to total change divided by time taken to bring that change. since we have to calculate the instantaneous velocity. we assume instant to mean " an infinitesimally short interval of time for which particle can be assumed to be moving with constant velocity and denote this infinitesimal time interval by $dt$. Now particle cannot go more than an infinitesimal distance $dx$ in an infinitesimally small time. therefore we define instantaneous velocity as
$v=\frac{dx}{dt}$  i.e. as ratio of two infinitesimal changes.
This also helps us to get correct units for velocity since for change in position it will be $m$ and for change in time it will be $s$.
While defining pressure at a point, acceleration, momentum, electric current through a cross section etc. we assume ratio of two infinitesimal quantities.
So I think for practical purposes you may assume it to be a ratio but what actually it is has been well clarified in other answers.
Also from my knowledge of mathematics, when I learnt about differentiating a function also provides slope of tangent, I was told that Leibniz assumed the smooth curve of some function to be made of infinite number infinitesimally small lines joined together ,extending any one of them gives tangent to curve and the slope of that infinitesimally small line = $\frac{dy}{dx}$ =  slope of tangent that we get on extending that line.
Even I learnt that we would need a magnifying glass of "infinite zoom in power" to see those lines which may not be possible in real.


Calculating area between two zero points I have the following function $$y=\cos(2x-1)$$ and I am supposed to find the area between two zero points.

I have calculated the first zero points which is $$x_1=0.285$$, but I do not know how to calculate the second point, $$x_2=?$$

If by ,,zero point'' you mean $x$ such that $f(x)=\cos{(2x-1)}=0$, then:

We know that $\cos(t)=t$ for $t=k\frac{\pi}{2}$, where $k$ is odd. And so we need $2x-1=k\frac{\pi}{2}$, and from this $x=k\frac{\pi}{4}+\frac{1}{2}$. You'd like to calculate the area between
$$x_1=\frac{\pi}{4}+\frac{1}{2}\text{ and }3\frac{\pi}{4}+\frac{1}{2}$$.

Then we have that area:
$$A=\int_{x_1}^{x_2}\cos{(2x-1)}\,\mathrm{d}x $$

You can for example substitute $u=2x-1$, then $du=2dx$ and $dx=\frac{du}{2}$ and our integral becomes:

$$ \frac{1}{2}\int_{x_1}^{x_2}\cos{u}du=\frac{1}{2}\left[\sin{u}\right]_{x_1}^{x_2}=\frac{1}{2}\left[\sin{(2x-1)}\right]_{x_1}^{x_2}$$

and so:
$$A=\frac{1}{2}\left(\sin(\frac{\pi}{2})-\sin(3\frac{\pi}{2})\right)=\frac{1}{2}(1-(-1))=\frac{1}{2}\cdot 2=1$$.

Through this MATLAB commands:

evalin(symengine,'solve(cos(2*x-1),x)')

you find all the roots:

Dom::ImageSet(pi/4 + (pi*k)/2 + 1/2, k, Z_)

$$
x=\frac{\pi}{4}+k\frac{\pi}{2}+1/2, k \in \mathbf{Z}
$$

Over the $cos$ function, you only then need to integrate the first two of them for getting the area:

eval(int(cos(2*x-1),pi/4+1/2,pi/4+pi/2+1/2))

Resulting in:

A=1


Ordered Arrangements: number of ways of arranging $1-9$ in a circle so that sum of 3 adjacent is divisible by 3 Find the number of ways of arranging the numbers $1, 2, 3, ... 9$ in a circle, so that the sum of any three adjacent numbers is divisible by $3$. (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)

I worked out that if I have the numbers in ascending/descending order, it will be divisible by $3$ (because $x-1 + x + x+1$ would be equal to $3x$).
I'm not sure how to use that information to work out the total number though (or if there are other ways to place them).

Please provide an answer to my question and help me with how to proceed with my approach.

Hint: $a_1 \equiv a_4 \equiv a_7 \pmod 3$.

Likewise for the other indices.

This gives us a constraint, dividing up into 3 sets of 3. Under this constraint, all permutations work.

Hence $ \dfrac{3!\times (3!)^3}{9} = 144$ ways.

First, when we look at this problem we must think what does every set of 3 numbers contain. Since, $1 + 2 + 3 \equiv  0 \mod 3$ and $2 + 3 + 4 ≡  0 \mod 3$ then $1 + 2 + 3 \equiv 2 + 3 + 4 \mod 3$. So $1 ≡ 4 \mod 3$.

Then we use the same tactic to find that $4 ≡ 7 \mod 3$. So $1 ≡ 4 ≡ 7 \mod 3$. Then we also find that $2 ≡ 5 ≡ 8 \mod 3$ and $3 ≡ 6 ≡ 9 \mod 3$. And since $1 + 2 +3 ≡  0 \mod 3$, we realize that a set of 3 must have one number from each set.

We should start by putting $1$ on top to account for rotational symmetry. Then we have $2$ choices for which set to be to the right of the one. Then we have $3$ choices for which number from the set shall go there. Then there are $3$ choices for the number to the right of that one because there are $3$ numbers from the set that we did not use so far. Then there are $2$ choices for the number that goes to the right of that because we already used $1$ from that set. Then another $2$ for the next. and $2$ for the next. then there is only one option for how to arrange the rest of the numbers because there is only one number left from each set.

