Closed form expression for $e^{-ne^{-at}}$ I am solving an equation, and I have come across the following expression which I am unable to solve. I will really appreciate if someone can help me with this.

$$\int_{0}^{\infty} e^{-ne^{-at}}dt$$

The original expression was...

$$\int_{0}^{\infty} t e^{-ne^{-at}}dt$$

Solving by by-part method, $u = t$, $v=-ne^{-at}$.

$$\int u,v \ dt= u\int v\ dt- \int u' (\int v \ dt) \ dt$$

Now the second part of by-part equation i.e.
$$\int v \ dt= \int -ne^{-at} \ dt= \frac{1}{na}e^{-ne^{-at}} $$

At this point I am stuck and need some hint on how to solve this

Concerning the problem of the antiderivative $$I = \int e^{-ne^{-at}}dt$$ change variable $$ne^{-at}=u \implies t=-\frac{\log \left(\frac{u}{n}\right)}{a}\implies dt=-\frac{du}{a u}$$ This makes $$I=-\frac 1a\int\frac{e^{-u}}{ u}\,du=-\frac{\text{Ei}(-u)}{a}$$ where appears the  exponential integral function.

Back to $t$, this will give $$I=-\frac{\text{Ei}\left(-n\,e^{-a t} \right)}{a}$$ If there is no problem for $t \to 0$, as user1952500 already answered, there is a major one for $t \to \infty$.

If you consider the case $n=a=1$, you would notice that $I$ almost behaves as $t$

$$I = \int_{0}^{\infty} e^{-ne^{-at}}dt$$

Let
$$y =e^{-ne^{-at}} \Rightarrow I = \int_{e^{-n}}^{1}y\cdot dt$$
$$\Rightarrow \ln y =-ne^{-at}$$
$$\Rightarrow \frac{dy}{y} = nae^{-at} \cdot dt = -a\,(-ne^{-at}) \cdot dt = -a\,(\ln{y}) \cdot dt$$
$$\Rightarrow -\frac{dy}{na \ln{y}} = y \cdot dt$$

$$\Rightarrow I = -\frac{1}{n \cdot a} \int_{e^{-n}}^{1} \frac{dy}{\ln y}$$
$$\Rightarrow I = -\frac{1}{n \cdot a} \bigg(\int_{0}^{1} \frac{dy}{\ln y} - \int_{0}^{e^{-n}} \frac{dy}{\ln y}\bigg)$$

The above integral does not converge since $$\int_{0}^{1} \frac{dy}{\ln y} = \infty$$ That is, the first integral diverges.

Side notes:

We have
$$\int_{0}^{x} \frac{dy}{\ln y} = li(x)$$

where $li(x)$ is the Logarithmic Integral Function.

Hence the above integral can be rewritten for the limits as:
$$I = \boxed{-\frac{1}{na}\cdot \big(li(1) - li(e^{-n})\big)}$$

However, note that $li(1) = \infty$ and hence the above integral diverges.


Homeomorphism definition I was told by my professor that homeomorphisms are continuous maps with continuous inverse, but do those conditions also imply that the map is bijective?

It was inconvenient that your professor worded it this way. From the get-go, saying that a homeomorphism is continuous function with a continuous inverse assumes that we have some function $f$, and that its inverse, $f^{-1}$, exists. Off-the-bat, just cause we have a function $f$, it does not mean that it has an inverse.

In addition, the way your professor phrased it doesn't make it clear whether we are talking about a subset of the range, or the whole range itself: in a homeomorphism, we must have the whole range, which is something your professors phrasing neglected to capture.

If I were you, I would just forget about what your professor said. It is kind of circular. Just to make it clear. By definition, a homeomorphism is a function $h$, from a topological space $X$, to a topological space $Y$, such that the following hold:

$h$ is 1-1
$h$ is onto
$h$ is continuous
$h^{-1}$ is continuous

Note, we do not say $h^{-1}$ exists here: this is a consequence of $h$ being 1-1, as we can always create $h^{-1}$ in such a case. ALSO, we imply that the image of $h$ is all of its range: this is captured by saying that $h$ is a bijection.

A function that has an inverse is injective, or one-to-one.  Every one-to-one function is onto some set.  So a continuous function with a continuous inverse is a homeomorphism onto some set even if there is some larger target set.

However, if $f:A\to B$ is continuous and has an inverse and its inverse is also continuous, that does not mean $f$ is a homeomorphism from $A$ to $B$.  Rather, $f$ is a homeomorphism from $A$ to some subset of $B$, which may or may not be all of $B$.

Among the simplest instructive examples of a continuous bijection whose inverse is not continuous is $\theta\mapsto e^{i\theta}$ from $[0,2\pi)$ to the circle.


Vertical asymptotes of a given non-rational radical funtion We have that $f$ is a function $f(x) = x\sqrt{x+4}$.

Hence, $f'(x) = \dfrac{3x+8}{2\sqrt{x+4}}$.

Then, $\lim_{x \to -4^+}f'(x) = -\infty$.

This means that $f$ has a vertical slope at $f(-4)$.

It was my first thought to say, by this, $f$ has a vertical asymptote at $x = -4$; I wonder, though, is it so that an asymptote can necessarily only exist when a function is rational?

Rational functions with a zero in the denominator are common causes of vertical asymptotes, but they are not the only ways this can occur.

For example, $f(x)=\tan(x)$ has vertical asymptotes.  Of course you can argue that the reason this occurs is that there is a zero hiding in the denominator since $\tan(x)=\frac{\sin(x)}{\cos(x)}$.

Another example, $f(x)=\ln(x)$ has a vertical asymptote at $x=0$.  I do not see any way (reasonably easy and meaningful) to write $\ln(x)$ as a fraction with a zero appearing in the denominator when $x=0$.

One can construct more functions like $\ln(x)$ (in terms of vertical asymptotes) by taking the inverse function of any one-to-one function with a horizontal asymptote.  In the case of $\ln(x)$, we found the inverse of $e^{x}$ which had a horizontal asymptote at $y=0$.

If the denominator contains a factor going to zero it has infinite value at an asymptote for any function.


Why is mean curvature extrinsic? I believe that an intrinsic property is dependent on a surface itself (not on how it is parameterized), and an extrinsic property may vary depending on the parameterization.

Why is mean curvature extrinsic then? It's a measure of the curvature of a surface, and curvature is intrinsic, right? Perhaps I'm confusing mean curvature for Gaussian curvature actually. I don't have a good intuitive grasp of what mean curvature is.

At a point on a surface, mean curvature is the mean of the two principal curvatures. The intrinsic curvature that you refer to is the Gaussian curvature, which is the product of the two principal curvatures, and can be shown to be invariant under isometry. To get a feeling for this, consider a flat piece of paper. Intuitively, it has zero mean as well as Gaussian curvature. But if you roll the paper into a cylindrical shape (let's say of unit radius), the principal curvatures are along the axis and the circular cross sections, and are $0$ and $1$ respectively, so the surface now has mean curvature $1/2$, while the Gaussian curvature is still $0$. In general, mean curvature is a property of an embedding, and depends on how the surface (or in general, manifold) is embedded into the ambient Euclidean space (or manifold). However, the Gaussian curvature depends on the surface itself (which is to say, depends solely on the metric), and does not depend on the embedding. To get more insight, you can look at the books of do Carmo or O'Neill. You will find that both the curvatures are defined in terms of the shape operator, which apriori depends on the embedding. If defined this way, neither appears to be intrinsic, but then one proves the intrinsic nature of the Gaussian curvature (this is the Theorema Egregium of Gauss).

the local values of curvature are necessarily dependent on the parametrization; what is invariant is the global property of curvature (e.g. when you integrate over the whole manifold etc.)


Understanding of multi-index notation - specific example I'm having trouble to really come to terms on how to understand this multi-index notation.

Given $x=(x_1, \dotsc, x_n) \in K^n$ and $\alpha \in \mathbb{Z}^n$, we write $x^{\alpha} = \prod_{i=1}^n {x_i}^{\alpha_i}$. We consider the multiplicative seminorm, for $\rho \in \mathbb{R}^n \cup \{ - \infty \}$,

$$\delta(\rho) : K[x_1, \dotsc x_n] \rightarrow \mathbb{R}_{\geq 0}, \\
 \sum_{\alpha} c_\alpha x^\alpha \mapsto \max_{\alpha} \{ |c_\alpha| \cdot \exp(\sum_{i=1}^n \rho_i \alpha_i) \}$$
where $|.|$ denotes the absolute value on $K$.

I tried to better understand this by setting $n=3$. This should give me, for $\alpha=(\alpha_1, \alpha_2, \alpha_3)$, $\sum_{\alpha} c_\alpha x^\alpha = c_{\alpha_1} x^{\alpha_1} + c_{\alpha_2} x^{\alpha_3} + c_{\alpha_3} x^{\alpha_3}= c_{\alpha_1} \prod_{i=1}^3 {x_i}^{\alpha_{i1}} + c_{\alpha_2} \prod_{i=1}^3 {x_i}^{\alpha_{i2}} + c_{\alpha_3} \prod_{i=1}^3 {x_i}^{\alpha_{i3}}$, with each $c_{\alpha_i} \in K$. BUT: Logically, there are no $\alpha_{i1}$ etc.??? So maybe I missunderstand the index $\alpha$ below the sum? (Similarly, I thought that the $\max$ on the RHS then runs over the coordinates of the $\alpha \in \mathbb{Z}^n$...is this right?)

It would be also great if someone could show me how a sum and a product of two sums of the form $\sum_{\alpha} c_\alpha x^\alpha$ would have to look like! (I'm working on actually proving that the map above is a multiplicative semi-norm, which doesn't work out yet, probably due to a mistake in the understanding of the index...)

Thank you very much!

If $\alpha$ is an $n$-dimensional vector all of whose coordinates are  integers, then $z^\alpha$ means the product $z_1^{\alpha_1}\cdots z_n^{\alpha_n}$; the notation $|\alpha|$ means $\alpha_1+\cdots+\alpha_n$. In this multi-index notation, a multivariable series can be written in the form $\sum_{\alpha}c_{\alpha}z^{\alpha}$ as abbreviation for $\sum_{\alpha_1\in\mathbb{Z}}\cdots\sum_{a_n\in\mathbb{Z}} c_{\alpha_1\ldots\alpha_n}z^{\alpha_1}\cdots z^{\alpha_n}$.

  In the three-dimensional case we have
  \begin{align*}
\sum_{\alpha}c_{\alpha}x^{\alpha}
=\sum_{\alpha\in\mathbb{Z}^3}c_{\alpha}\prod_{j=1}^3x_j^{\alpha_j}
=\sum_{\alpha_1\in\mathbb{Z}}\sum_{\alpha_2\in\mathbb{Z}}\sum_{\alpha_3\in\mathbb{Z}}c_{\alpha_1\alpha_2\alpha_3}x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}
\end{align*}

For example, you have the function":

$f(x,y,z)=x^2yz^3 $

With $\alpha=(1,0,2)$:

$\partial^\alpha f=\frac{\partial}{\partial x}\frac{\partial^2}{\partial z^2}f=6xyz$


How prove there exist prime numbers $P_{1},P_{2},\cdots,P_{n}$ such $P_{k}\mid c+k$ Question:

Let $c,n\in\mathbb{N}$, such that, $c>n^{n-1}$.
Show that: there exist distinct prime numbers $P_{1},P_{2},\dots,P_{n}$ such that:
$$P_{k}\mid c+k,k=1,2,3,\dots,n$$

My idea: if $c+k$ have greater than $n$ factor,this problem It is clear .
But other case,I can't

As others pointed out, what I wrote earlier did not work, and I spent several hours failing to fix it. But a correct solution turns out to be already published.

@JeppeStigNielsen [Comment under question] pointed out that this is related to Grimm's Conjecture. Grimm's Conjecture is similar but with the requirement that $c>n^{n-1}$ replaced by the requirement that the $n$ numbers $c+1,\dots,c+n$ be composite.

But in his original paper (American Mathematical Monthly 76 (1969) p1126), he proves the result above. The proof is quite short:

Consider the prime factoring of $c+1,\dots,c+n$. If $c+k$ is divisible by more than $n-1$ distinct primes, then no matter what primes are factored from the other $n-1$ numbers $c+j$ there would be a prime to factor from $c+k$. Hence we assume $c+k$ has fewer than $n$ distinct prime factors $p_{1k},\dots,p_{jk}$ (where $j<n$).

Write $$p_{1k}^{\alpha_{1k}}p_{2k}^{\alpha_{2k}}\dots p_{jk}^{\alpha_{jk}}=c+k>c>n^{n-1}$$ Since there are at most $n-1$ factors $p_{ik}^{\alpha_{ik}}$, at least one of them must be $>n$. Thus for each $c+k$ with fewer than $n$ prime factors we can choose at least one factor $p_{ik}^{\alpha_{ik}}>n$. If it were possible to choose the same prime power in two cases, say for $c+k$ and $c+j$ with $k>j$, then the smaller of the two powers $p^{\lambda}>n$ would divide the difference $c+k-(c+j)=k-j<n$, an impossibility. Thus a different prime can be associated with each number $c+k$ in all cases and the proof is complete.


