Finding the MLE estimate from a log function If I have the following function:
$$\ln L(\mu,\sigma^2)= -n \ln (\sqrt{2 \pi \sigma^2})+\sum_{i=1}^n \frac{-(x_i-\mu)^2}{2\sigma^2}$$
How can I find the MLE estimate $σ^2_\text{MLE}$?
I have already shown that $\mu = \bar{x}$

$$
\ln L(\mu,\sigma^2)= -n \ln \sqrt{2 \pi \sigma^2} - \frac 1 {2\sigma^2} \sum_{i=1}^n (x_i-\mu)^2
$$
You report that you have shown that the value of $\mu$ that maximizes this is $\overline x = (x_1+\cdots+x_n)/n.$
One could say that this is with $\sigma$ fixed, but more to the point, this result is the same regardless of $\sigma.$ And for any here who have not noticed this before, it is of interest to note that this is just the value of $\mu$ that minimizes $\sum_{i=1}^n (x_i-\mu)^2,$ and that the best way to see that is via the identity in algebra that says
$$
\sum_{i=1}^n (x_i-\mu)^2 = n(\mu-\overline x)^2 + \sum_{i=1}^n (x_i-\overline x)^2.
$$
What is sought is the pair $(\mu,\sigma^2)$ that maximizes $L(\mu,\sigma^2),$ and that is the same as the pair that maximizes $\ln L(\mu,\sigma^2),$ and that expression is used because it requires less work to get to the answer.
The maximizing value of $\mu$ with $\sigma^2$ fixed is the same regardless of $\sigma^2,$ and consequently we can just find the value of $\sigma^2$ that maximizes $L(\overline x,\sigma^2)$.
\begin{align}
\ln L(\overline x,\sigma^2) & = -n \ln \sqrt{2 \pi \sigma^2} - \frac 1 {2\sigma^2} \sum_{i=1}^n (x_i-\overline x)^2 \\[8pt]
& = -n\ln \sqrt{2\pi \sigma^2} - \frac{ns^2}{2\sigma^2} \\[8pt]
& = \text{constant} -\frac n 2 \ln (\sigma^2) - \frac{ns^2}{2\sigma^2}. \tag 1
\end{align}
I prefer to write that last line as $\text{constant}-n\ln\sigma - \dfrac{ns^2}{2\sigma^2}$ and differentiate with respect to $\sigma,$ but I've noticed various authorities preferring to treat $\sigma^2$ as a simple thing, so here I will do it that way.
The derivative of $(1)$ with respect to $\sigma^2$ is
$$
- \frac n {2\sigma^2} + \frac{ns^2}{2(\sigma^2)^2} = n\cdot\frac{s^2-\sigma^2}{\sigma^4} \quad \begin{cases} >0 & \text{if } 0<\sigma^2<s^2 \\ =0 & \text{if } \sigma^2=s^2, \\ <0 & \text{if } \sigma^2 > s^2. \end{cases}
$$
So this function of $\sigma^2$ increases on the interval $(0,\sigma^2]$ and decreases on the interval $[\sigma^2,+\infty).$ Therefore it has a glabal maximum at $\sigma^2 = s^2.$

Take the derivative wrt $\sigma^2$:
$$-\frac n {\sqrt{2\pi\sigma^2}}\cdot \frac{2\pi}{2\sqrt{2\pi\sigma^2}}+\frac{\sum -(x-\mu)^2}{2}\cdot \frac{-1}{(\sigma^2)^2}=0\\
=\frac{-2n\pi}{4\pi\sigma^2}+\frac{\sum(x-\mu)^2}{2(\sigma^2)^2}\\
\frac{n}{2\sigma^2}=\frac{\sum(x-\mu)^2}{2(\sigma^2)^2}\\
\sigma^2=\frac{\sum(x-\mu)^2}{n}$$
Elaborate verification
To be rigorous, we need to show that this value of $\sigma^2$ maximizes the density. The first derivative is $$\frac{-n}{2\sigma^2}+\frac{\sum(x-\mu)^2}{(\sigma^2)^2}?0\\
-\sigma^2+\frac{\sum(x-\mu)^2}{n}?0$$
where the question mark  represents an unknown relationship. We can see that the derivative is 0 at $\sum(x-\mu)^2/n$. When $\sigma^2$ is greater than this, the first derivative is negative, and when it is less than this, the first derivative is positive. So we can conclude that $\sigma^2$ is a global maximum. For fun, I also take the second derivative.
$$\frac{\partial^2}{(\partial \sigma^2)^2}\log l(\sigma^2|\mathbf x)=\frac{n}{2(\sigma^2)^2}-\frac{\sum(x-\mu)^2}{(\sigma^2)^3}\\
\sigma^2-\frac{2\sum(x-\mu)^2}{n}?0$$
This is zero when $\sigma^2$ is $\frac{2\sum(x-\mu)^2}{n}$, positive when it is greater than that, and negative when it is less than that. So there is an inflection point at $\frac{2\sum(x-\mu)^2}{n}$, the log likelihood is concave up when $\sigma^2$ is greater than that, and concave down when it is less than that.

Conclusion
$\sigma^2_{MLE}=\frac{\sum(x-\bar x)^2}{n}$


Proof of binomial coefficient formula. How can we prove that the number of  ways choosing $k$ elements among $n$ is    $\frac{n!}{k!(n-k)!} = \binom{n}{k}$    with $k\leq n$?
This is an accepted fact in every book but i couldn't find a proof.

Notice that given $n$ objects $\{a_i\}$ and $n$ positions, there are $n$ possible positions for $a_n$. Once we have placed $a_n$, there are $n-1$ remaining positions for $a_{n-1}$. Once we have placed $a_n$ and $a_{n-1}$, there are $n-2$ positions left for $a_{n-2}$, and so forth. Thus, there are
$$
n(n-1)(n-2)\cdots3\cdot2\cdot1=n!
$$
possible ways to arrange the $n$ items.

Suppose we divide the $n$ items into $k$ white objects and $n-k$ black objects. For each arrangement of the white and black objects that looks the same, there are $k!$ rearrangements of the white objects and $(n-k)!$ rearrangements of the black objects. Since there are $n!$ rearrangements of the $n$ objects, there must be
$$
\frac{n!}{k!(n-k)!}=\binom{n}{k}
$$
arrangements of the $k$ white and $n-k$ black objects that look different.

If you have 5 people (A, B, C, D & E) but wants to build a team of three people, the most number of ways you can do is 5x4x3= 60 ways. But realize that team ABC and CBA is basically same team.
So for three people, you can have 3x2x1=6 different combinations.
Meaning that, making a team with 5 poeple with 3 positions, you can have total (5x4x3) / (1x2x3) = 10 different combinations. This number 10 is the 3rd number on 5th row of Pascal's triangle. This basically tells us the number of combination we can have.
And (5x4x3) / (1x2x3) = 5! / (3! x 2!) = C (5, 3)


Proof that $a_n < b_n \implies \lim(a_n) \le \lim(b_n)$ First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $\limsup$ or $\liminf$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.

The problem is as follows:

Given $a_n < b_n$ prove that $\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$. The proof is then done by contradiction, assuming that $a = \lim_{n\to \infty}(a_n) > b =\lim_{n\to\infty}(b_n)$.

We take an $\epsilon = \frac{a-b} 2$, so that the $\epsilon$-neighborhoods of $a$ and $b$ are disjoint. From the definition of limits, we now know that there is such a $N$, so that $\forall n > N : |a_n-a|<\frac\epsilon2$ and $|b_n-b|<\frac\epsilon2$.

The next step is absolutely always confusing. Two variants I've found are either:

$a_n>a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon>b_n$

In which I do not understand why any two terms of that (in)equality are like that, or it is said that if $a > b$, there must be such an $\epsilon$ so that $a - \epsilon > b + \epsilon$. Then,
$a - \epsilon > b + \epsilon > b_n, a_n > a - \epsilon > b + \epsilon > b_n$, which contradicts $a_n \le b_n$. Here I simply cannot comprehend how we came to the conclusion that $a_n > a - \epsilon$ and $b + \epsilon > b_n$. The definition of the limit uses absolute values everywhere, so surely the values depends on the signs of $a_n$ and $a$, and $b_n$ and $b$. Please help me understand what is it that I'm missing here. Thanks in advance!

$\newcommand{\eps}{\varepsilon}$If $\eps > 0$ is real, then for every real number $x$,
$$
\text{$|x| < \eps$ if and only if $-\eps < x < \eps$.}
$$
Particularly,
\begin{align*}
  |a_{n} - a| < \eps &\quad \text{if and only if}\quad -\eps < a_{n} - a < \eps, \\
  &\quad \text{if and only if}\quad a - \eps < a_{n} < a + \eps.
\end{align*}

Second, if $a$ and $b$ are real numbers such that $b < a$, then $b < \frac{1}{2}(b + a) < a$ (the mean/midpoint lies between). A bit of algebra shows that if $\eps = \frac{1}{2}(a - b)$, then $\eps > 0$, and
$$
b + \eps = \tfrac{1}{2}(b + a) = a - \eps.
$$

In each case, it's a pleasant and instructive exercise to sketch a number line and see the obvious geometric fact these inequalities express.

$a_n=0$ for all $n$ and $b_n=\frac{1}{n}$


Endomorphism and Matrices In the context of linear maps I need help to understand the notation used in this exercise and the general method to solve it.

In the Euclidean space $\mathbb R^3$ with the standard inner product, consider the endomorphism $f : \mathbb R^3 \rightarrow \mathbb R^3$ defined by
$$f(1,0,1) = (1,4,1), \,\, f(1,0,-1) = (-3,0,3), \,\,f(0,1,0)=(2,-1,2).$$
Find the associated $f^*$.

Since I'm translating directly from greek notes I'm not sure if the word "associated" is correct in this context. Nowhere in my notes does the author gives the definition of $f^*$. This is the major problem that I have and therefore, this is my first question:
$(Q1)$ What is the precise definition of $f^*$? Specifically, what does the exercise ask us to find? Also, if the word "associated" is not correct, what is the correct one?
The author gives the following quick solution to the exercise:
Solution. The matrix of $f$ in the standard basis $B$ of $\mathbb R^3$ is
$$A = M_B^B(f) = \begin{pmatrix}-1 & 2 & 2\\2 & -1 & 2\\2 & 2 &-1\end{pmatrix}.$$
Since the matrix $A$ is symmetric it follows that $f = f^*$.
Therefore, the associated $f^* : \mathbb R^3 \rightarrow \mathbb R^3$ of $f$ is exactly $f$.

$(Q2)$ At the moment the above solution doesn't make any sense to me. I can see that the matrix $A$ is symmetric but I have no idea how to find it, what it actually represents, how it is connected to the exercise and what to conclude from it.
I'm looking for answers to $(Q1)$ and $(Q2)$ which will allow me to understand the exercise and solution.

As a linear transformation, $f$ can be represented through a matrix, whose columns are the images of the standard basis vectors. By linearity \begin{align}f(1,0,0)&=f\left(\frac12(1,0,1)+\frac12(1,0,-1)\right)=^{\text{linearity of $f$}}\\[0.2cm]&=\frac12f(1,0,1)+\frac12f(1,0,-1)=\frac12(1,4,1)+\frac12(-3,0,3)=(-1,2,2)\end{align} and similarly for $(0,1,0)$ and $(0,0,1)$. That is how he derived matrix $A$. Then $f^*$ is just the conjugate of $f$, the word "associated" seems unnecessary. Given the matrix $A$, the conjugate is found through the relation $$A^*=\overline{A}^T=A^T$$ since $A$ is real. For any $x,y \in \mathbb R^3$ it satisfies the relation $\langle Ax,y\rangle=\langle x,A^*y\rangle$ where $\langle \cdot, \cdot \rangle$ stands for the inner-product in $\mathbb R^3$, but most likely you do not need that in this question.

Basically this explains how to find a matrix representation of a function
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).The matrix representation is a transformation matrix which depends upon the basis vectors chosen. So for simplicity we choose the standard basis (1,0,0), (0,1,0) and (0,0,1).Then express the three given functions ( the function vector as taken together) in linear combinations of the basis vectors. So we find the functional values
 f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
This expressed as a matrix becomes A = M(f) as above. The matrix is symmetric because of linearity of the function f


proof by induction, two basic questions about the procedure I'm currently learning to prove something via induction. I have two questions:

1) In the inductive step, assuming that the statement $P(k-1)$ holds, so that you have to show that $P(k)$ follows, how exactly has the induction hypothesis to be formulated? Like this:

"The statement $P(k)$ holds for some $k\le n-1.$" ? I guess there should be no difference between this 'variant' and this one where you assume $P(k)$ and then have to show that 'P(k+1)' in the inductive step, because it is

$$\big(P(k)\Rightarrow P(k+1)\big)\;\iff \big(P(k-1)\Rightarrow P(k)\big),$$right?