So there are $2 * 3 * 3 * 2 * 2 * 2 * 1 = 144$ total ways to arrange the numbers $1-9$ in a circle if the sum of every $3$ adjacent numbers must be divisible by $3$.


check if estimation is unbiased? Assume we that we calculate the expected value of some measurements $x=\dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=\dfrac {x_1 + x_2 + x_2 + x_2} 4$.

How do I know if $v$ is a unbiased estimation of $x$?

I am not sure how to approach this problem, any ideas are appreciated!

[EDIT: Assumed $x_k$ are random variables.]

We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[\cdot]$ is expectation of random variables.

Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=\frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=\frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.

So $v$ is an unbiased estimator if $E[x]-E[v]=0 \Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$

Let $\theta$ be some parameter. Let $X$ be an estimator.

$X$ is called an unbiased estimator for $\theta$ if $E[X] = \theta$.

Note that $X$ is a random variable (or random vector) while $\theta$ would be a scalar (or vector).

Example

Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = \mu$. Then estimator $y = \frac{1}{5} x_1 + \frac{4}{5} x_2$ would be an unbiased estimate of $\mu$ since $E[y] = \mu$.


Fast algorithm for solving system of linear equations I have a system of $N$ linear equations, $Ax=b$, in $N$ unknowns (where $N$ is large).

If I am interested in the solution for only one of the unknowns, what are the best approaches?

For example, assume $N=50,000$. We want the solution for $x_1$ through $x_{100}$ only. Is there any trick that does not require $O(n^{3})$ (or $O$(matrix inversion))?

The fastest matrix multiplication algorithm is an important open mathematical question

The Wikipedia page https://en.wikipedia.org/wiki/Matrix_multiplication_algorithm#Sub-cubic_algorithms shows how the fast known asymptotic algorithm has evolved through time:

This begs the question: how close can we get to O(n^2), which is of course a lower bound since we have to read 2 * n^2 inputs at least once?

Proving the upper lower bound, even non constructively, would make you instantly famous. Wikipedia lists it as an "Unsolved problem in computer science".

The constant factor for the better algorithms is so large though that it makes them impractical for most matrix sizes found in practice today, and I doubt this will change if new algorithms are found.

Therefore any practical real world answer will come down to optimizing against a given machine model, which is an important software / hardware engineering optimization question, but not of much mathematical interest.

Personal guess about solving for single variables: I don't think you can reduce complexity like that in general, since the entire system is coupled. How would you know that your solution for some variables also satisfies the entire global solution?

Im suggesting you take a look into what quantum computing can do for you.

If you are able to make api calls form your product - then this might be the most performant approach by a lot.

This youtube video for example is a lecture which discusses problems that appear with large nr of equations : Quantum algorithm for solving linear equations

For quantum computer api you can use a service like: http://rigetti.com/forest

I personally did not tested this services, but this answer is just for pointing out the posibility. Just to show is something that can be used today, not science fiction.


Limit of square root where x approaches infinity I have to calculate the following limit, and I wondered if my solution to the question was true. Here it is:

$$\lim _{x \to -\infty} (\sqrt{(1+x+x^2)}-\sqrt{1-x+x^2})$$

Now I divide by $x^2$ and get:

$$\lim _{x \to -\infty} (\sqrt{\frac{1}{x^2}+\frac{x}{x^2}+\frac{x^2}{x^2}}-\sqrt{\frac{1}{x^2}-\frac{x}{x^2}+\frac{x^2}{x^2}})$$

I know that $$\lim_{x \to -\infty}\frac{1}{x}=0$$ so I get the following:

$$\lim _{x \to -\infty} (\sqrt{0+0+1}-\sqrt{0-0+1})$$
So we get the following:

$$\lim _{x \to -\infty} (\sqrt{1}-\sqrt{1})=0$$

Is my solution correct?

Thanks.

No, it isn't at all.

What you have shown is that

$$\lim_{x\to-\infty}\frac{\sqrt{x^2+x+1}-\sqrt{x^2-x+1}}{x}=0$$

because you have divided the function of that you were computing the limit by $x$.

To compute limits like this, it is customary to multiply and divide by an expression with the same roots, but added.

$$\lim_{x\to-\infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\frac{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}$$

No. First, you divided by $x^2$ inside the square roots, which means you divided the whole expression by $|x|$. So you have shown that the difference of the square roots, divided by $x$, goes to zero.

Here's a hint: Multiply and divide the expression by the sum of the two square roots.


Most students asked why is that ${a\over b}\div{c\over d}={ad\over bc}$ Most students asked why is that

$${a\over b}\div{c\over d}={ad\over bc}$$

I just told them: inverse the second fraction and multiply. Why? They ask me. I have no idea.

Any logical answers to them kids?

These day teachers just told students at secondary to memorise and no why is allowed in lessons. I think that is wrong. Putting people of studying maths.