Line integral depending on a parameter is entire Suppose you have a continuous function:

$$\phi:[0,1]\rightarrow \mathbb{C}$$

define the complex function:

$$f(z)=\int_0^1\phi(t)e^{itz}dt$$

prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.

Also, how can I apply those results to the sequence of functions:

$$f_n(z)=\int_0^n \sqrt{t}e^{-tz}dt$$

It's obviously entire as the derivative is $$ \int_0^1 ite^{izt}  \phi(t) \, dt$$ which exists (is convergent) for all $z$  as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ \int_0^1 (it)^n \phi(t)\, dt.$$


Your expression for $a_n$ is correct. So $$a_n=2+\cdots+2^n+2^{n+1}.$$ Now note that this is a sum of $n+1$ terms in geometric progression with first term $a=2$ and the common ratio $r=2$. The sum $S$ can be found by $$S=\frac{a(r^{n+1}-1)}{r-1}$$ So substituting $a=2$ and $r=2$ we get $$a_n=\frac{2(2^{n+1}-1)}{2-1}=2(2^{n+1}-1)$$ as required.

You can also do this in base 2 fairly easily:

$$
\begin{align*}
a_0 &= 10_2 \\
a_1 &= 2a_0 + 2 = 100_2 + 10_2 = 110_2 \\
a_2 &= 1100_2 + 10_2 = 1110_2.
\end{align*}
$$

We can see that by induction we will have.
$$a_n = \underbrace{11\cdots1}_{n+1}0_2 = 2(\underbrace{11\cdots1_2}_{n+1}).$$
Now note that
$$\underbrace{11\cdots1_2}_{n+1} + 1 = 1\underbrace{00\cdots0_2}_{n+1} = 2^{n + 1}.$$
Therefore
$$a_n = 2(2^{n + 1} - 1).$$


Series Expansion Of An Integral. I want to find the first 6 terms for the series expansion of this integral:

$$\int x^x~dx$$

My idea was to let: $$x^x=e^{x\ln x}$$

From that we have: $$\int e^{x\ln x}~dx$$

The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$

Then we have:

$$\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\frac{(x\ln x)^n}{n!}dx$$

I integrated the first 3 terms (since it was the easiest the work with):
$$x+\frac{1}{4}x^2(2\ln x-1)+\frac{1}{54}x^3(9(\ln x)^2-6\ln x+2)+\cdots$$

Is there any other efficient way of getting the first 6 terms of the series expansion of the integral?

Try taking a Taylor series for $\log x$, multiplying it by $x$, and plugging the result into the Taylor series for $e^x$. You can easily compute the first few terms.

Your choice of series for $\log x$ will affect where the result converges, so nothing will be valid everywhere. If this were a definite integral, it might be better.


Faithful Group Actions and Normal Subgroups Let $G\curvearrowright X$. Show that $K=\{g\in G:g\cdot x=x,\text{ for all }x\in X \}\trianglelefteq
G$. If $\phi\colon G\to Sym(X)$ is the homomorphism given by the
action, show that $K=\ker(\phi)$.

Let $G\curvearrowright X$ and let $K$ be as above. Show that
$\frac{G}{K}$ acts faithfully on $X$.

Attempt at solution: We want to show that $K = ker(\phi)$ The $ker(\phi)$ has already been show to be a normal subgroup. Also, by definition above we can see that $K$ is not faithful as it has more than one element that acts as the identity.

Take any element $g \in G$ and $x \in X$ Then by definition $g \cdot x = x$ but if we choose different elements $g_1 \in G$ and $x_1 \in X$ then we also get $g \cdot x_1 = x_1$ This is where I am stuck with writing the proof.

In part B when we kill the kernel of $G$ this makes only the identity map to the identity, as this group action now becomes faithful.

First of all, $G$ may act faithfully on $X$. It was not stated that $K$
had more than one element. Next $g\in K\iff gx=x,\forall\,x\in X\iff \phi(g)=\text{id}_X\iff g\in\ker(\phi)$.

Hints:

For any $\;k\in K\,,\,g\in G\;$ :

$$k\in K\implies k(gx)=gx\;,\;\;\forall\,g\in G\,,\,x\in X\implies $$

$$\implies g^{-1}kg(x)=g^{-1}(kg(x))=g^{-1}gx=1\cdot x= x\implies g^{-1}kg\in K\implies K\lhd G$$

As for (b):

$$(gK)x:=g(x)=x\iff g\in K\iff gK=\overline 1\iff\;\text{the induced action's faithful}$$


Stationary points of a family of functions which converge uniformly. Here's the motivation of my question:

Consider the family of functions $\mathrm{f}_n(x) := \frac{1}{n}\sin(nx)$. It is easy to show that $\mathrm{f}_n$ converges uniformly to the zero function.

To prove this, we need to show that for all $\varepsilon > 0$, there exists a positive integer $N$ such that $|\mathrm{f}_n(x) - 0| < \varepsilon$ for all $x \in \mathbb{R}$ and $n > N$.

$$\left|\frac{1}{n}\,\sin(nx)-0\right| = \left|\frac{1}{n}\,\sin(nx)\right|
=\left|\frac{1}{n}\right| \cdot \left|\sin(nx)\right| \le \frac{1}{|n|} \cdot 1=\frac{1}{n}$$

Hence $|\mathrm{f}_n(x) - 0| < \varepsilon$ if $\frac{1}{n} < \varepsilon$ and so $|\mathrm{f}_n(x) - 0| < \varepsilon$ for all $n > \frac{1}{\varepsilon}$.

This example is used often to show that derivatives are not inherited from uniformly convergent sequences. For example, $\mathrm{f}_n{'}(0) = 1$ or all positive integers $n$, and yet the zero function has derivative zero everywhere.

Here's my build-up to my question:

I'm interested in when the derivatives of the limiting functions agree with the limit function.
$$\mathrm{f}_n{'}(x) = \cos(nx)$$
The limit function, i.e. the zero function, has zero derivative everywhere. Let $$D_n := \{ x \in \mathbb{R} : \mathrm{f}_n{'}(x) = 0 \}$$
For a fixed $n$, we have $\mathrm{f}_n{'}(x) = 0$ if, and only if,
$$nx \in \left\{\frac{\pi}{2} + \pi k : k \in \mathbb{Z} \right\}$$
It would seem to me then that
$$D_n = \left\{ \left(\frac{1+2k}{2n}\right)\!\pi : k \in \mathbb{Z} \right\}$$

Here's my question:

What can be said about the set
$$\lim_{n \to \infty} D_n$$

My instinct is that it will be countable and dense in the reals, just like the rationals are.

EDIT: For example, for any $x \in \mathbb{R}$ and any $\varepsilon > 0$ can we find an $N \in \mathbb{Z}^+$ for which an element of $D_n$ is within $\varepsilon$ of $x$ for all $n > N$?

Counterexample: $f_n(x) = x/n \to 0$ uniformly on $[0,1],$ but $f_n(x)$ has no stationary points.


The Number Of Integer Solutions Of Equations The Number Of Integer Solutions Of Equations

An approach is to find the number of distinct non-negative integer-valued vectors $(x_1,x_2,...,x_r)$ such that $$x_1 + x_2 + ... + x_r = n$$

Firstly, considering the number of positive integer-valued solutions.

An approach to solving this problem for positive integer-valued solutions is to imagine that you have $n$ indistinguishable objects lined up and that you want to divide them into $r$ nonempty groups. To do so, you can select $r-1$ of the $n-1$ spaces between adjacent objects as the dividing points. See the diagram below for a visual representation.

$$0_\wedge0_\wedge0_\wedge..._\wedge0_\wedge0$$

$$n\,\, objects\,\,0$$
$$Choose\,\,r-1\,\,of\,\,the\,\,spaces\,\,_\wedge.$$

For instance if you have $n=8$ and $r=3$ and you choose the 2 divisors so as to obtain $$000|000|00$$ then the resulting vector is $x_1 = 3. x_2 = 3, x_3 = 2.$ As there are $n-1\choose r-1$ possible selections, you have the following proposition.

Proposition 1: There are $n-1\choose r-1$ distinct positive integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$x_1 + x_2 + ... + x_r = n, \,\,\, x_i > 0,\,\,\, i = 1,...,r$$
Finally, from Proposition 1 you can obtain the following proposition
Proposition 2: There are $n+r-1\choose r-1$ distinct non-negative integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$ x_1 + x_2 + ... + x_r = n$$
Question:   I understand all the steps taken prior to Proposition 2, so what I want to know is how is Proposition 2 derived from Proposition 1? I have drawn multiple diagrams using the spaces between objects analogy by adding $r$ extra possible positions for a divider, but none of them hold for all possible vectors.

For concreteness, let us work with the specific numbers $8$ and $3$ mentioned in the post, though the argument is general.

We have $8$ identical candies, and we want to distribute them among $3$ kids, with some kid(s) possibly getting no candy. Call this Task A.

Task B goes as follows. Distribute $8+3$ candies among the kids, with each kid getting at least $1$ candy. Then take away a candy from each kid.

It is clear that there are just as many ways to carry out Task B as there are to carry out Task A. And by the analysis of Proposition 1, there are $\binom{8+3-1}{3-1}$ ways to carry out Task B.

Another way: Imagine $8+3-1$ slots in a row, like this:
$$\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square$$
We will put place $8$ candies ("stars") in these slots, with the other $2$ slots serving as separators ("bars"), possibly adjacent. The number of ways of placing the candies is $\binom{8+3-1}{8}$. It is the same as the number of ways of choosing the $2$ slots to be left blank.

So the number of solutions is $\binom{10}{8}$, or equivalently $\binom{10}{2}$.

In general, the same analysis shows that the number of ways to distribute $n$ candies among $r$ kids is $\binom{n+r-1}{n}$, or equivalently $\binom{n+r-1}{r-1}$.

Well, this particular problem can be solved by thinking in following way:

Step 1: First find the total number of solutions in which none of the variables from x1, x2, and x3 is zero.

Step 2: Then find the number of solutions in which only one from x1, x2, and x3 is zero.

Step 3: Then find the number of solutions in which only two from x1, x2, and x3 is zero.

Step 4: Finally, add all the number of solutions found in the abovementioned steps. This is the final answer.

This method can be generalized and we will arrive at the same formula.

PS: I don't have any idea how to write math equations here, and therefore my contribution may rightly look ugly to some of the contributors here.


Cancellation Law for Addition Proof This is a question I have regard the proof of the cancellation law for addition in Apostol's Calculus. We are told that the sum of two real numbers x and y is x+y and that it is uniquely determined. We are also given a few axioms. The ones that matter are the commutative law, associative law and the existence of negatives.

Cancellation law for addition:
If $a + b = a + c$, then $b = c$.

We assume that $a + b = a + c$. By the Existence of Negatives Axiom, we know that there is a number $y$ such that $y + a = 0$.

My question regards this step:

Since sums are uniquely determined we have $y + (a + b) = y + (a + c)$. Why is it allowed to do this step? What is the reasoning behind it?

Cancellation law for addition:-
If $x,y,z\in\mathbb{R}$ and $x+y=x+z$, then $y=z$

Proof: $$x+y= x+z \qquad \qquad (1)$$

Since, $x\in\mathbb{R}$ (existence of additive inverse). Now adding $-x$ to both sides of $(1)$, we get $(-x)+(x+y) = (-x)+(x+z)$

$\Rightarrow (-x+x)+y =(-x+x)+z$       (by associativity of addition)

$\Rightarrow 0+y = 0+z$               ( by property of additive inverse)

$\Rightarrow  y = z$                        ( by property of additive identity)


Identity element of the operation $x*y = \min\{x+1, y+1\}$ on $\mathbb R$? The following is an operation $*$ on $\mathbb R$.

$$x*y= \min\{x+1, y+1\}$$

Explain whether or not $\mathbb R$ has an identity element with respect to $*$.

This question also had two other parts, asking to explain whether it is commutative and associative. I said that it is commutative but not associative.

As for the identity element, I am confused because $\mathbb R$ goes to $-\infty$ so won't we always be able to find an element that is smaller than $x+1$?

I feel like there cannot be a neutral element because the operation $*$ is defined on $\mathbb R$ and I feel like you will always be able to find a smaller element than $x+1$.

Plus, the neutral element has to be unique so how could this operation have a unique neutral element?

Thank you for any help!

The operation $x*y = \min(x+1, y+1)$, for $x, y \in \mathbb R$, is equivalent to the following:

$$x*y = \min(x, y) + 1$$

The neutral element, say $e \in \mathbb R$ must be such that $e*x = x*e = x$, that is $\min(e, x) + 1 = \min(x, e) +1 = x$, for all $x \in \mathbb R$.

But let's just pick a specific $x_1 \in \mathbb R$ to text.  $\min(e, x_1) + 1 = \min(x_1, e) = x_1 \iff e = x_1 - 1$.

Let's try another distinct element, $x_2 \neq x_1$ in $\mathbb R$.  We need $e' \in \mathbb R$ such that $e'*x_2 = x_2*e' = x_2 \iff e'= x_2 - 1$.