2) In the induction base, I'm still not sure in which cases I have to do verify the statement for more than one n, (i.e. not only n=0, but also n=1) and so on.. can you explain me typical situations in which I have to verify the statements for several n's for the induction base?

Thank you for any help.

First of all I applaud you for the username.

I'll do an induction proof for you, and show you why I do everything. Hopefully this will answer all your questions.

Suppose we have the following:

$$
f(n) = \left\{\begin{array}{lr}
        n, & \text{for } 0\leq n\leq 2\\
        3f(n-2)+2f(n-3), & \text{for } n>2\\
        \end{array}\right\}
$$

Let's prove $f(n)<2^n$ for all $n\in\mathbb{N}$

Let's define a predicate.

For all $n\in\mathbb{N}$, let $P(n):f(n)<2^n$

Pf of $\forall n, P(n)$.

Base case(s)

First of all, notice the when we recurse ($n>2$), we have to "go back" at most $3$ numbers, hence the $n-3$. This should suggest that we have $3$ base cases.

As for which ones, well definitely we must prove $P(0)$, and after that, we prove 2 more, so we prove $P(1),P(2)$. So in the inductive step, when we "go back", we have these base cases to work with, so we are okay. If you only proved $P(0)$, then in the inductive step you would get stuck, because you don't know anything regarding $P(n-2)$.

Let $n=0$

Then $f(n)=0$, by definition

$< 2^0 = 1$

$\therefore P(0)$ holds.

Now do $P(1)$ and $P(2)$

Induction Step:
For of all, since we proved $P(0-2)$, now we let $n>2$, or equally $n\geq 3$, so we enter the induction step.

Let $n>2$

Suppose $P(j)$ holds where $0\leq j\leq n-1, j\in\mathbb{N}$. [IH]

What does this say? This says let $j$ be an arbitrary natural number between $0$ and $n-1$, both inclusive. This says that suppose $P(0)\land P(1)\land\dots\land P(n-1)$ hold. This is how I write my IH, it's very clear. Moreover, you can also write this:

Suppose $P(j)$ holds where $0\leq j\lt n, j\in\mathbb{N}$. [IH]

We want to prove that $P(n)$ holds. It's much easier to prove $P(n)$ holds, rather then assuming $P(n)$ and proving $P(n+1)$.

$f(n)=3f(n-2)+2f(n-3)$, by definition of recursive step

$<3\cdot 2^{n-2}+2\cdot 2^{n-3}$, by IH, since $0\leq n-3<n-2<n$

$=6\cdot 2^{n-3}+2\cdot 2^{n-3}$

$=8\cdot 2^{n-3}$

$=2^n$

$\therefore P(n)$ holds.

We have showed it holds for all $\mathbb{N}$, and so our proof is done. For proving how many base cases, think about how many steps the recursion "goes back", and proceed from there. For writing your induction hypothesis, it becomes much easier with a predicate and proving $P(n)$. If you can understand the way I wrote mine, just use that. Assume the predicate holds within that range, where $0$ is your first base case, and $n-1$ is what you assume it holds to. Then prove $P(n)$. Hope you found this answer useful.

Assume you want to prove that

$$\forall n\ge n_0 \;\;P_n $$

In the induction base, you check that the first  $P_{n_0} $ is true.

In the inductive step , you start like this :

Let $n\ge n_0$ such that $P_n $.
then you prove $P_{n+1} $.


Confusion regarding totally bounded metric spaces having cauchy subsequences of every sequence. My book says "A metric space $(X,d)$ is totally bounded if and only if every sequence in $X$ contains a cauchy subsequence."

Let us take a sequence $\{x_i\}$ such that for every prime number $p$, when $i$ is of the form $p^k$, $x_i$ or $x_{p^k}$ is part of the subsequence converging to $p$. For example, for $p=3$, $\{x_3,x_9,x_{27},\dots\}$ forms a cauchy subsequence converging to $3$.

Clearly, this sequence contains infinite cauchy subsequences. But is it totally bounded? Any help with proving this sequence is totally bounded would be greaty appreciated. For a given $\epsilon\in\Bbb{R}$, selecting the primes as part of the finite set isn't an option, as there are an infinite number of them.

Thanks in advance!

In principle, the problem was already pointed out in the comments - let me just clarify this a little more:

The term "totally bounded" refers, as stated, to the metric space $(X,d)$, i.e. it is a property of the metric space and not of a single sequence in the metric space.
Now the statement reads that if a metric space is totally bounded, then every sequence contains a Cauchy-subsequence and conversely, if every (not only one!) sequence contains a Cauchy subsequence, then the space (not the sequence) is totally bounded.
Intuitively, totally bounded means that you cannot diverge with your sequences. Either you converge (within the topological closure), or you somehow oscillate (meaning that eventually every point of your sequence is close to some point in the space and the set of these points is bounded). But this behaviour cannot be decided upon looking at a single sequence.

For your example (you have not really specified the metric space), if you take the subsequence $\{x_{i_j}\}_j$ where $i_j=p_j^{k_j}$ with $p_j$ the j-th prime number and $k_j$ sufficiently large, such that $|x_{i_j}-p_j|<\varepsilon$ for some prechosen $\varepsilon>0$, which is certainly a subsequence of your sequence and hence a sequence of the metric space, then this sequence does not contain a Cauchy-subsequence - therefore your space cannot be totally bounded.

This is a very useful statement, since it readily follows that a metric space is compact iff it is complete and totally bounded. So I'll give a proof, even if that wasn't explicitly asked for.

Suppose $E$ is a totally bounded metric space. Let $\{p_n\}$ be a sequence in $E$ and $\varepsilon>0$. Then there exists a finite set of points $\{x_1, \ldots, x_m\}\subset E$ such that $$E=\bigcup_{j=1}^N B\left(x_j,\frac\varepsilon2\right).$$
Then one of these balls must contain $p_n$ for infinitely many $n$, which gives us a subsequence $\{p_{n_k}\}$ with $$d(p_{n_l}, p_{n_m}) \leqslant d(p_{n_l},x_j) + d(x_j, p_{n_m}) < \frac\varepsilon2 + \frac\varepsilon2=\varepsilon $$
for all $l, m$.

Conversely, if $E$ is not totally bounded, then there exists $\varepsilon>0$ such that for any finite subset $\{x_1,\ldots, x_N\}\subset E$, there exists $p\in E$ with $$p\notin\bigcup_{j=1}^N B(x_j,\varepsilon). $$
Let $p_1\in E$. Given $p_1,\ldots, p_N\in E$, choose $p_{N+1}\in E$ with $$p_{N+1}\notin\bigcup_{n=1}^N B(p_n,\varepsilon). $$
Then for all $N$, $d(p_n, p_N)\geqslant\varepsilon$ for $n>N$, and so $\{p_n\}$ cannot have a Cauchy subsequence.


Image of closed unit ball under a compact operator Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:

$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.

I have proven this when $X$ is a reflexive space.

Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.

In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.

Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.

On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.

But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.

It's easily proved that if the image is compact, the operator is also compact.

But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?

It is false in general that the image of the closed unit ball under a compact operator is closed (and hence compact). Here is an easy example:

Consider $X = C[0,1]$ with the uniform norm, and the compact operator $A \in B(X)$ defined by the formula:

$\displaystyle\qquad Af(x) = \int_0^x f(t)\,dt$.

Compactness of $A$ is easily proven using Arzelà–Ascoli. Our operator $A$ produces an anti-derivative of any input given to it, and the image of the closed unit ball of $X$ under $A$ is the set

$\displaystyle\qquad \{f \in C^1[0,1] \mathrel: f(0)=0,\ \lVert f'\rVert \leq 1\}$

which certainly is not a closed subset of $X$.

This is much easier. Use the definition of compact operator. An operator is compact if and only if the image of a bounded set is precompact. The unit ball is bounded and any other bounded set is inside some $\lambda$ times the unit ball.


Sum of reciprocals of primes factorial: $\sum_{p\;\text{prime}}\frac{1}{p!}$ The series
$$\sum_{p\;\text{prime}}\frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots$$
diverges as is well known. How about the following?
$$\sum_{p\;\text{prime}}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{11!}+\cdots$$
Its convergence is easily obtained by comparing with the series $e=\sum_{n=0}^\infty\frac{1}{n!}$. Numerically, we get
$$\sum_{p\;\text{prime}}\frac{1}{p!}\simeq 0.675198$$
Can we find the specific value?

I calculated your number to 15 decimal places and entered it into the Inverse Symbolic Calculator. The only match was for the sum $\sum_{p}\frac{1}{p!}$.

Therefore, I would venture to guess that this number is unlikely to admit a simple description in terms of any known constants.

Interesting question, disappointing answer. Sorry!

I have a formula which equals this, an infinite sum to be precise, but it is given in terms of a prime counting function.

Here it is:

If pi[n]= 'the number of primes up to and including the number n',

Then the sum of reciprocals of primes factorial equals:

∑pi[n]*n/(n+1)!,

where n goes from 2 to infinity.


How to calculate the time to travel around the ellipse? A particle moves along the ellipse $3x^2 + y^2 = 1$ with positions vector $\vec{r(t)} = f(t)\vec i + g(t) \vec j$. The motion is such that the horizontal component of the velocity vector at time $t$ is $-g(t)$. How much time is required for the particle to go once around the ellipse?

Now I found that the particle travels counterclockwise, though I suspect it doesn't matter. I also found that the position vector is $(x, ±\sqrt{1-3x^2} )$ and hence the velocity vector is $(\mp \sqrt{1-3x^2} , 3x)$.

I am not supposed to use arc length, so I am quite confused as to how to solve this problem. I am guessing that I am supposed to integrate something, but what, how, and why....

Lets paremertize the elipse

$f(t) = \frac {1}{\sqrt 3}\cos (\omega(t))\\
g(t) = \sin (\omega (t))$

and it is given that:
$f'(t) = -g(t)$

$f'(t) = -(\frac {1}{\sqrt 3}\cos \omega)\omega' = \sin (\omega (t))\\
\omega' = \sqrt 3\\
\omega(t) = \sqrt 3 \ t$

It will take $\frac {2\pi}{\sqrt 3}$ to complete one orbit.

Or you could say

$(x,y) = (\omega(t), \sqrt {1-3\omega^2})$

But this only gets you half way around.... then you need to get back along the path

$(x,y) = (\omega(t), -\sqrt {1-3\omega^2})$

$\omega' = -\sqrt {1-3\omega^2}$

and we solve the differential equation and get the same trig equations we have above.

Hint: The examiner wants you to use a parametrisation of the ellipse such that the derivative has a horizontal component which satisfies the given condition. That is, such that $f'(t)=-g(t).$

Well, one such parametrisation is got by setting $ x =f(t)=\frac{1}{\sqrt 3}\cos(\sqrt 3 t)$ and $y=g(t)=\sin( \sqrt 3 t).$


Compute $E[ \sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable. Let  $Q$  be non-central chi-square random variable of degree $k$ and centrality parameter $\mu$.

How to compute  the following expectation
\begin{align}
E\left[ \sqrt{Q} e^{-tQ} \right], t>0
\end{align}

I was thinking this can be computed by using the fact that $Q= \sqrt{ \sum_{i=1}^k Z_i^2 }$ where $Z_i \sim \mathcal{N}(\mu_i,1)$ and $\mu= \sqrt{\sum_{i=1}^k \mu_i^2}$, and $Z_i$'s are independent.

Using this observation we get that
\begin{align}
E\left[ \sqrt{\sum_{i=1}^k Z_i^2 } e^{-t\sum_{i=1}^k Z_i^2 } \right]= E\left[ \sqrt{\sum_{i=1}^k Z_i^2 }  \prod_{i=1}^k  e^{-tZ_i^2 } \right].
\end{align}
However, it is not clear to me how such exressions can be  manipulated or expanded.

I also tried direct integration but expression seem very long, so I stopped.

$$\mathbf E\left[ \sqrt{X} e^{-tX}\right] =
e^{-\lambda/2}
(1+2t)^{-(k+1)/2} \sqrt{2}
\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}
{}_1F_1 \left( \frac{k+1}{2}; \frac{k}{2};  \frac{\lambda}{2(1+2t)}\right)
$$

We proceed to show this over a number of steps; first I outline the main claims, and then I will come back to prove the first of these (the rest are referenced below).

We consider a $\chi^2$ variable parameterised by $\lambda > 0$, $k > 0$ with PDF

$$
f(x) = \frac12 e^{-(x+\lambda)/2} \left( \frac{x}{\lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\lambda x}),$$
where $I$ denotes the modified Bessel function of the first kind.