By definition, $x \div y$ should be a number $z$ such that $x = y  z$.
So you just verify: $$ \dfrac{ad}{bc} \times \dfrac{c}{d} = \dfrac{a}{b}$$
Ultimately this works because of the associative and commutative laws of multiplication, but you don't have to tell the students that.

"Why" should always be allowed.

$\frac{a}{b}=\frac{\frac{ad}{b}}{d}$.  So $\frac{ad}{bc}$ is how many times $\frac{c}{d}$ goes into $\frac{\frac{ad}{b}}{d}$.


Almost sure convergence for expected value of indicator function Let $Z$ a random variable which is $\mathscr{G}$-measurable.
If $\mathcal{C}$ is a class of events, closed under intersection which generates the $\sigma$-field $\mathscr{G}$ (i.e $\mathscr{G}=\sigma(\mathcal{C})$) and $\mathbb{E}[{Z \mathbf{1}_B}]=\mathbb{E}[Z]=0$ for every $B \in \mathcal{C} $
and  $\mathbb{E}[{Z \mathbf{1}_B}]=0$ for every $B \in \mathcal{G}$ then how show that :
$Z = 0 $ almost sure.
Proposal
If $Z\geq 0$, then we may argue that
$$
\mathscr{G}\ni \mathcal{C}\mapsto \mathbb{E}[Z \mathbf{1}_{B}]
$$
are measures that agree on the $\pi$-system $\mathscr{G}$ and thus also on the sigma-algebra $\sigma(\mathcal{C})=\mathscr{G}$.
Now for a general $Z\in L^1(\Omega,\mathcal{C},P)$ we have
$$
\mathscr{G}\ni \mathcal{C} \mapsto \mathrm{E}[Z^{+}|B]+\mathrm{E}[Z^-| B]
$$
are measures that agree on the $\pi-\lambda$-system.

Since $Z$ is $\mathcal G$ measurable an $EZ1_B=0$ for all $B \in \mathcal G$ we have $EZ1_{Z>0}=0$ and $EZ1_{Z<0}=0$. But this implies that $P(Z>0)=0$ and $P(Z<0)=0$ so $Z=0$ almost surely.


What are some good math specific study habits? What are/ were some of your good mathematician's study habits that you found really worked for you?  I'm a CS major at a respected school and have a solid GPA...  However, I definitely lack when it comes to getting great grades in relatively higher level maths (300 and 400 level).  I feel that even though I study a lot, I perhaps don't study effectively enough.  I would like to ask for any tips/ tricks as to how you keep yourself focused, motivated etc that could perhaps help me or others.  Thanks in advance.

Here's something I didn't do until grad school, but wish I had started earlier. Depending on my schedule each quarter I find one hour every single morning that I set aside for "basics."

I find that I absolutely must specify this hour for that and only that if I'm going to psychologically get myself to do this. Otherwise I'll just keep putting it off for later and later in the day until I just don't get around to it. Everything else will seem more important, but I've found this time to be extremely useful.

Here's what I use the time for. Take a textbook on a subject you're learning (presumably the assigned textbook, but certainly others work to get a variety of perspectives). Start going through it. By going through it I mean you have a huge stack of paper (usually from the recycling next to a public printer). You read the book, and after every single sentence make sure you can justify why it is true.

Paragraphs from the book will sometimes transfer to many pages of your own writing. Also, don't let the topics get too fancy. This time is set aside for basics.

What I find odd in math compared to practically everything else is that this type of thing isn't emphasized. Even the top musicians in the world set aside time for scales everyday. It is easy to forget while struggling to play a hard piece of music that the basics are what makes it easy to put together (what am I talking about again?).

Maybe everyday is excessive for an undergrad class, but I think a few times a week will really get your brain in gear to understand the harder topics, and it will certainly help you fill in the details of proofs on quizzes/tests if you've already carefully thought through why the steps of the big theorems are true.

If you have always thought of your role in math as that of a calculator, then this is the point where that view of math fails. No wonder so many students think math skills are irrelevant to life. Unfortunately, or fortunately, depending on if you are at it to learn something or if you want a piece of paper that lets you go to work, you will now need to learn to think abstractly in terms of solving problems. This is a good thing in my opinion; suddenly math will become relevant and related to everything else in life. The thought "I can just grab a calculator for this, so I don't need to learn it, or spend any mental energy on it, suddenly becomes obsolete."


Toss two coins until two Heads and two Tails come up You play a game where you toss two fair coins in the air. You always win $1. However, if you have tossed 2 heads at least once, and 2 tails at least once, you surrender all winnings, and cannot play again. You may stop playing at anytime. What’s your strategy?

My thoughts were that this seems similar to the coupon collector. We have two bad events (2H and 2T). So after the occurence of the first bad event the second one will occur in an expected number of 4 turns. So my strategy is to stop after 3 tosses from the moment the first bad event occured.
However, I can't prove it.