But we see that $e'\neq e$, because $e = x_1 -1$, and $e'= x_2 - 1$, and since $x_1\neq x_2$, we know that $e' \neq e$.

So  we see quickly, that there is a different "neutral element" dependent on and defined for, each distinct element $x\in \mathbb R$.

So no, there does not exist a unique neutral (identity) element $e$ in $\mathbb R$ such that for all $x \in \mathbb R$, $e*x = x*e = x$.


How many ways are there to place $2$ identical kings on an $8\times 8$ chessboard so that the kings are not in adjacent squares? This problem needs also to be extended to $n*m$ chessboard. I tried to think like this:

First I choose a place for the first king in $64$ ways. Then I have a  choice $64-5 = 59$ squares for the second king . But this solution is not right because this is not the case if I place the first king in the sidemost layers of squares. Then I have $64-4 = 60$ squares for the other king. How can I solve this problem?

The first King could be on a corner square($4$ ways), leaving $60$ other squares for the next King.

The first King could be on a edge square($24$ ways), leaving $58$ other squares for the next King.

The first King could be on a central square($36$ ways), leaving $55$ other squares for the next King.

This will double count the configurations, so we have $(4 \times 60 + 24 \times 58 + 36 \times 55)/2 =\color{red}{1806}$.

We will solve the problem by cases.
1) We put the first king on the corner square. Then there are 4 ways to do it, and for the second king we have 64-3=61. Then there are $4\cdot61=244$ ways to do it.
2) We put the first king on the edge square. Then for the first king $6+6+6+6=24$ ways to do it, then for the second it is $64-4=60$ ways to do it. So it is $24\cdot60=1440$ ways to do it.
3)We put king neither in the corner, nor on the edge. So we have $6\cdot6=36$ ways for the first king and for the second $64-5=59$ ways to do it. So it is $36\cdot59=2124$ ways.
So together it is $(2124+1440+244)/2=1904$ ways to do it.  We are dividing by 2 because we are double-counting when we analyze cases, because the order of putting kings doesn't matter.
And now, I think, it is easy to generalise it for $n×m$ case.


Why do axiomatic systems for propositional calculus include IFF axioms? I am reading : https://en.wikipedia.org/wiki/Propositional_calculus#Axioms, and the following three axioms seemed unnecessary to me:
$$
 IFF-1 : ( \phi \iff \chi ) \implies (\phi \implies \chi ) \\
 IFF-2 : ( \phi \iff \chi ) \implies (\chi \implies \phi ) \\
 IFF-3 : ( \phi \implies \chi ) \implies ( ( \chi \implies \phi ) \implies ( \phi \iff \chi ) )
$$
I understand how they would define an $ \iff $ operator, but wouldn't you always be able to substitute $(\phi \implies \chi) \land (\chi \implies \phi)$ for $ (\phi \iff \chi) $ in any proof? From $(\phi \implies \chi) \land (\chi \implies \phi)$ and the three $\land$ axioms you can recover the three $\iff$ axioms. In this scheme $ (\phi \iff \chi) $ in proofs would be notation for ($\phi \implies \chi) \land (\chi \implies \phi)$.
You would end up with an equivalent notion with three fewer axioms and one fewer operator in the system.

Firstly, let us make clear that there are many different but equivalent sets of axiom schemata for propositional calculus, which serve the same purpose: from them, we can prove the same theorems. Some of them include only the least possible number of axiom schemata. Others have redundant axiom schemata, that can be proved from the rest and thus, are not really necessary. However, there is a subtle point here: you have to introduce more axioms, in order to include in a formal language, more primitive symbols, like logical connectives. For example, see the following set of axioms, from Jan Łukasiewicz:

$p\rightarrow\left(q\rightarrow p\right)$
$\left[p\rightarrow\left(q\rightarrow r\right)\right]\rightarrow\left[\left(p\rightarrow q\right)\rightarrow\left(p\rightarrow r\right)\right]$
$\left(\neg p\rightarrow\neg q\right)\rightarrow\left(q\rightarrow p\right)$

From these three schemata and using modus ponens as the only inference
rule, one can prove all the tautologies of propositional calculus.
There is no redundant axiom schemata here. If any of them is removed,
the system will not work. Note that only two logical operations are
used in the system: negation "$\neg$" and material implication
"$\rightarrow$". The other usual logical operation symbols, could
be seen as abbreviations:

Conjunction "$p\wedge q$" is an abbreviation for "$\neg\left(p\rightarrow\neg q\right)$".
Disjunction "$p\vee q$" is an abbreviation for "$\neg p\rightarrow q$".
Biconditional "$p\leftrightarrow q$" is an abbreviation for "$\neg\left[\left(p\rightarrow q\right)\rightarrow\neg\left(q\rightarrow p\right)\right]$".

Etc. A crucial point here, in respect of your question, is that abbreviations do not belong to the formal language. One can use these symbols in order to make an expression or a proof more readable and shorter. However, a strictly formal expression, does not include them.
For another example, lets take J. Barkley Rosser's system:

$p\rightarrow p\wedge p$
$p\wedge q\rightarrow p$
$\left(p\rightarrow q\right)\rightarrow\left[\neg\left(q\wedge r\right)\rightarrow\neg\left(r\wedge p\right)\right]$

Again, modus ponens is the only inference rule and there are not any redundant axiom schemata. We see three logical
operations here. However, in Rosser's system, only negation and conjunction
are primitive symbols, while material implication
is informal:

"$p\rightarrow q$" is an abbreviation for "$\neg\left(p\wedge\neg q\right)$".
"$p\vee q$" is an abbreviation for "$\neg\left(\neg p\wedge\neg q\right)$".
"$p\leftrightarrow q$" is an abbreviation for "$\neg\left(p\wedge\neg q\right)\wedge\neg\left(q\wedge\neg p\right)$".

Rosser's axiom schemes, in the formal language of his system, should look like this:

$\neg\left[p\wedge\neg\left(p\wedge p\right)\right]$
$\neg\left[\left(p\wedge q\right)\wedge\neg p\right]$
$\neg\left[\neg\left(p\wedge\neg q\right)\wedge\neg\neg\left[\neg\left(q\wedge r\right)\wedge\neg\neg\left(r\wedge p\right)\right]\right]$

Also, modus ponens should look like this:
$$\dfrac{\begin{array}{c}\neg\left(p\wedge\neg q\right)\\p\end{array}}{q}$$
Now, if you need more primitive logical connectives, you need more
axiom schemata also. Stephen Cole Kleene proposes the following 13
axiom schemata for all the usual connectives $\neg$, $\wedge$, $\vee$,
$\rightarrow$ and $\leftrightarrow$:

$p\rightarrow\left(q\rightarrow p\right)$
$(p\rightarrow q)\rightarrow\left(\left[p\rightarrow(q\rightarrow r)\right]\rightarrow(p\rightarrow r)\right)$
$(p\rightarrow q)\rightarrow\left[(p\rightarrow\neg q)\rightarrow\neg p\right]$
$\neg\neg p\rightarrow p$
$p\rightarrow(q\rightarrow p\wedge q)$
$p\wedge q\rightarrow p$
$p\wedge q\rightarrow q$
$p\rightarrow p\vee q$
$q\rightarrow p\vee q$
$(p\rightarrow r)\rightarrow\left[(q\rightarrow r)\rightarrow(p\vee q\rightarrow r)\right]$
$(p\rightarrow q)\rightarrow\left[(q\rightarrow p)\rightarrow(p\leftrightarrow q)\right]$
$(p\leftrightarrow q)\rightarrow(p\rightarrow q)$
$(p\leftrightarrow q)\rightarrow(q\rightarrow p)$

From this Kleene's system, if you don't need the biconditional as
a primitive symbol, you can remove schemata 11-13, which are the 3 IFF schemata that you mention in your question. So, you get a shorter
set of 10 axiom schemata, for the connectives $\neg$, $\wedge$,
$\vee$ and $\rightarrow$. Then again, if you don't need disjunction
as a primitive symbol either, then you can also remove schemata 8
and 9. Now, you have a set of 8 axioms for the connectives $\neg$,
$\wedge$ and $\rightarrow$. Further more, if you don't need conjunction
as a primitive symbol either, then you can also remove schemata 5-7.
Now, you have a set of 4 axioms for the connectives $\neg$ and $\rightarrow$.
This is as far as you can go, with this system.
A second crucial point, is that, as far as modus ponens is the only
rule of inference, substitutions of composite parts of a formula, which contain logical operations, with other formulas, are not allowed in a completely formal proof. Since abbreviations
are informal anyway, most writers feel free at any time to replace
a formal expression with the abbreviation, or vice-versa. But, in
a strictly formal proof, every formula is either an axiom (or an axiom schema, in propotitional calculus), a hypothesis, a repetition of a previews formula or it follows from two previews formulas by application of modus ponens. Thus, unless you have substitution as an inference rule, you can not use it.
However, in order to make things simpler, one could prove the substitution theorem: Let $p_{1},p_{2},...,p_{n},q,r$ be statements and let $w$ be built
up out of some or all of the previews, occurring one or more times,
by means of the primitive connectives only. Let $v$ be the result
of replacing some or all occurrences of $q$ in $w$ by $r$. Then,
substitution theorem states that:
$$\text{If }\vdash w\text{ and} \vdash q\leftrightarrow r\text{, then }\vdash v\text{.}$$
It will take to much space to include the whole analysis here, but the proof of substitution theorem is such that allows you, by following specific rules, to transform a proof which uses the substitution theorem, to a completely strict formal proof, where every formula is either an axiom, a hypothesis, a repetition of a previews formula or follows from two previews formulas by application of modus ponens. Then, one feels confident to write shorter proofs with the help of substitution theorem, knowing that he can get the strict form in any time, with an algorithmic procedure.

"I understand how they would define an ⟺ operator, but wouldn't you always be able to substitute (ϕ⟹χ)∧(χ⟹ϕ) for (ϕ⟺χ) in any proof?"
No.  In intuitionistic logic, such substitution is not necessarily allowed, since defining (ϕ⟺χ) in terms of $\land$ and ⟹ is not possible necessarily.  Definiability of distinct connectives is not allowed in at least some intuitionistic logical systems.
Also, using definable connectives is sometimes disallowed in proofs.  Definition-free proofs would disallow using ((ϕ⟹χ)∧(χ⟹ϕ)) for (ϕ⟺χ) in any proof.


What is the fractal dimension of a cauliflower? I tried to calculate it's dimension using mass. Here's how I did,
I took a small cauliflower and measured it's mass = 105g, then I divided it into 12 similar looking branches and calculated the mass of one of the branch (scaled down version) and got 15g, Then I found 15/105 = 1/7. Then since I scaled it down by 1/12 so it put the number in the log formula which is log base 12 of 7 and got 0.78.
So the fractal dimension of cauliflower = 0.78 however I think this is wrong cause on wiki, fractal dimension of cauliflower is given as 2.88..And also for a cauliflower to be a fractal it's dimension should exceed it's topological dimension of 3 which is clearly not the case here. Where am I wrong? Please correct me!

The version of dimension that you seem to be looking to compute is the similarity dimension.  Roughly speaking, if a set $F$ is the (disjoint) union of $n$ sets, each of which is congruent to the original set scaled down by some factor $r$, then the similarity dimension is given by
$$ \dim_s(F) = -\frac{\log(n)}{\log(r)} \tag{1}$$
(note that this will be positive, since we are assuming that $r<1$ so that we are scaling down).  This comes from the idea that the "volume" (or measure) of $F$ is given by
$$ \operatorname{vol}(F) = n \operatorname{vol}(rF) = nr^d \operatorname{vol}(F)
\implies \log(\operatorname{vol}(F)) = \log(n) + d \log(r) + \log(\operatorname{vol}(F)),$$
which we can solve for $d$ in order to get the formula above.  The point is that $d$ is the magic exponent that tells us how to scale the radius of a set, and $n$ is like some kind of "normalizing constant".  If this doesn't make sense, don't worry about it---the key result is the formula at (1).

More generally, if $F$ is made up of self-similar pieces, not all of which are the same size, then the similarity dimension is the unique real solution $d$ to the Moran equation:
$$ 1 = \sum_{j=1}^{J} r_j^d, $$
where $F$ is the disjoint union of $J$ copies of itself, scaled by the factors $r_j$.

Now, this is an abstract mathematical definition of dimension, built to deal with ideal mathematical objects, which don't really exist anywhere in the world.  When you try to compute the dimension of a cauliflower, there are a lot of variables that are going to make the computation more difficult to deal with.  When you break the plant into 12 similar pieces, how accurate is your measurement of the scaling factor between the original plant and the smaller pieces?  Is the mass really a good proxy for volume?  Or do you want to measure the dimension with respect to mass, which is a different kind of measure (though, perhaps, related---especially if the density is constant; is it?)?  Do you really need to compute the mass/volume at all?  Does it matter that the cauliflower does not take a consistent dimension (for example, I would imagine that the stalk is genuinely three dimensional, while the florets are lower dimensional)?  These kinds of experimental errors and physical phenomena can make it hard verify mathematical results empirically.

