Basis for a subspace I need to calculate the basis for
$$W = \lbrace (a,b,c,d) \: : \:  a+b+c = 0 \rbrace.$$
I find it hard to understand how does the fact that d is not part of the equation effects the basis.

Thanks in advance.

$W$ is the kernel of a non zero linear form. Therefore the dimension of $W$ as a subspace of $\Bbb{R}^4$ is $4-1=3$. So it suffices to find three independent vectors belonging to $W$

$(0,0,0,1)$ is an obvious element of $W$

$(1,-1,0,0)$ and $(0,-1,1,0)$ are two other vectors.

The three vectors are independent therefore a basis.

Note that in the definition of $W$ you've given, $a+b+c = 0$ is best understood as a constraint on the 4-tuple $(a,b,c,d)$.

That is, there aren't any explicit constraints on $d$, so conceivably it can be whatever it wants.

Usually, linear algebra deals with real numbers, so it's my guess that $d\in R$, but that's not in the definition you've given, and so is just a guess on my part.


On geodesics in Schwarzschild spacetime I am required to show that a circular lightlike geodesic exists in the Schwarzschild spacetime, and to find its radius. What's the best way to start this?

You can start with the system of geodesic equations together with the Schwarzschild metric expression divided by $d\tau^2$ (formulated in terms of $u^\mu := dx^\mu /d\tau$). Coordinates $x^\mu := (x^0 := ct, r, \theta, \varphi)$. Then you can obtain geodesics by assuming as $u^2 \equiv 0$ ($\theta = \pi/2$) due to the spherical symmetry. Solutions in $u^0$ and $u^3$ are evident thanks to this symmetry.


Why can there be an infinite difference between two functions as x grows large, but a ratio of 1? I learned in grade school that the closer $a$ and $b$ are to one another, the closer $\frac{a}{b}$ is going to be to $1$. For example, $\frac{3}{\pi}$ is pretty close to 1, and $\frac{10^{100}}{42}$ isn't even close to 1.

So, why is:

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

But:

$$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$$ ?

Seems pretty counterintuitive. What's going on here?

Vadim's answer handles the math (and I've upvoted it), so I will try to provide intuition.

The idea is the word "closer" is relative.  That is, in some sense, $100{,}000$ is closer to $100{,}010$ than $1$ is to $0$.  Of course, in an absolute sense, $1$ is $1$ away from $0$ while $100{,}000$ is $10$ away from $100{,}010$.

But, say you had $\$1$ and I took away $\$1$ and contrast that with the scenario where you have $\$100{,}010$ and I take $\$10$.  In the second scenario, I take more money.  But in the first scenario, you care a lot more about the theft.

In your particular example, something similar happens.  Yes, $x^2 + x$ and $x^2$ have a very large difference when $x$ is large.  But the relative difference is tiny.  For example, when $x = 1{,}000$, then $x^2 + x = 1{,}001{,}000$ while $x^2 = 1{,}000{,}000$.  So, yes, the difference is large (it's $1000$), but the relative difference is not that big - these two numbers are separated by a mere $0.1\%$  And this percentage shrinks as $x$ gets even bigger.

Trying to develop an intuitive idea using images and, doing so, cutting a lot of corners.

Suppose Bill Gates gets a raise of 1 dollar an hour. And suppose he decides to work forever. He will make 1$ more every hour. So an infinite amount of money more than without the raise. But do you think this raise is significant? No. The amount of money he makes with or without the raise is almost the  same. The ratio between the fortunes he amasses is pretty close to one.


Does $\int_{-\infty}^\infty f(x) dx < \infty$ where $f \ge 0$ implies $\sup_{x \in \mathbb R} f(x)<M $? Question. Does $\,\int_{-\infty}^\infty\, f(x)\, dx < \infty,\,$ where $\,f \ge 0,\,$ imply that $\,\,\mathrm{ess}\sup_{x \in \mathbb R}\, f(x)<\infty\,? $

My attempt

Since $f \ge 0$ and $\int_{-\infty}^\infty f(x) dx < \infty$  we get $\lim_{x \to \infty} f(x)=0$ and $\lim_{x \to -\infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $\sup_{x \in \mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?

I am trying to show that if $f,g$ are non-negative  and the integral exists then $f\cdot g(x)$ exists for almost all $x$.

If not can you give a counterexample and explain the problem in my reasoning?

No. A very simple counter-example can be constructed as follows.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by
$$
f(x)=\begin{cases}
x^{-\frac{1}{2}}, & \mbox{ if }x\in(0,1]\\
0, & \mbox{ otherwise}
\end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $\int f<\infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that $\{x\mid|f(x)|>M\}=(0,\frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.

Neither if $f$ is continue, let $\rho:\mathbb R\rightarrow [0, +\infty[$ a continue non identically zero function such that $\rho(x)\neq 0$ only inside $[-1, 1]$ and
$$
\int^{+\infty}_{-\infty}\rho(x)dx = 1
$$

Then let
$$
f(x)=\sum^{+\infty}_{i=1}2^i\rho\left(i+4^ix\right)
$$
$f$ is continuous and
$$
\int^{+\infty}_{-\infty}2^i\rho\left(i+4^ix\right)dx=\int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^i\rho\left(i+4^ix\right)dx=2^i4^{-i}\int^{1}_{-1}\rho(y)dy\leq 2^{-i}
$$
then $\int fdx<+\infty$ but $f$ isn't bounded almost everywhere: let $k=\max_{x\in\mathbb R}\rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $\rho$ is continue exists an open set $U\subseteq\mathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $x\in U$
$$
M < 2^i\rho(i+4^ix)\Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.

You need uniform continuity on $\mathbb R$ because statement
$$
\lim_{x\rightarrow \infty}f(x)=0
$$
is true if $f$ is uniformly continue.

Let $f$ uniformly continue on $\mathbb R$ and suppose exists $\epsilon>0$ and increasing sequence $x_n$ such that $x_n\rightarrow +\infty$ and $f(x_n)>\epsilon$.

From uniform continuity exists $\delta>0$ such that for every $x, y\in\mathbb R$ such that $\lvert x-y\rvert <\delta$ then $\lvert f(x)-f(y)\rvert<\frac{\epsilon}{2}$. If $x\in ]x_n-\delta, x_n+\delta[$ then
$$
f(x)\geq f(x_n)-\lvert f(x)-f(x_n)\rvert>\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
\int^{+\infty}_{-\infty}f(x)dx\geq\sum^{+\infty}_{k=0}\frac{\epsilon}{2}\left(x_{n_k}+\delta-x_{n_k}+\delta\right)=\sum^{+\infty}_{k=0}\epsilon\delta=+\infty
$$
absurd.

Your last assertion is true: let $f:\mathbb R\rightarrow [0, +\infty]$ such that integral is finite let $K=\{x\in\mathbb R : f(x)=+\infty\}$ then
$$
\int^{+\infty}_{-\infty}f(x)dx\geq\int_Kf(x)dx=\begin{cases}
+\infty & \text{ if }\lvert K\rvert >0\\
0 & \text{ if }\lvert K \rvert =0
\end{cases}
$$
so $f(x)<+\infty$ for almost every $x\in\mathbb R$.


How many cards out of a standard deck would you need to draw to ensure two of them are different suits? I believe that we are determining, in a worst case scenario, how many cards must we pick to draw a two different suit. My thought process is that we must get rid of all 13 cards to deplete an entire suit before we would be able to ensure the next card drawn is of a different suit. If this is so, then would the correct answer be to draw 14 cards(13 same suit and 1 different suit) to ensure a different suit?

Followup Question for Clarification

What is the probability that we will pick two cards with two different suits?  I know we are able to pick any card for first one so I believe that means we have a 52/52 possibility for the first card. For the second card would we then have a 39/51 possibility (where 39 is the remaining number of cards that are not the suit)?

Your reasoning in both parts is perfectly fine. For the first problem we need to exhaust one suit completely to be sure that we'll draw a card from a different suit. As each suit has $13$ cards we need at least $13+1=14$ draws.

For the second part you could think like this: The first card is only required to determine the suit we don't want and hence the good combination would be drawing a card from those $3$ suits in the second draw. As you've noticed that probability is $\frac{39}{51}$

We ll take the worst case that we have selected 13 cards but they are all of same suit. 14th draw will guarantee the draw of two different suits.
For the second part:
Two cards randomly can be chosen in C(52,2) ways.
But two suits can be selected in C(4,2 ) ways. And two cards of those selected suits can be chosen in C(13,1)×C(13,1) ways. So it becomes 39/51.


$u$ continuous and the weak derivative $Du$ continuous $\Rightarrow$ $u \in C^1$? Supose we have $u \in W^{1,p}$ (i.e $u$ has weak partial derivatives, which we denote by $Du$), and that both $u$ and $Du$ are continuous (More precisely, there is a continuous representative in the equivalence class of $u$ and $Du$ that is continuous).

  Can we ensure that $u \in C^1$?

I mean, Can we ensure that $u$ (The continuous representative) has classical partial derivatives and that they are continuous?

I am working with the Variational Approach and at a certain step it is required to show that $u$ is $C^2$ and the results I have prove that there is a continuous representative for all the derivatives of $u$ and $u$, which seems to be not enough for me.

Let $\Omega \subset \mathbb{R}^n$ open. If $u \in \mathscr{C}(\Omega)$, and the distributional derivatives of $u$ are given by integration against continuous functions, that is, there exist $v_1,\dotsc, v_n \in \mathscr{C}(\Omega)$ such that for all $\varphi \in \mathscr{D}(\Omega)$ we have

$$- \int_{\Omega} u(x)D_k\varphi(x)\,dx = \int_{\Omega} v_k(x)\varphi(x)\,dx$$

for $1 \leqslant k \leqslant n$, then $u \in \mathscr{C}^1(\Omega)$.

Weak derivatives, when they exist, coincide with distributional derivatives, so this includes your setting.

Proof sketch: Let $\psi \in \mathscr{D}(\mathbb{R}^n)$ with $\psi \geqslant 0$, $\operatorname{supp} \psi \subset \overline{B_1(0)}$ and $\int_{\mathbb{R}^n} \psi(x)\,dx = 1$. For $\varepsilon > 0$, define $\psi_{\varepsilon}(x) := \varepsilon^{-n}\psi(x/\varepsilon)$. For $\delta > 0$ and $0 < \varepsilon < \delta$, let

$$u_{\varepsilon}(x) := (\psi_{\varepsilon} \ast u)(x) = \int_{\Omega} u(x-y) \psi_{\varepsilon}(y)\,dy = \int_{B_1(0)} u(x-\varepsilon y) \psi(y)\,dy$$

for $x \in \Omega_{\delta} = \{x \in \Omega : d(x,\partial \Omega) > \delta\}$.

Then $u_{\varepsilon} \in \mathscr{C}^{\infty}(\Omega_{\delta})$, and

$$D_k u_{\varepsilon} = D_k (\psi_{\varepsilon}\ast u) = (D_k \psi_{\varepsilon}) \ast u = \psi_{\varepsilon} \ast v_k. \tag{1}$$

Since $u, v_1,\dotsc, v_n \in \mathscr{C}(\Omega)$, it follows that $u_{\varepsilon} \to u$, and $D_k u_{\varepsilon} \to v_k$ locally uniformly on $\Omega_{\delta}$ as $\varepsilon \to 0$.

Then fix $x^{(0)} \in \Omega_{\delta}$, and note that for $0 < \eta < d(x^{(0)}, \partial \Omega) - \delta$ we have

\begin{align}
u(x^{(0)} + he_k) - u(x^{(0)}) &= \lim_{\varepsilon \to 0} u_{\varepsilon}(x^{(0)} + he_k) - u_{\varepsilon}(x^{(0)})\\
&= \lim_{\varepsilon \to 0} \int_0^h D_ku_{\varepsilon}(x^{(0)} + te_k)\,dt\\
&= \int_0^h v_k(x^{(0)} + te_k)\,dt
\end{align}

for $\lvert h\rvert < \eta$, which shows that $u$ is partially differentiable at $x^{(0)}$ with $\dfrac{\partial u}{\partial x_k}(x^{(0)}) = v_k(x^{(0)})$.

Since $\delta > 0$ was arbitrary, it follows that $u$ is partially differentiable in the classical sense on all of $\Omega$, and the partial derivatives are continuous by assumption, so $u \in \mathscr{C}^1(\Omega)$.

Pseudo proof: Suppose $f,g$ are continuous functions on $\mathbb R,$ with $g$ the weak derivative of $f.$ Let $0<x.$ Define piecewise linear functions $\varphi_n$ such that $\varphi_n$ connects $(0,0), (1/n,1), (x-1/n,1), (x,0)$ with line segments, $\varphi_n =0$ elsewhere. Now  $\varphi_n$ is not a smooth test function, but pretend it is. Then

$$\tag 1 -\int_0^x f \varphi_n' = \int_0^x g\varphi_n.$$

The left side of $(1)$ equals

$$-n\int_0^{1/n} f  + n\int_{x-1/n}^x f \to f(x) - f(0)$$

by the continuity of $f$ at $0,x.$ The right side of $(1)$ tends to $\int_0^x g.$ Thus $f(x)-f(0) = \int_0^x g.$ This gives $f\in C^1.$


Convergence test for series Let  $\space \displaystyle\sum_{n=1}^{\infty}  \space \left | \frac{\cos(n\pi)}{n+1}\right |$. Does this series converge or not?