After the first bad event one has $x\geq1$ dollars in the pocket. The question is whether one shall go on playing. Denote by $E(x)$ the expected total win under the best strategy in this situation. If we decide to quit we have won $x$, and if we make a next turn we have (with probability ${3\over4}$) won $x+1$  and the possibility of further play, or (with probability ${1\over4}$) we have won $0$. This shows that
$$E(x)=\max\left\{x, \ {3\over4}E(x+1)\right\}\ .\tag{1}$$
I'm assuming (without proof) that there will be an $n$ where we definitely won't play once more; hence $E(n)=n$. From this and $(1)$  we get
$$E(n-1)=\max\left\{n-1,\ {3\over4}n\right\}=n-1\qquad(n\geq4)\ .$$
By downwards induction it follows that
$$E(n)=n\qquad(n\geq 3)\ .$$
Using $(1)$ one then computes
$$E(2)=\max\left\{2,\ {3\over4}E(3)\right\}={9\over4}\ ,\qquad E(1)=\max\left\{1,\ {3\over4}E(2)\right\}={27\over16}\ .$$
This means that one should play on when $1\leq x\leq2$, and quit otherwise.

If while playing the game,  the latest 3 tosses are HHT or TTH, then in the next move, you have a $50\%$ chance of losing. If you take the chance and manage to stay in the game, you again have a $50\%$ chance of losing. However the total probability of the game going in your favour will be $25\%$ by the second move itself. So, it somewhat depends on the amount of money you have already won. If the game goes something like HTHTHHTor HTHHHHHT and you have already won $7-8\$$, then you should not take the $50\%$ chance. However, if you have won only $3-4\$$, then you may take the chance as you have little to lose, but after that, it is best to stop playing as a $25\%$ chance clearly doesn't seem optimal.


Applications of $f(x_0)=f'(x_0)$ If a function $f(x)$ has a derivative $f'(x)$ then where $f'(x_0) = 0$ there is an extreme point at $x=x_0$.
And where $f''(x_0)=0$ there is an inflection point at $x=x_0$.

I am asking are there any significance or applications where $f(x_0)=f'(x_0)$ at some point $x=x_0$?

I am referring to a case like $f(x)=x^2$ and $f'(x)=2x$, they intersect at $(0,0)$ and $(2,4)$ what is special or can be derived from these points?

If I'm interpreting this correctly, I believe you are asking if there is anything special about this happening locally, i.e at a certain point $x_0$, does the fact that $f(x_0)=f'(x_0)$ mean anything? I would say no; note that by just translating the function appropriately, you can make this happen at any point. To see this, for any differentiable $f$, and any point $x_0$ define $g_{x_0}(x)=f(x)+(f'(x_0)-f(x_0))$, and this function will have that property at the point $x_0$.

Well, $f(x) = f'(x)$, for every $x \implies f(x) = Ce^x$, C is a constant that can be determined if you have a inicial condition.

There are many applications, such as radioactive decay or growth of some population!


What is wrong with $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$ The title says it all, $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$ cannot possibly be true, so what am I missing?

I think there is nothing wrong with $x=x^{2/2}$. The order of operatins say you should evaluate $2/2$ first. So this is the same as $x=x^1$.
The problem is with $x^{2/2}=\sqrt{x^2}$. You have changed the order of execution. The right side is the same as $\left(x^2\right)^{1/2}$. So this comes down to whether $x^{(2/2)}$ is the same as $\left(x^2\right)^{1/2}$. You are familiar with the identity $\left(x^a\right)^b=x^{ab}$, which you might think applies here, but that identity only holds when $x$ is non-negative and/or when both $a$ and $b$ are integers.
For example, $(-1)^{2/2}=(-1)^1=-1$, but $\left((-1)^2\right)^{1/2}=\left(1\right)^{1/2}=1$.

$x^{\frac{2}{2}}=\sqrt{x^2}$ only if $x\ge0$,
in fact $x^{\frac{n}{m}}$ (where $n,m\in \mathbb{N}-\left\{0\right\}$) is defined as $\sqrt[m]{x^n}$ only for any $x\ge0$,
so $x=|x|$ is not wrong because $x$ cannot be negative if we use the equality $x^{\frac{2}{2}}=\sqrt{x^2}$.


About conditional expectation Can someone give me some hints on the following problem? Many thanks!!

Let $x$, $y$, and $z$ be integrable random variables on a probability space $(X,\Sigma, \mathbb{P})$. Show that if both $x$ and $y$ are independent of $z$, then $\mathbb{E}(x\,|\,y)=_{\text{a.s.}}\mathbb{E}(x\,|\,y,z)$.(Here $\mathbb{E}(x\,|\,y,z)$ stands for $\mathbb{E}(x\,|\,\sigma\{y,z\})$, where $\sigma\{y,z\}$ is the smallest $\sigma$-algebra generated by $y$ and $z$.)

If the hypothesis is that $x$ and $z$ are independent and that $y$ and $z$ are independent, the result does not hold and the usual counterexample applies: assume that $x$ and $y$ are i.i.d. symmetric Bernoulli $\pm1$ random variables and that $z=xy$.

Then $x$ and $y$ are independent hence $E[x\mid y]=E[x]=0$, while $x=yz$ hence $E[x\mid y,z]=x$.

Hint: the conditional expectation is the (almost) unique function satisfying...