That being said, there does seem to be an error in your computation.  You said that you divided the cauliflower into 12 pieces (are they all about the same size?), then weighed them.  But the formula deals with the scaling ratio, not the original or resulting volume.  If the original cauliflower was 6 inches tall, and each of your 12 pieces is 2 inches tall, then the scaling ratio is $\frac{1}{3}$ (not $\frac{1}{7}$, which is what you got by measuring the weight).  That is, you need to compare the lengths, not the volumes.

It depends on what you are looking for in the fractal dimension...

Let me explain. Suppose you are looking for the fractal dimension of a cube (since cubes are nice and simple). Then the length of the cube gets halved 2 and the mass gets divided by 8 (scaled down by 8). Hence the fractal dimension of the volume of the cube is log$_2$8=3. However, the fractal dimension of the surface area is different. Instead, we scale down to half the size but the surface area (of a single cube) reduces to a quarter of the large cube (scaled down by 4). Hence the fractal dimension if log$_2$4=2, as expected.

Now onto your question. The number of shapes you have divided into is 12 (so it has been scaled down by 12). Assuming that the shape is indeed similar, we have (length$^3$ $\propto$ mass) so the length is scaled down by roughly the cube root of 7 $\approx$ 1.912. log$_{1.912}$12 is roughly 3.833.

EDIT: In retrospect, 12 similar-looking branches should mean that each one has a mass of 105/12 grams. This is bad. I think, instead of mass, you measure the length of each one compared to the length of the original. Also, looking it up online yields a fractal dimension of 2. The thing is, the topological dimension of a 3d object is not 3, but rather 2.


Find $\lim\limits_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$ $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$$

I'm surprisingly struggling with this limit, could you give me a hint how to handle it (no L'Hospital and no prior knowledge what the limit value is)?

Generally what are some basic methods to handle $\infty \cdot 0$?

Hint:

Set $\dfrac1{\sqrt n}=2h,n=\dfrac1{4h^2}$

as $n\to\infty, h\to0^+$

and use  $$\cos2h=1-2\sin^2h$$

Enforcing the change of variables  $t^2 =1/n $ then   $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)= - \lim_{t \rightarrow 0}\frac{1-\cos(t) }{t^2}=-\frac12 .$$

This last limit is classical for trigonometric function


nCr and nPr: Order of Operations Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

Where would a counting concept like nCr and nPr fall into this mix?

There is no strong convention for these notations.

In most cases the $C$ and $P$ notations are self-delimiting, because the arguments are typeset as subscripts or superscripts. They can be written as ${}_nC_r$ or ${}^nC_r$ or $C^n_r$, where in each case it is unambiguous that the argument expression is whatever is written in smaller type -- or as $C(n,r)$ where the parentheses and comma make the structure explicit.

There are a few cases where ambiguity does arise, such that if you write the functions as ${}_nC_r$ and want to multiply two of them. In that case it is strongly advisable to use parentheses to disambiguate rather than try to rely on a convention: $({}_aC_b)({}_dC_e)$.

In any case, in more complex situations it is often preferred to use binomial-coefficient notation $\binom nr$ instead of ${}_nC_r$, and that is completely self-delimiting. The counts that are written ${}_nP_r$ in some elementary combinatorics texts appear to be rare enough in advanced material that it is generally feasible to write $\frac{n!}{(n-r)!}$ for them.

$_nC_r$ and $_nP_r$ are more like counting functions that map $\mathbb N^2\to \mathbb N$. What this means is they aren't operations more or less. they have an equation related to them $P(n,r):= \frac{n!}{(n-r)!}$ and $C(n,r):= \frac{n!}{r!(n-r)!}$.

So in a sense they use multiplication, since $n!= 1*2*3*\dots*(n-1)*n$ . However algebraically, there is no relationship to the order of operations.


Need to check if a t- test should be used instead of a z -test here! For this question shouldn't they be using a t test and the test statistic should be t and not z as the sample is small? Is this a mistake in the mark scheme?

Deciding between a z-test and a t-test has to do with whether or not the population standard deviation is known.  In this case, you are told that the standard deviation is .8.  When performing a t-test, the population standard deviation is unknown and thus must be estimated with the data via the sample standard deviation.

Since it is given that the distribution is normal and the standard deviation is $0.8$ mg, then the $z$-test is correct.  If the location were unknown and were being inferred from the data, a $t$-test would be correct.

A $t$-test incorporates the simultaneous uncertainty in both the mean and standard deviation of a distribution.  A fluctuation in the mean can be caused by a sample with the correct mean, but an unlucky set of deviations.  Thus, the fluctiation in the sample conflates both the location and variance of the distribution.  Since we absolutely know the mean, there is no possibility of conflation.

Edited for cross-wiring between fingers and head:  "mean" -> "s.d.".


If $\sum A_n$ converges, does $\sum A_n x^n$ converge uniformly on $[0,1]$? Question: If $\sum_{n=0}^\infty A_n$ converges, does the series $\sum A_n x^n$ converge uniformly on $[0,1]$?

Pointwise convergence is easy enough to see, and intuitively I think the series should converge uniformly as well, but I'm having trouble showing that in proof. Indeed, if we fix $\varepsilon > 0$ and procure a $N$ by the convergence of $\sum A_n$, it is not universally true that $|\sum_{n=q}^p A_n x^n| \leq |\sum_{n=q}^p A_n| < \varepsilon$ whenever $p \geq q \geq N$. I also tried to use the statement $|\sum_N^\infty A_n x^n| < |\sum_N^\infty x^n|$ past some point where $|A_n| < 1$, but this only gives pointwise convergence.

Perhaps we can break up the interval in $[0, \xi]$ and $[\xi, 1]$, the first interval converging uniformly using the geometric property and the second somehow else?

The answer is yes. We use partial summation (used to prove Abel theorem) to get an estimate of Cauchy sum.

Let $B_n=\sum_{k=m}^n A_k$. So by Cauchy Criterion, for any $\epsilon>0$, there is a $N$ such that
$$
|B_n|<\epsilon \quad\text{whenever }\quad n,m>N\tag1
$$
We have
\begin{align}
\sum_{k=m}^n A_kx^k&=\sum_{k=m}^n (B_k-B_{k-1})x^k
\\
&=\sum_{k=m}^n B_kx^k -\sum_{k=m}^n B_{k-1}x^k
\\
&=\sum_{k=m}^{n-1} B_k(x^k-x^{k+1})+B_nx^n\tag{2}
\end{align}
Note: $B_{m−1}=0$.

Since $\lim_{n\to\infty}x^n$ exists and $\:x^n \downarrow$ on $[0,1]$,  for all $k>0$ and $x\in[0,1]$ we have
$$
x^k-x^{k+1}\geqslant0\:
$$
Since for all $k>m$, $-\epsilon<B_k<\epsilon$
$$
|B_k(x^k-x^{k+1})|<\epsilon(x^k-x^{k+1})\tag3
$$
So for all $n,m>N-1$ and $x\in[0,1]$, by $(1)$, $(2)$ and $(3)$ there is
\begin{align}
\left|\sum_{k=m}^n A_kx^k\right|&\leqslant\sum_{k=m}^{n-1} |B_k(x^k-x^{k+1})|+|B_nx^n|
\\
&\leqslant\sum_{k=m}^{n-1} \epsilon\:(x^k-x^{k+1})+\epsilon \:x^n
\\
&=\epsilon \:(x^m-x^n+x^n)
\\
&=\epsilon \:x^m
\\
&\leqslant \epsilon
\end{align}
So by Cauchy Criterion, $\sum_{k=1}^{\infty} A_kx^k$ converges uniformly on $[0,1]$.

We may use Abel's theorem itself here.

$ \sum A_n $ is uniformly convergent (as it is free from 'x').

And sequence {$ {x^n} $} is monotonically decreasing and bounded in (0,1).

Therefore, by Abel's theorem, $\sum A_nx^n$ is uniformly convergent.


Two restaurants competing in the same region On a square area $[0,a]\times[0,b]$ exists a continuous function $f(x,y)$. For any $2l$ points in this area and we divide them into two sets, $K=\{k_{1},k_{2},\cdots,k_{l}\},M=\{m_{1},\cdots,m_{l}\}$.
Now define
$$M(x,y)=\begin{cases}
0&\text{$d((x,y),M)>d((x,y),K)$}\\
f(x,y)&\text{$d((x,y),M)<d((x,y),K)$}\\
\displaystyle\frac{1}{2}f(x,y)&\text{$d((x,y),M)=d((x,y),K)$}
\end{cases}$$
If we arrange the set $K$ first, then $M$, can we guarantee that
$$\iint M(x,y)\text{d}\sigma\geq\frac{1}{2}\iint f(x,y)\text{d}\sigma$$
Or in a vivid way, we may regard $f(x,y)$ as the density of customers. The points are restaurants. Where to have dinner is only determined by the distance. May the initiative restaurant dominate this area or not?

Yes, one can choose $M$ such that the inequality is true.
For a given set $K=\{k_1,\dots,k_l\}$ we choose
$m_i:=k_i$.
Then, according to the definition of $M(x,y)$, we have
$M(x,y)=\frac12 f(x,y)$ for all $(x,y)$.
This implies the inequality (in this situation, it is even an equality).


Taylor series for $x\sin(x^2)$, looking for $a_n$ I had to find the coefficient $a_n \in \mathbb{R}$ so that $\sum\limits_{n=0}^\infty a_nx^n=x\sin(x^2)$,$\forall x \in \mathbb{R}$

Hmm I thought since:

$\sin(x)= \sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

$x\sin(x^2)=x\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{2(2n+1)}}{(2n+1)!}=x\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{4n+2}}{(2n+1)!}=\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}(x)^{4n+3}$

But can I get this in term of only $x^n$? Or is this the solution?

I took some derivatives before and figured that indeed for $x_0=0$ as our center the first non-zero derivative is the 3rd.

So I dont see how we could express our function as a series with $x^n$ instead of $x^{4n+3}$

Could someone tell me if the solution for $a_n$ is indeed $\frac{(-1)^n}{(2n+1)!}$

As you found, for every $x \in \Bbb{R}$,
\begin{align}
x \sin(x^2) &= \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}x^{4n+3}
\end{align}
This can of course be written as $\sum_{k=0}^{\infty}a_k x^k$ for specifically chosen coefficients $a_k$, but I doubt doing so explicitly is very illuminating. Anyway, here it is: define for each integer $k \geq 0$,
\begin{align}
a_k := \begin{cases}
0 & \text{if $k \not\equiv 3 \mod 4$} \\\\
\dfrac{(-1)^{\frac{k-3}{4}}}{\left(\frac{k-3}{2} + 1 \right)!}&\text{if $k \equiv 3 \mod 4$}
\end{cases}
\end{align}
Then, $x \sin(x^2) = \sum\limits_{k=0}^{\infty}a_k x^k$. All I did was to do a case by case definition of what $a_k$ is, and in the second part, I simply set $k = 4n+3$, and solved for $n$, then plugged it into $\dfrac{(-1)^n}{(2n+1)!}$. So, again, while this is a complete expression for the various coefficients, I'm not sure how helpful you'll find it.


Does $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{} k+1}}$ converge? Does the following series converge? $$\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{|} k+1}}$$

I tried using the ratio test and the comparison test but I wasn't able to solve this.

I think I should try manipulating the denominator to use comparison test but I can't figure out how?

Note that $\frac{1}{k\sqrt{k+1}} < \frac{1}{k \sqrt{k}} = k^{-1.5}$.

Hence, by the comparison test, $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}} < \displaystyle\sum_{k=1}^n k^{-1.5} < \infty$ for all $n$. Hence $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}}$ converges, and in fact it is somewhere  close to $2.04288$ by Wolfram Alpha.

Using the fact that square root of x+1 > square root of (x), for x>0 and p test, it converges


To what value will the series converge? I have done a Fourier series expansion and get $$\frac{12}{\pi(2n-1)}\sin((2n-1)x)$$
How to find the value it converges at $x=\frac{\pi}{2}$?
isn't it divergent?
Please show me the correct way step by step

Thanks


Excision property false when set is not measurable. From Royden & Fitzpatrick's Real Analysis, Problem 19 section 2.4:

"Let E have finite outer measure. Show that if E is not measurable, then there is an open set O containing E that has finite outer measure and for which
m*(O/E) > m*(O) - m*(E)."

My process goes like this:

1) m(E) is finite => E is bounded => There exists open set O containing E such that for all e1>0, m(O)-m(E)

2) E not measurable => there exists some e2>0 such that there exists an open set Q containing E and m(Q/E)>e2.

If I could somehow show that for the same e>0, Q=O then I would have the property immediately. But I cannot see any way that I can justify equating O with Q, and I do not see any other angle of approaching the problem, anyone can offer any hints or answers, I'd much appreciate it.