Claim 1. (proof below) Let $X \sim \chi^2(\lambda,k)$ and $\tilde X \sim \chi^2(\tilde \lambda, k)$, where $\tilde \lambda = \lambda/(1+2t)$. Then

$$ \mathbf E\left[ \sqrt{X} e^{-tX}\right] = \exp\left(- \frac{t}{1+2t}\lambda\right) (1+2t)^{-(k+1)/2} \mathbf E \left[ \sqrt{ \tilde X}\right]  $$

So in particular we have reduced the problem to calculating the expectation of the square root of a non-central chi-squared distribution. To do this we will want to relate the non-central $\chi^2$ distribution to the centralised one with $\lambda = 0$.

Claim 2. (Muirhead, 2005. Corollary 1.3.5) Let $P \sim \text{Poi}(\lambda/2)$ be Poisson distributed, then
$$\chi^2(\lambda, k) = \chi^2(0,k+2 P).$$
Where the equality is in distribution.

In particular this claim gives us a way to calculate $\mathbf E[\sqrt{\tilde X}]$ by conditionining on $P$:

$$ \mathbf E[ \sqrt{\tilde X}]  = e^{-\tilde \lambda/2} \sum_{p=0}^\infty \frac{(\tilde \lambda/2)^p}{p!}  \mathbf E \left[\sqrt{X^{(k+2p)}}\right]$$
where I have denoted $X^{k} \sim \chi^2(0,k)$ for a centralised chi-squared distribution with $k$ degrees of freedom.

To tie things up we will require one further claim.

Claim 3. (Forbes, Evans, Hastings, Peacock 2011. Chapter 11.3)
$$ \mathbf E \left[ \sqrt{X^{(k)}} \right] = \sqrt{2} \frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}, $$

Piecing all of this together we have the main result:

Theorem
$$\mathbf E\left[ \sqrt{X} e^{-tX}\right] =
e^{-\lambda/2}
(1+2t)^{-(k+1)/2} \sqrt{2}
\sum_{p=0}^\infty \frac{\left(\frac{\lambda}{(2(1+2t)}\right)^p}{p!}
\frac{\Gamma\left(p + \frac{k+1}{2}\right)}{\Gamma\left(p + \frac{k}{2}\right)}
$$

The power series can be recognised as a confluent hypergeometric function, so we have the following corollary.

Corollary
$$\mathbf E\left[ \sqrt{X} e^{-tX}\right] =
e^{-\lambda/2}
(1+2t)^{-(k+1)/2} \sqrt{2}
\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}
{}_1F_1 \left( \frac{k+1}{2}; \frac{k}{2};  \frac{\lambda}{2(1+2t)}\right)
$$

Claims 2 and 3 are referenced above. Therefore I prove only Claim 1.

Following Raskolnikov's suggestion we turn to the PDF of the distribution, defined above. And note

$$ \begin{aligned}
\mathbf E[ \sqrt{X} e^{-tX} ]& =
\int x^{1/2} e^{-tx}f(x) dx \\
& =  \int x^{1/2} \frac12 e^{-((1+2t)x+\lambda)/2} \left( \frac{x}{\lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\lambda x}) dx.
\end{aligned}
$$
Consider the change of variables $\tilde x = (1+2t)x$ and $\tilde \lambda = \lambda / (1+2t)$, then in particular note that the term in the exponential becomes

$$(1+2t)x + \lambda = \tilde x + (1+2t)\tilde \lambda,$$
whilst
$$\left(\frac{x}{\lambda}\right)^{k/4 - 1/2} = \left(\frac1{(1+2t)^2} \frac{\tilde x}{\tilde \lambda}\right)^{k/4 -1/2} =(1+2t)^{1-k/2} \left(\frac{\tilde x}{\tilde \lambda}\right)^{k/4 -1/2} $$
and
$$ \sqrt{\lambda x} = \sqrt{\tilde \lambda \tilde x}$$

Putting these together we have:
$$ \begin{aligned}
\mathbf E[ \sqrt{X} e^{-tX} ]& =
(1+2t)^{1-k/2} \int x^{1/2} \frac12  \exp\left(-(\tilde x + (1+2t)\tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) dx \\
& = (1+2t)^{1-k/2} \exp(-t \tilde \lambda) \int x^{1/2} \frac12  \exp\left(-(\tilde x + \tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) dx
\end{aligned}
$$
We still have two terms which remain in terms of $x$, rather than $ \tilde x$:

$$ x^{1/2} = (1+2t)^{-1/2} \tilde x,$$
and to complete the change of variables formula

$$ dx = (1+2t)^{-1} d \tilde x.$$

Including these changes of variables we have

$$ \begin{aligned}
\mathbf E[ \sqrt{X} e^{-tX} ]& =
(1+2t)^{-(k+1)/2} \exp(-t \tilde \lambda) \int \tilde x^{1/2} \frac12  \exp\left(-(\tilde x + \tilde \lambda)/2\right) \left( \frac{\tilde x}{\tilde \lambda} \right)^{k/4 -1/2} I_{k/2 -1}(\sqrt{\tilde \lambda \tilde x}) d \tilde x \\
& = (1+2t)^{-(k+1)/2} \exp(-t \tilde \lambda) \int \tilde x^{1/2} \tilde f(\tilde x) d \tilde x
\end{aligned}
$$
where $\tilde f$ denotes the PDF of the $\chi^2$ with parameters $\tilde \lambda$, and $k$. The result now follows on writing
$$ \exp(-t \tilde \lambda) = \exp(-t\lambda/(1+2t))$$

Finally we check the above using a few different methods in R.

First we set some arbitrary choices of the parameters

t <- 1
lambda <- 2
k <- 2

In the first method, we sample $X$ directly as a $\chi^2(\lambda, k)$ distribution.

# Sample chi-squared.
x <- rchisq(10e06, df = k, ncp = lambda)
mean( sqrt(x) * exp(-t *x) )
[1] 0.1436569

In the second we apply Claim 1, and sample the $\chi^2(\lambda/(1+2t,k)$ distribution

y <- rchisq(10e06, df = k, ncp = lambda/(1+2*t))
mean(sqrt(y)) * exp(- t * lambda/(1+2*t) ) * (1 + 2 *t)^{-(k + 1)/2}
[1] 0.1436606

In the third we apply Claim 2 and check that the Poisson weighting formula is correct.

df <- data_frame(p = rpois(10e06,lambda/(2*(1+2*t))) ) %>%
  mutate(
    z = rchisq(n(), df = k + 2 * p,ncp=0)
  )

mean(sqrt(df$z)) * exp(- t * lambda/(1 + 2 *t) ) * (1 + 2 *t)^{-(k + 1)/2}
[1] 0.1436255

Finally we compute the formula presented in the theorem statement.

library(CharFun) # Required for hypergeom1F1 function
exp(-lambda/2) *(1+2*t)^{-(k+1)/2} * sqrt(2) * gamma((k+1)/2) / gamma(k/2) * hypergeom1F1(lambda/(2*(1+2*t)), (k+1)/2, k/2)
[1] 0.1436613

Mathematica 11.3.0 produces

Mean[TransformedDistribution[Sqrt[x]*Exp[-t*x],x \[Distributed]
 NoncentralChiSquareDistribution[k, \[Mu]],  Assumptions -> t > 0]]

  $$\sqrt{2} e^{-\frac{\mu }{2}} (2 t+1)^{-\frac{k}{2}-\frac{1}{2}} \Gamma \left(\frac{k+1}{2}\right) \, _1\tilde{F}_1\left(\frac{k+1}{2};\frac{k}{2};\frac{\mu }{4 t+2}\right) $$

Edit. Thanks to @owen88 who paid my attention to the typo in my code:x*Exp[-t*x] instead of Sqrt[x]*Exp[-t*x]. Fixed.


Proving a basis for inner product space V when $||e_j-v_j||< \frac{1}{\sqrt{n}}$. Suppose $(e_1,e_2,...,e_n)$ is an orthonormal basis of the inner product space $V$ and $v_1,v_2,...,v_n$ are vectors of $V$ such that $$||e_j-v_j||< \frac{1}{\sqrt{n}}$$ for each $j \in \left\{1,2,...,n \right \}$. Prove that $(v_1,v_2,...,v_n)$ is a basis of $V$.

I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?

Note that it's enough to show that the $v_i$ are linearly independent. Suppose that there are scalars $c_1,\dots,c_n$ not all zero such that $\sum_ic_iv_i=0$. Then
$$ 0=\sum_{i}c_iv_i=\sum_{i}c_i(v_i-e_i)+\sum_ic_ie_i$$
hence
$$ \Big|\Big|\sum_{i}c_i(v_i-e_i)\Big|\Big|=\Big|\Big|\sum_{i}c_ie_i\Big|\Big|$$
However,
$$\Big|\Big|\sum_{i}c_i(v_i-e_i)\Big|\Big|\leq \sum_i|c_i|\cdot ||v_i-e_i||<\frac{1}{\sqrt{n}}\sum_{i=1}^n|c_i|\leq \Big[\sum_{i=1}^n|c_i|^2\Big]^{\frac{1}{2}} $$
(with the last step using Cauchy-Schwarz), while since the $e_i$ are orthonormal we have
$$ \Big|\Big|\sum_{i}c_ie_i\Big|\Big|^2=\sum_{i=1}^n|c_i|^2 $$
so that
$$ \Big|\Big|\sum_{i}c_ie_i\Big|\Big|=\Big[\sum_{i=1}^n|c_i|^2\Big]^{\frac{1}{2}} $$
This contradiction shows that the $v_i$ are linearly independent.

Start with $\sum_i \lambda_iv_i = 0$ and assume one $\lambda_i\neq 0$ and denote $d_i = e_i -v_i$

$\Rightarrow \sum_i \lambda_i(e_i +d_i) =0$

$\Rightarrow \sum_i \lambda_ie_i =- \sum_i \lambda_id_i$

$\Rightarrow ||\sum_i\lambda_i e_i ||= ||\sum_i \lambda_id_i||$

(Take $d=\max\{||d_i||\}$)

$\Rightarrow ||\sum_i\lambda_i e_i ||^2= ||\sum_i \lambda_id_i||^2
\leq (\sum_i |\lambda_i|\cdot ||d_i||)^2
\leq (\sum_i |\lambda_i|\cdot d)^2
< (\sum_i |\lambda_i|\cdot \frac{1}{\sqrt{n}})^2
\leq \frac{1}{n} (\sum_i |\lambda_i| )^2
\leq \frac{1}{n}\cdot n (\sum_i |\lambda_i|^2)
$

The last inequality uses Cauchy–Schwarz inequality.

Now, since {e_i} is orthonormal basis we know that $||\sum_i\lambda_i e_i ||^2= \sum_i|\lambda_i|^2$ and thus:

$\sum_i|\lambda_i|^2= ||\sum_i\lambda_i e_i ||^2< \frac{1}{n}\cdot n (\sum_i |\lambda_i|^2) = \sum_i|\lambda_i|^2$.

Contradiction.


A system of 1st-order ODEs I am trying to solve $\begin{aligned} \dfrac{dx}{dt} &= -2 x - 6 y \\ \dfrac{dy}{dt} &= x - 7 y \end{aligned}$.

Let $P = \begin{bmatrix}-2&-6\\1&-7\end{bmatrix}$. My approach is to take $\operatorname{det}(P - \lambda I) =0 \implies \lambda = - 5 \vee -4$.

So then I get $\left[\begin{array}{c}
2\cr
1
\end{array}\right]$ as an eigenvector corresponding to $\lambda = -5$ and $ \left[\begin{array}{c}
3\cr
1
\end{array}\right]$ as an eigenvector corresponding to $\lambda = -4$.

So it seems to me that a solution should be:
$\begin{aligned} x_1 = A2e^{-5t}+3Be^{-4t} \\ x_2 = Ae^{-5t}+Be^{-4t}\end{aligned}$ however this is apparently incorrect. Any help anyone can give on where I went wrong is appreciated.

From second equation
$$x=y'+7y$$
derive it to get
$$x'=y''+7y'$$
substitute in equ.1
$$y''+7y'=-2(y'+7y)-6y$$
$$y''+9y'+20y=0$$
the general solution is
$$y=Ae^{-5t}+Be^{-4t}$$
now derive this equation and substitute in equ 2 to get
$$x=2Ae^{-5t}+3Be^{-4t}$$

I guess there is an $e$ missing in $x_2$ (that is, in the solution for $y$). If you take $x_2=Ae^{-5t}+Be^{-4t}$, your solution is correct. Just derive both soolutions and check that they satisfy the system.


Constructive vs computable real numbers I find it confusing that all of the following statements are true :

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

Perhaps I am misunderstanding something.
I suppose I really have two questions. In constructive analysis :

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

It is consistent with the constructive analysis that every real is computable. (In fact, it is consistent with intuitionistic ZF $\mathsf{IZF}$.) This follows from that the consistency of Church's thesis, which claims every total function from $\mathbb{N}$ to $\mathbb{N}$ is computable.
However, it does not mean constructive analysis can prove every real is computable. We know that classical analysis is a superset of constructive analysis (as a theory), that is, every statement that is provable from the constructive analysis is also a theorem of the classical analysis. And classical mathematics proves not every real is computable. Hence it would be accurate to say whether every real is computable is independent of constructive analysis.
To add some comment, I think it does not mean Bishop is simply wrong. Bishop's constructive analysis is minimal in the sense that it is contained in classical analysis, Brouwer's intuitionistic mathematics, and Recursive constructive analysis (also known as Russian constructivism.) The latter one reflects the behavior of recursive mathematics, so Bishop's claim would be true in that sense.