Is there an alternative definition of finite-coproduct categories? In order theory, there's two possible definitions of the term unital join-semilattice.

A unital join-semilattice is a poset $P$ with a least element $0$, such that for any two $x,y \in P$, there exists a least upper bound $x \vee y \in P$.
A unital join-semilattice is a set $S$ together with a distinguished element $0 \in S$ and a distinguished function $\vee : S \times S \rightarrow S$ satisfying the axioms for an idempotent commutative monoid.

Is there anything like this in category theory? In particular, I want to replace unital join-semilattices with finite-coproduct categories. The problem then becomes the second dot point. I was thinking maybe we can replace $S$ with a groupoid equipped with a symmetric monoidal structure, together with some coprojection and codiagonal maps. The details aren't clear to me. For example, can we speak of the "underlying category" of such a thing?

A poset with a least element is generalised to a category with an initial element:
just pretend the arrows are inclusions to see this.

A poset with a lub for each pair of items is generalised to a category with sums:
just pretend the arrows are inclusions within the sum universal property and from
that you can obtain the lub properties.

For your second point, the idea is essentially the same:
you want a category with an inital object and a functor that assisgns pairs of objects
to their sum within the category.

Categories can be thought of as coherently constructive lattices!

Here are some possible definitions of a finite coproduct category, essentially generalizing the definitions you had.

A finite coproduct category is a category with an initial object such that any pair of objects has a coproduct.
A finite coproduct category is a category $\mathcal{C}$ equipped with functors $\mathcal{C}\times\mathcal{C}\to \mathcal{C}$ and $1\to \mathcal{C}$ that are left adjoints to the diagonal functor $\mathcal{C}\to \mathcal{C}\times\mathcal{C}$ and the unique functor $\mathcal{C}\to 1$ respectively. This was already mentioned in the comments.
A finite coproduct category is a symmetric monoidal category $(\mathcal{C},\otimes,I,\alpha,\rho,\lambda,\sigma)$ equipped with codiagonals, i.e. a monoidal natural transformation with components $A\otimes A\to A$ and a monoidal natural transformation with components $I\to A$ such that certain diagrams commute. These diagrams are essentially the unit-counit equations from the previous definition, so in a sense we're only doing a change of language here. On the other hand, those same equations could be interpreted as equipping each object with the structure of a monoid, which leads to our final definition.
A finite coproduct category is a symmetric monoidal category $(\mathcal{C},\otimes,I,\alpha,\rho,\lambda,\sigma)$ such that the forgetful functor $cMon(\mathcal{C})\to\mathcal{C}$ is an isomorphism, where  $cMon(\mathcal{C})$ is the category of commutative monoids in $\mathcal{C}$.

Assuming enough choice these amount to the same thing.


Clarify about primary ideals in a commutative ring In a commutative ring, if one takes a primary ideal $I$, then $\sqrt I$ is prime. It is not true in general that an ideal with such property is primary. For example, given a prime ideal $\mathfrak p$, one has that the radical of $\mathfrak p^n $ is $\mathfrak p$, clearly, but a $\mathfrak p^n $ is not always primary. The notes from which I'm studying prove that, in $\mathbb Z$, every ideal $I$ such that $\mathfrak q:=\sqrt I$ is prime is a power of $\mathfrak q$. Then they conclude noticing that every power of a prime ideal is primary; however, it seems to me that they only proved that in $\mathbb Z $, for an ideal, being a power of a prime is equivalent to have the property that its radical is prime. This proves that every primary is a power of a prime (in $\mathbb Z$) but not the converse, which is what we need. What am I missing? Thanks

Take $n=p_1^{a_1}p_2^{e_2}\dotsm p_k^{e_k}$. Then it's not difficult to show that $\sqrt{\smash[b]{(n)}}=(p_1p_2\dots p_n)$, so the radical is prime if and only if $n$ is a prime power.
Conversely, $(p)^e=(p^e)$ being a $(p)$-primary ideal is easy to show.
It is generally false that the power of a prime ideal is primary, but it can hold for particular prime ideals. Indeed, if $p$ is a prime element of a domain, then $(p^e)=(p)^e$ is $(p)$-primary.

We can prove that any power of a prime in $\mathbb{Z}$ is primary by hand. Let $I=(p^n)$ for a prime $p\in\mathbb{Z}$. Suppose $xy\in I$, but $x\not\in I$. Since $xy\in I$, $p^n m = xy$ for some $m$. $p^n$ divides the LHS, so $p^n$ must divide the RHS. Since $x\not\in I$, we have $p^j\mid x$ and $p^k\mid y$ for some $j$ strictly less than $n$ and some $k$ strictly greater than 0. Thus $y^n\in (p^{kn})\subset (p^n)$.


How to denote "powers" of a function? I'm working with functions themselves, and I have learned that functional powers mean composition so:

$f^3 = f \circ f \circ f$

But I'm looking for something that means $fff$. So $(fff)(x) = (f(x))^3$
Is there a de facto standard notation I can use when mixing the two? Preferably without having to write out the $(x)$ part?