The serie is valid for the natural numbers, so it can be writen as $\space \displaystyle\sum_{n=1}^{\infty}  \space  \frac{\left |\cos(n\pi)\right |}{n+1}$.

One knows that $\space 0\leq\left |\cos(n\pi)\right |\leq 1$. By using the asymptotic concept, one can say that the leader terms of the top and bottom of the fraction are $\displaystyle \frac{1}{n}$.

And so, the original serie is equivalent to this one
$$\space \displaystyle\sum_{n=1}^{\infty}  \space \frac{1}{n}$$

that is a $p$ series, where $p\leq 1$ and so diverges, and also the original series. This is correct? Thanks.

No. Since the p-series you've considered was only an upper bound for the series, the fact that it diverges to infinity tells you nothing.

Since $|\cos(n\pi)| = 1\;\; \forall n,\;$ your series is equivalent to:

$$\sum_{n = 1}^{\infty}\frac{\left|(-1)^n\right|}{(n + 1)} = \sum_{n = 1}^{\infty} \frac {1}{n+1} $$

You're "spot on" with respect to comparing your given series to the harmonic series: $\;\;\displaystyle \sum_{n = 1}^\infty \dfrac 1n,\;$ but since $\;\dfrac 1{n+1} \lt \dfrac 1n,\;$ you can note that

$$\sum_{\color{blue}{\bf n = 1}}^\infty \frac 1{n+1} \quad = \quad \sum_{\color{blue}{\bf n = 2}}\frac 1n$$ and so, save for one term, your series is the harmonic series, and diverges just as does the "strict" harmonic series.

If you want a tighter argument, you can use the limit comparison test of your series and the "strict" harmonic series to prove your series diverges.

$$\text{Put, say}\;\;a_n = \dfrac 1{n+1},\;\;b_n = \dfrac 1n \iff
\dfrac{a_n}{b_n} = \dfrac{1/(n+1)}{1/n} = \dfrac n{n+1}$$

$$\text{Then we have}\quad\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac n{n+1} = 1 < \infty$$

Since the harmonic series given by $\sum b_n$ diverges, so too does your series.


Logic problem about set of disjunction forms(Fixed) I will call the disjunction of literals a disjunction form. Let $\Sigma$ be a set of disjunction forms, and $\alpha$ is a well-formed formula satisfying $\Sigma \vDash \alpha$. For all propositional variable $p$, assume that $p$ and $\neg p$ do not occur in element of $\Sigma$ at the same time.

Prove that $\exists \sigma \in \Sigma$ such that $\sigma \vDash \alpha$.  And the above proposition is true if we change all  disjunctions to  conjunctions.

Can you help me?

Neither statement is true.

Let $\Sigma = \{P_1\lor P_2,  Q_1\lor Q_2\}$, and $\alpha = (P_1\lor P_2) \land (Q_1\lor Q_2)$.

Then $\Sigma\models \alpha$, but there is no $\sigma\in \Sigma$ such that $\sigma\models \alpha$.

Similarly, for "conjunction forms" instead of "disjunction forms", let $\Sigma = \{P_1\land P_2,  Q_1\land Q_2\}$, and $\alpha = (P_1\land P_2) \land (Q_1\land Q_2)$.

Edit: In response to the comments, I'll prove the following.

Suppose $\Sigma$ is a nonempty set of disjunctions of literals, such that for every propositional variable $p$, at most one of the literals $p$ and $\lnot p$ occurs in $\Sigma$. And suppose $\alpha$ is a disjunction of literals. If $\Sigma\models \alpha$, then there is some $\sigma\in \Sigma$ such that $\sigma\models \alpha$.

Case 1: There is some propositional variable $p$ such that both of the literals $p$ and $\lnot p$ occur in $\alpha$. Then $\alpha$ is a tautology, so $\sigma\models \alpha$ for any $\sigma\in \Sigma$.

Case 2: There is some $\sigma\in \Sigma$ such that every literal occuring in $\sigma$ occurs in $\alpha$. Then we are done, since $\sigma\models \alpha$.

We will show that if we assume we are not in Case 1 or Case 2, we have a contradiction. Since we are not in Case 2, for every $\sigma\in \Sigma$,  we can pick some literal $l_\sigma$ which appears in $\sigma$ but not in $\alpha$.

Let $M$ be a model/interpretation such that the literals $\{l_\sigma\mid \sigma\in \Sigma\}$ are all true, and the literals appearing in $\alpha$ are all false. We can do this, because:

We have arranged that no literal $l_\sigma$ appears in $\alpha$.
Since we are not in Case 1, no literal and its negation both appear in $\alpha$.
By hypothesis, no literal and its negation both appear in $\{l_\sigma\mid \sigma\in \Sigma\}$, since all these literals appear in $\Sigma$.

But then for each $\sigma\in \Sigma$, we have $M\models l_\sigma$, so $M\models \sigma$, and hence $M\models \Sigma$. But $M\not\models \alpha$. So $\Sigma\not\models \alpha$, contradiction.

Hint: For every propositional formula $\phi$ there is an equivalent formula $\psi$ in conjunctive normal form, that is,

$$ \psi = \bigwedge^{m}_{i=1} (\bigvee^{m_i}_{j=1} \ell_{ij}) $$

where the $\ell_{ij}$'s are literals. And there is a similar result for disjunctive normal forms.


Determining the Average number of Attempts before a result occurs (using defined Probabilities) I'm trying to solve this problem but I'm completely unsure of what sort of concepts/formulas/proofs I need in order to tackle it. I'm just lost on where to start. Here's an example story problem as it relates to item drops off a boss in an MMORPG...

Loot Table:

Pants - 30%
Helm - 20%
Shoulders - 15%
Chest- 10%
Arms - 25%

Given these values and that the boss always drops a single item and the probabilities of those drops are not influenced by previous drops, what the average number of boss kills one would need to get a Chest piece? What's the average number of attempts one would need to get 3 pants, 2 Shoulders and an 1 Arm piece?

Any help pointing me in the right direction for what I should be studying or reading up on to solve this would be most helpful.

As has been noted in comments, the expected number of attempts to get one item that has probability $p$ per attempt is $1/p$.

To find the expected number of attempts for a collection of items, we can use inclusion-exclusion. Let $A_n$ denote the event that the loot from $n$ kills doesn't contain $3$ pants, $B_n$ the event that it doesn't contain $2$ shoulders and $C_n$ the event that it doesn't contain $1$ arm piece. Then by inclusion-exclusion the probability that it doesn't contain the entire collection is

$$
P(A_n\cup B_n\cup C_n)=\\P(A_n)+P(B_n)+P(C_n)-P(A_n\cap B_n)-P(A_n\cap C_n)-P(B_n\cap C_n)+P(A_n\cap B_n\cap C_n).
$$

The individual terms are

\begin{align}
P(A_n)&=\sum_{k=0}^2\binom nk0.3^k0.7^{n-k}\;,\\
P(B_n)&=\sum_{k=0}^1\binom nk0.15^k0.85^{n-k}\;,\\
P(C_n)&=\sum_{k=0}^0\binom nk0.25^k0.75^{n-k}\;,\\
P(A_n\cap B_n)&=\sum_{k=0}^2\sum_{l=0}^1\binom n{k,l,n-k-l}0.3^k0.15^l0.55^{n-k-l}\;,\\
P(A_n\cap C_n)&=\sum_{k=0}^2\sum_{l=0}^0\binom n{k,l,n-k-l}0.3^k0.25^l0.45^{n-k-l}\;,\\
P(B_n\cap C_n)&=\sum_{k=0}^1\sum_{l=0}^0\binom n{k,l,n-k-l}0.15^k0.25^l0.6^{n-k-l}\;,\\
P(A_n\cap B_n\cap C_n)&=\sum_{k=0}^2\sum_{l=0}^1\sum_{m=0}^0\binom n{k,l,m,n-k-l-m}0.3^k0.15^l0.25^m0.3^{n-k-l-m}\;.
\end{align}

The expected number of kills required to obtain the entire collection is the sum over $P(A_n\cup B_n\cup C_n)$ for all n, which we can get by summing the individual terms. For the first three terms, this just yields the known individual expectations:

\begin{align}
\sum_{n=0}^\infty P(A_n)&=\frac3{0.3}=10\;,\\
\sum_{n=0}^\infty P(B_n)&=\frac2{0.15}=\frac{40}3\;,\\
\sum_{n=0}^\infty P(C_n)&=\frac1{0.25}=4\;.
\end{align}

For $P(A_n\cap B_n)$, we get

\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^2\sum_{l=0}^1\binom n{k,l,n-k-l}0.3^k0.15^l0.55^{n-k-l}\\
={}&\sum_{k=0}^2\sum_{l=0}^1\binom{k+l}k0.3^k0.15^l\sum_{n=0}^\infty\binom n{k+l}0.55^{n-(k+l)}\\
={}&\frac1{0.45}\sum_{k=0}^2\sum_{l=0}^1\binom{k+l}k\frac{0.3^k0.15^l}{0.45^{k+l}}\\
={}&\frac{0.45^3+0.15\cdot0.45^2+0.3\cdot0.45^2+2\cdot0.3\cdot0.15\cdot0.45+0.3^2\cdot0.45+3\cdot0.3^2\cdot0.15}{0.45^4}\\
={}&\frac{200}{27}\;,
\end{align}

for $P(A_n\cap C_n)$

\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^2\sum_{l=0}^0\binom n{k,l,n-k-l}0.3^k0.25^l0.45^{n-k-l}\\
={}&\sum_{n=0}^\infty\sum_{k=0}^2\binom nk0.3^k0.45^{n-k}\\
={}&\sum_{k=0}^20.3^k\sum_{n=0}^\infty\binom nk0.45^{n-k}\\
={}&\frac1{0.55}\sum_{k=0}^2\left(\frac{0.3}{0.55}\right)^k\\
={}&\frac{4460}{1331}
\end{align}

and for $P(B_n\cap C_n)$

\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^1\sum_{l=0}^0\binom n{k,l,n-k-l}0.15^k0.25^l0.6^{n-k-l}\\
={}&\sum_{n=0}^\infty\sum_{k=0}^1\binom nk0.15^k0.6^{n-k}\\
={}&\sum_{k=0}^10.15^k\sum_{n=0}^\infty\binom nk0.6^{n-k}\\
={}&\frac1{0.4}\sum_{k=0}^1\left(\frac{0.15}{0.4}\right)^k\\
={}&\frac{55}{16}\;.
\end{align}

Finally, for $P(A_n\cap B_n\cap C_n)$,

\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^2\sum_{l=0}^1\sum_{m=0}^0\binom n{k,l,m,n-k-l-m}0.3^k0.15^l0.25^m0.3^{n-k-l-m}\\
={}&\sum_{n=0}^\infty\sum_{k=0}^2\sum_{l=0}^1\binom n{k,l,n-k-l}0.3^k0.15^l0.3^{n-k-l}\\
={}&\sum_{k=0}^2\sum_{l=0}^1\binom{k+l}k0.3^k0.15^l\sum_{n=0}^\infty\binom n{k+l}0.3^{n-(k+l)}\\
={}&\frac1{0.7}\sum_{k=0}^2\sum_{l=0}^1\binom{k+l}k\frac{0.3^k0.15^l}{0.7^{k+l}}\\
={}&\frac{0.7^3+0.15\cdot0.7^2+0.3\cdot0.7^2+2\cdot0.3\cdot0.15\cdot0.7+0.3^2\cdot0.7+3\cdot0.3^2\cdot0.15}{0.7^4}\\
={}&\frac{7300}{2401}\;.
\end{align}

Putting it all together, we obtain the expected number of kills required:

$$
\sum_{n=0}^\infty P(A_n\cup B_n\cup C_n)=10+\frac{40}3+4-\frac{200}{27}-\frac{4460}{1331}-\frac{55}{16}+\frac{7300}{2401}=\frac{22334578793}{1380555792}\approx16.178\;.
$$

Thanks to normalization the probability of looting a chest piece in N successful boss fights is:
$$P_{N}(Chest)+P_{N}(No Chest)=1$$
The probability of getting no chest piece in N trials is much easier to write down:
$$P(NoChest)= (0.9)^N$$
So, $$P(Chest)=1-P(NoChest)\rightarrow 1-(0.9)^N$$

The number of times you would expect an event is given by:
$$<n_{Chest}>=\sum_{j=1}^N j*P_N=\sum_{j=1}^N j*(1-0.9^j)$$

Since you want at least one chest piece, then you want to know when for what value of capital N does the number of expected chest pieces raise above 1. Hope this helps.


What is the practical difference between a differential and a derivative? I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:

So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get.

Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?

And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.

Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle.

I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.

Any assistance?

Originally, "differentials" and "derivatives" were intimately connected, with derivative being defined as the ratio of the differential of the function by the differential of the variable (see my previous discussion on the Leibnitz notation for the derivative). Differentials were simply "infinitesimal changes" in whatever, and the derivative of $y$ with respect to $x$ was the ratio of the infinitesimal change in $y$ relative to the infinitesimal change in $x$.

For integrals, "differentials" came in because, in Leibnitz's way of thinking about them, integrals were the sums of infinitely many infinitesimally thin rectangles that lay below the graph of the function. Each rectangle would have height $y$ and base $dx$ (the infinitesimal change in $x$), so the area of the rectangle would be $y\,dx$ (height times base), and we would add them all up as $S\; y\,dx$ to get the total area (the integral sign was originally an elongated $S$, for "summa", or sum).