$\mathbb{R}$ is uncountable in the sense that there is no bijection between $\mathbb{N}$ and $\mathbb{R}$. The proof is available from Bishop's Constructive analysis. (Theorem 2.19 of Bishop and Bridges.) Here is a rough proof:

Theorem. There is no bijection from $\mathbb{N}$ to an interval $[x_0,y_0]=\{z\in\mathbb{R}\mid x_0\le z\le y_0\}$.
(Caution: $x\le y$ is not ($x< y$ or $x=y$.))
Proof. The proof uses diagonalization argument. Assume that $f:\mathbb{N}\to\mathbb{R}$ is a function. We will find a Cauchy real $x\in [x_0,y_0]$ such that $f(n)\neq x$ for all $n$.
We will construct sequences of natural numbers $(x_n)$ and $(y_n)$ recursively such that

$x_0\le x_n\le x_m<y_m\le y_n\le y_0$ if $1\le n\le m$,
$x_n>f(n)$ or $y_n<f(n)$, and
$y_n-x_n<1/n$.

Assume that $x_i$ and $y_i$ are given for all $i<n$.
Then we have either $f(n)>x_{n-1}$ or $f(n)<y_{n-1}$. (This follows from the following constructively valid theorem: if $x<y$ are reals, then either $z<y$ or $x<z$.)
Assume that we have $f(n)>x_{n-1}$. (The remaining case is analogous.)
Choose a rational number $x_n$ and $y_n$ such that $x_{n-1}<x_n<\min(a_n,y_{n-1})$ and $x_n<y_n<\min(a_n,y_{n-1},x_n+1/n)$. Then the mentioned inequalities are satisfied.
Hence $(x_n)$ forms a Cauchy sequence of rational numbers. Let $x$ be a limit of $(x_n)$, then $x$ satisfies the desired properties.

Note that the above proof makes use of the countable choice that Bishop accepted. As far as I know, it is open whether $\mathbb{R}$ is countable or not without the countable choice. Also, note that $\mathbb{R}$ can be subcountable, i.e. $\mathbb{R}$ can be an image of a subset of $\mathbb{N}$. (Unfortunately, I forgot whether the subcountability of $\mathbb{R}$ is compatible with $\mathsf{IZF}$, although I believe it is true.)

1.Because there is at least one real number that is not computable.
2.
Step 1 : 0.0 , 0.1
Step 2 : 0.00 , 0.01 , 0.10 , 0.11
Step 3 : 0.000 , 0.001 , 0.010 , 0.100 , 0.011 , 0.101 , 0.110  , 0.111
step n : 0.(all combinations 1,0 of length n)
continue above indefinitely in binary to get all values in the interval 0,1.


When is $\gcd(a+b,c)=\gcd(a,c-b)$? Let $a,b,c$ be three positive integers. Under which conditions do we have $\gcd(a+b,c)=\gcd(a,c-b)$ ?

Set $d= a+b$. Then the question becomes: When is $\gcd(d,c) = \gcd(d-b,c-b)$?

Let $g=\gcd(d,c)$. Then $d=sg$ and $c=tg$ (with $s,t$ coprime). So there are two cases when the sought-for equality will not hold:

$g$ does not divide $b$ (equivalently, $b$ is not a multiple of $g$).
$b=rg$ such that $s-r$ and $t-r$ have a common factor, so the resulting $\gcd$ is a multiple of $g$. This is harder to test for.

Sample case: $(a,b,c) = (40, 6, 16)$. $\gcd(a+b,c) \ne \gcd(a,b) \ne (gcd(a,b-c)$


Formula for calculating volume discount for bulk product purchases I'm trying to work out a basic formula for calculating a volume discount for volume product purchasing (i.e: the per-unit price decreases more and more as the number of units purchased increases). The simple way to go is to set pricing tiers (i.e: <100 = $40.00 per unit, 100-199 = $30.00 per unit, 200-299 = $25.00 per unit etc.), but if this were to be plotted on a graph displaying per-unit prices as the number of units purchased increased, it would show a stepped trend, i.e:

Therefore, a similar graph showing total cost as the number of units purchased would also show a stepped trend.

I was hoping there is a formula I can use that dispenses with the lazy tiering method, and can provide a smooth trend at a per-unit resolution, suiting the below example:

What would be a good way to achieve this?

Any help would be much appreciated.

You would need to use the X to the Y power function on the calculator and use a decimal for the power, .95 or something like that to represent the discount for volume discounts. For example X($40)to the .95 power. "X" would be the number of units they are  purchasing and the $40 is your starting price and the .95 power would do the discounting (cheaper per unit the more more you buy). Apologize for the autoformatting I can't seem to fix it so I included a written equation in the link.
Click on link to view equation written out
x($40) to the .95power


Why are absolute values involved in functions of random variables? From a textbook:

  If $X$ is a continuous random variable, then so too is the new random variable $Y = Y (X)$. The
  probability that $Y$ lies in the range $y$ to $y + dy$ is given by
  $$g(y)=\int_{dS}f(x)dx,$$
  where $dS$ corresponds to all values of $x$ for which $Y$ lies in the range $y$ to $y +dy$.
  The simplest case occurs when $Y (X)$ possesses a single-valued inverse
  $X(Y)$. In this case, we may write
  $$g(y)dy=\lvert\int_{x(y)}^{x(y+dy)}f(x')dx'\rvert=\int_{x(y)}^{x(y)+\lvert           \frac{dx}{dy}\lvert dy}f(x')dx',$$
  from which we obtain
  $$g(y)=f(x(y))\lvert\ \frac{dx}{dy} \lvert.$$

What I know:

For this case, $x$ and $y$ has one to one correspondence, ie for each
value of $x$, $Y(x)$ gives a unique value of $y$. Similarly $X(y)$ give
unique $x$.
What the integral does is to integrate a range of infinitesimal probability $f(x)$. That range is all $x$ that gives that particular value of $y$ by $Y(x)$.

What I don't know:

Why is the integral needed to be "absoluted"? I think $f(x)$ is always positive because it is a probability distribution and by definition $0\le f(x)\le 1$, so there is no need to make sure the integral gives non-negative values.
How is the absolute sign moved to the upper limit of the integral?
How to evaluate the equation to give the final result? I think the text just divide the whole equation by $dy$ to give the final result and ignore what happens to the lower and upper limit of the integral. Again I don't how the absolute sign appears.

Attempts to answer:
I searched wikipedia and google and found several online lecture notes. None of them explained the source of the absolute sign.

Find a graph of the normal distribution function.  Compare the left half as a PDF and the right half as  PDF.  (Each half is monotone, so each half is invertible.)  Pick a narrow range of $y$ values; there will be $x$s in each half that contribute mass to the PDF in terms of $y$.  It doesn't matter whether the derivative is positive or negative, they contribute positive mass.

Now the integral you have has the derivative appearing at one endpoint.  You think of it as the upper endpoint, but that is only true if the derivative is positive.  If the derivative is negative, your endpoints of integration are reversed and the result of the integral is negative.  One way to correct this is to take the absolute value of the integral -- you don't care whether the PDF in $x$ was increasing or decreasing, you only care how widely it was distributing mass among the $y$s.  This is an answer to your question 1.

For your question 2, again we only care about the magnitude of the slope (how rapidly we move through the $y$s receiving mass for each unit of $x$).  Further, we may consider the linear approximation (in particular, by Taylor series: $x(y + \mathrm{d}y) \approx x(y) + \dfrac{\mathrm{d}x}{\mathrm{d}y} \mathrm{d}y$) as running from the upper bound of integration down or from the lower bound of integration up.  If your derivative is negative, swap the order of the endpoints and switch from which endpoint the linear approximation proceeds.  This will give you the absolute value on the derivative.

The fundamental theorem of calculus with the chain rule is the answer to your third question.  (The upper endpoint is moving $\left| \dfrac{\mathrm{d}x}{\mathrm{d}y} \right|$-times as fast as $\mathrm{d}y$, which constant pops out by the chain rule.)

Sorry, I'm on a smartphone so I'll answer shortly.

1) $f>0$ but $x(y)$ could decrease, the integral must be taken from the lowest $x$ so, using the absolute value, you are just swapping the integration limits.

2) same as above, but we don't swap the limits but force the upper limit to be greater than the lower (here we abuse the $dx$ smallness).

3) suppose you know the primitive, the integral is a difference that, for $dx$ going to $0$, becomes again the $f$ function.


Finding the 3rd axis in 3d space I want to draw 3 perpendicular axis in 3d space, for example like this

I figured out this much:

start by drawing the first axis (x). Any line will do
draw the second axis (y). Any line that intersects the x axis should do (because in 3d space we can find an infinite ammount of lines perpendicular to x, and it is impossible to know if a line is perpendicular or not without the 3rd axis)
make a ritual sacrifice to the gods of math
???

The first two axis can be basically any lines that intersect, as there is an infinite ammount of 2d planes in 3d space. But the third axis has to be perpendicular to the first two, and there is exactly ONE such line in 3d space. I just don't know how to find it.

You can use the cross product.

Given two vectors

$$v_1=(x_1,x_2,x_3)\;,\;\;\;v_2=(y_1,y_2,y_3)$$

the perpendicular vector is

$$v_3=v_1\times v_2:=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2\,,\,x_3y_1-x_1y_3\,,\,x_1y_2-x_2y_1)$$

the direction of $v_3$ is given by the right hand rule.

Done a lot of research and came up with this:

Draw any line. This will be our x axis.
Draw any line intersecting with x. This will be our y axis.

This is where it gets interesting. I thought that two lines fully define a plane. No, when projecting from 3d to 2d space they actualy don't. This is because the plane can also be tilted. If you only projected the axis onto the picture plane, then there can be an infinite ammount of planes with different degrees of tilt that offer the same projection, for example

and

(purple is an example rectangular shape on the infinite planes)

okay so here's how we go about it:
3. Draw a point on each axis and connect them. The distance from the center of the coordinate system and a given point will represent one unit of length on that axis

IT IS IMPORTANT to make sure that the angle shown on the image is at least 90*. If it isn't you have to put your points elsewhere.

Draw a line going through the center of the coordinate system and perpendicular to the line you've just drawn

And there you have it! That's your z axis.

Now, you may ask "but what about the z axis metric? We have a metric for x and y, we could use a metric for z". Here's how:

BONUS STEP:
Draw a line going through one of the points you've marked earlier and perpendicular to either x or y. Wherever that line intersects z is your metric. You can double check by drawing the other line and making sure that it intersects in the same place.


How to solve linear differential equations such as $y''-7y'+10y=0$? The question asks to solve the linear equation

$$y''-7y'+10y=0$$

I'm not sure where to even start.

I'm new to this so it would be nice if I got an in depth explanation ( with rules too )

Thank you!

I'm sure others will rush in and give you the solution to your particular equation.

Answering this question the way you want requires a whole lecture in ODEs. I recommend watching this good one from MIT on the exact topic, Solving second order linear ODEs with constant coefficients:

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-9-solving-second-order-linear-odes-with-constant-coefficients/

or equivalently

https://www.youtube.com/watch?v=vP-oRQqmeg4#t=30

Your differential equation has characteristic equation $t^2-7t+10=0$, which forces $t=-5,-2$. Therefore the solutions are of the form $f(x)=Ae^{5t}+Be^{2t}$.


Working with C++ for GF(2) Pardon me if it is off topic.But, is there anyone who could suggest me some basics with how to get started with working with C++ for GF(2)??

I am new in C++.I am learning to working with arrays and pointers. I have no idea if there is any in built function like MATLAB. Would be so helpful if anyone could suggest me anything. Any kind of suggestion is appreciated.

Thanks in Advance

You could define your own operators for + and * by operator overloading. That means that you define the rules of GF(2) yourself, and you can then compute with it using the + and * operators.

C++ is a general purpose programming language. It comes with a standard library, but nothing comparable to a dedicated mathematics package like Matlab.

Usually you look for external libraries that offer the desired functionality, in case you should really need unique Matlab functionality, you should investigate about interfaces between Matlab and C or C++. (Example)


Proving that $\mathbb{Q}$ is neither open or closed in $\mathbb{R}$ I need to figure out if $\mathbb{Q}$ is either open or closed in $\mathbb{R}$ and prove my assertions. I don't think it's either. I would like a review/critique of my proofs to make sure that there's not an easier way to do it.