It seems we have a good deal ambiguity here. Right at the question title: The power of a function may mean two things. Sometimes even the circle is omitted to make it look like multiplication.

I've already seen the following, which I adopted: $$\begin{align} f^n(x) &= (f(x))^n \\ f^{(n)}(x) &= \frac{{\rm d}^nf}{{\rm d}x^n} \\ f^{\circ n}(x) &= (f \circ f \circ \cdots \circ f)(x)\end{align}$$
However $f^{\circ n}$ does not seem to be standard, so you should always warn the reader when using it.

  The power of a function may mean two things. Sometimes even the circle is omitted to make it look like multiplication.

That's because the collection of all bijections from a set to itself form a group with composition of functions - and the operation in a group is usually seen as multiplication.

The way i read it, raising a number gives a power of the measure, so

$\sin^2 x$ corresponds to x sin square, and one is supposed to read down the tower of powers, not up them.

Multiple application, like $\sin \sin x$ = x, sin, sin, is not that common, and is best served by sequencing the terms from inner right to outer left.


$\int_{0}^{2008}x|\sin\pi x| dx$ Evaluate:

$$\int_{0}^{2008}x|\sin\pi x| dx$$

That modulus sign is causing problems. How do I handle it?
I am trying integration by parts

I have even evaluated: $\int_0^1 {|\sin \pi x|}= \frac 2 \pi$. Not sure how to utilise it in the problem.

I just need help with the modulus part.

Let $I$ denote the given integral.

Using,

$\int_{a}^b f(a+b-x)dx = \int_{a}^{b} f(x)dx$

we get:

$2I = \int_0^{2008} 2008 |\sin \pi x|dx$

Using the periodicity of $|\sin \pi x|$, we obtain:

$I = \dfrac{2008^2}{\pi}$

Not sure how to bring graphs into answers, but here is a link to your function.

http://www.wolframalpha.com/input/?i=graph+%7Csin(pix)%7C

As you can see, the period is $1$, i.e. if

$f(x)=|\sin(\pi x)|$ then $f([0,1])=f([1,2])$.

So $\int_0^{2008}f(x) dx = 2008\int_0^1 f(x) dx$

${}{}{}$


Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$ What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?

All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am very much stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?

We'll find sums at each place (units, tens, etc.) and add them.
At thousands place each of $1,2,3$ occurs $6$ times. Sum is  $6(1+2+3)\cdot 1000=36000$.
For hundreds place, we first look at numbers with thousands digit $1$. These are six numbers in which each of $2,3,0$ occurs twice. Hence in the $18$ nos, each of $1,2,3$ occurs $4$ times while $0$ occurs $18-3\cdot 4=6$ times as expected. Sum at hundreds place is $4(1+2+3)\cdot 100=2400$.
Sum at tens and units places are similar - $240$ and $24$ respectively.
Desired sum is $36000+2400+240+24=38664$.

The repetition case is exactly similar and probably easier.
We have $3\cdot 4^3$ nos in all, with $4^3$ starting with $1,2,3$ each. Sum at thousands place is $4^3(1+2+3)\cdot 1000$
For hundreds place, lets fix any digit as $$\square \, 0 \,  \square \, \square$$ There'll be $3\cdot 4^2=48$ occurences of the fixed digit. Thus sum at hundreds place is $48(0+1+2+3)\cdot 100$.
Similarly at tens and units place - $48(0+1+2+3)\cdot 10$ and $48(0+1+2+3)\cdot 1$ resp.
Desired sum is $$(1+2+3)(64000 + 48\cdot 111) = 415968$$

It would be easy to write down four blanks side by side, and then try to see how many possible choices could be filled in each block that would satisfy the condition.
      _____   ______  _____ _____

Check the different cases, for which repetition is allowed, and for repetition not being allowed.
For case 1, repetition being allowed, there are 3 choices for the first blank(0 not incuded), 4 for the next three.
So total number of choices, as per counting principle, total= 3 * 4 * 4 * 4=192
For case 2, repetition not being allowed, there are 3 choices for the first blank, 3 for the next, 2 for the third, and 1 for the last.
so total number of choices, total=3 * 3 * 2 * 1=18.


Finding sum of a finite series. Consider the series

$$\frac {q_1} {p_1} + \frac {q_1 q_2} {p_1 p_2} + \cdots + \frac  {q_1q_2 \cdots q_n} {p_1 p_2 \cdots p_n}$$ where $p_i + q_i = 1$ and $0 < p_i < 1$ and $0 <  q_i < 1$  for all $i=1,2, \cdots , n$.

How can I find the sum of this series? Please help me in this regard.

Thank you very much.


How to simplify $\frac{(5+7i)}{(2+i)}*\left(\frac{(1+2i)}{(3+4i)}\right)^{-2}$ I´d like to simplify the following term. But there is something I haven´t understood yet on how to deal with the negativ exponential $^{-2}$.

$$\frac{(5+7i)}{(2+i)}*\left(\frac{(1+2i)}{(3+4i)}\right)^{-2}$$

First I tried to divide both terms hence:
$$\frac{(5+7i)}{(2+i)} = \frac{(17+9i)}{5}$$
and
$$\frac{(1+2i)}{(3+4i)} = \frac{(11+2i)}{25}$$
therefore
$$\frac{(17+9i)}{5}*\frac{1}{\left(\frac{11+2i}{25}\right)^2}$$

But this looks weird to me. First I don´t know if this is even the right approach? My next step would be to square the bracket and then simplify further but my results seem to be wrong. Did I overlook some important rules? I´m thankfull for any hints!