Infinitesimals, however, cause all sorts of headaches and problems. A lot of the reasoning about infinitesimals was, well, let's say not entirely rigorous (or logical); some differentials were dismissed as "utterly inconsequential", while others were taken into account. For example, the product rule would be argued by saying that the change in $fg$ is given by
$$(f+df)(g+dg) -fg = fdg + gdf + df\,dg,$$
and then ignoring $df\,dg$ as inconsequential, since it was made up of the product of two infinitesimals; but if infinitesimals that are really small can be ignored, why do we not ignore the infinitesimal change $dg$ in the first factor?  Well, you can wave your hands a lot of huff and puff, but in the end the argument essentially broke down into nonsense, or the problem was ignored because things worked out regardless (most of the time, anyway).

Anyway, there was a need of a more solid understanding of just what derivatives and differentials actually are so that we can really reason about them; that's where limits came in. Derivatives are no longer ratios, instead they are limits. Integrals are no longer infinite sums of infinitesimally thin rectangles, now they are limits of Riemann sums (each of which is finite and there are no infinitesimals around), etc.

The notation is left over, though, because it is very useful notation and is very suggestive. In the integral case, for instance, the "dx" is no longer really a quantity or function being multiplied: it's best to think of it as the "closing parenthesis" that goes with the "opening parenthesis" of the integral (that is, you are integrating whatever is between the $\int$ and the $dx$, just like when you have $2(84+3)$, you are multiplying by $2$ whatever is between the $($ and the $)$ ). But it is very useful, because for example it helps you keep track of what changes need to be made when you do a change of variable. One can justify the change of variable without appealing at all to "differentials" (whatever they may be), but the notation just leads you through the necessary changes, so we treat them as if they were actual functions being multiplied by the integrand because they help keep us on the right track and keep us honest.

But here is an ill-kept secret: we mathematicians tend to be lazy. If we've already come up with a valid argument for situation A, we don't want to have to come up with a new valid argument for situation B if we can just explain how to get from B to A, even if solving B directly would be easier than solving A (old joke: a mathematician and an engineer are subjects of a psychology experiment; first they are shown into a room where there is an empty bucket, a trashcan, and a faucet. The trashcan is on fire. Each of them first fills the bucket with water from the faucet, then dumps it on the trashcan and extinguishes the flames. Then the engineer is shown to another room, where there is again a faucet, a trashcan on fire, and a bucket, but this time the bucket is already filled with water; the engineer takes the bucket, empties it on the trashcan and puts out the fire. The mathematican, later, comes in, sees the situation, takes the bucket, and empties it on the floor, and then says "which reduces it to a previously solved problem.")

Where were we? Ah, yes. Having to translate all those informal manipulations that work so well and treat $dx$ and $dy$ as objects in and of themselves, into formal justifications that don't treat them that way is a real pain. It can be done, but it's a real pain. Instead, we want to come up with a way of justifying all those manipulations that will be valid always. One way of doing it is by actually giving them a meaning in terms of the new notions of derivatives. And that is what is done.

Basically, we want the "differential" of $y$ to be the infinitesimal change in $y$; this change will be closely approximated to the change along the tangent to $y$; the tangent has slope $y'(a)$. But because we don't have infinitesimals, we have to say how much we've changed the argument. So we define "the differential in $y$ at $a$ when $x$ changes by $\Delta x$", $d(y,\Delta x)(a)$, as $d(y,\Delta x)(a) = y'(a)\Delta x$. This is exactly the change along the tangent, rather than along the graph of the function.  If you take the limit of $d(y,\Delta x)$ over $\Delta x$ as $\Delta x\to 0$, you just get $y'$. But we tend to think of the limit of $\Delta x\to 0$ as being $dx$, so abuse of notation leads to "$dy = \frac{dy}{dx}\,dx$"; this is suggestive, but not quite true literally; instead, one then can show that arguments that treat differentials as functions tend to give the right answer under mild assumptions. Note that under this definition, you get $d(x,\Delta x) = 1\Delta x$, leading to $dx = dx$.

Also, notice an interesting reversal: originally, differentials came first, and they were used to define the derivative as a ratio. Today, derivatives come first (defined as limits), and differentials are defined in terms of the derivatives.

What is the practical difference, though? You'll probably be disappointed to hear "not much". Except one thing: when your functions represent actual quantities, rather than just formal manipulation of symbols, the derivative and the differential measure different things. The derivative measures a rate of change, while the differential measures the change itself.

So the units of measurement are different: for example, if $y$ is distance and $x$ is time, then $\frac{dy}{dx}$ is measured in distance over time, i.e., velocity. But the differential $dy$ is measured in units of distance, because it represents the change in distance (and the difference/change between two distances is still a distance, not a velocity any more).

Why is it useful to have the distinction? Because sometimes you want to know how something is changing, and sometimes you want to know how much something changed. It's all nice and good to know the rate of inflation (change in prices over time), but you might sometimes want to know how much more the loaf of bread is now (rather than the rate at which the price is changing). And because being able to manipulate derivatives as if they were quotients can be very useful when dealing with integrals, differential equations, etc, and differentials give us a way of making sure that these manipulations don't lead us astray (as they sometimes did in the days of infinitesimals).

I'm not sure if that answers your question or at least gives an indication of where the answers lie. I hope it does. Added. I see Qiaochu has pointed out that the distinction becomes much clearer once you go to higher dimensions/multivariable calculus, so the above may all be a waste. Still...

Added. As Qiaochu points out (and I mentioned in passing elsewhere), there are ways in which one can give formal definitions and meanings to infinitesimals, in which case we can define differentials as "infinitesimal changes" or "changes along infinitesimal differences"; and then use them to define derivatives as integrals just like Leibnitz did. The standard example of being able to do this is Robinson's non-standard analysis Or if one is willing to forgo looking at all kinds of functions and only at some restricted type of functions, then you can also give infinitesimals, differentials, and derivatives substance/meaning which is much closer to their original conception.

The limit is the value corresponding to a function (say $f(x)$) at a specific point (say $x = x_0$). The differential of $f$ is a formula in which the limit-notion is used.


How many ways? - difficult combinatorics These are from an archival local math contest:

The city purchased train rails in two colors: gray and black. The only difference between them is the color. The project assumes that the rails will be installed in series of 10. We say that the series is elegant when no two black rails are directly adjacent to each other. How many different elegant series of 10 rails are possible?

I made some observations:

In 3. I know the answer is supposed to be 144. If we have 6 black rails or more we cannot make an elegant series. For 1 black rail we can make 10 different series (the order matters), for two it is 8 + 7 + ... + 1 = 36 (i placed the first rail and looked where I can place the second one). For 5 rails there are two elegant series. However I cannot calculate the numbers for 3 or 4 black rails, is there any nice and clean way?

I will answer the third one first.
For 1,2 and 5 black rails you have already calculated it so I would not bother calculating again.
We will use the gap method
Case 1: 3 black rails
First, you lay down 7 grey rails so there are 8 gaps
Out of those 8 gaps, you select 3 because there are 3 black rails.
So $\dbinom{8}{3} = 56$
Case 2: 4 black rails
You lay down 6 grey rails so 7 gaps are left and you have to select 4 for 4 black rails
.
So $\dbinom{7}{4} = 35$
You have made an error in the case where there are 5 black rails. The answer should be 6 for that case(6C1)
Total = 10 + 36 + 56 + 35 + 6 +1= 144

A good way to count 3's and 4's is to "fix" the first rail, and the second, variate the third, then variate the second and so on. You could try complementary counting, but to my knowledge, casework is the best approach.


What are the last two digits of $3^{100}11^{50}+7^{518}$? What are the last two digits of $3^{100}11^{50}+7^{518}$?

I know that to find the last two digits implies that you are essentially finding the remainder when divided by $100$.

Can I simplify $3^{100}$ to $3$ mod $1$ by dividing both the exponent and $100$ (which are the same in this case) by $100$?

Also, how would I simplify $7^{518}$ if that were the case?

Any help is appreciated; thanks!


Norm of linear operator. Let $T$ be an invertible linear operator on $\mathbb{R}^2$. Is it true that if $T$ has determinant $\pm 1$ , then $T$ and $T^{-1}$ have the same norm (the usual norm operator)?

If you define the matrix norm as spectral norm or 2-norm, which is the largest singular value $\mu$, then $T$ and $T^{-1}$ have the same norm if $|det(T)|=1$, because the singular values of the $T^{-1}$ is just the inverse of singular values of the $T$, and $|det(T)|=\mu_1 \mu_2 = 1$, so $T$ and $T^{-1}$ have the same norm $|\mu|$.


Probability of emails can anyone can enlighten me on how to solve this?
Are there any good approaches to tackle this kind of problem?

  The sizes of e-mail messages received could be 1K bytes, 2K bytes, 3K bytes, 4K bytes or 5K bytes. All sizes are equally likely. What is the probability that 6 incoming emails comprise one 3K byte message, three 4K byte messages and two 5K byte messages?

Thanks in advance.

You can use the Multinomial Distribution.

  For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories.

In your case: $n=6$, $k=5$, and $P(K=k)=\frac{1}{5} \forall k$.

I believe that is the correct answer,

 - **Probability getting different email:**

       Based on your question given,
       P(1k)=P(2K)=P(3K)=P(4K)=P(5K)=1/5
       (But it is not important issue)

       I believe the problem is related to the combination (nCr) to be used

Total 6 emails
= one 3Kbyte email + three 4Kbytes email + two 5Kbytes email
= 6C1 x 6C3 x 6C2 *(or =COMBIN(6,1) * COMBIN(6,3) * COMBIN(6,2) from excel)*
= 6 x 15 x 20
= 1800

Total combination of 6 emails for 5 kinds of emails
= 5 ^ 6
= 5 * 5* 5* 5* 5 * 5
= 15625

  So the actual probability is
      =1800 / 15625
      =0.1152


If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$ For $\{x,y\}\subset  \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$

Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.

Let $x = \tan\alpha$ and $y = \tan\beta$ for some $\alpha,\beta\in(-\pi/2,\pi/2)$. Then $\sqrt{x^2+1} = \sec\alpha$ and $\sqrt{y^2+1} = \sec\beta$, so the condition can be rewritten as
$$(\tan\alpha+\sec\alpha)(\tan\beta+\sec\beta) = 1. $$
Multiplying by $\cos\alpha\cos\beta$ yields
$$(\sin\alpha+1)(\sin\beta+1) = \cos\alpha\cos\beta. $$
Multiplying by $(1-\sin\alpha)(1-\sin\beta)$ yields
\begin{align} (1-\sin^2\alpha)(1-\sin^2\beta) &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies\cos^2\alpha\cos^2\beta &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta).
\end{align}
Thus
\begin{align} (1+\sin\alpha)(1+\sin\beta) = \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta) \\
\implies 1+\sin\alpha+\sin\beta+\sin\alpha\sin\beta &= 1-\sin\alpha-\sin\beta+\sin\alpha\sin\beta \\
\implies 2(\sin\alpha+\sin\beta) &= 0.
\end{align}
This implies that $\alpha = -\beta$, and hence $x = \tan\alpha = \tan(-\beta) = -\tan\beta = -y$, as desired.


If a map between separable Banach spaces has closed graph, does it have a point of continuity? It is well known that the closed graph theorem does not directly extend to nonlinear maps: even for functions from $\mathbb{R}$ to $\mathbb{R}$, having closed graph does not imply continuity. But let's consider the following reformulation of the closed graph theorem:

  If a linear map between Banach spaces has closed graph, it has a point of continuity.

This does not essentially change the meaning, since a linear map is either everywhere continuous or everywhere discontinuous. But now there is a better chance of nonlinear generalization. Hence, my question:

  Suppose that a nonlinear map between separable Banach spaces has closed graph. Does it necessarily have a point of continuity?

Nonseparable counterexample

Since  the above seems too good to be true,   I tried to find a counterexample. So far, found it only in nonseparable setting.

Let $X$ the space of all bounded functions $x:(0,1]\to\mathbb R$ with the supremum norm. Let $(q_n)_{n=1}^\infty$ be an enumeration of the rationals. Define the function $y=F(x)$ separately on each subinterval $(2^{-n},2^{1-n}]$, $n=1,2,\dots$ as
$$
y(t) = \begin{cases} 1 \quad &\text{if   $f(2^nt-1)>q_n$} \\
0 \quad &\text{if  $f(2^n t-1)\le q_n$}
 \end{cases}
$$
I claim that the map $F:X\to X$ satisfies $\|F(x_1)-F(x_2)\|=1$ whenever $x_1\ne x_2$.
Hence, it is nowhere continuous and its graph is a discrete (in particular closed) set.

Indeed, since $x_1\ne x_2$, there is a point $s\in (0,1]$ and a number $n\in\mathbb N$  such that $q_n$ is strictly between $x_1(s)$ and $x_2(s)$; according to the definition of $F$ this implies that the functions $F(x_1)$ and  $F(x_2)$ take on different values at the point $t=2^{-n}(s+1)$.

Finite-dimensional case

In finite dimensions, a map with a closed graph is continuous outside of a closed set with empty interior. The proof is not very interesting, so I link to it instead of adding it to the post.