The contrary to say $E$ is measurable if for a set $A$
$$m^\*(A)\geq m^\*(A\cup E)+ m^\*(A\cup E^c)$$
is that for every set we have
$$m^\*(A)< m^\*(A\cup E)+ m^\*(A\cup E^c)$$
In specific for a open set $O$ covering $E$, with $O\cup E=E$, we must have
$m^\*(O)< m^\*(O\cup E)+ m^\*(O\cup E^c)     .....$
$m^\*(O)< m^\*(E)+ m^\*(O~E) ....$
$m^\*(O)-m^\*(E) <  m^\*(O\sim E)$


Joint distribution of $(X,\min(X,Y))$ for $X$ and $Y$ i.i.d. uniform on $(0,1)$ Take two independent random variables $X_1, X_2$, identically uniformly distributed over $[0,1]$. Now take $U= \min (X_1, X_2).$ There's a standard procedure to evaluate the distribution of $U$: $p_U(u) = 2(1-u).$ What I would like to find is the joint distribution of $X_1$ and $U$. It's probably a standard question, and reference is fine for me: I just couldn't find it by myself. What I tryed, is something informal: $P(X_1 = x, U = u) = P(X_1 = x | U = u)2(1-u)$. But this is something strange, since, from simmetry, I would say that $P(X_1 = u | U = u) = 0.5,$ and so my guess for $P(X_1 = x | U = u)$ is $\frac{1}{2} 1_{\{u\}}(x) + \frac{1}{2(1-u)} 1_{(u,1]}(x),$ which looks somehow wrong.

A measure-theoretic solution:

Let $F$ denote the distribution function of $X$. Intuition gives that the conditional probability of the event $[U > u]$ given the $\sigma$-field $\sigma(X)$ is given by:
$$P(U > u | X)(\omega) = I_{[X > u]}(\omega)(1 - F(u)) \tag{$*$}.$$
(A rigorous proof for $(*)$ will be given at the end of this answer.)
Intuitively, for fixed $u$, given the observation of $X(\omega)$, if $X(\omega) \leq u$, there is no hope that $U(\omega)$, which is at most $X(\omega)$, can exceed $u$, this explains the existence of the first term of $(*)$; if $X(\omega) > u$, then $U(\omega) > u$ if and only if $Y(\omega) > u$, by independence of $X$ and $Y$, this gives $1 - F(u)$.

Now for any $x \in \mathbb{R}^1$, since $[X \leq x] \in \sigma(X)$, it follows by $(*)$ that
\begin{align}
& P(U > u, X \leq x) \\
= & \int_{X \leq x} P(U > u|X)(\omega) dP \qquad \text{by the definition of conditional probability} \\
= & \int_{X \leq x} (1 - F(u))I_{[X > u]}(\omega) dP \qquad \text{by $(*)$} \\
= & \int_{(-\infty, x]}(1 - F(u))I_{(u, +\infty)}(v) dF(v) \qquad \text{change of variable formula}\\
= & \begin{cases}
0 & \text{if $x \leq u$}; \\
(1 - F(u))(F(x) - F(u)) & \text{if $x > u$}.
\end{cases}
\end{align}
Together with that $P(X \leq x) = F(x)$, it follows that
\begin{align}
& P(U \leq u, X \leq x) \\
= & P(X \leq x) - P(U > u, X \leq x) \\
= & \begin{cases}
F(x) & \text{if $x \leq u$}; \\
F(x) - (1 - F(u))(F(x) - F(u)) & \text{if $x > u$}.
\end{cases} \tag{$**$}
\end{align}

Proof of $(*)$: Clearly, the right hand side of $(*)$ is $\sigma(X)$-measurable, in view of $\{X > u\} \in \sigma(X)$. In addition, for every $H \in \mathscr{R}^1$, by the change of variable formula,
\begin{align*}
& \int_{X \in H} I_{[X > u]}(\omega)(1 - F(u))dP = (1 - F(u))\int_H I_{(u, +\infty)}(x)dF(x) = P(Y > u)P([X \in H] \cap [X > u]) \\
= & P([Y > u] \cap [X > u] \cap [X \in H]) = P([U > u] \cap [X \in H]),
\end{align*}
where we used $X$ and $Y$ are independent in the first equal sign of the second line above. Since $[X \in H]$ is the general element of $\sigma(X)$, this proves $(*)$.

Some poster was curious about why I didn't report the joint density (i.e., pdf) instead of the joint cdf. Well, the fact is: this is an example that a random vector doesn't have joint density!

Let $A = [(x, u): x = u]$ and let $\mu$ be the probability measure on $\mathbb{R}^2$ induced by $(X, U)$. It can be seen that
$$\mu(A) = P(X = U) = P(X \leq Y) = \frac{1}{2}.$$
Therefore there is a positive probability mass concentrated on a set which has Lebesgue measure $0$, so it is impossible for $(X, U)$ to have a joint density $f$ with respect to the planar Lebesgue measure $\lambda_2$ (otherwise, $\mu(A) = \iint_A f(x, u) dx du = 0$, contradiction!). Therefore the most clear way to describe the joint distribution of $(X, U)$ is still through (**).

Conditioning on sets of measure zero makes symmetry arguments misleading.
Bulletproof approach deals with CDF:

$$P(X_1> x,U> u)=P(X_1>\max(x,u))P(X_2>u)=(1-\max(x,u))(1-u)$$
Differentiating we get:
$$f_{X_1,U}(x,u)=\frac {\partial^2P}{\partial x\partial u}=\frac{\partial}{\partial u}(-I(x>u)(1-u))=I(x>u)+(1-u)\delta(x=u)$$


Linear rotation spanning 4d Let $M$ be a real $4 \times 4$ matrix  Let $x$ be the time-dependent 4-dimensional vector evolving according to the linear ODE

$$ \frac{ d x }{dt} = M x $$

Can you give an example of a matrix $M$ such that the trajectory will span all 4 dimensions for any initial condition?

In other words, find an $M$ such that starting from any vector $x_0$, the subsequent trajectory will not be confined in a plane, or even in a certain 3d space, but will evolve periodically in the 4d space.

I tried to answer this by starting from the fact that any two eigenvectors can be non-orthogonal. But how the pairs of eigenvectors corresponding to imaginary eigenvalues? If the 4d matrix has two pairs of imaginary eigenvalues, won't the two cyclic behavior of the ODE be uncoupled? And if so, we can always choose an initial condition orthogonal to one plane, so the system will rotate exclusively in the other plane?

But then finding an M as described above is impossible. Please enlighten me!

Let $Mu=\lambda u$. So, $x(t) = e^{\lambda t} u$ is a solution to $\dot{x} = Mx$. This means if you start on $u$, you will stay on $u$ forever. If $\lambda$ is real, then $x(0) = u$ selection can only span a 1d space. If it is complex then $x(0) = \alpha \operatorname{Re}u + \beta \operatorname{Im}u$ selection can only span a 2d space for any real $\alpha, \beta$.

In summary, there exists an initial condition where the solutions can span at most 2d space.


If $T$ is a bounded linear map and $\sum x_n$ is an absolutely convergent series, then $T(\sum x_n) = \sum T(x_n)$ Is the following true? If so, how to prove it?

If $T:X \to Y$ is a bounded linear map between the Banach space $X$ and the normed vector space $Y$ and $\sum x_n$ is an absolutely convergent series, then $T(\sum x_n) = \sum T(x_n)$.

I need this claim to finish a proof that if $X$ is a Banach space and $M$ is a proper closed subspace, them $X/M$ is a Banach space, but I am not able to show it.
Thanks in advance and kind regards.

On a Banach space $X$ a linear operator $F$ is continuous if and only if it's bounded.
For your case, if it is bounded you have that $\exists L$ such that for any $x\in X$ $\Vert F x\Vert \leq L \Vert x \Vert$.
Hence you have that if $x_n\to x$, fixed any $\epsilon$, for $n$ large enough $\Vert x_n-x\Vert\leq \epsilon$.
Then $\Vert Fx_n - F x\Vert = \Vert F(x-x_n)\Vert \leq L\Vert x-x_n\Vert \leq L\epsilon$.
Now an absolutely convergent series is a converging sequence (of the partial sums) so your statement follows.

You can find detailed proofs of both facts as Lemma 4.4 and Theorem 4.5 in:
http://www.pitt.edu/~hajlasz/Notatki/Functional%20Analysis2.pdf


Faithfullness of Weyl group action Let $I$ be a finite indexing set, and $A \in \operatorname{Mat}_I(\mathbb{Z})$ be a generalised Cartan matrix, i.e.

$a_{ii} = 2$,
$a_{ij} \leq 0$ for $i \neq j$, and
$a_{ij} = 0 \iff a_{ji} = 0$.

Then there is an abstract Weyl group $W_A$ associated to the matrix $A$, defined on the generators $\{s_i \mid i \in I\}$ subject only to the following relations:

$s_i^2 = 1$ for all $i \in I$, and
$(s_i s_j)^{m_{ij}} = 1$ for $i \neq j$, where $m_{ij} = 2, 3, 4, 6$ or $\infty$ depending on whether $a_{ij} a_{ij} = 0, 1, 2, 3$ or $\geq 4$ respectively. (If $m_{ij} = \infty$, we mean that $s_i, s_j$ generate an infinite dihedral group).

A based root datum of type $A$ is the data $R = (X, X^\vee, \langle - , - \rangle, (\alpha_i)_{i \in I}, (\alpha_i^\vee)_{i \in I})$ of:

Two $\mathbb{Z}$-modules $X, X^\vee$ of free finite rank in a perfect pairing $\langle -, - \rangle \colon X \times X^\vee \to \mathbb{Z}$, and
Choices of elements $\alpha_i \in X$ and $\alpha_i^\vee \in X^\vee$ such that $\langle \alpha_j, \alpha_i^\vee \rangle = a_{ij}$ for all $i, j \in I$.

Any time we have such data, we may define the simple reflections
$$ r_i \colon X \to X, \quad r_i(\lambda) = \lambda - \langle \lambda, \alpha_i^\vee \rangle \alpha_i, $$
$$ r_i^\vee \colon X^\vee \to X^\vee, \quad r_i^\vee(\mu) = \mu - \langle \alpha_i, \mu \rangle \alpha_i^\vee $$
and with a little work it is possible to show that the map $W_I \to \operatorname{GL}_{\mathbb{Z}}(X)$ defined by $s_i \mapsto r_i$ is in fact a group homomorphism, and similarly for $W_A \to \operatorname{GL}_{\mathbb{Z}}(X^\vee)$. Note that the perfect pairing is $W_A$-invariant, meaning that $\langle r_i \lambda, r_i^\vee \mu \rangle = \langle \lambda, \mu \rangle$, and so the $W_A$ action on $X$ is faithful iff the action on $X^\vee$ is faithful.

Question: What are necessary and sufficient conditions on the based root datum $R$ such that the abstract Weyl group acts faithfully on $X$ and $X^\vee$?

A sufficient condition is that both the sets $\{\alpha_i\} \subseteq X$ and $\{\alpha_i^\vee\} \subseteq X^\vee$ are linearly independent. Then one may do some chamber geometry on $X \otimes \mathbb{R}$ to show that $\langle r_i \mid i \in I \rangle$ is an abstract Coxeter group, as is done in Kac' book Infinite dimensional Lie algebras. However, this condition is not necessary, as one can see by taking the affine $\mathfrak{sl}_2$ Cartan matrix
$$\begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix}$$
and taking a root datum where $X \cong \mathbb{Z}^2 \cong X^\vee$ and let $\alpha_1^\vee, \alpha_2^\vee$ be a basis of $X^\vee$, forcing $\alpha_1 + \alpha_2 = 0$ to be linearly dependent roots. It is not hard to see that $r_1$ and $r_2$ then have no relation, so the action here is faithful.
On the other hand, we could also take a rank-one realisation of that same Cartan matrix by choosing any $\alpha_1 \in \mathbb{Z}$ and $\alpha_1^\vee \in \mathbb{Z}$ satisfying $\langle \alpha_1, \alpha_1^\vee \rangle = 2$ and then setting $\alpha_2 = -\alpha_1$ and $\alpha_2^\vee = - \alpha_1^\vee$. In this case, both $r_1$ and $r_2$ are just multiplication by $-1$, and so we have the relation $r_1 r_2 = 1$.

The progress I made on this shortly afterwards is that a sufficient condition is for either the simple roots or coroots to be linearly independent. Assuming the simple roots are linearly independent, one finds during the proof of (rirj)mij=1 that mij is the smallest integer for which this holds, meaning that the Coxeter matrix coincides with the matrix [ord(rirj)]ij. Together with the fact that the ri satisfy the exchange condition (3.11 in Kac), the ri define the same Coxeter group as the si. – Joppy


Introduction to Proof via Linear Algebra Many universities offer a transition course from computational courses like Calculus to proof-oriented courses like Abstract Algebra. Such courses often go by a name like "Introduction to Proof" or "Transition to Higher Mathematics". They typically contain an introduction to first-order logic (conditionals, conjunctions, negations, quantifiers, etc.) as well as various methods of proof (contradiction, induction, etc.).

I'm hoping to find a text for a first course in linear algebra that fills the role of a "transition course" by deliberately incorporating first-order logic and proof techniques as part of the instruction.