To see that it's not open, take some $\frac{m}{n}$ with both $m$ and $n$ perfect squares so that $\left(\frac{m}{n}\right)^{\frac{1}{2}}$ is rational. We have that $x \mapsto x^{\frac{1}{2}}$ is continuous. Hence for any $\epsilon$-neighborhood of $\left(\frac{m}{n}\right)^{\frac{1}{2}}$, there exists a $\delta$-neighborhood of $\frac{m}{n}$ that maps into it. but

$$
\begin{eqnarray*}
\frac{m}{n} - \frac{cm}{cn + 1} < \delta &\Leftrightarrow& \frac{m}{n(cn + 1)} < \delta n(cn + 1) \\
&\Leftrightarrow& \frac{m}{\delta n^{2}} < \left(nc^{2} + 1\right)^{2} \\
&\Leftrightarrow& \frac{m}{\delta n^{4}} < c^2 + \frac{2}{n}c + \frac{1}{n^{2}} \\
&\Leftrightarrow& \frac{m}{\delta n^{4}} < \left(c + \frac{1}{n}\right)^{2} \\
&\Leftrightarrow& \left(\frac{m}{\delta n^{4}}\right)^{\frac{1}{2}} - \frac{1}{n} < c
\end{eqnarray*}
$$

So we can make $\frac{cm}{cm + 1}$ as close to $\frac{m}{n}$ as we like by taking $c$ sufficiently large. If $cm$ is a perfect square, then $c$ must be a perfect square because $m$ is so we can just choose a $c$ that's not a perfect square and hence $\left(\frac{cm}{cn +1}\right)^{\frac{1}{2}}$ is irrational. Because we can place an irrational arbitrarily close to at least one positive rational, $\mathbb{Q}$ is not open.

To show that $\mathbb{Q}$ is not closed, simply pick some irrational $\alpha$ and let $\epsilon_n = \frac{1}{n}$. Let $x_n$ be chosen from between $\alpha - \epsilon_n$ and $\alpha$. Then the sequence $x_n$ converges to $\alpha$ which shows that the complement of $\mathbb{q}$ in $\mathbb{R}$ is not open as we can place a rational in any $\epsilon$-neighborhood of any irrational.

Proving that the rationals aren't open was so much harder than proving that they're not closed because my book hadn't proved that the irrationals are dense in $\mathbb{R}$ and so I couldn't use that. Is there some other way that I could do this?

You are certainly right that all you need to show that $\mathbb{Q}$ is not open is to show that given a rational $q$, you can find irrationals arbitrarily closed to $q$. A very simple way of doing this is to take your favorite irrational, say $\sqrt{2}$, and then consider $\sqrt{2}/n$ with $n$ an integer. These numbers are never rational, and get arbitrarily small; so if $q$ is a given rational, then $q+\frac{\sqrt{2}}{n}$ will be as close as you want to $q$ by taking $n$ large enough.

To show Q is not open you can use the fact that Q is not open if and only if the complement of Q is not closed.....take a sequence Un in Q complement (R\Q)such that Un = (n+√2)/(n+1) .......
.lim Un = p....(p∈R) then p ∈Q so p∉Q complemet ....then Q complement is not closed therefore Q is not open


Could the Riemann hypothesis be provably unprovable? I don't know much about foundations and logic, so I ask forgiveness if my question is just plain stupid.

Assume we have a statement of the form:

  There exist no $x\in X$ such that $P(x)$.

where $X$ is some kind of set (or class, or similar stuff) and $P$ is a set of properties. An example of such a statement could be the Riemann hypothesis:

  There exist no $x\in\mathbb{C}$ such that $\Re(x)\neq\frac{1}{2}$ such that $x$ is not a negative even integer and $\zeta(x)=0$.

Can such a statement be provably unprovable?

Intuitively, I would say no, because to show that it is unprovable we should show that we cannot find $x\in X$ such that $P(x)$ (else finding such an $x$ would be a proof that the statement is false), but doing so would prove the statement to be true.

Is this correct, or am I missing something?

Edit: Please read the question correctly: it is not properly a question on the RH, but more a question on logic!

You can check out the answers to this related MO question:

"Can the Riemann hypothesis be undecidable?"

No. The problem is in the definition of zeros. If we would not have 1,2,3,4,5,6... on the left side of the equation, which is the sum

we could argue that we do not have enough information about zeros, hence some of them are not defined in some particular sense, hence Riemann cannot be decided. However, although we cannot calculate all of the zeros, there is nothing ambiguous about them. They are all perfectly defined objects on their own accord, so if the Riemann hypothesis is false there is the first zero off 1/2 mark.

Although I said this, hypothetically speaking it could be true that the set of zeros is not defined completely, by simply stating that they are zeros of the extended sum as above. This might mean that we would need to add something to the definition of natural numbers, some minuscule and yet not discovered condition that would change their nature in such a way that with one condition Riemann is true, yet with another false.

This last is not likely because we can already define the way of calculating any zero anywhere, although we do not have time and space to collect all of them.
So it is definitely true or false with Riemann.


what does it mean equal in distribution Consider a discrete countable probability space $V$ and consider two probability measures such that, for any finite set
$A \subset  V$,
$$
P_1(A) = P_2 (A).
$$

How to prove that, for any bounded measurable function $f : V^{\mathbb{N}} \rightarrow \mathbb{R}$ ?
$$
E_1(f(x)) = E_2(f(x)),
$$
where $E_1( \, \cdot \,)$ is the expectation with respect to $P_1$ and
$E_2( \, \cdot \,)$ is the expectation with respect to $P_2$? Is the boundedness assumption necessary just to ensure that the expectations are well defined?

We know that $E_1(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_1(x)$ and $E_2(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_2(x)$

$P_1(A) = P_2(A)$ for $A \subset V$, then we have $P_1(x) = P_2(x)$ for $x \subset V^{\mathbb{N}}$.

So the two expectations are equal.

I believe that the boundedness is necessary. If $f(x)$ is infinite for some $x$, then the corresponding term in $E_1$ or $E_2$ would be infinite if the $x$ has non-zero probability (since the distribution is discrete), and thus the expectation would not be well defined.


What's 4 times more likely than 80%? There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.

What's the new probability as a percentage and how do you work it out?

As I remember it the question was posed like so:

  Suppose there's a student, Tom W, if you were asked to estimate the
  probability that Tom is a student of computer science. Without any
  other information you would only have the base rate to go by
  (percentage of total students enrolled on computer science) suppose
  this base rate is 80%.

  Then you are given a description of Tom W's personality, suppose from
  this description you estimate that Tom W is 4 times more likely to be
  enrolled on computer science.

  What is the new probability that Tom W is enrolled on computer
  science.

The answer given in the book is 94.1% but I couldn't work out how to calculate it!

Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.

The most reasonable way to match the answer in the book would be to define the likelihood to be the ratio of success over failure (aka odds):
$$
q=\frac{p}{1-p}
$$
then the probability as a function of the odds is
$$
p=\frac{q}{1+q}
$$
In your case the odds are $4:1$ so $4$ times as likely would be $16:1$ odds which has a probability of
$$
\frac{16}{17}=94.1176470588235\%
$$
This matches the $3\%$ to $11.0091743119266\%$ transformation, as well.

Bayes' Rule

Bayes' Rule for a single event says that
$$
O(A\mid B)=\frac{P(B\mid A)}{P(B\mid\neg A)}\,O(A)
$$
where the odds of $X$ is defined as earlier
$$
O(X)=\frac{P(X)}{P(\neg X)}=\frac{P(X)}{1-P(X)}
$$
This is exactly what is being talked about in the later addition to the question, where it is given that
$$
\frac{P(B\mid A)}{P(B\mid\neg A)}=4
$$

The only way I see to make sense of this is to divide by $4$ the probability it does not happen. Here we obtain $20/4=5$, so the new probability is $95\%$.


Is this proof claiming that P(A union B) does not equal P(A) union P(B) correct? Let $A$ and $B$ be finite sets and $\mathbb{P}(A)$ the power set of $A$. Then $\mathbb{P}(A \cup B) \neq \mathbb{P}(A) \cup \mathbb{P}(B)$ for some $A$ and $B$.

Let $A$ be the set $\{1,2,3\}$ and $B$ the set $\{4,5,6\}$. Then $|A \cup B| = |A|+|B| = 6$ since A and B are disjoint. And so  $|\mathbb{P}(A \cup B)| = 2^6$. On the other hand the size of $\mathbb{P}(A) \cup \mathbb{P}(B)$ is at most $|\mathbb{P}(A)|+|\mathbb{P}(B)| = 2^3+2^3=2^4$. Since the two sets don't have the same size, they cannot equal.

Is this correct?

What you said is correct but does not prove the theorem. You only proved that the equation is not in general true. The theorem is that the equation is always false. You need a more general cardinality argument.


How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$? How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$ without a proof by exhaustion?

$$ 4b+3m=17 \implies 4b=17-3m $$

$$\implies 4b=16+(m+1)-4m = 4(4-m)+ (m+1) $$

Thus $m+1$ must be a multiple of $4$ where $m<4$ to make  both sides positive.

The only choice is $m=3$ which implies $b=2$

If $b\geq 9$ then (since $m\geq 1$) $$4b+3m>18+3m>17$$
If $m\geq 6$ then (since $b\geq 1$) $$4b+3m>4b+18>17$$

Thus there are only finitely many solutions, $1\leq b< 9$, $1\leq m< 6$, and they can be checked directly.


Bound on the sup norm for derivatives of a particular $C^\infty$ function I'm reading textbook "A Primer of Real Analytic Functions" and on page 86 the following "obvious" claim is made:

  Let $|| \cdot ||$ be the sup norm on $[0, 2 \pi]$ and define function  $f$ to be $e^{\frac{1}{x(x-2\pi)}}$ on $(0, 2 \pi)$ and $0$ at points $0$ and $2 \pi$. Note that $f \in C^\infty([0,2\pi])$. Show that there exists a constant $C>0$ s.t. for all $n \in \mathbb{N}$ we have $$ || f^{(n)}|| \leq C \cdot 2^{2^n}$$

I suspect that this bound can be obtained by noticing that $f$ is in some class that can be represented by some integral kernal - however I know of no such thing.  The book mentions a couple facts from Fourier analysis so perhaps that is relevant - I've included it as a tag.

This is only a hint.

Try the following problem first: Define $g(t) = e^{1/t}$ on $(0,1]$ and $g(0) = 0.$ Can you show that $g(t) \in C^\infty([0,1])?$ What estimates can you make on the sup norm of the derivatives of $g.$

Now the problem at hand. To show $f$ is $C^\infty$ on an interval like $[0, \epsilon]$ with $\epsilon < 2\pi,$ write $f(t) = g(t)^{1/(t-2\pi)}.$ You can use your solution from the above part, to obtain the conclusion. For $f$ near $2\pi,$ repeat the same argument with the two singularities switched around.


Equation of tangent and normal for translated conics Suppose I have a translated parabola
$(y-k)^2=4a(x-h)$
and I need to find the equation of its tangent, can I just replace $x$ and $y$ in
$y=mx+a/m$
by $(x-h)$ and $(y-k)$?
similarly can I do that with its equation of normal?
And if not, what will be the correct equations of normal and tangent?

P.S. I would appreciate it if you explained it using the theory of conics rather than calculus as I haven’t studied calculus yet.

What basically happens during translation is the shift of the origin.(0,0) shifts to (h,k).Thus, every y coordinate changes to y-k and x to x-h.Think like this-
"You have a circular rubber-band, and on it you have a net.When you 'move' the 'net' keeping the circle's position under it fixed, you actually shift the origin in terms of coordinate geometry.With reference to the previous position of the net, you have shifted it to a new position, by a distance 'h' horizontally and 'k' vertically.Thus the new origin with respect to the old has 'shifted' to (h,k).And so changes the x and y coordinates, which, too, 'shifts' by h and k...


Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:

$\sqrt{4x-1} = \sqrt{x+2}-3$
$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$

How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.

Note that if your equation includes $\sqrt{4x-1}$, this means that $4x-1\ge0$. Similarly, you are only looking for solutions among numbers that fulfill $x+2\ge0$, since the equation contains the expression $\sqrt{x+2}$.

So you can write $\sqrt{(4x-1)^2}=|4x-1|=4x-1$. But do not forget in the end include only those solutions where $4x-1\ge0$, i.e., $x\ge\frac14$. (And the same for the condition $x+2\ge0$.)

After thinking a bit about some of what @Daniel Hast typed, I realized that the reason why, say, $\sqrt{(4x-1)^2} = 4x-1$ was that more generally, given that $\sqrt{x} = r$ such that $r^2 = x$, squaring $r$ in $\sqrt{x} = r$ produced $(\sqrt{x})^2 = r^2 = x$ and subsequently $(\sqrt{x})^2 = x$.


Convergence of $\sum _{n=1}^{\infty \:}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$ $$\sum _{n=1}^{\infty \:}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$$

Upon research I know this series diverges because of the comparison test but I do not know what to compare the series to as I am having trouble finding a series smaller that diverges. And I know when I use the comparison test, my comparison series will diverge because of the $p$-series test.