A variation on the non-separable counter-example.
Let $X$ as above, be the space of all bounded functions $x:(0,1]→\mathbb(R$ with the supremum norm.
Since $\#(\ell_2(\mathbb N)) = \#(0,1]$, there is a bijection $g: \ell_2(\mathbb N) \rightarrow (0,1]$. Define $F: \ell_2(\mathbb N) \rightarrow X$ as
$
F(\phi) = I_{(0,g(\phi)]}
,$
the indicator function of the interval $(0,g(\phi)]. $ As abobe, $\|F(\phi_1)-F(\phi_2)\|=1 $ if $ \phi_1 \ne \phi_2 .$


solving word problems of triangles/bearings This is a long word problem so be prepared.

A group of computer game junkies decided to act out their sky rim adventures in real-life. Till Whompson, Mampton Cunninghah, Baaron Bew, Mas Darnol, and Whephan Stidener all set out from Mas Darnol's house on a bearing of N 62 degrees E for 3 miles. After stopping at the local restaurant called the Boar's Nest for some grub, they decided to head on a new bearing of S 25 degrees W for 5 miles. Unfortunately, they were ambushed by the deadly water bubble monsters. Mampton immediately was devoured by one of the monsters, and Till, Baaron, and Wephan all knocked themselves out when they turned to run away. Fortunately Mas escaped and headed back home. What bearing should Mas take to get back to his house and how far will he have to travel?

I already calculated how far he will have to travel and got $3.32$ miles using law of cosines. This could be wrong, and if it is feel free to correct me but what I am having so much trouble with is the bearing. I cannot figure it out for the life of me. Please help!

Here's a nice diagram.

Using the Law of Cosines:

$$\begin{align}
c^2&=a^2+b^2-2ab\cos C\\
c^2&=3^2+5^2-2\cdot3\cdot5\cos37\\
c^2&=34-30\cos37\\
c&=\sqrt{34-30\cos37}\\
c&\approx3.16874
\end{align}$$

$\theta$ can be found with the Law of Sines:

$$\begin{align}
\sin\theta&\approx\frac{\sin37}{3.16874}\cdot3\\
\sin\theta&\approx0.56977\\
\theta&\approx34.73
\end{align}$$

The direction is North $(34.73-25)$ degrees West

Two ships left the same port; one going in the direction N 70 E and the other is sailing directly East. The first ship traveled at the rates of 15 mph. After 30 minutes, the second ship was observed to be directly South of the first. What is the speed per hour of the second ship?


Geometric Problem. Find an angle. In△ABC , AB=AC ∠ACB=72∘. D is a point in △ABC such that ∠DBC=42∘,∠DCB=54∘.Find an angle∠BAD. Is there any geometric method?

Let, $AB=AC=a$ in isosceles $\triangle ABC$

Using Sine rule  in isosceles $\triangle ABC$  $$\frac{\sin \angle BAC}{BC}=\frac{\sin \angle ABC}{AC}$$ $$\frac{\sin 36^\circ}{BC}=\frac{\sin 72^\circ}{a}\implies BC=2a\sin 18^\circ$$ Similarly applying Sine rule in $\triangle BCD$
$$\frac{\sin \angle DCB}{BD}=\frac{\sin \angle BDC}{BC}$$  $$\frac{\sin 54^\circ}{BD}=\frac{\sin 84^\circ}{2a\sin 18^\circ}$$  $$\implies BD=\frac{2a\sin 18^\circ\sin 54^\circ}{\cos 6^\circ}$$
Let, $\angle BAD=\alpha$
Applying Sine rule in $\triangle ABD$
$$\frac{\sin \angle BAD}{BD}=\frac{\sin \angle ADB}{AB}$$  $$\frac{\sin \alpha}{\frac{2a\sin 18^\circ\sin 54^\circ}{\cos 6^\circ}}=\frac{\sin (180^\circ-(\alpha+30^\circ))}{a}$$ $$\cos 6^\circ\sin \alpha=2\sin 18^\circ\sin 54^\circ\sin (\alpha+30^\circ)$$  $$\frac{\sin(\alpha+30^\circ)}{\sin \alpha}=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$ $$\frac{\sin \alpha\cos 30^\circ+\cos \alpha\sin 30^\circ}{\sin \alpha}=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$
$$\frac{\sqrt 3}{2}+\frac{1}{2}\cot \alpha=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$
$$\frac{1}{2}\cot \alpha=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}-\frac{\sqrt 3}{2}$$

$$\cot \alpha=\frac{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}{\sin 18^\circ\sin 54^\circ}$$

$$\tan \alpha=\frac{\sin 18^\circ\sin 54^\circ}{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\alpha=\tan^{-1}\left(\frac{\sin 18^\circ\sin 54^\circ}{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}\right)=24^\circ}}$$

This problem is contradictory and has no solution. Here are the details: Let x=angle BAd, y=angle DAC. Extend DA to E and let EDC=t,EDB=z. Then we have z=x+30,t=y+18,t+z=42,x+y=36. We get the following matrix

1   0   -1  0   -30
0   1   0   -1  -18
0   0   1   1   42
1   1   0   0   36
                If we use Gauss elimination we get the last row being
0 0 0 0 42
This means 0=42 which is absurd.


A slight modification of Beta function The Beta function is defined as follows: for real values $a,b>0$
$$B(a,b) = \int_0^1 w^{a-1}(1-w)^{b-1} dw.$$

It is known that
$$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$
where $\Gamma$ is the Gamma function.

Is there a way to find a closed expression or bound for the following integral?
$$I:=\int_0^1 w^{a-1}(1-w)^{b-1} e^{-cw} dw,$$
where $c>0$. Of course, one has
$$|I| \leq B(a,b),$$
but I would like to keep an expression in terms of the exponential of $c$. ($c$ will be some $x^2$ which should then be an integrable function on $\mathbb{R}$, that is why).

Any ideas or suggestions? Thanks a lot!! :)

Use Taylor series for the exponential and get

$$\int_0^1 w^{a-1}(1-w)^{b-1}\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}w^k\ \text{d}w$$

Which becomes

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\int_0^1 w^{a-1+k}(1-w)^{b-1}\ \text{d}w$$

The integral can be computed easily, and gives

$$\frac{\Gamma(b)\Gamma(a+k)}{\Gamma(a+b+k)}$$

which is nothing but the Beta

$$B(b, a, k)$$

hence in the end you have

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\ B(b, a, k)$$

Further Manipulations

Since $B(a, b, k)$ is expressed as we wrote above, you can take out the first Gamma term which does not depend on $k$ and get a familiar result:

$$\Gamma(b)\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\frac{\Gamma(a+k)}{\Gamma(a+b+k)}$$

The series can be summed into The so called HyperGeometric Function:

$$\sum_{k = 0}^{+\infty}\frac{(-c)^k}{k!}\frac{\Gamma(a+k)}{\Gamma(a+b+k)} = \frac{\Gamma (a) \, _1F_1(a;a+b;-c)}{\Gamma (a+b)}$$

Hence the final result is:

$$\Gamma(b)\frac{\Gamma (a) \, _1F_1(a;a+b;-c)}{\Gamma (a+b)}$$

We again recognize a Beta, so finally:

$$\boxed{ B(a, b)\, _1F_1(a;a+b;-c)}$$

According to Maple, it is

$$ \eqalign{ \Gamma \left( a \right) &{{\rm e}^{-c/2}}{c}^{-a/2-1-b/2}\Gamma
 \left( b \right)  \cr&\frac{\left(  \left( 1+a \right)  \left( b+a \right) {
{M}_{a/2-b/2+1,\,b/2+a/2+1/2}\left(c\right)}+b{{ M}_{a/2-b/2,\,
b/2+a/2+1/2}\left(c\right)} \left( c+a+b \right)  \right) }{\Gamma
 \left( b+a+2 \right) }}
$$

Where $M$ is the Whittaker M function.


What is the Fourier Transform of $f(x) = x^2$? What is the Fourier transform of $f(x) = x^2$ ?
I am unsure how the problem is tackled.

Recall that for the function $f_0(x) := x^0 = 1$, we have
$$ \def\F{\mathscr F}\F(f_0) = \sqrt{2\pi} \delta $$
Moreover, if $f_1(x) = x$, then for each $u \in \mathscr S'(\mathbf R)$, we have
$$ \F(f_1 u) = i\F(u)' $$
Hence, for $f = f_1^2f_0$, we have
$$ \F(f_1^2f_0) = i^2\sqrt{2\pi}\delta'' = -\sqrt{2\pi} \delta''$$

Using the following definition of the FT
\begin{equation}
\mathcal{F}\{\rho(x)\} \equiv s(k)= \int_{-\infty}^{\infty}\rho(x)e^{-i2\pi kx} dx
\end{equation}
the derivative property of the FT is
\begin{equation}
\mathcal{F}\{\rho'(x)\} = i2\pi k s(k).
\end{equation}
Then, recalling that $\mathcal{F}\{1\} = \delta(k)$ where $\delta(\cdot)$ is a Dirac delta function, we have that
\begin{aligned}
\mathcal{F}\{1\} &= i2\pi k \mathcal{F}\left\{x\right\} \\
\mathcal{F}\left\{x\right\} &= \left(\frac{1}{i2\pi k}\right)\delta(k).
\end{aligned}
Since $\frac{d}{dx} \frac{1}{2}x^2 = x$,
\begin{equation}
\mathcal{F}\{x\} = i2\pi k \mathcal{F}\left\{\frac{1}{2}x^2\right\}.
\end{equation}
Substituting in $\mathcal{F}\left\{x\right\}$ from the previous equation and simplifying, we have
\begin{aligned}
\mathcal{F}\{x^2\} &= 2 \left(\frac{1}{i2\pi k}\right)^2 \delta(k) \\
&= -\frac{1}{2\pi^2 k^2} \delta(k).
\end{aligned}
Repeating the process, it can be shown that
\begin{aligned}
\mathcal{F}\{x^m\} &= m! \left(\frac{1}{i2\pi k}\right)^m \delta(k).
\end{aligned}


More elegant proof of that this diagram commutes Define an $\mathbb R$-linear isomorphism $f_n : \mathbb C^n \to \mathbb R^{2n}$ by $(x_1 + iy_1, x_2 + iy_2, \dots) \mapsto (x_1, y_1, x_2, y_2, \dots )$. Let $A \in M^n (\mathbb C)$ and define $R_A(v) = v \cdot A$. Furthermore define an injective homomorphism $\rho_n : M^n(\mathbb C) \to M^{2n}(\mathbb R)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -b_{ij} & a_{ij} \end{array}$ if $A_{ij}=(a_{ij} + i b_{ij})$. It is an exercise in Tapp's book to show that the following diagram commutes:
$\require{AMScd}$
\begin{CD}
    \mathbb C^n @>\displaystyle f_n>> \mathbb R^{2n}\\
    @V \displaystyle R_A V V\# @VV \displaystyle{R_{\rho_n(A)}} V\\
    \mathbb C^n @>>\displaystyle f_n> \mathbb R^{2n}
    \end{CD}

I proved it by calculating

$f_n(R_A(v))_i = (\sum_k u_k a_{ki} -w_kb_{ki}, \sum_k u_kb_{ki} + w_k a_{ki})$

and

$R_{\rho_n(A)}(f_n(v))_i  = (\sum_k u_k a_{ki}-w_k b_{ki}, \sum_ku_k b_{ki}+w_ka_{ki})$

(by slight abuse of notation)

Is there a nicer way to show this? I don't quite get the point of the exercise if it boils down to a battle of computing indices.

HINT:
\begin{eqnarray}
(x+iy)\cdot (a+ib) &=& (x a - y b) + i (x b + y a) \\
 \left(
\begin{array}{cc}
x,&\!\!\!\!y\\
\end{array} \right)\cdot \left(
\begin{array}{cc}
a&b \\
-b&a \\
\end{array} \right)&=& \left (\begin{array}{cc}
x a - y b,&\!\!\!\!x b + y a
\end{array} \right)
\end{eqnarray}


Prove that the greatest integer function: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$ Statement: the greatest integer function int: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$

Proof: let $x \in \mathbb{R}$, then $int(x) \leq x$ and is an element of $\mathbb{Z}$. Since $\mathbb{Z}$ is an element of $\mathbb{R}$, the greatest integer function maps onto $\mathbb{Z}$. However, it is not one-to-one, because $int(0.2)=int(0.3)=0$

Is my proof valid?

You'd want to more clearly show how given $x \in \mathbb{Z}$ you could find a real number $r \in \mathbb{R}$ such that $\text{int}(r)=x$. Then you have onto. As for $1:1$, you have shown that it cannot be $1:1$. Note that a function that can be plotted in $\mathbb{R}^2$ can only be $1:1$ if every horizontal line drawn in the plane intersects the graph at most one point--clearly not the case here.

Yes, your reasoning is correct, but you should say $\Bbb Z$ is a subset of $\Bbb R$. The element relation does not hold between $\Bbb Z$ and $\Bbb R$. (Some students confuse containment and membership early on.)

It looks like you are using the fact that integers map to themselves to demonstrate the map is onto, and that's good. However, it might strengthen what you wrote if you stated that a little more explicitly.

(Incidentally, the question should also be more like "determine if this function is 1-1 and/or onto," but I can tell that's really what you were interested in because of your solution.) (Obsolete due to intervening edits.)