The text should be accessible to students with two semesters of Calculus (roughly the basics of single-variable differentiation, integration, and infinite series). In particular, the overwhelming majority of students will have never written a formal proof and will have extremely limited exposure to logic and set theory.

Ideally, the author would discuss these topics just as they are needed in the treatment of linear algebra (as opposed to supposing the reader is familiar with them already). For example, the author might have a digression on proof by contradiction just prior to using it in some proof about linear independence.

Less ideal (but still acceptable) would be a text that at the very least makes use of all the relevant ideas from first-order logic and proof techniques that one expects from a transition course. Hopefully, the progression of such a text would be such that the instructor could use a supplemental text to discuss, say, proof by contradiction just as it is about to make its first appearance in the text.

Linear Algebra - As an Introduction to Abstract Mathematics (direct link to pdf here) seems like a good shout. I haven't looked through it thoroughly, but the following bold claim is made in the intro:

  In the setting of Linear Algebra, you will be introduced to abstraction. As the theory
  of Linear Algebra is developed, you will learn how to make and use definitions and how
  to write proofs.

Indeed, at the end of each chapter are special "Proof-Writing Exercises".

My favourite book in linear algebra, as the first course, would be Schaum's Outline of Linear Algebra by Seymor Lipschutz.

Pros: Starts with an intense use of algorithms: vector, matrices, elimination, linear equations, etc. This comprehends the first three chapters. The rest of the book is the basic material for a first time course in linear algebra. It starts with axiomatic definition of vector spaces, then linear transformations, then connections between linear transformations and matrices, then inner products, diagonalization, Jordan's canonical form, etc. It has a plethora of examples.

Cons: It is highly devoted to algorithm, so mathematicians don't like that a much (This methods, though, are essential: imagine a mathematician incapable of solving a $3 \times 3$ ODE or being unable to diagonalize a small size matrix). Proofs are delayed to the end of each chapter, so the theorems are stated, then there are plenty of examples and then, at the end, the proofs.


About Differential Equation. How can I prove $y(x)\neq 0$? This is a familiar differential equation.
\begin{cases}
\dfrac{dy}{dx}=y \\
y(0)=1 \\
\end{cases}
When we solve this equation, we will do the following operation.
From $\dfrac{dy}{dx}=y,$ we get $\dfrac{1}{y} \dfrac{dy}{dx}=1.$
But when we do this, we have to mention that $y(x)\neq 0$ for all $x$.
I wonder how I can prove $y(x)\neq 0$ for all $x$.
This is my attempt.
Suppose there exists $x_1$ such that $y(x_1)=0$.
Because $y(0)=1,$ $x_1\neq 0$.
And we can say that $\dfrac{dy}{dx}(x_1)=y(x_1)=0.$
I couldn't proceed from this.
I think that I have to use "the uniqueness of solution of differential equations". But I don't know where I can use this.
I would like you to give me some ideas.

If you allow knowledge of the exponential function, $e^x$ then you can use an "integrating factor".  You have the equation $\displaystyle \frac{dy}{dx} = y$.  Write it as
$$
\frac{dy}{dx}-y = 0
$$
and multiply by $e^{-x}$ to get,
$$
e^{-x} \frac{dy}{dx} - e^{-x} y = 0.
$$
Apply the product rule to see that the left side is now a differential, and,
$$
\frac{d}{dx}\Big( e^{-x}y \Big) = 0.
$$
You only need now to know that the derivative of a function is zero if and only if the function is constant, and the constant is set from your initial condition.  Thus $y = e^x$ which we know to be non-zero.
Alternatively, if you want to avoid using knowledge of the exponential function but you do know that such differential equations have a unique solution given any initial condition $y(x_0)$ at some $x=x_0$ you can argue that if $y(x)$ is a solution that ever equals zero at a point, $x_0$ say, it must be zero for all $x$ because the zero function is the unique solution satisfying the given condition. Therefore a solution cannot also satisfy $y(0)=1$.

as you said, if:
$$\frac{dy}{dx}=y\Rightarrow \int\frac1y\frac{dy}{dx}dx=\int dx\Rightarrow \ln|y|=x+C_1$$
$$y=C_2e^x$$
and so for $y=0$ at any point $x$ would require $C_2=0$ but $C_2=e^{C_1}$ and so: $C_1\to-\infty$. All of this gives us that if $|x|<\infty,y\ne0$ or it will contradict $y(0)=1$


Computing cyclic cubic splines Cubic splines (piecewise interpolator with $C_2$ continuity) are well-known to be computable from a tridiagonal system of equations that give estimates of the second derivatives at the interpolated points.
Different endpoint conditions are allowed, such as natural or constrained. The resolution is easy and fast.
The cyclic case (equal derivatives at endpoints) is more difficult, as the matrix is cyclic-tridiagonal (nonzero coefficients in the corners). Despite Web search, I didn't find a clear explanation of the modified resolution in this case.
Can you help ?

I haven't looked at that case specifically, but Wilf's "generatingfunctionology" (available for free at the link) gives a delightful treatment of cubic splines using generating functions to solve the equations. Worth a read.


Formula for a geometric series Helping my brother with his math homework and I am hung up on this one.

Find a formula for $a_n$ for the geometric sequence:
$a_1=2$, $a_{k+1}=-3a_k$

If anyone here can help that would be great. Thanks. My thought would be that I just need to turn the formula above into $a_k = ____ a_{k-1}$

I think you are searching for $a_n = 2\cdot(-3)^n$


How to deal with three equations. We had a lesson in chemistry when we learned this equation  $R=K(A)^n*(B)^m$.
They give you a table with the value of R,a and b in three different situations(different value) to get the rest (k,n and m)
I can't find a way to make one equation once for n and m and other one for k so
I can find its value with only a,b and r

for example

   A= B= R=

2          3        216
4          2        256
3          4        1152

so what is n, and m and k
by the equation I will find that
n=2
m=3
k=2

So you are given that
$$\begin{align} K\cdot 2^n\cdot 3^m&=\hphantom{0}216=2^33^3\\
K\cdot 4^n\cdot2^m&=\hphantom{0}256=2^8\\
K\cdot 3^n\cdot4^m&=1152=2^73^2\end{align}$$
Assuming for simplicity that $K=2^u3^v$ for some $u,v$ and nothing messy happens, we obtain the following equations by considereing the exponents of $2$ and $3$ in the given equations:
$$u+n=3\qquad v+m=3 $$
$$u+2n+m=8\qquad v=0 $$
$$u+2m=7\qquad v+n=2 $$
which readily leads to $v=0,n=2,m=3,u=1$. The latter system of six equations is overdetermined, so we need to check all equations and are somewhat lucky that this simple trick works out (this is not guaranteed with more "random" numbers as input).

(The right method for the general case is taking logarithms and solving three linear equations in three unknowns $n,m,\ln K$, as Yves Daoust suggests)

Take the logarithm and set $k=\ln K$:
$$n\ln2+m\ln3+k=\ln216,\\n\ln4+m\ln2+k=\ln256,\\n\ln3+m\ln4+k=\ln1152.$$
Then by pairwise subtraction
$$n\ln2+m(\ln2-\ln3)=5\ln2-3\ln3,\\n(\ln3-2\ln2)+m\ln2=2\ln3-\ln2.
$$

By inspection (or by Cramer), $n=2$ and $m=3$. $k$ follows.


demostration of interger part integration. I need help for solving this demostration, I  appreciate your suggestions very much.

$$\begin{array}{rclr}
\int ^{n}_{0}[x] dx= \frac{n(n-1)}{2}
\end{array}$$

Pd. If you have any suggestion of a book that deepens on the subject please communicate it.

Hint: $\lfloor x \rfloor = k$ for $x \in [k,k+1)$ for $k = 0,1,2,\ldots,n-1$. Hence
$$\int_0^n \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k \; dx,$$
where the integrand is a constant with respect to $x$.

For any positive integer n,  $[x]=n$ for $n\le x< n+1$
$$\int_0^n[x]dx=0+1+2+3+4+\ldots +(n-1)=\frac{n(n-1)}{2}$$


$ \lim_{n\rightarrow \infty}[a_n-a]=0 $ implies $ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0 $? Suppose that
$$
\lim_{n\rightarrow \infty}[a_n-a]=0
$$
Does this imply
$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0
$$
? Does the answer change if in place of $a$ we have another sequence $b_n$?

Yes. We have that $\lim_{n\to\infty}[a_nb_n]=\lim_{n\to\infty}a_n\lim_{n\to\infty}b_n$ if the limits on the right hand side exist. Hence,
$$
\lim_{n\to\infty}\frac1n[a_n-a]=\lim_{n\to\infty}\frac1n\lim_{n\to\infty}[a_n-a]=0\cdot0=0.
$$
If we replace $a$ with some sequence, the limit might not be $0$ anymore (take, for example, $a_n=1/n$ and instead of $a=0$ take $b_n=n$).

For constant $a$:

$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]
$$

$$
=\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-a]=(0)(0)
$$

For sequence $b_n$:

$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-b_n]
$$

$$
=\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-b_n]=(0)(0)
$$

Remark: Wherever they exist, limits of sequences behave like limits of functions, i.e. We can split of limits by products into products of limits. This is because a sequence is a function whose domain is countable.


Geometry of images of maps $f: \mathbb R \to \mathbb C$? I am having trouble seeing what a continuous map $f: \mathbb R \to \mathbb C$ might look like. If it was linear it would look like a line but it's not clear to me what happens if it's any map.

I started to wonder about this after I was trying to work out what a convolution of two such maps might look like.

Then I ended up thinking about characteristic functions. These of course do look like just a line of the length of the set of which they are the characteristic function of. But are there maps for which the image is not a line but a plane?

  What can images of continuous maps $f: \mathbb R \to \mathbb C$ look
  like?

The Hahn-Mazurkiewicz theorem says (among other things) that a subset of $\mathbb{C}$ is a continuous image of $[0,1]$ iff it is compact, connected, and locally connected.  This includes, for instance, any closed disk, as well as many other more complicated sets that don't look anything like a "curve".  If you consider continuous images of $\mathbb{R}$ rather than $[0,1]$, you get exactly those sets which are path-connected and a union of countably many compact, connected, and locally connected sets (see this answer on MO, for instance).  In particular, for instance, $\mathbb{C}$ is the continuous image of some function $\mathbb{R}\to\mathbb{C}$.

Probably the best-known explicit construction of a curve in the plane whose image has nonempty interior is the Peano curve which is described nicely on Wikipedia.  The key technical tool in its construction is the fact that a uniform limit of continuous functions is continuous; the curve is constructed as the limit of a sequence of polygonal curves that get increasingly "dense" in a square.

Note that all of this is about continuous maps.  Given the tags you chose for the question, it sounds like you might actually be more interested in differentiable maps.  In that case, the story is quite different.  In particular, for instance, if $f:\mathbb{R}\to\mathbb{C}$ is $C^1$ (or more generally, locally Lipschitz), then it is not hard to show the image of $f$ must have Hausdorff dimension $\leq 1$.  In particular, this means it must have measure zero.

Every continuous map $ f: \mathbb R \to \mathbb C $ is a curve in complex plain, in particular a map $f:[a,b] \to \mathbb C$ " $a$ and $b$ are real numbers", is called a path,  which is used in complex integration theory. But about your second question I have to say that no! No such a map has a plain as its image. A simple example of a such map is $$f: [0,2\pi] \to \mathbb C$$                      $$f(t)=e^{it}$$ which its image is a circle.


PDE: Guess non-polynomial solution of $\Delta u=\frac{\delta^2u}{\delta x^2}-\frac{\delta^2u}{\delta y^2}=0$ I need to verify that the function $$u(x,y)=a\cdot ln(x^2 + y^2)$$ is a solution of the two-dimensional Laplace equation $$\Delta u=\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \qquad (1)$$ and later I have to use the result to guess one non-polynomial (non-trivial) solution of the equation $$\Delta u=\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0 \qquad (2)$$ and verify my result.
I did first part of the problem where I calulated first and second derivative of u and inserted it into equation. Later I made a guess that the solution for (2) can be: $$u(x,y)=a\cdot ln(x^2 - y^2) \qquad (3)$$ and I calulcated derivaties and so on. My question is, is my guess correct?

Your guess

$$u(x,y)=a\cdot ln(x^2 - y^2)  $$

does  satisfy the $$\frac {\partial ^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0  $$

Thus your guess is as good as gold.

The function $$u(x,y)=a\cdot ln(x^2 - y^2)$$ is a solution of the equation $(2)$, as you successfully guessed.
$$\Delta u=\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0 \qquad (2)$$

Note that a lot of solutions can be found on the general form :
$$u(x,y)=f(x+y)+g(x-y)$$
with any differentiable functions $f$ and $g$.

For example, taking $\quad \begin{cases}f(x+y)=a\:\ln(x+y) \\g(x-y)=a\:\ln(x-y) \end{cases}\quad $
$$u(x,y)=a\:\ln(x+y)+a\:\ln(x-y)=a\ln((x+y)(x-y))=a\ln(x^2-y^2)$$
is solution of Eq.$(2)$.

But other functions of $(x^2-y^2)$ are not solution of Eq.$(2)$.