Try multiplying the top and the bottom by $\sqrt{n+1}+\sqrt{n}$. After simplifying we get
$$\frac{1}{\sqrt{n}(\sqrt{n+1}+\sqrt{n})}\geq \frac{1}{\sqrt{n+1}(\sqrt{n+1}+\sqrt{n+1})}=\frac{1}{2(n+1)}$$

If you are stuck, I'd try doing a little algebra first. Note that $$\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}} = \frac{\sqrt{n+1}}{\sqrt{n}}-\frac{\sqrt{n}}{\sqrt{n}} \\ = \sqrt{\frac{n+1}{n}}-1 $$ Now you can compare $$\sqrt{\frac{n+1}{n}}-1 \geq \sqrt{\frac{0+1}{n}}-1 \\ = \frac{1}{\sqrt{n}}-1$$ and hence $$-1+\left(\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\right) \leq \sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$$


Does finite projective dimension localize? Let $R$ be a commutative (but not necessarily Noetherian) ring with unity.  Let $M$ be an $R$-module.  Suppose that, for all $\mathfrak p \in\text {Spec}(R),$ $\text{pd}_{R_{\mathfrak p}}M_{\mathfrak p}< \infty $.  Is it the case that $\text{pd}_RM < \infty$?

Let $R$ be a ring with $\operatorname{gldim}(R)=\infty$. Then there exist $R$-modules of arbitrary large projective dimension. By taking their direct sum we can find an $R$-module $M$ of infinite projective dimension. Now, if the global dimension of all localizations of $R$ is finite, that is, $R_{\mathfrak p}$ is a regular local ring for all $\mathfrak p\in\operatorname{Spec}(R)$, it follows that the projective dimension of all localizations of $M$ is finite.

Such an $R$ is the classical example of Nagata of a noetherian ring of infinite Krull dimension.

This is wrong, I'm leaving it up in case it helps anyone else avoid the mistakes I just made :) Thanks to QiL and YACP for the pointers!

Okay yeah you can carry out my first comment for affine schemes: namely, if you take an arbitrary product of rings $R_i$, $$Spec \; \prod R_i \simeq \coprod Spec \; R_i$$

Where $\coprod$ is disjoint union. With that, if we take a countable product of say $$R_i  = \mathbb{C}[x_1, \ldots, x_i]$$just so that $gldim \; R_i = i$, and take $M$ to be literally $(x_1, \ldots, x_i)$, inside $Spec \; R_i$ it has projective dimension $ i$ at the origin and is 0 elsewhere.


How to classify the opinions to get the best prediction? Assume I have four friends, each of them will predict the result of each football match (based on history data, weather data, etc). They will tell me their decisions.

I also have the set of their data history data and weather data. I can also do the similar prediction. However, I want to classify which of my four friend make the right decision instead of predict the football match. (The reason I said that because I do NOT want to put all the data including my friends' decisions into a classify to make the decision. It is because their decisions will be the same weight of the history and weather data. My friends' decisions should weight much more  )

How I can maximize the correct prediction?

Your model is highly inefficient. Each of the players prediction has no influence on the other players predictions' truthfulness. So instead of inputting them all at once, you should input them sequentially. So:

friend1 + other data = output
friend2 + other data = output
friend3 + other data = output
friend4 + other data = output

This makes your network smaller, making the training happen in less iterations and in a less amount of time.

But still. You shouldn't input your friends prediction at all. The only thing that your network is doing is comparing it's own prediction with your friends prediction internally.

In what way should your friends' predictions add anything useful to the (each others) output?


What can I guess for this differential equation? What can I guess as a particular solution for $y''+25y = -x\sin(5x)$

I tried $(Ax+B)(C\cos(5x)+Dsin(5x))$, but that didn't work.

We need to use something similar to the right hand side i.e.
$${y_p} = Ax\sin (5x) + Bx\cos (5x).$$

Do you understand why?

If you do not understand why this is the way we guess, then try another but longer method. Take all the functions that appear in the product on the right hand side. Multiply them together and then simply put undetermined coefficients in from of each term.

If one of the functions is $\cos (ax)$ or $\sin (bx)$ then $\cos (ax) + \sin (bx)$ should appear in the product.

Here is a list of different combinations of functions for particular solutions.


Upper and lower bounds in rounding to 1 Decimal place I have a measurement $a=8.3$cm to 1 d.p. I'd like to find the upper bound of this. Why is the upper bound $8.35$cm? not $8.3\dot{4}$cm? Surely 8.35 rounds to 8.4cm?
And the lower bound $8.25$cm not $8.24\dot{9}$cm?

The upper bound is 8.35 cm because anything less than 8.35 will round to 8.3, but anything 8.35 or greater will round to 8.4. 8.35 itself does not round to 8.3, but it is the smallest such number, or the least upper bound.

Similarly, the lower bound is 8.25 because anything less than 8.25 rounds to 8.2, and 8.25 rounds to 8.3, so 8.25 is the lower bound (And also the minimum).

Also, your question about $8.25$ and $8.24\dot9$ doesn't really make much sense -- those two decimals represent the same real number.

Its because the upper bounds of 8.3 is 8.349999999999.... So you round it up to 2 decimal places which is 8.35.


Do gradients existing almost everywhere for a pointwise maximum function Consider the following function: $f: \mathbb{R}^n\rightarrow\mathbb{R}$
     $$f(x)=\max_{p\in\mathcal{P}} x^Tp$$
where $\mathcal{P}\in\mathbb{R}^n$ is a nonempty and compact set.

It is easy to see that $f(x)$ is a piecewise convex function but is it true that $f(x)$ has gradients almost everywhere? Any references and counterexamples are very welcome.

It is not actually the case that $f$ is always a piecewise convex function.  It depends completely on the set $\mathcal{P}$.

Onto your question, I assume by "almost everywhere" you mean that the set of non-differentiable points has measure $0$.

So, if $\mathcal{P}$ has measure $0$, then $f$ is differentiable "almost everywhere".

If $\mathcal{P}$ has non-zero measure, then it may be the case that $g$ is non-differentiable on a set with non-zero measure (some subset of $\mathcal{P}$ that has non-zero measure, perhaps $\mathcal{P}$ itself).


Convergence of $\sum_n^\infty (-1)^n\frac{\sin^2 n}n$ Could anyone give a hint how to prove the convergence of the following sum?

$$\sum_n^\infty (-1)^n\frac{\sin^2 n}n$$

I tried writing it like this instead:

$$\sum_n^\infty \frac1n (-1)^n \sin^2 n.$$

From here, it is easy to see that $\frac1n$ is a bounded and strictly decreasing sequence. It would be sufficient to prove that the sequence of partial sums of $(-1)^n\sin^2 n$ is bounded.

From here, I get that $(-1)^n\sin^2 n = (-1)^n\frac{1 - \cos 2n}2 = \frac{(-1)^n}2 - \frac{(-1)^n \cos2n}2$, where the sequence of partial sums of $\frac{(-1)^n}2$ is bounded as well as the sequence of partial sums of $\frac{\cos 2n}2$. Unfortunately, I cannot tell anything about $\frac{(-1)^n\cos 2n}2$.

Thank you.

We will use Dirichlet's test: http://en.wikipedia.org/wiki/Dirichlet%27s_test

Write $\frac 1 n (-1)^n\sin(n)^2 = \frac 1 n  (-1)^{n+1} \frac{e^{2in}-2+e^{-2in}}{4}$, which follows from Euler's formulas.

Our series naturally splits in a sum of $3$ series. Set $a_n = \frac 1 n$. Set $b_n =(-1)^{n+1} \frac{e^{2in}}{4}$. Then $\sum_{n=1}^M b_n $ is a geometric sum, with quotient $-e^{2i}$, that can be evaluated as $\frac{e^{2i}}{4} \frac{(-1)^M e^{2iM}-1}{-e^{2i}-1}$. We see that $|\sum_{n=1}^M b_n|<2$ for all $M$. By Dirichlet's test $\sum a_n b_n$ converges. The remaining two series can be handled similarly.

Using Alternative convergence theorem.
As long As $b_n$ has limit when $n$ approaches infinity, and the limit equals to $0$,
the serise $a_n=(-1)^n b_n$ is convergence.


Is aleph-$0$ a natural number? Would I be right in saying that $\aleph_0 \in \mathbb N$?

Or would it be a wrong thing to do?

$\aleph_0$ is not a natural number. It is the cardinality of the set of natural numbers - each individual natural number is finite, but the set of all natural numbers is infinite.

Based on the comments below, let me share an anecdote: I was once part of a medical survey on Things Neurological. One of the tasks they had us do was come up with short definitions of common words, on the spot. Beforehand, I (talk about tempting fate) joked about how easy this would be. The very first word they gave me: "Number." For the life of me, I couldn't come up with anything. The doctors gave me very weird looks.  Finally I sputtered something like "An element of a formal system," which didn't help with the weird looks but did let us move on to other, simpler words.

The Moral: Man, number is weird.

No.

1 is a natural number.
A number made by taking a natural number and adding 1 to it is a natural number.
There are no other natural numbers than the ones whose existence is implied by 1 and 2.

$\aleph_0$ is not equal to 1. Therefore it can only be a natural number if there is some natural number $x$ such that $x+1 = \aleph_0$. If you think it is a natural number, you have to be able to say what $x$ is.

In fact, there is no such $x$.


Cube question without coordinates, cross sections and planes We have a cube with corners $ABCDA_1B_1C_1D_1$. The points $A_1B_1C_1D_1$ lie above the square $ABCD$. We also have a plane that goes through the middle of line $BC$, through the middle of square $ABA_1B_1$, through the middle of $A_1B_1C_1D_1$. At what ratio does the plane cut line $AB$.
What I did, I've put the cube into the coordinate system $B(0,0,1), A_1 (0,1,0) \text{ and } A(0,0,0)$, $D(1,0,0)$ and then it was pretty easy to solve. I just got the equation of the plane and then solved for intersection of the line that goes through AB, to get the point. Then I knew the ratio.
My question is, how to solve this problem without putting the cube into the coordinate system?
SOLUTION: solution should be that the plane cuts AB in ratio 3:1

Fairly quickly:
(1) the plane (call it $\Pi$) also passes through the centre $M_{ABC_2D_2}$ of square $ABC_2D_2$ ($C_2,D_2$ are the reflection of $C,D$ in line $AB$).  This is just extending $M_{A_1B_1C_1D_1}M_{ABB_1A_1}$ to meet the bottom plane $ABCD$.
(2) So the line $M_{BC}M_{ABC_2D_2}$ lies in $\Pi$.  This line intersects AD at $X$ where $AX=\frac32 AD$ and $X$ is opposite to $D$.
Now look at the configuration $A,B,X,M_{BC}$ and you have similar (right-angled) triangles $\triangle AXY$ and $\triangle YBM_{BC}$ where $Y$ is the intersection of diagonals, i.e., intersection of $AB$ with $\Pi$.  So $AY:YB=AX:BM_{BC}=3:1$.

In response to the question how I solved with coordinate system:
Let's say:
$$AA_1 = (0,0,1) = c$$
$$AB = (0,1,0) = a$$
$$AD = (1,0,0) = b$$
Now we get all three points as mentioned above:
$$P_1 = c + (1/2)a + (1/2)b = (1/2,1/2,1/2)$$
$$P_2 = (1/2)c + (1/2)b = (0,(1/2),(1/2))$$
$$P_3 = a + (1/2)b = ((1/2),1,0)$$
Now we get two vectors $P_2P_3$ and $P_2P_1$. When we do cross product we get: $$n = (1,-2,-1)$$
The equation of the plane is thus: $$x - 2y - z = \frac{-3}{2}$$
Because: $\text{We use the second point and get:}0*1 + (1/2)*(-2)+(1/2)(-1) = \frac{-3}{2}$
Now we know that $AB$ is on  $y$ axis , or the second coordinate as we may call it is a parameter $t$. So the equation of the line on which $AB$ is on is $y = t, x =0, z=0$, where $t$ is the parameter.
Now we input $y =t$ into the plane equation and get $$(-2t = \frac{-3}{2}\iff t =\frac{3}{4} )\implies \text{ratio is: } 1:3$$


Using the product rule when differentiating I am asked to find $dy/dx$ if $y = x(2x + 5)^7$ using the product rule.

I have made $U = X$ and differentiated this to get $1$.

And $V = (2x + 5)^7$ and differentiated to get $7(2x + 5)^6$.

However the solutions say that $V'$ should be $14(2x + 5)^6$.

Should it be $7$ and not $14$ ? Has there been an error in the solutions or am I incorrect ?

By the way I got $U$ and $V$ from:
$$
\frac{dy}{dx} = U \cdot \frac{dV}{dx} + V \cdot \frac{dU}{dx}
$$

The problem here, is that you have not used the chain rule.