Proof that $\sum\limits_{j,k=1}^N\frac{a_ja_k}{j+k}\ge0$ I found this inequality holds:

  Given a sequence of real numbers,
  $a_1,a_2,...,a_N$, if
  $$S=\sum_{j,k=1}^N\frac{a_ja_k}{j+k}$$ then $$S\ge0$$

Is it a known result? How is it possible to prove it?

It strongly reminds me of the Hilbert matrix, although slightly different. And here is a simple and brutal proof:

$$ S
= \sum_{j,k=1}^{N} a_j a_k \int_{0}^{1} x^{j+k-1} \, dx
= \int_{0}^{1} \bigg( \sum_{j=1}^{N} a_j x^{j-1} \bigg)^2 x \, dx
\geq 0 $$

It is also interesting to notice that the equality holds if and only if $a_1 = \cdots = a_N = 0$.

In fact, no two real numbers $a_j,a_k$ will produce a negative result regardless of the numbers, so when it's broadened to $N$ numbers it won't be negative since it's just the sum of multiple sets of 2-variable solutions. $$\frac{a_j^2}{2j}+\frac{a_ja_k}{j+k}+\frac{a_k^2}{2k}$$ The above is the expanded version of a 2-variable solution. This function will never be negative as long as $j,k$ are positive; it's irrelevant how large or small they are.


What is the pdf of g(x,y) were x and y are two random variables from a uniform distribution? Is there a formula analogous to the single random variable case i.e the pdf of y(x) where x is generated from a pdf=f(x) is given by:
$g(y)=f(x(y))|\frac{dx}{dy}|$ ?

Specifically, I have the following problem:
X,Y are two uniform distributions with $a<x<b$ and $c<y<d$. Find the pdf of $Z=\frac{1-r^2}{1+\rho^2}$ where $r=\sqrt2\rho$ and $\rho=\sqrt{(x^2+y^2)}$

There is no general formula (that I know of), (unless, you mean the this one, which is the definition):
$$P(z)=\int\delta\Big(z-g(x,y)\Big)P(x,y)dx\,dy$$
$\delta$ is the Dirac delta distribution, $P(x,y)$ is the (joint) PDF of $X$ and $Y$ and $P(z)$ is the PDF of $Z$.

However, your problem is relatively simple as you can first calculate the PDF of $\rho$ and then use the formula you cited for $Z$. The PDF of $\rho$, unfortunately, is not very nice. You can get the answer for $a=c=0$ and $b=d=1$ in this question, and you can probably generalize it to other cases, but it will not be nice.


There is not such function in $\mathbb{R}$ such that $f>0$, $f'>0$ and $f''<0$. Prove that there isn't a function $f \colon \mathbb{R} \to \mathbb{R}$  such that
$f>0$, $f'>0$ and $f''<0$.

Any suggestion will be appreciated.

Since $f$ is strictly concave and non-constant, either $\lim_{x \to -\infty} f(x)=-\infty$ or $\lim_{x \to +\infty} f(x)=-\infty$. In both cases, $f$ can't be always positive.

$f(x)= \sqrt{x}+1$ is perhaps a counter example. It is always positive, has a positive derivative and concave down.


If $k \in \mathbb N$ and if $\sum_n a_n, \sum_n a_n^2, \dots, \sum_n |a_n|^k$ converges then $\prod_n (1+a_n)$ converges Suppose that $(a_n)_n$ is a sequence of complex numbers with the property that $\sum_n a_n, \sum_n a_n^2, \dots,\sum_n |a_n|^k$ all converge where $k\in \mathbb N$. I want to show that this implies that the product $\prod_n (1+a_n)$ converges. I read this in a paper without proof so it should be true. Di you know how to prove this?

The sums $p_k = \sum_n a_n^k$ are known as power sums.
Using elementary symmetric polynomials $e_j$, you have (see here) that
$$
\prod_n (1+a_n) = \sum_{j=1}^n e_j
$$
Further, Newton's identities allow to represent all  elementary symmetric polynomials again as finite sums of products of power sums, see here. So we have that
$$
\prod_n (1+a_n) = \sum_{j=1}^n \sum_h b_{j,h} \prod_{i =1}^j c_{j,i} p_i^{m_i}
$$
with a finite number of fixed constants $b_{j,h}$ and $c_{j,i}$. Hence if all power sums converge, so does $\prod_n (1+a_n)$.


Convolution powers tend to zero Let $\mu$ be a probability measure  on $[0,\infty)$ that is not degenerated ($\mu(0) < 1$) and $f$ be a bounded function on $[0,\infty)$.
Show that pointwise
$$f * \mu^{*n} \rightarrow 0$$
where $*$ denotes convolution
$$(f * \mu)(t) = \int_0^t f(t-u) d \mu(u)$$
and
$$\mu^{*n} = \mu * \dots * \mu\quad n\text{ times}$$

Edit: Added that $\mu$ is not degenerated.
Edit2: Clarified that $f$ is defined on the positive reals.

Okay, here is my attempt, please correct me if I'm wrong.
Let's assume that
$$\nu(t) := \lim_{k\rightarrow\infty} \sum_{n=0}^k \mu^{*n}(t)$$
defines a measure that is finite for bounded sets.

Then for every $t \ge 0$
$$\lim_{k\rightarrow\infty} \sum_{n=0}^k (f*\mu^{*n})(t)= (f*\nu)(t)$$
is finite, and hence
$$\lim_{n\rightarrow\infty} (f*\mu^{*n})(t) = 0$$
For the special case $f \equiv 1$ this means
$$\lim_{n\rightarrow\infty} \mu^{*n}(t) = 0$$


Find the maximum radius of a incircle? I was soving geometry problem.
And found this question , i tried to solve it .But i am unable to get the answer.

The hypotenuse of a right triangle has length of $5$ cm.

Determine its maximum possible area of its incircle.

I try to solve it by assuming that for maximum area of incircle is when the other two sides of triangle are equal.

But this i am unable to get the answer.
Where am i wrong?

Thanks in advance.

The inradius of a right triangle is equal to $\frac{b+c-a}2$ where $a$ is the hypothenuse. Therefore the maximal radius is when $b+c$ is maximal.

On the other hand you have the inequality
$$ b+c \leq \sqrt{2(b^2+c^2)}=a\sqrt{2} $$
with equality if and only if $c=b$.

Given the three sides of a triangle will result in only one incircle. Radius of incircle is the largest circle that can be inscribed in any triangle. Maximum radius of incircle can only occur if only two sides are given, because it will result into an infinitely number of third sides.

Suppose the question is this, "What is the maximum radius of incircle of a triangle with sides $3$ cm and $4$ cm?" You cannot assume that it is a right triangle and its maximum inradius is $1$ cm because there are so many oblique triangles with sides $3$ cm and $4$ cm with different radii of incircles.


Eigenvectors of real symmetric matrices are orthogonal Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.

For any real matrix $A$ and any vectors $\mathbf{x}$ and $\mathbf{y}$, we have
$$\langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle.$$
Now assume that $A$ is symmetric, and $\mathbf{x}$ and $\mathbf{y}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\mu$. Then
$$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$
Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$.

Now find an orthonormal basis for each eigenspace; since the eigenspaces are mutually orthogonal, these vectors together give an orthonormal subset of $\mathbb{R}^n$. Finally, since symmetric matrices are diagonalizable, this set will be a basis (just count dimensions). The result you want now follows.

Note a real symmetric matrix is a linear operator on Euclidean space with respect standard basis (orthonormal). So the fact that it equals to its conjugate transpose implies it is self-adjoint.

For two distinct eigenvalues $\lambda_1, \lambda_2$ and corresponding eigenvectors $v_2, v_2$, $$(\lambda_1-\lambda_2)<v_1, v_2>=<Tv_1,v_2>-<v_1,\bar{\lambda_2} v_2>=<Tv_1,v_2>-<v_1,T^* v_2>=0$$ where the 2nd last equality follows from properties of self-adjoint (thus normal) linear operator (Lemma below).

Lemma: Assume $T$ is normal. If $(\lambda, v)$ is eigenvalue and eigenvector of $T$, $(\bar{\lambda}, v)$ is eigenvalue and eigenvector of the adjoint $T^*$. (pf.) Trivial from definition of normality.


Product of polynomials with negative coefficients Given $a \in \mathbb{Z}$ with $|a| > 1$. Let $g(x) \in \mathbb{Z}[x]$ be a polynomial with $g(a)=\pm 1$. Let Let $h(x) \in \mathbb{Z}[x]$ be a polynomial with $h(a)=\pm p$, a prime. Let $g(x)$ and $h(x)$ have not all coefficients negative nor positive. Can $g(x)h(x)$ have only positive coefficients? Is there a way to generate such polynomials?

Answered here:
by Gerry Myerson https://mathoverflow.net/questions/121568/on-reducible-polynomials


Can a polynomial of degree 2 be non quadratic By wikipedia:
In algebra, a quadratic function, a quadratic polynomial, a polynomial of degree 2, or simply a quadratic, is a polynomial function in one or more variables in which the highest-degree term is of the second degree.

But the below polynomial is also of degree 2 and non quadratic:

xy + x + y

So, why do we say, polynomial of degree 2 is a quadratic function?

We say it because of conventions. It makes sense.

I think the biggest reason why is that whether a function on the plane is quadratic shouldn't depend on exactly which coordinate axes we impose on the plane. For instance, if I didn't like your $xy$-axes, and instead wanted $uv$-axes $45^\circ$ to yours (say my point $(1,0)$ is at your $(1,1)$ and my point $(0,1)$ is at your $(-1,1)$), then my expression for the same function would be
$$
\frac14u^2-\frac14v^2 + u
$$
which clearly has quadratic terms.

$xy+x+y =0$ (I added the = 0 to make your question complete) is not of degree two for x.

Rearranging: $y=\frac{-x}{x+1}$

Thus it is not quadratic.


Can a machine exist making more steps than the current record, which is no busy beaver? There are some machines in the busy-beaver-competition for turing machines with $5$ states and $2$ symbols for which it is not known whether they halt or not.

  Is it possible that among those machines, there is a halting machine beating the current record with respect to the number of steps, but not being a busy beaver beacuse it doesn't write down more ones than the currect record  ? Or do we know that such a halting machine would also be a new record holder with respect to the number of ones ?

I tend to believe that we cannot rule out this possibility. In fact, I am not sure whether we can even rule out any final output (except trivial ones like an empty tape, if the machine contains $1RH$ forcing at least one $1$ on the tape).

I would say that yes, this is perfectly possible. We know that the champion as far as the number of $1$'s written at the end does not need to be the same as the champion in terms of steps taken (here are some champions for the quadruple definition Turing machines, and the champion for $1$'s is not the same machine as the champion for steps)

So, the situation you describe is can certainly be true at the beginning of our investigations, when every machine is a holdout ... so why would that change later on, assuming we still haven't found neither the $1$-champion nor the step-champion?

I don't know the exact machines in the competition, but in general the answer could be "we have a machine which isn't known to halt or diverge, which we've been running for years but has written down fewer 1's so far than the best-known busy-beaver candidate". In that case, exactly one of the following is true, but we don't know which:

The machine halts, leaving the tape with more $1$'s than the current record holder. Then it is a busy-beaver record beater.
The machine halts, but does not leave the tape with more $1$'s than the current record holder. Then it is not a busy-beaver.
The machine diverges. Then it is not a busy-beaver candidate at all, let alone a record-beater.


Is there a general formula for the sum of a quadratic sequence? I tried Googling "formula for sum of quadratic sequence", which did not give me anything useful. I just want an explicit formula for figuring out a sum for a quadratic sequence. For example, how would you figure out the sum of $2+6+12+20+\dots+210$? Can someone please help? Thanks
For those of you who do not know, a quadratic sequence is a sequence where the differences of the differences between the terms are constant. Let's use $2+6+12+20+\dots$ as an example. The differences between the terms are $4$, $6$, $8$, etc. The difference between the differences of the terms is $2$. So the sequence will continue like $2+6+12+20+30+42+56+72+\dots$

Yes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I  mean? Consider this

$$
t_n = n+n^2
$$

Subsituiting $n=1,2,3,\cdots$ generates your terms. By the way, $202$ doesn't occur in this sequence, the 13th term is $182$ and the $14th$ term is $210$. I am assuming it was supposed to be $210$.

So we need to find

$$
\sum_{i=1}^{n}i+i^2 = \sum_{i=1}^{n}i+\sum_{i=1}^{n}i^2
$$

where $n=14$. There are well known formulas for $\sum_{i=1}^{n}i$ and for $\sum_{i=1}^{n}i^2$. Substituting them, we get,

$$\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$$
$$=\frac{n(n+1)}{2}\left(1+\frac{2n+1}{3}\right)$$
$$=\frac{n(n+1)}{2}\left(\frac{3+2n+1}{3}\right)$$
$$=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)$$
$$=\frac{n(n+1)(n+2)}{3}$$

where $n=14$. Thus our sum is $1120$.

I have a sum formula for quadratic equation.

$$s=\frac{n}{6} (3d(n-1)+(n-1)(n-2)c) +an $$

Where $n$ is the number of terms, $d$ is the first difference, $c$ is the constant difference or difference of difference, $a$ is first term. For calculation of sum first put value of $a$ $d$ $c$ then you get a formula like $an^2+bn+c$.