Just rationalize, compute the inverse of the square and rationalize again.

At the end you will get

$$\frac{477}{25}+\frac{61 i}{25}$$

Note

$$\left(\frac{A}{B}\right)^{-2} \equiv \frac{B^2}{A^2}$$


Let $H$ be a normal $p$ -subgroup of a finite group $G$ ,then $H$ is contained into each Sylow $p$ subgroup of $G$ Any good reference for the proof of this theorem "Let $H$ be a normal $p$ -subgroup of group $G$ ,then $H$ is contained into each Sylow $p$ subgroup of $G$",i could not find after googling ?

Also a question like this here - If H is normal p-subgroup of G, then H is contained in every sylow-p subgroup.

is not clear to me.

Any help!

Let $P$ be a Sylow-$p$-subgroup of $G$,

then we need to prove that $H \subset P $ and since $P$ is any arbitrary Sylow $p$ subgroup the result will follow,

So, From Sylow's theorem we know that $H$  is contained in some Sylow -$p$-subgroup say $P'$ , now from another theorem by Sylow we also know that both $P$ and $P'$  are conjugate to each other that is $gP'g^{-1} = P$ for some $g \in G$,

Now since $H \subset P' \implies gHg^{-1} \subset gP'g^{-1} = P$

that is $gHg^{-1} \subset P$ , but as $H $ is normal in $G$ so $gHg^{-1} = H$ ,

so we obtained $H \subset P$ as desirable thus a normal $p$ subgroup of $G$ is contained into each Sylow $p$ subgroup.

Let $H$ normal a normal $p-$group and let $K$ a $p-$subgroup. Suppose that $|H|\leq |K|$. In particular, by Sylow second theorem, $K$ has a subgroup $L$ of cardinality $|H|$ that is conjugate to $H$, i.e. there is $g\in G$ s.t. $L=gHg^{-1}$. But since $H$ is normal, you get $L=H$. If $|K|\leq |H|$, then you take $L$ as a subgroup of $H$ and do the same.


What are the group objects in the category of finite sets and bijections, and its functor category? An object $G$ in a category $\mathcal{C}$ is called a group object if, given any object $X$ in $\mathcal{C}$, there is a group structure on the morphisms $\operatorname{hom}\left(X,G\right)$ such that $X\mapsto \operatorname{hom}\left(X,G\right)$ is a (contravariant) functor from $\mathcal{C}$ to $\text{Grp}$.

Group objects of the category $\text{Set}$ are groups in the usual sense. Similarly, group objects of the category $\text{FinSet}$ are finite groups.

Assuming that good (non-tautological) descriptions exist,

What are the group objects of the category $\text{FinBij}$, the category whose objects are finite sets, and whose morphisms are bijections?
What are the group objects of the functor category of $\text{FinBij}$, the category whose objects are functors of $\text{FinBij}$, and whose morphisms are natural transformations?

There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.

The question uses a nonstandard definition of group object.

Under the ordinary definition of group object, the answer to the first question is

For the identity morphism $1 \to G$ and the multiplication morphism $G \times G \to G$ (if products exist...) to live in a category whose morphisms are bijections, $G$ must have cardinality $1$.

The definition in the question is equivalent to the usual only when the category has finite products.  Although your definition would seem to be more general, in a category with terminal object, the subcategory generated by that object has a unique structure of finite products (all of which are equal to the terminal object) and one would therefore expect that the terminal object fits any definition of "group object".  As written in Mariano's answer this is not the case for your definition.  Therefore a different phrase than "group object" should be used for this more general definition.


Polarity of $\frac{x + 3}{x - 4}$ This problem is giving me loads of confusion. I just need someone to walk through it because I have the answer and I can't get to it to save my life. I have been on it for days. Please help.

$$\frac{x + 3}{x - 4}\le 0$$

we have only two cases:
a) $$x\geq -3$$ and $$x<4$$
or
b) $$x\le -3$$ and $$x>4$$ and this is impossible.
Thus we have $$-3\le x<4$$

Hint:
Let "$+$" be any positive real number, and "$-$" be any negative real number. It should be clear that:

$$\frac{+}{+} > 0;\ \ \frac{-}{-} > 0$$
$$\frac{+}{-} < 0;\ \ \frac{-}{+} < 0$$

If you can figure out for which values of $x$ has $(x+3)$ positive\negative, and which values of $x$ make $(x-4)$ positive\negative, can you use the above rules to figure out where the fraction is negative?