Put simply, the chain rule says:

$\text{ If you have a function composed of another function }$

$\text{ you must take derivative of outside function $\times$ derivative of inside function }$

$$\frac{d}{dx}\left[ f(g(x))\right] = f'(g(x)) \cdot g'(x)$$

In our case here, we have $(2x+5)^7$. What you must recognize is that this is a combination of $(2x+5)$ and $x^7$.

So we get that the derivative of $(2x+5)^7$ is:

$$7*(2x+5)^6 \text { (product rule on the outside $x^7$ function)} \times (2) \text {(linear differentiation of 2x + 5)} $$

$$14(2x+5)^6$$

You can not apply the reasoning that $\frac{d}{dx}(2x+5)^7 = 7(2x+5)^6$. You have attempted to apply the power rule to a function that is not a polynomial term, and the power rule applies only to functions of the form $ax^n$. Instead, you may do two things:

Expand and apply power rule to each individual term (slow)
Apply the chain rule

The latter of these options you may not have been introduced to you, so here it is:

$$\frac{d}{dx}\left[ f(g(x))\right] = f'(g(x)) \cdot g'(x)$$


Phase difference of two signal of different frequency Currently, I have two signals, the main components of both signals are 60Hz, but both also have weaker response at 180Hz + small amount of noise. As shown in the photo below, I want to find the phase difference between the 60Hz component of first signal, and the 180Hz component of the second signal.

As you can see the second waveform of the figure, the phase lagg is of the 180Hz signal is 1.39ms, which coresponds to a 90 degree phase shift.

I am wondering, if I want to implment this phase measurement, could I simply take the fft of two signals, and then find the phase at 60Hz of signal 1, and subtract the phase at 180Hz of signal 2?

Or if there is better way to do so?

if im not wrong this figure is from "UNIQUE ASPECTS OF DISTRIBUTION SYSTEM HARMONICS DUE TO HIGH IMPEDANCE GROUND FAULTS"

Actually, measuring the phase difference between two different frequency signals does not make sense. imagining two rotating sticks, one is faster than another one, the phase difference keeps changing.


Averaging over Mahalanobis distance vectors of different clusters Given vectors from two different clusters (in particular in my example from two different experimental conditions, called "CS" and "US") where the Mahalanobis Distance is calculated according to:
$M^2 = (x-\mu)^T \Sigma^{-1}(x-\mu)$
the explanation reads as follows: "where x and $\mu$ are individual and mean population vectors for CS and US ensemble responses [that is n-dim vectors; my addition], respectively, and $x^T$ and $\mu^T$ are their transposes. We averaged the M over all points x in both sets". The source is here in the methods, under "Population vector analysis"
And the illustration looks like this:

My questions:

As I understand the take one cluster as the reference in turn and then average over vector-by-vector calculations of the mahalanobis distance? Is this a common and correct approach?
Does this lead to a distance measure which is no longer symmetric?

Preamble
Whenever all the variables under consideration are expressed in the same unit of measurement and therefore they appear to be directly comparable, the Euclidean scalar product or the associated Euclidean distance is usually used,
$ d^2(\mu,x)\equiv \langle \mu - x,\mu - x \rangle_{eucl}:= \sum_{i=1}^n (\mu_i - x_i)^2 $
Often, the collected data do not have homogeneous units of measurement, for example in the pair of weight-height variables, the unit of measurement of the first variable is mass, while the second variable expresses the measurement of a length. Weight and height cannot be added together, so it is necessary to express the data in dimensionless numbers. If the variables refer to different units of measurement, the Weighted Euclidean Distance is used in order to transforms the observable variables into dimensionless and therefore comparable numbers.The transformation is operated by a suitable multiplier or weight, $\lambda_i$, which makes the observed difference $(\mu_i - x_i)$ of a character directly comparable with the observed difference in one of the other characters $(\mu_j - x_j)$ where $i\neq j$ and $i,j=1,\cdots, n$.
If the measurements of the single variables have non-univocal values, i.e. in probabilistic terms they present non-zero variance, it is common to choose the reciprocal of the standard deviation as the multiplier: $\lambda = 1 / \sigma $ (Mahalanobis transformation), where $\sigma $ designates the standard deviation.
The Mahalanobis transformation that operates on vectors by multiplying them by the reciprocal of the corresponding standard deviation $ x \mapsto f (x):= \sigma^{-1}x$ is an isometry with respect to the usual Euclidean scalar product:
$\langle\mu - x,\mu - x\rangle_{mahal} = \langle f(\mu - x),f(\mu - x)\rangle_{eucl}$
If the n variables are statistically independent, therefore uncorrelated, the square of the Mahalanobis distance is written as:
$d^2(\mu - x,\mu - x)_{mahal}:= \sum_{i=1}^n \tfrac{(\mu_i - x_i)^2} {\sigma^2_{ii}} $
where $\sigma^2_{ii}$ is the variance referred to the i-th variable.
Otherwise, if the n variables are correlated, the square of the Mahalanobis distance is written as
$d^2(\mu - x,\mu - x)_{mahal}:= \sum_{i=1}^n \sum_{j=1}^n (\mu_i - x_i) \cdot \sigma^{-2}_{ij} \cdot (\mu_j - x_j) $,
where $\sigma^{2}_{ij} $ is the element of the inverse matrix of the variance-covariance matrix $\Sigma$.
In short

$M=(\mu - x)^t \Sigma^{-1} (\mu - x) $

Question 1
Once a distance $d$ able to express the notion of proximity for the observations has been adopted, it is possible to introduce a classification rule for the observations with respect to a representative of each class. The statistical character of the discriminant analysis derives from the fact that the average or the sample mean of the populations under examination is chosen to be the representative of each class.
In the example of the two classes (sets "US" and "CS"), indicated with $ \mu_1$ the average of the first set (population) and with $ \mu_2$ the average of the second population, the observation $x$ belongs to class 1 if and only if:

$d(\mu_1,x) < d(\mu_2,x)$

that is if $x$ is closer to $\mu_1$.
Question 2
Just consider $d^2(\mu,x)=\|\mu-x\|^2=\langle \mu-x , \mu-x \rangle$ to express the rule of belonging to set 1 in terms of the dot product as
$\langle \mu_1-x, \mu_1-x\rangle < \langle \mu_2-x, \mu_2-x\rangle$
The relationship just written, exploiting the bilinearity property of the scalar product, can be rewritten after simple steps such as
$\langle \mu_1-\mu_2, x-1/2(\mu_1+\mu_2)\rangle > 0$.
Indicated with $x_0=1/2(\mu_1+\mu_2)$ the classification rule for set 1 elements becomes:

$Set_1: \langle \mu_1-\mu_2, x-x_0 \rangle > 0 $


Generalisation of characterisation of continuous Fréchet-differentiability to infinite dimensions In the following $E,F$ are Banach spaces. A standard result from analysis is

  Let $f: \mathbb{R}^n \to F$ with $n \in \mathbb{N^{\times}}$. Then $f$ is continuously Fréchet-differentiable if and only if all the partial derivatives $\mathbb{R}^n\ni h \mapsto \frac{\partial f}{\partial x_j}(h) \in F$ for $j \in \{1,...,n\}$ exist and are continuous.

Does there exist a generalisation of this for maps with a general Banach space as a domain? A direct generalisation without appealing to basis vectors as in the finite-dimensional case would be

  Let $f: E \to F$. Then $f$ is continuously Fréchet-differentiable if and only if for every $e\in E$ the Gâteaux-derivative $E\ni h \mapsto Df(h)e\in F$ exists and is continuous.

However, I suspect that this is false, since the proof of the finite-dimensional statement pretty much leans on the fact that $\mathbb{R}^n$ has a finite basis. What would be a counter-example to this?

Can the finite-dimensional case still be generalised in some way? What about if $E$ is a (separable?) Hilbert space and we replace the basis vectors by a maximal orthonormal system? Finally, what about the following statement:

  Let $f: E \to F$. Then $f$ is continuously Fréchet-differentiable if and only if the map $E\oplus E \ni (h,e) \mapsto Df(h)e \in F$ exists and is continuous, where $Df(\cdot)$ denotes the Gâteaux-derivative.

Your first statement is true. If you have a continuous Gâteaux derivative,
then your function is Fréchet differentiable. Here, it is important that
you equip $\mathcal L(E,F)$ with the usual operator norm. The proof is not hard,
it uses the weak mean value inequality.

There is also a concept of partial derivatives: if $E = E_1 \times E_2$ is a product of Banach spaces,
Fréchet differentiability follows from partial differentiability with continuous partial derivatives,
see Theorem 3.7.1 in the book ``Differential calculus'' by Cartan.

Your last statement is false. For a counterexample consider $E = L^2(0,1)$ (usual Lebesgue measure)
and
\begin{equation*}
 f(u) := \int_0^1 \sin(u(t)) \, \mathrm dt.
\end{equation*}
Then, $f \colon L^2(0,1) \to \mathbb R$ is Gâteaux differentiable with
\begin{equation*}
 f'(u)\,h = \int_0^1 \cos(u(t)) \, h(t) \, \mathrm dt,
\end{equation*}
but this is not a Fréchet derivative.
Moreover, $(u,h) \mapsto f'(u)\,h$ is continuous.

In fact to define a Fréchet derivative for infinite dimensional spaces, you're facing two problem.

First for a fixed point $x_0 \in E$, $Df(x_0)$ needs to be continuous. In finite dimensional spaces, all norms are equivalent. So defining the continuity doesn't depend on the norm. This is no more true for infinite dimensional spaces. So you have to define $Df(x_0)$ and its continuity for a given norm... Knowing that $Df(x_0)$ might be discontinuous for another norm.

Second, and more complicated... $Df(x)$ is a function from $E$ to $\mathcal L(E,F)$ so to say that $Df(x)$ is continuous, you have to define a topology on $\mathcal L(E,F)$. Not an easy question.


probability that 5 square lie along a diagonal line If $5$ squares are chosen at random from a chess board, what is the probability that they lie on a diagonal line?

HINT:

There are $64$ squares altogether; how many ways are there to choose $5$ of them?
There are $15$ diagonals in each direction, but only $7$ of them are long enough to contain $5$ squares. Specifically, in each direction there are two diagonals of length $5$, two of length $6$, two of length $7$, and one of length $8$. How many $5$-element subsets are there of each of these diagonals?

The total in (2) is the number of sets of $5$ squares that lie on a diagonal. The total in (1) is the total number of sets of $5$ squares. Combine these two numbers to get the desired probability.

There is $16$ possible arrangements of $5$ squares so that they are on the diagonal lines and $64*63*62*61*60$ possible ways to put squares on the chessboard so the probability is $16\over(64*63*62*61*60)$


prove that $(n+3)^{2}=\Theta(n-5)^{2}$ I need to find $c_1,c_2,n_0>0$ so for all $n\geq n_0$:
$$c_{1}(n-5)^{2}\leq(n+3)^{2}\leq c_{2}(n-5)^{2}$$
I have problem with the right side of the inequality. I can't find c2.
thanks for any help :)

Write
$$\sqrt{c_1}\le\frac{n+3}{n-5}\le\sqrt{c_2}$$
or
$$\sqrt{c_1}\le1+\frac8{n-5}\le\sqrt{c_2}.$$
Now choose some $n_0>5$.


Viewing integration as a $\mathbb{Z}$ action on function spaces Note: This question has been edited thanks to a comment by Chris Culter. Thank you for this comment.

The question:

I was fooling around and thought of a perspective which i thought to be quite curious:

Consider the space $X = C^\infty(\mathbb{R})/\{\textrm{polynomials}\}$, quotiented via addition (quotient is a quotient of abelian groups and not rings). Given $\bar{f} \in X$, we may lift it to a $C^\infty$ function $f: \mathbb{R} \to \mathbb{R}$, we define $(\int)^n \bar{f}$ to be:

(1) For $n$ positive, the integral $\overline{\int \int ... \int f}$, ($n$ times).

(2) For $n \le 0$, $\overline{\textrm{the $n$'th derivative of $f$}}$.

One can check that this is well defined on the quotient space, and:

Then, one can easily see that $\int^{m+n}\bar{f} = \int^n (\int^m \bar{f})$, and $\int^0 \bar{f} = \bar{f}$. Thus, we see that $n \mapsto \int^n$ defines a $\mathbb{Z}$ action on $X$.

Questions:

(1) Has anybody come across this perspective? Is this perspective useful?

(2) One question one might ask about this is what are the orbits (and finite orbits). For example, $\{ \sin, \cos, -\sin, -\cos \}$ is an interesting orbit of size $4$. Can anyone think of some nice results along these lines?

$\mathbb{Z}$ is amenable so we have:
Fixed-point property. Any action of the group by continuous affine transformations on a compact convex subset of a (separable) locally convex topological vector space has a fixed point. For locally compact abelian groups, this property is satisfied as a result of the Markov–Kakutani fixed-point theorem.
This might be interesting


find a value in pascal triangle given row and column How can I find a value from this pascal triangle given row and column number without calculating $^nC_r$?