Laplacian of a function restricted to a submanifold Let $f$ be a sufficiently smooth function on a manifold $S$. Let $M$ be a sub manifold of $S$. Can someone show how I can write  $\Delta^M f$ in terms of  $\Delta^{S}f$ ? ((i.e the relation between them))

PS: $\Delta^M$ and   $\Delta^{S}$ represent the Laplace-Beltrami operator the sub-manifold $M$ and the manifold S respectively.

Thank you

Let $e_1,\cdots e_m$ be local orthonormal basis on $M$ and $e_{m+1} , \cdots e_s$ be a local orthonormal frame normal to $M$. Let $\overline \nabla$ and $\nabla$ be the covariant derivative on $S$ and $M$ respectively.

$$\begin{split}
\Delta^S f &= \sum_{k=1}^s \overline\nabla^2 f(e_k, e_k) \\
&= \sum_{k=1}^m \overline\nabla^2 f(e_k,e_k) + \sum_{k=m+1}^s \overline \nabla ^2 f(e_k,e_k)
\end{split}$$

The first term on the right can be written as

$$\begin{split}
\sum_{k=1}^m \overline  \nabla^2 f(e_k,e_k) &= \sum_{k=1}^m e_k e_k f - \overline\nabla_{e_k} e_k f\\
&= \sum_{k=1}^m e_ke_k f - \nabla_{e_k } e_k f - (\overline\nabla_{e_k}e_k)^\perp f \\
&= \Delta^M (f|_M) - \vec H f,
\end{split}$$

where $\vec H$ is the mean curvature vector of $M$ in $S$. So

$$(\Delta^S f)|_M = \Delta^M f|_M - \vec H f + \sum_{k=m+1}^s \overline \nabla ^2 f(e_k,e_k) .$$

It seems that not much can be said about the last term (which cannot be seem from $f|_M$).

If you are using $$\Delta f = \frac{1}{\sqrt{detg}}d _i (g^{ij}d_jf)$$
formula , where summation is occurring. Then, by picking a local orthonormal frame on $M$, and then extending to an orthonormal frame on $S$, I think you can simply ignore those derivatives in the above formula that involve directions other than those on $S$.

Since we are dealing with an intrinsic construction, this should work. Check with $R^2$ and $R^1$ case, for instance.


conditional expectation given 2 random variables Let's say there are random variables $A$ and $B$ being independent.
And random variable $X$.

Are there any properties to simplify $E(X | (A,B))$  : expectation of $X$ given $A$ and $B$ ?

In particular, do we have a "simplification", like  $E(X | (A,B)) = E( E(X|A) | B)$ ?

I think it's incorrect (take X = A*B) but it seems strange as intuitively this would seem to be correct, so I feel like there must be some formula linking $E(X | (A,B))$ and the conditional expectations given a single variable

thanks !

Notice that the random variable $\mathbb{E}(X|A)$ is $\sigma(A)$-measurable, so, being $\sigma(A)$ and $\sigma(B)$ independent:
$$\mathbb{E}\left({\mathbb{E}}\left(X|A\right)|B\right)=\mathbb{E}\left({\mathbb{E}}\left(X|A\right)\right)=\mathbb{E}(X),$$
so it is $\mathbb{P}$-a.e. constant.

On the other hand, for example, if $X$ is $\sigma(A,B)$-measurable and it is not $\mathbb{P}$-a.e. constant, then: $$\mathbb{E}\left(X|\sigma(A,B)\right)=X\neq\mathbb{E}(X).$$

So, where did your intuition fall? When you get $\mathbb{E}(X|A)$, you get the best prediction knowing $A$ of $X.$ Now, when you get $\mathbb{E}(\mathbb{E}(X|A)|B)$, you get the best prediction knowing $B$ of [the best prediction knowing $A$ of $X$] and not the best prediction knowing $A$ and $B$ of $X$. While the first prediction has to be constant (because it is a prevision made on information independent on the quantity you are estimating, and so, knowing useless information, the best prediction you can make is to take the best prediction you can make when you know nothing at all, i.e. the expectation of the quantity) the second could be very well non-constant.

$A$ and $B$ being independent is a common, but actually strong assumption. It implies one special property for the joint probability density function:

$$
f_{AB}(a,b)=f_{A}(a)f_{B}(b)
$$

If $X$ is a random variable of its own, the joint probability of $X$, $A$ and $B$ is :

$$
f_{XAB}(x,a,b)
$$

But the conditional probability of $X$ given $A$ and $B$ has the following property:
$$
f_{X|AB}(x|a,b) = \frac{f_{XAB}(x,a,b)}{f_{AB}(a,b)}
$$

And therefore:
$$
f_{X|AB}(x|a,b)f_{AB}(a,b) = f_{XAB}(x,a,b)
$$

And if $A$ and $B$ are independent:
$$
f_{X|AB}(x|a,b)f_{A}(a)f_{B}(b) = f_{XAB}(x,a,b)
$$

And from that it's possible to derive some of the identities you may need:
$$
E(X|(A,B))= \int_{\Omega_X} x f_{X|AB}(x|a,b) dx = \int_{\Omega_X} x  \frac{f_{XAB}(x,a,b)}{f_{AB}(a,b)} dx
$$

Now, in a general case I believe $E[X|(A,B)] \neq E[E[X|A]|B]$ Because when computing E[X|A] the result is a function $g(a)$, in the sense that is may algebraically depend on the assumed value of $A$, but the result would be no longer a function of the random variable $B$ (nor of an algebraic value $b$), thus:
$$
E[E[X|A]|B] = E[g(a)|B] = g(a) = \int_{\omega_{X}} f_{X|A}(x|a)\, dx
$$

The expectation given both $A$ and $B$ is a function $h$ of both algebraic values $a$ and $b$:
$$
E[X|(A,B)] = \int_{\Omega_{X|(A,B)}} f_{X|AB}(x|a,b) dx = h(a,b)
$$

If however, $X$ was assumed independent of both $A$ and $B$, then $E[X] = E[X|(A,B)] = E[E[X|A]|B]$ because the values of $A$ and $B$ wouldn't  matter.

If X is not a random variable with its distribution, but a function of $A$ and $B$, then, the principle above holds:
$$
X= w(A,B) \Rightarrow
$$
$$
E[X|(A,B)] = \int_{\Omega_{A,B}}w(a,b) f_{AB|AB}(a,b)\, da\, db = w(a,b)
$$
$$
E[X|A] = \int_{\Omega_{A,B}}w(a,b) f_{B|A}(b|a)\, db = g(a)
$$


Almost sure convergence of Bernoulli r.v. Assume we have an independent sequence of Bernoulli r.v. $(X_n)_{n=1}^\infty$
each $X_n$ gets 1 w.p. $\frac{1}{\sqrt{n}}$ and otherwise $0$.
How can we show that almost surely the following holds?
$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N X_n  = 0  $$

Let's start with a sequence that is known to converge $a.s.$:

Let $B_n(p)$ be a sequence of iid Bernoulli r.v.s with $P(B_n(p)=1)=p$

In this case, it is known that
$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N B_n(p) = p\; (a.s.)$$

For a given $p \in (0,1]$, we can let $M_p=\lceil \frac{1}{p^2}\rceil$ and we get:

$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \leq \lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N B_n(p) = p \;(a.s.)$$

This is due to the fact that both $X_n$ and $B_n(p)$ are Bernoulli, but $X_n$ has a strictly decreasing probability of being $1$. This means that $\frac{1}{N}\sum _{n=M_p}^{N+M_p} X_n$ is bounded in probability by $\frac{1}{N} \sum _{n=1}^N B_n(p)$; i.e.,

$$P\left(\frac{1}{N}\sum_{n=M_p}^{N+M_p} X_n > 0\right) \leq P\left(\frac{1}{N} \sum_{n=1}^N B_n(p) > 0\right)$$

From the above two arguments, we have:

$$\forall p \in (0,1]\;\;\exists M_p\in \mathbb{N}: \lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \leq p\; (a.s.) $$

So, each value of $p \in (0,1]$ represents an almost-sure upper bound on the limit of an appropriately initialized series in $X_n$ (i.e., a suitably large $M_p$).

What about the $M_p-1$ terms  we omit from $\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n$. We know that these terms can add up to at most $M_p-1$, so our revised upper bound is:

$$\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n \leq \lim\limits_{N\rightarrow \infty} \left[ \frac{M_p-1}{N} +  \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \right] \leq 0+p=p\; a.s.$$

We've shown that each $p \in (0,1]$ represents an almost-sure upper bound on the value of $\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n $

Recasting in terms of the definition of limits, we can see that:

$$\forall p \in (0,1], \exists M: \left|\frac{1}{N} \sum _{n=1}^{N} X_n\right|<p\;\forall N>M\; (a.s.) \implies \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n = 0 \; a.s.$$

$R_n=\frac{1}{N} \sum _{n=1}^N X_n=\frac{1}{N}(\sum _{n=1}^\sqrt{N}{X_n})+\frac{1}{N}(\sum _{n=\sqrt{N}+1}^N{X_n})$

the first summand is $ \leq \frac{\sqrt(N)}{N}$ (number of terms)

the second summand has less than N terms each of them having an expectation smaller than $\frac{1}{\sqrt{N}}$. As it is divided by N, its expectation is smaller than $\frac{1}{\sqrt{N}}$

So the expectation of $R_n$ is $\leq \frac{2}{\sqrt{N}}$


Examples of mathematical results discovered "late" What are examples of mathematical results that were discovered surprisingly late in history? Maybe the result is a straightforward corollary of an established theorem, or maybe it's just so simple that it's surprising no one thought of it sooner.

The example that makes me ask is the 2011 paper John Baez mentioned called "Two semicircles fill half a circle", which proves a fairly simple geometrical fact similar to those that have been pondered for thousands of years.

This proof of the irrationality of $\sqrt 2$ appears to have been discovered in 1892 by A.P. Kiselev:

If $\sqrt2$ is rational, let $\triangle ABO$ be the smallest possible isosceles right triangle whose sides are integers.
Construct $CD$ perpendicular to $AO$ with $AC = AB$.  $\triangle OCD$ is another isosceles right triangle.
$AC=AB$, therefore $AC$ is an integer, therefore $OC = OA - AC$ is an integer. $\triangle OCD$ is isosceles, so $OC=CD$ and CD is an integer.  $CD$ and $BD$ are equal because they are tangent to the same circle, so $BD$ and $OD = BO-BD$ are integers.  But then $\triangle OCD$ is an isosceles right triangle with integer sides, contradicting the assumption that $\triangle ABO$ was the smallest such.
I am amazed that this wasn't found by the Greeks, because it is so much more in their style than the proof that they did find.
(Source: http://www.cut-the-knot.org/proofs/sq_root.shtml#proof7)

An example of this that jumps to my mind is Fermats last theorem, for which was conjectured by him in 1637 but only proved in 1995 despite many years work by countless mathematicians. But this proof (obviously) is not simple nor straightforward. I don't know how important that criteria is on our answers, but it is an example of a result that many believed to be true for a very long time, yet to be unable to be proved.
Fermats last theorem is that,

For a,b and c being positive intergers

$a^n+b^n\neq c^n$
For n bigger than 2


Can a sequence which decays more slowly still yield a converging series? In Bergman's companion notes to Rudin, he says that "If a sequence of positive terms has convergent sum, so does every sequence of positive terms which decays more rapidly." So given a sequence $\{a_n\}$ of positive terms such that $\sum_n a_n$ converges, if $\{b_n\}$ is such that
$$
\lim_{n\to\infty} \frac{a_n}{b_n} = +\infty,
$$ then $\sum_nb_n$ converges. I can prove this given $\{a_n\}$ and $\{b_n\}$. However, can we find $b_n$ which decays more slowly, i.e.
$$
\lim_{n\to\infty} \frac{b_n}{a_n} = +\infty
$$ such that $\sum_n b_n$ converges?

Similarly, we have the claim "if a sequence of positive terms has divergent sum, then so does every sequence of positive terms which decays more slowly."

The question is the following:

  Show that for every converging series $\sum\limits_na_n$ with positive terms there exists a  converging series $\sum\limits_nb_n$ with positive terms such that $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=0$.

A hands-on approach is as follows: for every $n$, consider $$A_n=\sum_{k\geqslant n}a_k,$$ then, by hypothesis, $A_n\to0$ hence, for each $k\geqslant0$, there exists some finite $\nu(k)$ such that $$A_{\nu(k)}\leqslant2^{-k}.$$ Assume without loss of generality that the sequence $(\nu(k))$ is nondecreasing and define $(b_n)$ by
$$\color{red}{\forall k\geqslant0,\quad\forall n\in[\nu(k),\nu(k+1))},\quad \color{red}{b_n=k\,a_n}.$$
Then $\frac{a_n}{b_n}\to0$ when $n\to\infty$ and $$\sum_{n\geqslant\nu(0)}b_n=\sum_{k\geqslant0}k\sum_{n=\nu(k)}^{\nu(k+1)-1}a_n\leqslant\sum_{k\geqslant0}kA_{\nu(k)}\leqslant\sum_{k\geqslant0}k2^{-k},$$ which is finite, hence the series $\sum\limits_nb_n$ converges, as desired.

There are a couple of things... slightly off... in your understanding.

The theorem does not say, as you imply, that if $$\sum_{n=1}^\infty a_n$$ exists and if $$\lim_{n\to\infty}\frac{b_n}{a_n} = \infty,$$ then the sum $$\sum_{n=1}^\infty b_n$$ exists. I advise you to re-check what your notes say.