Other example, with $f(x+y)=\sin(x+y)$ and $g(x-y)=\sin(x-y)$ we obtain $u(x,y)=2\cos(x)\sin(y)$ which is solution of Eq.$(2)$.

One can found many other solutions with other functions $f$ and $g$ , even with $f\neq g$, including a lot of polynomial solutions in case of polynomial functions $f$ and $g$.


What does $\delta$ mean here? What does $\delta$ mean here? If you help me at least one formula, then I'll try to do the rest.

Here $\delta$ means finite change in the variable in question.


Polynomial interpolation to LP How I can convert next program to LP (linear programming) or to get some direction how to do this (references).
The problem is :

  I want to find a polynomial $P$ of degree at most $d$ that $||P(x)-y||_1$ is minimal.
  Where $x, y$ vectors in $\mathbb{R}^n$ and $P$ is vector of $(p(x_1), \ldots, p(x_n))$

Let $A$ be the Vandermonde Matrix of $x_1,x_2,...$ then if $z$ is the Vector of polynomial coefficients $z=(a_0,a_1,...)^t$ it is $P(x)=Az$.

For $\textrm{min }||Az-y||_2$ the solution is $z=(A^tA)^{-1}A^ty$.

The problem $\textrm{min }||Az-y||_1$ is equivalent to the LP

minimize $\sum \lambda_i$

subject to

$+Az-I\lambda<=+y$

and

$-Az-I\lambda<=-y$

with $\lambda:=(\lambda_1,...)$

In the following change $||Az-y||$ to $||Ax-b||$

Here's some matlab code that does polynomial regression

This is the output from the code:

I suppose that if you want to find $P$, this means that $x$ and $y$ are known vectors. If you also know $d$, than your polynomial becomes $p(x) = \sum_{k=1}^d a_k x^k$. You can formulate your linear programming problem as :

$$\min z = \sum_{i = 1}^n p_i + n_i$$
subject to
$$ p_i - n_i + p(x_i) - y_i, \forall i=1\dots n$$
$$ 0 \le p_i, d_i < \infty$$
$$ -\infty \le a_k < \infty$$

In this formulation,you can take $p_i$ and $n_i$ as the positive and negative parts of the absolute value. If $p(x_i) - y_i>0$, then $p_i >0$ and $n_i = 0$, and conversly.

Hope this helps.


A Time Calculation Problem: 3yrs or 3yrs and 1 day? Suppose, a person A has his birthday on 30 Sept 1994 and another person B has his birthday on 30 Sept 1997, then which of the following statement would be correct and why?

The difference between the ages of the two persons is

Three years correct to the number of Days
Three years and 1 day correct to the number of Days (because of 29th Feb 1996 that comes in between)?

It rather depends what you mean by a year, which is approximately 365.25 days. This article
 gives a huge range of options with more precision and specificity.

Really you should be using SI Units, they are designed to stop problems like this occurring.


Proving two sets of polynomials are equal How could I go about showing that A is equal to B?

$A=\{q\in \mathbb{P}_2 : q(x)\geq 0 \text{ for all }x\in \mathbb{R}\}$,

$B=\{p\in \mathbb{P}_2 : p(x)=a(x−x_0)^2+b, \text{ where } a,b,x_0 \in \mathbb{R}, \text{ and } a\geq 0, b≥0\}$.

Certainly $B \subset A$, as we note that $u^2 \geq 0$ for each $u$, so $(x - x_0)^2 \geq 0$. Multiplying by a positive number keeps it positive, and adding a positive number keeps it positive still.

For $A \subset B$, try writing the quadratic in its "vertex form" $a(x-h)^2 +k$. In fact, show that when $q(x) = ax^2 + bx + c$ has leading coefficient $a \geq 0$, the aforementioned form can be achieved.

For $B$ we must require:

Parabola is concave up.
Apex is at or above $0$.

Given
$$
 p(x) = a(x-x_{0})^{2} +b
$$

1 concavity

The concavity is given by the second derivative
$$
 p''(x) = 2a
$$
The concavity requirement is expressed as
$$
\boxed{
 a \ge 0
}
$$

2 minimum value

The $x$ coordinate of the apex is at $x=x_{0}$.
$$
 p(x_{0}) = b
$$
Therefore the requirement is
$$
 \boxed{
 b \ge 0
}
$$

Conclusion

All the polynomials created with $a\ge 0$, $b\ge0$: $B\subset A$.

The general quadratic form is
$$
 q(x) = \alpha x^{2} + \beta x + \gamma
$$

Recycle the two criteria from above.

1 concavity

$$
 q''(x) = 2 \alpha
$$
Again, the requirement that $q(x)\ge 0$ is
$$
 \boxed{
\alpha \ge 0
}
$$

2 minimum value

The apex is located using
$$
 q'(x) = 2\alpha x - \beta = 0 \qquad \Rightarrow \qquad x_{min} = -\frac{\beta}{2\alpha}
$$
The minimum function value is
$$
 q (x_{min} ) = -\frac{\beta^2}{4 \alpha} + \gamma
$$
For $q (x_{min} ) \ge 0$, we must have
$$
\boxed{
 \gamma \ge \frac{\beta^2}{4 \alpha}
}
$$


If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is? If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is ?
My approach:
The given expression can be written as $(y-1)(y)(y+1)(y+k-2) + 1$ where $y = x + 2$
This gives me $(y)(y^2 - 1)(y+k-2) = n^2 - 1$ something seems similar in LHS and RHS. But not sure how to proceed next? Should I put $(y)(y+k-2) = 1$ ? although it doesn't seem right for some reason also I dont think it leads anywhere.
Any hints/explanations are appreciated!

If you mean that the polynomial $f(x) = (x+1)(x+2)(x+3)(x+k)+1$ is a perfect square, i.e. $\exists g(x) \in \mathbb{Z}[x] | f(x) = g(x)^2$ then I'd try and figure out what $g(x)$ must look like. What is the degree of $f$? What then must the degree of $g$ be? Then from there you could compare the coefficients of $f$ and what $g^2$ must look like, come up with some equations, and then solve for $k$!
I.e. if $ax^4+bx^3+cx^2+dx+e = fx^4+gx^3+hx^2+ix+j$ then $a=f,b=g,c=h,d=i,e=j.$
Hint if you want to know what $k$ is:

 According to my work, $k$ should be 0 or 4.


“Distinguished element” in group theory I have asked that given a short list of triples of a set, a binary operation on the set, and a distinguished element, whether they forms a group. For example, the set is Z, the operation is addition and distinguished element is 1. In this context, what does distinguished element mean ?

I know that the integer set with addition is a group with identity element 0, but if distinguished element means identity element, they don’t form a group.

I've seen these exercises before.  This is what your book is asking you:  given a trio of data as:

a set $G$
an operation $\ast$ on $G$
an element $x_0 \in G$ (your "distinguished element")

does $\ast$ give a group operation on $G$ with identity $x_0$?  Thus, for example, the answer for the data $(\mathbb{Z}, + , 1)$ is "no" but for the data $(\mathbb{Z}, + , 0)$ is "yes."

A relic of abstractionist parlance.  "Distinguished" just means "specific".  So, what they mean is that, in a group, the identity element is a specific element (with the required properties).

Therefore, the answer to your question is: the triple $(\mathbb{Z}, +, 0)$ is a group, while the triple $(\mathbb{Z}, +, 1)$ is not.

Some more background to guard against future difficulties arising from the abstractionist approach--which some will find tangent and ranty--but, in the end, I will re-address your question in terms of what follows.

This misunderstanding (and many others) would not even arise if groups were introduced properly; i.e., as groups of transformations.

Definition:  A transformation of a set $X$ is a 1-to-1 correspondence from $X$ to $X$.

Corollary:  Composition of transformations of $X$ gives transformations of $X$, and this composition is associative.

This Corollary can be proved by direct inspection of composition.

Definition: A nonempty collection of transformations of $X$ is called a group of transformations of $X$ if this collection contains all compositions and inverses of its members.

Examples: i) $X$ = the Cartesian $xy$-coordinate plane, the transformations are the rotations of $X$ about the origin. ii) $X$ is a finite set, and the transformations are all permutations of $X$. iii) $X$ is any set, and the collection of transformations contains just one transformation: the one that maps every $x$ in $X$ to itself.  This transformation is called the identity transformation of $X$ will be denoted by $1_{X}$.

There is only one identity transformation: if there were two, multiply them, and the product equals each of them, hence they are the same.

Theorem:  If $G$ is a group of transformations of a set $X$, then $G$ contains the identity transformation $1_{X}$.

Proof:  If $g$ is a transformation in $G$, then so is $g^{-1}$ (by the latter definition), and therefore so is $g \circ g^{-1} = 1_{X}$.  Done.

If now we define an abstract group (the uninformative formalist way it's usually done: a set, a binary operation, a "distinguished" identity element, etc.), then:

Every abstract group is isomorphic to a group of transformations of some set.

(Why?  If $G_{abs}$ is an abstract group, then each $g \in G_{abs}$ gives us the transformation $x \mapsto gx$ of the set of the elements of $G_{abs}$.)

Therefore, groups of transformations really give us ALL the groups that exist.

For finite groups, the version of this last boldfaced statement is Cayley's theorem.

To address your question in terms of groups of transformations:

Consider the shift of all integers by some constant integer (positive, negative, or zero). Such shifts are transformations of the set $\mathbb{Z}$. Among these, the identity transformation is the one that leaves all the integers in place; i.e., the shift by 0.  The shift by 1 is not the identity transformation.

That's why $(\mathbb{Z},+, 0)$ is a group, and $(\mathbb{Z},+,1)$ is not.


Prove $\neg (p \land q) \vdash \neg p \lor \neg q$ by natural deduction I have tried solving this by negating $\neg p \lor \neg q$ to get to a pbc, but apparently you need to use a few lem's (law of excluded middle) to solve the problem. Where (and how) have I gone wrong in the following deduction?

$\neg(p \land q)$ .............................hyp

$\neg(\neg p \lor \neg q)$ .................hyp

$\neg p$ ........................hyp

¬p ∨ ¬q ....... vIL, $3$
$F$ ...................$\neg$E, $2$, $4$

$\neg q$ ........................hyp

$\neg p \lor \neg q$ .......$\lor$ER, $6$
$F$...................$\neg$E, $2$, $7$

$F$ .................................$\lor$E, $2$, $3-5$, $6-8$

$\neg p \lor \neg q$ .............................pbc $2-9$

Any help would be greatly appreciated!

Fairly close.    Always keep an eye on the prize.   You want to be able to contradict the first premise, $\lnot(p\land q)$, so that requires deriving $p$ and $q$, then introducing a conjunction.

Thus the purpose of those subproof are to be proofs by contradiction too.

$$\def\bot{\mathcal F}\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}
\fitch{~~1.~\lnot(p\land q)}{\fitch{~~2.~\lnot(\lnot p\lor\lnot q)}{\\\fitch{~~3.~\lnot p}{~~4.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 3\\~~5.~\bot\hspace{9ex}\lnot\mathsf E,2,4}\\~~6.~p\hspace{13ex}\mathsf {PBC},3{-}5\\\\\fitch{~~7.~\lnot q}{~~8.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 7\\~~9.~\bot\hspace{9ex}\lnot\mathsf E,2,8}\\10.~q\hspace{13ex}\mathsf {PBC},7{-}9\\\\11.~p\land q\hspace{9ex}\land\mathsf I, 6,10\\12.~\bot\hspace{12ex}\lnot\mathsf E,1,11}\\13.~~\lnot p\lor\lnot q\hspace{9ex}\mathsf {PBC},2{-}12}$$

PS: The PBC steps contain the LEM, in the form of double negation elimination (DNE).
$$\begin{array}{|l}\fitch{~~3.~\lnot p}{~~4.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 3\\~~5.~\bot\hspace{9ex}\lnot\mathsf E,2,4}\\~~6.1.~~\lnot\lnot p\hspace{10ex}\neg\mathsf I, 3{-}5\\~~6.2.~p\hspace{13ex}\lnot\lnot\,\mathsf E,6.1\end{array}$$

The following proof is patterned on the one provided in section 19.6 of the forallx text as a proof of the first De Morgan rule.

Note the use of the law of the excluded middle (LEM) on line 10 which references the two cases for $P$ on lines 2-7 and $¬P$ on lines 8-9.  The links to the proof checker and the textbook are provided below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf


Find the average of $\sin^{100} (x)$ in 5 minutes? I read this quote attributed to VI Arnold.

  "Who can't calculate the average value of the one hundredth power of the sine function within five minutes, doesn't understand mathematics - even if he studied supermanifolds, non-standard calculus or embedding theorems."

EDIT Source is "A mathematical trivium" A book of 100 problems that university students "should be able to solve". The statement asks for calculation within 10% accuracy.

So the average value over the entire domain should be the same as the average value over  $[0,\pi/2]$

$$\langle\sin^{100} (x)\rangle= \frac{\int_0^{\pi/2} \sin^{100}(x) dx}{\int_0^{\pi/2} dx}.$$

So here's what I did:

First, this graph would be a train of highly sharp peaks. The integrand would assume values close to zero a up till before it sharply rises to 1.