For example, for row=$4$, column=$3$: value is $10$,

For row=$3$, column=$5$: value is $15$.

Is there any way to get this value without using $^nC_r$ explicitly?

I needed the following values modulo P (a prime):
$$^xC_1, ^{x+1}C_2, ^{x+2}C_3, ..., ^{x+y-1}C_y$$
where
$$0<x<10^{15}, 0<y<10^5, y\le x$$
Now you can consider each $^nC_r$  value as a fraction of numerator and denominator.

So $^xC_1$ = $\frac{x}{1}$, $^{x+1}C_2$ = $\frac{x.(x+1)}{1.2}$, $^{x+2}C_3$ = $\frac{x.(x+1).(x+2)}{1.2.3}$, ...

You can get modulo value of numerator and denominator separately.

Ultimately, $^nC_r = \frac{A}{B} \mod P = (A.B^{-1}) \mod P$ ......... (i)

Now, from Fermat's little theorem,
$$a^P \equiv a \mod P$$
$$a^{P-1} \equiv 1 \mod P$$
$$a^{P-2} \equiv a^{-1} \mod P$$
So we need $B^{P-2} \mod P$ in equation (i) which can be calculated by modular exponentiation

Gerry Myerson, can you edit this answer to format properly and add some details if you want?

~~I'm editing in a note at the top of my response here, because defining row and column in the way that you have leads to a different sort of problem than what I have addressed. Apologies~~

~~The value you are looking for is (C+R-2)choose(R-1) where C and R are the row and column values in your examples.~~

The triangle diagram you have included is incorrectly labelled. The rows do not increase on the RHS in the same way that the columns increase on the LHS. Consider that the first row is the 0th row and the first column is the 0th column. So the 0th row has only 1 column. The 1st row has only 2 columns. The nth row has n+1 columns (because the first column of each row is the 0th column). That is, the rows go horizontally (starting from the top) and the columns go vertically (starting from the left); they are not angled as is shown in your diagram.

In your first example, the 3rd element on the 4th row has a value of 4, counting the row and column from 0. Using natural numbers, this would appear to be the 4th element of the 5th row.

In your second example, the 3rd row does not have a 5th column, and so the result of 15 is incorrect.

To answer your question, the best way to solve for any location in the triangle is by drawing it out or by calculating n!/r!(n-r)! Keep in mind that the row and column start at (0,0) and also that the triangle is horizontally symmetrical.

Here is my source for this answer: https://en.wikipedia.org/wiki/Binomial_coefficient


$n^{th}$ derivative of a tetration function I stumbled upon this very peculiar function last summer, namely: $f(x)=x^{x^{x^{...^{x}}}}$, where there is a number $n$ of $x$'s in the exponent, I tried to find the derivative for the function and I was successful, it turned out not to be the most elegant formula but it worked. (Firstly, I invented a new notation, namely, a function such as $f(x)$ we can write it as the following: $f(x) =x^{\langle x \vert n\rangle}$ where $x$ is the exponent that is getting "powered" up $n$ times.) The formula I obtained by pattern matching was: $$f^{\prime}(x)=x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -1}\left[1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j\right]\tag{$n\geqslant 2$}.$$ I know this looks like a mad mess and I am aware that people like this have done it more elegantely, but now for the question. This is only the first derivative of the function, is there a way, or rather is there a general derivative i.e a $n^{th}$ derivative of this function?

Update: December 23th

I have tried to approach the problem myself since I asked the question and I have not gotten to a stage to say if it is impossible or possible to do, however I think I am on the right track. At first, I thought of distributing the factor $x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ to all the terms in the parentheses, but I quickly realized I had to deal with at least derivatives of triple products. Now I have come to realize that the easiest way is to differentiate the function just as it is and get a normal product and thus I must use the following formula: $$(f \cdot g)^{(n)}=\sum_{k=0}^{n}{n\choose k}f^{(k)}\cdot g^{(n-k)}$$ where $f =x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ and $g=1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j$. Since $k$ and $n-k$ are arbitrary numbers this leads us to find the general derivative for $f$ and $g$, this is where I am right now. (I do realize that I am trying to find the $n^{th}$ derivative of the first derivative but that is easily fixed later). Please come with suggestions on how to tackle this problem.

Update: December 24th

I have made progress with the help of Maple 17, namely, I have found a repeating pattern in at least a part of the general derivative, but there is still a part of it I cannot yet explain. Nonetheless, I present to you the part of the general derivative I have found: $$D_x^{\xi}f(x) = x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -\xi} \Big[(-1)^{\xi}\cdot\xi! +O(x)\Big]$$

I renamed the degree of the derivative as $\xi$ since $n$ is taken for the number of $x$s. The $O(x)$ is the (perhaps) series which I am currently working on finding, I do think I am on the right track though. The approach above with the product rule turned out to be less successful.

I found a method of doing the first derivative but it's relatively messy and makes you go all the way down the "n" chain as it were. Hopefully my ideas can give you some inspiration or spark a secondary idea.

I am using $^nx$ as the nth tetration of x.

What I did to reach my answer was start taking the derivatives of $^nx$ with increasing values of n using $e^{lnx}$. So for n=1 you obviously get
$$\frac{d}{dx}(^1x) = \frac{d}{dx}(x) = 1$$

For n=2 you get
$$\frac{d}{dx}(^2x) = \frac{d}{dx}(x^x)$$
$$=\frac{d}{dx}(e^{lnx^x})$$
$$=\frac{d}{dx}(e^{xlnx})$$
$$=e^{xlnx}\frac{d}{dx}(xlnx)$$
$$=(^2x)\Bigl(lnx\frac{d}{dx}(x)+x\frac{d}{dx}(lnx)\Bigl)$$
We already know the value of $\frac{d}{dx}(x)$ from the last problem, so we can plug it right in.
$$=(^2x)\Bigl(lnx+\frac{x}{x}\Bigl)$$
$$=(^2x)(lnx+1)$$

For n=3 you get
$$\frac{d}{dx}(^3x) = \frac{d}{dx}(x^{x^x})$$
$$=\frac{d}{dx}(e^{ln(x^{x^x})})$$
$$=\frac{d}{dx}(e^{(x^x)(lnx)})$$
$$=\frac{d}{dx}(e^{(e^{ln(x^x)})(lnx)})$$
$$=\frac{d}{dx}(e^{(e^{xln(x)})(lnx)})$$
$$=(^3x)\frac{d}{dx}(e^{xln(x)})(lnx)$$
$$=(^3x)\Bigl((lnx)\frac{d}{dx}(e^{xln(x)})+(e^{xln(x)})\frac{d}{dx}(lnx)\Bigl)$$
Note here that $e^{xln(x)}$ equals $^2x$. This means that we can substitute in values we already know, just like in the last problem, and a pattern starts to emerge.
$$=(^3x)\Bigl((lnx)\frac{d}{dx}(^2x)+(^2x)\frac{d}{dx}(lnx)\Bigl)$$
$$=(^3x)\Bigl((lnx)\frac{d}{dx}(^2x)+\frac{^2x}{x}\Bigl)$$
We don't need to plug in the end result of $\frac{d}{dx}(e^{xln(x)})$ here, because the form the equation is in now will end up fitting our generalization later.

For now, let's check by plugging in n=4
$$\frac{d}{dx}(^4x) = \frac{d}{dx}(x^{x^{x^x}})$$
$$=\frac{d}{dx}(e^{ln(x^{x^{x^x}})})$$
$$.$$
$$.$$
$$.$$
$$=\frac{d}{dx}(e^{(e^{(e^{xln(x)})(lnx)}(lnx)})$$
$$=(^4x)\frac{d}{dx}(e^{(e^{xln(x)})(lnx)}(lnx)$$
$$=(^4x)\Bigl((lnx)\frac{d}{dx}(e^{(e^{xln(x)})(lnx)})+(e^{(e^{xln(x)}(lnx)})\frac{d}{dx}(lnx)\Bigl)$$
Here again, $e^{(e^{xln(x)})(lnx)}$ is the same as $^3x$.  So if we rewrite our equation as
$$(^4x)\Bigl((lnx)\frac{d}{dx}(^3x)+\frac{^3x}{x}\Bigl)$$
we can see that a general form of the first derivative can be written as
$$\frac{d}{dx}(^nx) = (^nx)\Bigl((lnx)\frac{d}{dx}(^{n-1}x)+\frac{^{n-1}x}{x}\Bigl)$$

Obviously this has the problem of relying on the derivatives down the power tower, but I think this could have interesting applications. Hope these inane ramblings of a 17 year old help!

I think y'all are over complicating it a bit... Is it not the following?

This isn't based on hard scientific fact, this is just the pattern which seems to appear as it goes along.

Based on the following ( "X" is multiplication, as asterisk makes italics ) :

g( x ) = x^f ( x )

g'( x ) = x^f ( x ) X ( f ( x ) / x + ln( x ) X f'( x ) )

If applied to f ( x ) = x^^( k - 1 ), this pattern seems to follow.

(I'm not sure how to input the notation into the text thing, so here's a picture)

Additionally, in the actual picture I use tetration notation similar to an exponent, but the number of times the exponent is repeated in to the top-left of the base.

NOTE: X is multiplication, PI is product, SIGMA is summation, "^^" is tetration, and parenthesis following product and summation are first the bounds, then the function being evaluated ------------------------------------------------------------------------------------------------------------------------------------------------------------->>> d/dx( x^^k ) = PI(i = 2, k ) ( x^^i ) X SIGMA ( i = k - 2, k - 1 ) ( ln( x )^i ) + ( x^( -1 ) ) X ( SIGMA ( a = 2, k-1 ) ( ln( x )^( k - 1 - a ) X PI ( i = a, k )( x^^i ) ) )


which of the following statements are true?? ( Calculus and Differential equations)(NBHM-$2014$) Which of the following statements are ??

a. Let $\phi$ be a non-negative and continuously differentiable function on $(0,\infty)$ such that $\phi'(x)\le\phi(x)$ for all $x$ $\in (0,\infty)$. Then

$$lim_{x\to \infty}\phi(x)=0$$

b. Let $\phi$ be a non-negative function continuous on $[0,\infty)$ and differentiable on     $(0,\infty)$ such that $\phi(0)=0$ and such that   $\phi'(x)\le\phi(x)$ for all $x$ $\in   (0,\infty)$. Then $\phi=0$.

c. Let $\phi$ be a non-negative function continuous on $[0,\infty)$ and such that
      $$\phi(x) \le \int_{0}^{x}\phi(t) dt$$ for all $x \in [0,\infty)$. Then $\phi=0$.

Hint(Grownwall's Inequality)

  If
  $$f(t)\leq K+\int_a^tf(s)g(s)ds,$$
  then
  $$f(t)\leq K\exp(\int_a^tg(s)ds)$$

  And $K\geq 0$; $f,g$ is nonnegative continuous in $[a,b]$.

All of them are wrong.

For statement 1, consider $\phi(x) = e^x$. For statement 2, consider $\phi(x) = e^x - 1$. For statement 3, consider $\phi(x) = e^x$.


How to show that $R/I$ is Artinian when R is PID I'm working through some of Hungerfords "Algebra", and having trouble with Excercise VIII 1.2.:

  Show that if $I$ is a non-zero ideal in a principal ideal domain (PID) $R$, then the ring $R/I$ is both Noetherian and Artinian.

I know that $R$ is Noetherian since it is a PID (this follows from Lemma III. 3.6 ). To show that $R/I$ is Noetherian I have then noted that since $I$ is a submodule of $R$ (viewed as an $R$-module) and since $R$ is Noetherian it follows that $R/I$ is Noetherian (by Corollary VIII 1.6).

My problem is then how to show that $R/I$ is Artinian.

Can someone give me a hint?

We know ideals of $R/I$ are of the form "ideal of $R$ containing $I$" mod $I$. So a descending chain of ideals looks like $I_1/I \supseteq I_2/I \supseteq I_3/I \supseteq \cdots$ where $I_j$ are ideals of $R$ such that $I \subseteq I_j$.

Next, $R$ is a PID so there exists $a_j \in R$ such that $I_j=(a_j)$ and $a \in R$ such that $I=(a)$. Don't forget $a \not=0$ because $I \not= \{ 0 \}$.

What does $I \subseteq I_j$ say? What does $I_{j+1} \subseteq I_j$? Notice that $I \subseteq \cap_{j=0}^\infty I_j$.
Use unique factorization into primes to see that an infinite chain of proper divisors is impossible.

We know that a ring $R$ is Artinian iff it is Noetherian and every prime ideal is maximal. Now since $R$ is PID, so $R/I$ is PID. On the other hand, in a PID ring every prime ideal is maximal.So the Noetherian ring $R/I$ is Artinain.