Discontinuities of $f(x)=\sum \limits_{n=1}^{\infty}\frac{\{nx\}}{n^2}$ Let $f(x)=\sum \limits_{n=1}^{\infty}\dfrac{\{nx\}}{n^2}$ where $\{\}$ is fractional part. Find all discontinuities of function $f(x)$.

I think that $f(x)$ is discontinuous at every rational point.

Can anyone show how to prove strictly that $f$ for example is discontinuous at $0$ or $1/2$?

Expanding the comment:

Let, $x = \dfrac{p}{q} \in \mathbb{Q}^{+}$ (the case of negative rationals can be dealt with similarly), and for a $n \in \mathbb{N}$, we have:

Either $nx \notin \mathbb{N}$, in which case for any positive $\epsilon$-nbd of the point $x$, where, $\displaystyle 0 < \epsilon < \frac{1}{n}$,

$$\left\{\frac{np}{q}\pm n\epsilon\right\} = \left\{\frac{np}{q}\right\} \pm n\epsilon$$

Else, if $nx \in \mathbb{N}$, for points in the right $\epsilon$-neighborhood of $x$,

$$\left\{\frac{np}{q} + n\epsilon\right\} = n\epsilon$$

and, for points in the left $\epsilon$-neighborhood of $x$,

$$\left\{\frac{np}{q}- n\epsilon\right\} = 1- n\epsilon$$

If we let $\epsilon \to 0$, the left and right limits when $nx \notin \mathbb{N}$ are the same, but they are not the same when $nx \in \mathbb{N}$.

Thus, the $N^{th}$ partial sum function: $\displaystyle f_N(x) := \sum\limits_{n=1}^{N} \frac{\{nx\}}{n^2}$, is discontinuous at all $x \in \mathbb{Q}$ for sufficiently large $N$ (we need only find a $N$ large enough so that $nx \in \mathbb{N}$ for some $n < N$).

This jump discontinuity at any $x \in \mathbb{Q}$ is a positive number, call it:

$\displaystyle J_N(x) = \left|\lim\limits_{\epsilon \to 0^+} f_N(x) - \lim\limits_{\epsilon \to 0^-} f_N(x)\right| > 0$ is bounded,

and since the tail $\displaystyle f(x) - f_N(x) = \sum\limits_{n=N+1}^{\infty} \frac{\{nx\}}{n^2} \to 0$ as $N  \to \infty$, we may let $N$ large enough to ensure that $J_N(x) > |f(x) - f_N(x)|$, so that the contribution from the tail  can't make up for the jump at a discontinuity. Thus, $f(x)$ is always discontinuous at rational points.

Edit:

To prove uniform-continuity of $f_N$ at $x \in \mathbb{R \setminus Q}$,

If, $x \in \left(\frac{m}{N},\frac{m+1}{N}\right)$, then for any positive $\epsilon$ such that, $(x - \epsilon, x+\epsilon) \subset \left(\frac{m}{N},\frac{m+1}{N}\right)$ and $y \in (x - \epsilon, x+\epsilon)$,

$$|f_N(y) - f_N(x)| = \left|\sum\limits_{n=1}^{N} \frac{\{nx\} - \{ny\}}{n^2}\right| \le \epsilon \sum\limits_{n=1}^{N}\frac{1}{n^2} < 2\epsilon$$

Start with $x = a$ and conclude that $f(x) = f(a)$.  What if we are not sure of the value of $x$.  Is it true that $x \approx a$ implies $f(x) \approx f(a)$.  This is the basis of our concept of "continuity" - often we are able to guess the value of a function based on "nearby" or "similar values".

Continuous functions behave nicely under limits: $\displaystyle \lim_{x \to a} f(x) = f(a)$

Let's use this definition to try to check your function is continuous for irrational $x \not \in \mathbb{Q}$ such as $x = \sqrt{\color{#0055FF}{2}}, \sqrt{3}$.  If I just write out the equations.

$$ \lim_{x \to \sqrt{\color{#0055FF}{2}}}\; \sum \limits_{n=1}^{\infty}\frac{\{nx\}}{n^2} \stackrel{?}{=} \sum \limits_{n=1}^{\infty}\frac{\{n \sqrt{\color{#0055FF}{2}}\}}{n^2} $$

I have no idea about the number on the right side.  Is it even finite?  It's smaller than another number I am familiar with.  Even if the function is discontinuous or not, at east point it has a finite value.

$$ \sum \limits_{n=1}^{\infty}\frac{ \{ n \sqrt{2}\}}{n^2} \leq \sum \limits_{n=1}^{\infty}\frac{ 1}{n^2}  = \frac{\pi^2}{6} < \infty $$

What about nearby points?  Can we show that for $x \approx \sqrt{2}$ this function attains close values?

$$ \sum \limits_{n=1}^{\infty}\frac{ \{ n (\sqrt{2} \pm \color{red}{\epsilon})\}}{n^2} \stackrel{?}{\approx}  \sum \limits_{n=1}^{\infty}\frac{ \{ n \sqrt{2} \}}{n^2} $$

Let $\epsilon \ll 1$ be a really small number.  How about $\epsilon = \frac{1}{1000}$ or $\frac{1}{M}$.  We can bound the error in one of two possible ways.

$ \{ n \sqrt{2}  \} - n \epsilon < \{ n \sqrt{2} \pm \epsilon \} < \{ n \sqrt{2}  \} + n \epsilon$
the error here is smaller than $\frac{1}{n^2} \times n \epsilon = \frac{\epsilon}{n} \leq \epsilon$

Jumping does not occur before the $M = \frac{1}{\epsilon}$ term. For $n > M$:  $ \displaystyle 0 \leq  \frac{ \{ n \sqrt{2} \pm \epsilon \}}{n^2} \leq \frac{1}{M^2} = \epsilon^2$.

It seems we have established uniform convergence in the terms of the series and therefore our limit is correct for $x \notin \mathbb{Q}$.

Here's the plot I found.

It certainly looks discontinuous at $x = \frac{1}{2}$ as we expected.  It looks like a kind of "swoosh" shape repeated many times,  but a similar jump discontinuity should appear at all $x \in \mathbb{Q}$.

It's a totally reasonable-looking function. It's defined at all points in $[0,1]$, $\int_0^1 f(x) \, dx$ makes perfect sense and yet this function has horrible discontinuities everywhere!


Roots of $x^3+5x-18=0$ using Cardano's method Given that $x^3+5x-18=0$. We have to solve it using Cardano's method.

Using trial $x=2$ will be a root. Dividing the equation by $x-2$ we shall get the other quadratic equation and solving that one, we shall obtain all the roots.

But when I am trying to solve the equation using Cardan\rq s method, the calculation is becoming very difficult. I don\rq t know why. Please help.

Here is how did I proceed.

Let $x=u+v$. Then $x^3=u^3+v^3+3uvx$ i.e.  $x^3-3uvx-(u^3+v^3)=0$. So $-3uv=5$ and $u^3+v^3=18$. Clearly $u^3, v^3$ are the roots of
\begin{align}
&t^2-(u^3+v^3)t+(uv)^3=0\\
\Rightarrow &t^2-18t-\frac{125}{27}=0\\
\Rightarrow &27t^2-(27\times 18)t-125=0
\end{align}

and from here when we are getting the roots of $t$, they are very complicated. Hence I do not know how to simplify them so that $x=2$ finally be achieved along with the other two roots.

Please help me

Cardono's formula says that one root is$$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}+\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}.$$But$$\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}=1+2\sqrt{\frac23}\text{ and }\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}=1-2\sqrt{\frac23}.$$Adding these two numbers, you'll get $2$.


Proof relating to Measurement of angle The question asks to prove that radian is a constant angle.

First, I didn't understand the question and what is way to prove that the radian is a constant angle.

Your question is not so clear.

Anyway you can proceed from figure similarity comparisons bringing in other figures that are implied here.

Draw a series of circular arcs enclosed between two radial lines making an angle $\theta $ between the lines or arms.  Join the chords. When triangles or sectors are placed so that circle centers are same, it will be noticed that the ratio of corresponding sides or arcs is same. That is,

$$ \frac{Arc_1}{Arc_2} = \frac{Arm_1}{Arm_2}$$

This ratio is called proportion in its simplest form. It is a constant.It is saying the same thing as:

$$ \frac{Arc_1}{Arm_1} = \frac{Arc_2}{Arm_2}$$

This ratio is called angle $\theta$ here,  which must be also... and likewise be constant.

Let $O$ be the centre of a circle and radius $OR = r$. If we take an arc $AB = OA = r$, then by definition, $∠AOB =1$ radian.
Let $AO$ be produced to meet the circle at the point $C$. Then the length of the arc $ABC$ half the circumference and $∠AOC$, the angle at the centre subtended by this arc $= a$ straight angle $=$ two right angles.
Now if we take the ratio of the two arcs and that of the two angles, we have
arc $AB/$arc $ABC = r/(1/2 × 2\cdot\pi\cdot r) = 1/ \pi$
$∠AOB/∠AOC = 1$ radian$/2$ right angles
But in geometry, we can show that an arc of a circle is proportional to the angle it subtends at the centre of the circle.
Therefore, $∠AOB/∠AOC =$ arc $AB/$arc $ABC$
or, $1$ radian$/2$ right angles $= 1/\pi$
Therefore, $1$ radian $= 2/\pi$ right angles
This is constant as both $2$ right angles and $\pi$ are constants.
The approximate value of $\pi$ is taken as $22/7$ for calculation


Does $ W $ form a vectors subspace of the vector space of polynomial with real coefficients. $\displaystyle W=\{p(x) \in V: p(x)=p(1-x)\} , x\in R$.}

Does $ W $ form a  vectors subspace  of the vector space  of polynomial with real coefficients?

My attempts:  $p(x) $ = $p(1-x)$   as  $p(x) - p(1-x)  =  p(x+1-x)$

now   $c .p(x+1-x)  =  cp(x) - cp(1-x)$ so my answer is yes.......

Is that correct?

Let $p(x),q(x)\in W$ and $\alpha,\beta\in\mathbb{R}$ then
$$r(x):=\alpha p(x)+\beta q(x)=\alpha p(1-x)+\beta q(1-x)=r(1-x)$$
hence $r(x)\in W$ so $W$ is a subspace.


Maximum number of edges in a non-Hamiltonian graph I need to show that the maximum number of edges of non-Hamiltonian, simple graph, on $n$ vertices noted by $t(n,H_n)$ is $\binom{n-1}{2} + 1$. It's essential to show the upper and lower bounds for this question.

Now, I can see why I need to show the upper bound - for $t(n,H_n) > \binom{n-1}{2} + 1$ there's a clique on $n-1$ vertices and one other vertex with at least 2 edges. Surely, a Hamiltonian cycle exist in such graph.

What about the lower bound, though? Am I need to prove that for every $t(n,H_n) \leq \binom{n-1}{2} + 1$ there is no Hamiltonian cycle? By negation, I seem to not advance anywhere, because I don't know of any necessary conditions for a Hamiltonian cycle to exist, beside the cut-vertex one. Can someone give me a clue on that?

Thanks in advance.

Found a lot easier proof for the harder part, using Ore's theorem: If $G$ is complete, there's a Hamiltonian cycle. So suppose it's not. We'll take some non-neighbor vertices $v,w$, and delete them from the graph. For the resulting graph $G'$ - $|E(G')| \geq \binom {n-1}{2} + 2 - (d(v)+d(w))$. In the complete graph on $n-2$ vertices there are $\binom {n-2}{2}$ edges. So, lastly, we get that $$\binom{n-2}{2} \geq \binom{n-1}{2} + 2 - (d(v)+d(w))$$ and after some easy operations, we get that $$d(v)+d(w) \geq n$$ which is exactly Ore's condition for a Hamiltonian graph to exist.

First see that you can have a complete graph on n-1 vertices where the number of edges is n-1 C 2 and then you just need to consider how many edges you can add to a new incoming vertex such that the resulting graph is Non- Hamiltonian. Hamiltonicity of the complete graph implies that only one edge can be added .


Produce an explicit bijection between rationals and naturals? I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and the natural numbers, but can anyone produce an explicit formula for such a bijection?

We will first find a bijection $h_{+}:\mathbb Z^+\to \mathbb Q^+$. From there, we easily get a bijection $h:\mathbb Z\to \mathbb Q$ by defining: $$h(n)=\begin{cases}h_{+}(n)&n>0\\
0&n=0\\
-h_{+}(-n)&n<0
\end{cases}$$

From there, we can use any of the bijections $\mathbb N\to\mathbb Z$ to get our bijection between $\mathbb N$ and $\mathbb Q$. (We'll need a specific such bijection below, $s$.)

Now, every positive integer can be written uniquely as $p_1^{a_1}p_2^{a_2}\cdots$, where the $p_1=2,p_2=3,p_3=5,\dots$ is the sequence of all primes, and the $a_i$ are non-negative integers, and are non-zero for only finitely many $i$s.

Similarly, every positive rational number can be written uniquely as $p_1^{b_1}p_2^{b_2}\cdots$ where the $b_i$ are integers and only finitely many of the $b_i$ are non-zero.