So up till some $\epsilon \in [0,\pi/2]$ we will have $\sin x \approx x$ and for the remaining $\pi/2 - \epsilon$ interval I could find the area of triangle with base $\pi/2 - \epsilon$ and height $1$

$$\langle \sin^{100} (x)\rangle \approx \frac{2}{\pi} \left(\int_0^\epsilon x^{100} dx + .5 (\frac{\pi}{2}-\epsilon)\right).$$

I believe in principal it should be possible to find an $\epsilon$ such that the above expression yields the exact answer. So I try to approximate it, no good. Then I try mathematica and it is looking like there is no $\epsilon$ for which the value I am expecting is even close to the actual value. I plot the original and find that my approximation is hopeless.

Not to mention that my 5 minutes were over. So I admit I do not understand mathematics and humbly ask if someone could:

Point out my mistake (Other than that $\epsilon$ is probably incomputable within 5 mins)
How the hell is this done in 5 minutes?

The picture below has the $\sin^{100} x$ in blue (bottom) and my approximation of it plotted against $\epsilon$ (pink). Although there is no reason for them to be together, the upper graph has a minima quite above the exact value of the integral.

EDIT
Just realized

Let $$u=\cos x.$$

$$\int_0^{\pi/2} \sin^{100}(x) dx = \int_0^1 (1-u^2)^{99/2}du\approx \int_0^1 \left(1 - \frac{99}{2} u^2\right) du $$

I tended to the five minutes side, less the accuracy.  As shown by others, if you are on top of your complex variables the exact answer (unevaluated) can be found in under five minutes, and computers can handle the problem in a whiz.

It was fun to consider the intuitive shape of the equation.  The average over a large region is simply $1/2$; near zero, it is essentially zero.  However, an accuracy to a percent takes a little bit of work.

Approximate the integral as an area of height 1 and width from where $ \sin^{100} (x)=.5$ to $ \pi /2 $.  In the region of interest,
$$\sin(x)=1-(\pi/2-x)^2/2 $$ Sure, the sloping curve does not quite have rotational symmetry, but at 10% a rectangle should be fine.

If $\sin^{100}(z)=.5$, then $\sin(z)=\sqrt[100]{.5} $, which I am not used to finding without a calculator.  I took $\sqrt{.49}=.7$ and $\sqrt{.7} \approx 1-.3/2 -.01 = .84$ because I was not quite near enough to one to assume the number was constant, that is, $(1-a)^n \approx 1-n\cdot a$,

Similarly, $\sqrt[25]{.84} \approx 1-\dfrac{\frac{16}{100}}{25}$, and for the width of my rectangle, y, $$\frac{1}{2} y^2 =  \dfrac{\frac{16}{100}}{25}$$ $$(10 y)^2 = \frac{32}{25}$$ $$10y\approx 1+ \frac{1}{2} \frac{7}{25}$$ $$10y\approx 1+.04\cdot 3.5$$ or $y\approx .114$, giving an average of $1\cdot y\frac{2}{\pi}\approx .075$

I assume the primary error comes from the approximation finding the 25th root, which should be less than the estimate, so $y$ should be larger than found.  I partially accommodated this in my lazy division by $\pi$ in my head.


proof of 1) $|a| = 0 \iff a = 0$ and 2) $|a| \ge 0$ On the internet there are many proofs but very summary

Definition
$|a| = $
$a$ if $a \ge 0$
$-a$ if $a < 0$

Proposition 1 $|a| = 0 \Leftarrow \Rightarrow a = 0$

$\Rightarrow$)

Suppose that $|a| = 0$
This implies that $a\ge 0$ xor $a<0$
Case $a<0$
$|a| = -a$ by hypothesis $-a= 0$
As $a<0$
$a + (-a)<0 + 0$
$0<0$ (!contradiction)
Case $a \ge 0$
$|a| = a$ by hypothesis $a = 0$

$\Leftarrow$)
Suppose that $a= 0$
$|a| = a$ by hypothesis $|a| = 0$

proposition 2
$|a| \ge 0$
case $a \ge 0$
$|a| = a$ by hypothesis $|a| \ge 0$
case $a < 0$
$|a| = -a$
As $a<0$
$a+(-a)<0+(-a)$
$0<-a$
$0<|a| \Rightarrow 0 \le |a|$

are valid my proofs?

I think you do a little bit complicate. $$|a|=0\iff (a=0\quad \text{or}\quad  -a=0)\iff a=0.$$

Moreover, you don't need that fact that $a\geq 0$.

By the definition,

$$|a|=0\implies ( 0=a\land a\ge0)\lor(0=-a\land a<0)\iff ( 0=a\land a\ge0)\iff 0=a.$$

And obviously,

$$a=0\implies |a|=0.$$

For the second proposition,

$$a\ge0\implies|a|=a\ge0,\\
a<0\implies |a|=-a>0$$
so that $a\ge0$ holds as all $a\in\mathbb R$ have been covered.


Showing that solutions to a differential equation are non-negative for all time We've been given the following question:

Let $a$ and $p$ be real-valued piecewise continuous functions defined on $[t_0,t_1]$ and let $E \subset [t_0,t_1]$ be the (finite) set of points at which $a$ or $p$ fail to be continuous. Assume that $x:[t_0,t_1] \rightarrow \mathbb{R}$ is continuous and satisfies
$$ x(t) = x(t_0) + \int_{t_0}^t a(s)x(s)+p(s) ds$$
Show that if $x(t_0) \ge 0$ and $p(t) \ge 0$ for all $t \in [t_0,t_1]$, then $x(t) \ge 0$ for all $t \in [t_0,t_1]$.

I've little idea of how to go about showing this. I've tried going about it by contradiction, assuming that $x(t)$ was negative at some point (and so in some small interval by continuity) but this got me nowhere. I'd appreciate a pointer in the right direction.

Suppose we divide $[t_0,t_1]$ into intervals $[b_1,b_2],[b_2,b_3],\dots$ in each interval I can differentiate and write the equation as $x'(t)=a(t)x(t)+p(t)$

We have, by the integrating factor method, $$x(t)\exp(\int_{b_i}^t-a(y)dy)=\int_{b_i}^tp(y)\exp(g(y))dy+C$$, where $g(y)=\int_{b_i}^y-a(t)dt$ .Now assume $x\geq 0$ at some $b_i$. As the term inside the integral is $\geq 0$(right hand side) and the entire right hand side is $\geq 0$ at $t=b_i$.(exponentials are positive),we have that $x$ is always $\geq 0$ in $[b_0,b_i]$(As the right hand side is an increasing function) By induction it is $\geq 0$ everywhere.


How should the Calculus of Variations deal with $\delta(t-t_0)$ variations? I'm familiar with using the Calculus of variations to find the condition for which first order variations of a functional wrt a function are zero:

We start with a functional $J[x]= \int_{t_f}^{t_i}L(x(t),\dot x(t),t)dt$, where $t_i, t_f$ are constants and $\dot x(t)= dx/dt$
If $J[f]$ is stationary at $f$, we add to $f$ another continuous function $\epsilon\eta(t)$ where $\eta(t)$ is arbitrary and $\epsilon$ is a positive number
For any small number $\epsilon$ close to 0, the first order variation of the functional is zero around $f$, giving the Euler-Lagrange equations as the necessary condition.

If $\delta(t)$ is what physicists/engineers call the Dirac-Delta function, even though mathematicians wouldn't define it as a function:

Can the above procedure be applied to $\delta(t-t_0)$ variations of f at $t = t_0$ to again yield the Euler-Lagrange equations?

I've had a go by adding $\epsilon\delta(t-t_0)$ to $f$ as the variation, getting the differential of $L$ in terms of the usual partial derivatives and differentials of the independent variables. But the differential of $x$, for example, then becomes $dx = \epsilon\delta(t-t_o)$ which appears to be infinite for any $\epsilon$, making my expression for $dL$ nonsense.

Again, according to an answer given on PSE that motivated me to ask my question here, the change in $L$ to an $\epsilon\delta(t-t_0)$ variation in $\dot x$ is:

$$dL= {\partial L \over \partial \dot{x_1}} \delta \dot{x_1} = {\partial L \over \partial \dot{x_1}}\epsilon\delta(t-t_0)$$

Is this correct?

You can't proceed naively because $\epsilon \delta(x)$ is not continuos for any finite $\epsilon$.


Determine the convergence of $\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{n(n-1)}$ So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test.
$$
\sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)}
$$

You could do a limit comparison with $\dfrac{1}{n^{3/2}}$

Or we could be a bit witty. Write $\displaystyle \sum_{n = 2} \frac{\sqrt{n +1}}{n(n-1)} = \sum_{n = 1} \frac{\sqrt{n + 2}}{(n+1)n} < \sum_{n = 1} \frac{\sqrt{3n}}{n^2} = \sqrt 3 \sum_{n=1} \frac{1}{n^{3/2}}$

So that's just basic comparison, no limits or anything.

$\dfrac{\sqrt{n+1}}{n(n-1)}\le\dfrac{{n+1}}{n-1}=(1+\dfrac{1}{n})(1-\dfrac{1}{n})^{-1}=(1+\dfrac{1}{n})(1+\dfrac{1}{n}-\cdots)=1+\dfrac{2}{n}+O(\dfrac{1}{n^2}).$

$\displaystyle\sum\dfrac{{n+1}}{n-1}$ is convergent and so is the given series.


Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$ Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$ where

$f(x) = \frac{1}{x^3}$ if $x$ be rational $\geq 1$ and

$f(x) = \frac{-1}{x^3}$ if $x$ be irrational $>1$.

The solution given is straight forward $\int_{1}^{\infty}|f(x)|dx= \int_{1}^{\infty}\frac{1}{x^3}dx$ so its convergent, since is $f(x)$ absolutely convergent so is $\int_1^{\infty} f(x) dx$.

But what I don't understand is, the set of points of discontinuity of  $f(x)$ is infinite and has infinite number of limit points. If so its not even riemann integrable in the first place. How then can we comment on its convergence ?

We aren't integrating f(x), so we don't have to worry about f(x) being continuous. Rather, we are integrating g(x) = (1/x)^3, which is continuous.

Since |f(x)| = g(x), we can say that
Since we know g(x) > f(x) and -g(x) < f(x).
If g(x) approaches C, then f(x) is between C and -C. Therefore, it approaches a finite value.

Basically, we avoid having to worry about integrating f(x) by integrating a different function, g(x), instead and comparing the two.

Edit because my answer was unsatisfactory: Basically squeeze theorem is what proves the correctness of this method. Because squeeze theorem does not require a function to be continuous, this method works.


Probability - point inside disk $X, Y \sim \mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $\mathbb{P}[(X,Y)  \in D_r ] = 0.3$.

My attempt: $(X, Y) \in D_r$ means $X^2 + Y^2 \le r^2$. What should I do next?

HINT

For any event $A$, you have
$$
\mathbb{P}[A] = \iint_{\mathbb{R}^2} \mathbb{I}_{A(x,y)}f(x,y)dxdy,
$$
and in your case, since $X$ and $Y$ are independent, $f(x,y) = \phi(x) \phi(y)$.

Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...


Stirling Numbers of the First Kind and $S_n$. I know that, on the one hand, if $s(n, p)$ denotes the unsigned Stirling Numbers of the First Kind, then $(x)_n=\displaystyle\sum_{p=0}^n s(n, p)x^p$, where $(x)_n=x(x-1)\cdots(x-n+1)$. It follows that $n!=\displaystyle\sum_{p=0}^n s(n, p)n^p$.

But I also know that $s(n, p)$ counts the number of permutations in $S_n$ that can be written as the product of $p$ disjoint cycles, so shouldn't $n!=\displaystyle\sum_{p=0}^n s(n, p)$, instead?

The Stirling numbers of the first kind, $s(n,p)$, count the number of permutations in $S_n$ that are a product of $p$ disjoint cycles, but you are forgetting about sign! The relevant identities are:

$$n!=\sum_{p=0}^n s(n,p)n^p$$ and $$n!=\sum_{p=0}^n |s(n,p)|.$$ Often the symbol $\left[{n\atop p}\right]$ is used to denote $|s(n,p)|$, the unsigned Stirling numbers of the first kind. The reason why we define the Stirling numbers of the first kind $s(n,k)$ to be signed is because then the infinite matrices $(s(n,k))_{k,n\in\mathbb N}$ and $(S(n,k))_{n,k\in\mathbb{N}}$ (where $S(n,k)$ are the Stirling numbers of the second kind) which are lower and upper triangular respectively, form a change of basis matrix pair (they are inverses of each other) that allow us to move from the power series basis of polynomials $1,x,x^2,\dots$ to umbral calculus basis $(x)_0,(x)_1,(x)_2,\dots$.

You've got the sum wrong. It is:
$$
x^{\overline{n}}
  = \sum_{0 \le p \le} \genfrac{[}{]}{0pt}{}{n}{p} x^p
$$
(i.e., on the left hand side you have $x (x + 1) \ldots (x + n - 1)$).