Proof $\frac{\sin(ax)}{x} \rightarrow a$ as $x \rightarrow 0$ in the context of a Laplace Transformation I'm finding it difficult to get my head around

$\frac{\sin(ax)}{x} \rightarrow a$ as $x \rightarrow 0,$

however for the Laplace transform of this function,

$\mathcal{L} \{ \frac{\sin(ax)}{x} \} = \arctan(\frac{a}{s}),$

surely, $\arctan(\frac{a}{s}) \rightarrow 0$ as $s \rightarrow \infty$?

But to retain the behaviour seen in the prior limit, as

$\mathcal{L}^{-1}\{ \frac{a}{s} \} = a$,

we would require the nonsensical situation of

$\arctan(\frac{a}{s}) \rightarrow \frac{a}{s}$ as $s \rightarrow \infty.$

Does this create an issue for our description of this function as $t \rightarrow 0$?

Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.

The final value theorem for Laplace's transform actually states:

$$
\lim_{s\to \infty} sF(s) = \lim_{t\to 0^+} f(t)
$$

Thus:
$$
\lim_{s\to \infty} s \left[\text{sgn}(a) \frac{\pi}{2}-\arctan \left( \frac{s}{a} \right) \right] = \lim_{t\to 0} \frac{\sin(at)}{t}
$$

Check the result for the Laplace transform with Wolfram Alpha.

The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.


Is $k$ that satisfies $\alpha^{\frac{1}{k}}<1+ε$ polynomial in $\frac1ε$? I am doing an exercise in approximation algorithms course, and my
algorithm takes as input a constant number $\alpha$ and finds the
minimal $k_{0}$ s.t $\sqrt[k]{\alpha}<1+ε$ (this would promise
me a $1+ε$ approximation).

Clearly, such $k_{0}$ exist and I am not bothered with calculating
it within my algorithm. What I am bothered about is how my solution
behaves as $ε\to0$.

My question is this: Is $k_{0}$, which is the minimal solution for
$\sqrt[k]{\alpha}<1+ε$, polynomial in $\dfrac{1}{ε}$?

Since$$
α^{\frac{1}{k}} < 1 + ε \Longleftrightarrow \frac{1}{k} \ln α < \ln(1 + ε) \Longleftrightarrow k > \frac{\ln α}{\ln(1 + ε)},
$$
then$$
k_0 = \frac{\ln α}{\ln(1 + ε)} \sim \frac{\ln α}{ε} = \ln α · \frac{1}{ε}. \quad (ε \to 0^+)
$$

Taking logarithms on both sides of your equations we have
$$\ln(\alpha)\frac{1}{k}<\ln(1+\epsilon)$$
$$k>\frac{\ln(\alpha)}{\ln(1+\epsilon)}$$
As $\epsilon\to 0$, we can expand $\ln(1+\epsilon)$ using it's taylor expansion as $\ln(1+\epsilon)=\epsilon-\frac{\epsilon^2}{2}+\frac{\epsilon^3}{3}-...\approx \epsilon-\frac{\epsilon^2}{2}$ upto second order in $\epsilon$. Thus,
$$k>\frac{\ln(\alpha)}{\epsilon-\frac{\epsilon^2}{2}}=\frac{\ln(\alpha)}{\epsilon(1-\frac{\epsilon}{2})}$$
Now, for small epsilon, we also have $(1-\epsilon/2)^{-1} \approx 1+\epsilon/2$. Thus,
$$k>\frac{\ln(\alpha)}{\epsilon}(1+\frac{\epsilon}{2})$$


Differential equation, substitution By means of the substitute $y = v(x)Y (x)$, where $Y (x)$ is to be specified, solve the differential equation:

$$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}$$

with $y=2$ at $x=1$

Anyone can solve it for me with explaining the steps please, I have no idea how to do it. Thank you very much

$$Y(x)=x$$

$\therefore y=vx\implies \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$\therefore\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}\\\implies 2v+x\dfrac{dv}{dx}=v^2x\\\implies \dfrac{dv}{dx}+2\left(\dfrac{1}{x}\right)v=v^2$

Which is in Bernoulli's Form.

$$
y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x}
$$
if we re-write
$$
v'Y = -vY' -\frac{vY}{x} + \frac{v^2Y^2}{x} = -\left(Y' + \frac{Y}{x}\right)v + \frac{v^2Y^2}{x}
$$
set the brackets term to zero i.e.
$$
Y' = -\frac{Y}{x}\implies Y(x) = \frac{C}{x}
$$
then we obtain
$$
v' = \frac{1}{Y}\frac{v^2Y^2}{x} = \frac{Y}{x}v^2 = \frac{C}{x^2} v^2
$$
which is separable.
Thus
$$
v = \frac{x}{C+C_1x}\implies y(x) = \frac{C}{C+C_1x}
$$
put it back into the equation we find
$$
y' +\frac{y}{x} = \frac{-C C_1}{\left(C+C_1x\right)^2} + \frac{C}{\left(C+C_1x\right)x} = \\
\frac{-C}{\left(C+C_1x\right)^2x}\left[C_1 x -\left(C+C_1x\right)\right] = \frac{C^2}{\left(C+C_1x\right)^2x} = \left(\frac{C}{\left(C+C_1x\right)}\right)^2\frac{1}{x} = \frac{y^2}{x}
$$