So define $s:\mathbb N\to\mathbb Z$ (where we take $\mathbb N$ to include $0$):

$$s(n)=
(-1)^n\left\lfloor\frac{n+1}{2}\right\rfloor
$$

The sequence $s(0),s(1),s(2),s(3),\dots$ would be $0,-1,1,-2,2\dots$, and this is a bijection from $\mathbb N$ to $\mathbb Z$. The only properties we really need for $s$ is that $s$ is a bijection and $s(0)=0$.

Then for any $n=p_1^{a_1}p_2^{a_2}\cdots\in\mathbb Z^+$, define $$h_{+}(n)=p_1^{s(a_1)}p_2^{s(a_2)}\cdots $$

This then defines our bijection $h_{+}:\mathbb Z^+\to \mathbb Q^{+}$.

A potientially interesting feature of $h_+$ is that it is multiplicative - that is, if $\gcd(m,n)=1$ then $h_{+}(mn)=h_+(m)h_{+}(n).$

This is a bijection between the Stern-Brocot tree and the tree of Natural numbers.
Every left node is given by $ L_n = [2 P_n ] $ and every right one by $ R_n= [2 P_n +1 ]$ where $P_n $ is the value of the parent node and $P_0=[1]$.
We have the sequence of transformations $ P_n \rightarrow [ L_n  , R_n  ]$,    $L_n \rightarrow P_{n+1}$,  $R_n \rightarrow P^\prime_{n+1}$ .

In list notation for the tree (count the brackets) this is

$$ n = 1 \mapsto
[1,[2],[3]]
$$

$$n = 2  \mapsto [1,[2,[4],
      [5]],
   [3,[6],
      [7]]]
$$
$$
n = 3  \mapsto [1,[2,[4,[8],[9]],[5,[10],[11]]],[3,[6,[12],[13]],[7,[14],[15]]]]
$$
and so on.


$(x+2)/20=1/5$. What is $x$? $(x+2)/20=1/5$. I have been stuck on this problem and tried various solutions to no avail. Please help.

$$
\frac{x+2}{20}=\frac{1}{5} \tag*{ – multiply both sides by $20$}
$$
$$
\frac{x+2}{1}=\frac{20}{5} \tag*{  – simplify}
$$
$$
x+2=4 \tag*{  – substract $2$ from both sides}
$$
$$
x=2 \tag*{  – the answer}
$$

In general, whenever dealing with a fraction equality such as$$\frac{a}{b} = \frac{c}{d},$$ remember the extremes rule: $$a \times d = b \times c.$$

Applied to your case, this gives $$5 \times (x+2) = 20\times 1$$ which we now expand to $$5x + 10 = 20$$ leading to $$5x = 10$$ and thus $$x=2.$$


Version of iterated expectations conditioned on subsets: Simple proof? Thanks for any help with this. It is from the Stokey and Lucas (1988) Recursive Methods text (pg. 208) and uses notation from a Dynamic Modeling course taught at Carnegie Mellon and at Florida International University.

How do I sketch a detailed proof with the joint and conditional PDFs to show the following "general" version of the law of iterated expectations?

Let $I_t$ denote information available at time $t$ where $I_t \in I$. And let $\omega \subset I$ be a subset of the original information set. Then the expectation of a random variable, $x$ conditional on the realization $I_t \in I$ is denoted $E[x|I_t]$. The law of iterated expectations is

$$E[E[x|I_t]|\omega_t] = E[x | \omega_t]$$

The furthest I can get is:

$$\sum^{I_t} E[x|I_t] f(I_t|\omega_t) = \sum^{I_t}\sum^x x f(x|I_t)f(I_t|\omega_t) = \sum^{I_t}\sum^x x \frac{f(x,I_t)}{f(I_t)}\frac{f(I_t,\omega_t)}{f(\omega_t)}= ?$$

It's difficult to see how you end up with $f(x,\omega_t)$ (which you'd need to obtain $f(x|\omega_t)$) from the last term. It seems to come down to whether it's true that $f(I_t,\omega_t) = f(I_t)$ when $\omega \in I$.
Thanks for any help.

It seems that the question asked is the following:

  Let $X$ denote an integrable random variable defined on a probability space $(\Omega,\mathcal F,P)$, $\mathcal G$ a sub-sigma-algebra of $\mathcal F$ and $A\in\mathcal  G$ an event of positive probability, then $$E(E(X\mid \mathcal G)\mid A)=E(X\mid A).$$

To prove this, let $Y=E(X\mid \mathcal G)$, then $$E(Y\mid A)=\frac{E(Y\mathbf 1_A)}{P(A)},\qquad E(X\mid A)=\frac{E(X\mathbf 1_A)}{P(A)},$$ hence it suffices to show that $$E(Y\mathbf 1_A)=E(X\mathbf 1_A).$$
But the fact that this last identity holds (for every $A$ in $\mathcal G$) is (together with the fact that $Y$ is $\mathcal G$-measurable) exactly the definition of the conditional expectation $Y=E(X\mid \mathcal G)$, see the obvious reference.

Here is a possible answer, but the full answer still depends on whether $f(I_t,\omega_t) = f(I_t)$ when $\omega \in I_t$ ($\omega$ is a subset of $I_t$). If so, here's a proof in the other direction. I'm just using $E[x | I_t, \omega_t ]$ instead of $E[x |I_t]$. Of course, this could come from $f(I_t,\omega_t) = f(I_t)$.

$$E[x | I_t, \omega_t] = \sum^x x f(x|\omega_t) = \sum^x x \frac{f(x,\omega_t)}{f(\omega_t)}=\sum^x x \frac{ \sum^{I_t}f(x,I_t, \omega_t)}{f(\omega_t)}=\sum^x x \frac{ \sum^{I_t}f(x|I_t, \omega_t)f(I_t, \omega_t)}{f(\omega_t)}=\sum^{I_t}\frac{f(I_t, \omega_t)}{f(\omega_t)}\sum^x xf(x|I_t, \omega_t)=\sum^{I_t}{f(I_t |\omega_t)}E[x|I,\omega_t)]=E[E[x|I_t, \omega_t]|\omega_t]$$


How many positive integers $< 1{,}000{,}000$ contain the digit $2$? How many positive integers less than $1{,}000{,}000$ have the digit $2$ in them?

I could determine it by summing it in terms of the number of decimal places, i.e. between $999{,}999$ and $100{,}000$, etc.

Then to determine the number of numbers between $999{,}999$ and $100{,}000$ that have the digit $2$ in them would be $9^5$.

Is this correct, or am I miscounting?

Though not always the smartest way, such questions can mechanically be answered as follows.  (In this case the "smart" way to do it is Cameron's answer.  It is instructive to see that this mechanical procedure basically recovers Cameron's method.)  Let $a_n$ and $b_n$ be the amounts of $n$-digit numbers that do not and do have a $2$ in them.  So $a_0=1$ and $b_0=0$. These number satisfy the recurrence
$$
\begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix}=
\begin{pmatrix}9&0\\1&10\end{pmatrix}
\begin{pmatrix}a_n\\b_n\end{pmatrix}
$$

(Take a moment to understand what this recurrence expresses.)  Now
$$
\begin{pmatrix}9&0\\1&10\end{pmatrix}^6 \begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}531441\\468559\end{pmatrix}
$$

so the answer is $468559=10^6-9^6$.

the answer is $427608$ for the number with $2$ between $100000$ and $999999$. I create program to count the number with number two between $100000$ and $999999$.
Here is the code I used

for (int num = 100000; num < 999999; num++)
            {
                if (num.ToString().Contains("2"))
                {
                    numberWithTwo.Add(num);
                    count++;
                }
            }
            Console.WriteLine("Total number that contain number 2 is :"+ count);
            Console.ReadLine();


Answer not matching for calculating the value of an integral Consider the following integral $$\int_ {-\infty}^{\infty} \frac{x^2 - x + 2 }{x^4 + 10x^2 + 9} dx \!$$

I need to find its value using residue calculus. So, I considered the complex function $$\int_ {-\infty}^{\infty} \frac{z^2 - z + 2 }{z^4 + 10z^2 + 9} dz \!$$

The poles of above rational function are at $i , -i, 3i , -3i$. Now I calculated the residues at each singularity. In order to get the value of the integral I used $$\int_{\delta D}f(z)dz = 2 \pi i \sum_{j=1}^{m}Res[f(z), z_j]$$ where $z_j $'s are the isolated singularities and $\delta D$ is a piecewise smooth boundary. Using the above formula when I am summing up all the residues (and multiplying by $2 \pi i$) I am not getting the answer given in the book. Instead the answer given in the book has been calculated by adding just the residues at $z =i$ and $z = 3i$. I don't understand this. Why the sigularities which are in upper half plane are only considered?

Also I have a second question - How will the value of integral changes on changing the integral limits as $0$ to $\infty$ ?

It's about choosing the right piecewise smooth boundary $\delta D$.

Let $\delta D_r = \{z =re^{it}, t\in [0,\pi] \}\cup [-r,r]$ the boundary of an upper half-disk. With $r$ sufficiently large (i.e. $r>3$), we have that $$\int_{\delta D_r}f(z)dz = 2\pi i \sum_{z_j: \Re (z_j)  > 0}\operatorname{Res}(f,z_j).$$

Furthermore, $$\int_{\delta D_r}f(z)dz = \int_{\{z =re^{it}, t\in [0,\pi] \}}f(z)dz + \int_{[-r,r]}f(z)dz.$$

The left hand side summand (the integral over the upper arc) vanishes as $r\rightarrow \infty$ (why?) and the other summand converges to your real integral (why?).

How are you calculating the residues?
The residue at 3i is  $$ \frac{1}{16} - \frac{7i }{48} $$
The residue at i is  $$ -\frac{1}{16} - \frac{i }{16} $$
[In class, verify these by showing your work.]
Add these to obtain $$ -\frac{5i }{24} $$
These yield the integral value using the Res Theorem.

You should get the same answer if you perform the integration using Partial Fractions.


Proving that $\sum_{i=0}^n 2i=n^2+n$ using induction Prove that $$\sum_{i=0}^n 2i=n^2+n$$

I attempted to solve the question by proving the base case $n=0$, and assuming the inductive hypothesis $$\sum_{i=0}^n 2i=n^2+n$$ to be true, so $$\sum_{i=0}^{n+1}2i=\sum_{i=0}^n{2i}+2(n+1)=n^2+n+2(n+1)$$

I am confused as to what I should be getting. In the similarly solved examples, some expressions on the right were reduced to the original inductive hypothesis, while others worked on the right side to produce the left side. I worked on the right and got the following result:

$n^2+n+2n+1=n^2+2n+1+n+1=(n+1)^2+(n+1)$ which looks like the inductive hypothesis on the right side.

I am having issues with similar induction proves that involve the summation/sigma notation; one is an inequality and the other is an algebraic expression with 'i' in the denominator. I am not sure if I should include them or ask separately as they are very similar.

Based on your inductive hypothesis, you just need to confirm that:

$$\left(\sum_{i=0}^n 2i\right)+2n+2 \stackrel{?}{=} (n+1)^2 + (n+1)$$

i.e.

$$(n^2+n)+2n+2 = n^2+3n+2\stackrel{?}{=}\\ (n+1)^2 + (n+1) = n^2+2n+1+n+1=n^2+3n+2$$

They are equal, so that is the solution.

Observe that you can take the two out: then

$$\begin{align}
\sum_{i=0}^n 2i&=2\sum_{i=0}^n i \\
&= 2\left(\frac{n}{2}(n+1)\right) \\
&=n(n+1) \\
&=n^2+n,
\end{align}$$

since $\sum_{i=0}^n i=\frac{n}{2}(n+1)$ by induction.


How can I evaluate this improper integral? I have a problem: evaluate $$\int_{0}^{\infty}\frac{\cos(x)-\cos(2x)}{x}dx\,.$$ I am told this integral is not an elementary one and that's why I am stuck where to start. Thank you for helping me.

$$\int_a^{\infty}\frac{\cos(2x)}{x}dx = \int_{2a}^{\infty}\frac{\cos(x)}{x}dx$$ and so this integral equals $$\lim_{a\downarrow 0}\int_a^{2a}\frac{\cos(x)}{x}dx$$ which is easily found to be equal to $\log(2)$ using the inequalities $$1-\frac{x^2}{2}\leq\cos(x)\leq 1.$$

You have to use the limited developement of cos(x) and cos(2x).


What is the determinant of a binary matrix with exactly two ones in each row and column? Consider a binary matrix $A \in \{0,1\}^{n \times n}$. If it has exactly one $1$ in each row and column, is is called a permutation matrix and its determinant is $\pm 1$. What if I have a binary matrix with exactly two ones in each row and column? What is its determinant?

The $1$s form loops in the matrix.  Start with any $1$, go to the other $1$ in the same row, then the other $1$ in the same column, then same row, and so on.

Suppose they form one loop.  Then expanding along any row, it can be seen the determinant will be $\pm1\pm1$.  I think, it will be $0$ if there is an even number of rows, and $\pm2$ if there is an odd number.

Suppose they form several loops.  The determinant of the combination is the product of the $0$ and $\pm2$ from each individual loop.

So I think it will be $\pm2^k$, where $n$ can be written as the sum of $k$ odd numbers (each $3$ or more) and no even numbers; or it may be $0$.

(Playing at 4am)

If 2x2, 0.

If 3x3,
$
1 1 0\\
1 0 1\\
0 1 1
$
is -2.

An expansion by minors
should help,
but I can't think clearly
now.