Two seemingly different versions of the Egyptian Multiplication Algorithm - how do they relate? Here is a computer-based implementation of the Egyptian Multiplication algorithm which I believe is correct:
def multiply_egyptian(a, b):
    print(f"{a} is the multiplicand, {b} is the multiplier")
    print(f"a = {a}, b = {b}")
    result = 0
    while b != 0:
        if b & 1 != 0:  # if b is odd
            result += a
        a <<= 1  # Double it
        print(f"a = {a}, ", end="")
        b >>= 1  # Halve it
        print(f"b = {b}")
    print(f"result is: {result}")
    return result

multiply_egyptian(7, 5)

7 is the multiplicand, 5 is the multiplier
a = 7, b = 5
a = 14, b = 2
a = 28, b = 1
a = 56, b = 0
result is: 35
>>>

Here is the written version:
b    a
1  ✓ 7
2    14
4  ✓ 28
8    56

How do they relate?
The algorithm seems to go backwards in some sense. For one thing, the powers of 2 are chosen at the end in the written version.
b starts at 5, but a starts at 7, so contrary to the written version, the movement seems to be "diagonal" - b starts at the bottom, but a starts at the top.
I get that if b is odd, then it must lie between a power of 2, and thus give a remainder when divided by 2. so the remainder at that step of the algorithm is added to the result. What is the remainder at that step of the algorithm?
5//2 = 2 r 1, so I would have expected to add 28, maybe, at that stage, as much as I would expect anything from my current level of understanding - at least 28 is opposite the nearest power of 2 to 5.
Could someone please explain the relationship of the the written algorithm to the computer-based one given, in as much detail as possible?

There are two strands:
The product of two numbers, one of which is written in binary form is: $$A \times (c_n2^n+c_{n-1}2^{n-1}+\cdots+c_12+c_0)=c_n(2^nA)+\cdots+c_1(2A)+c_0A$$ where $c_i=0$ or $1$. Hence one needs to add the doublings of $A$ only for $c_i=1$.
To write a number in binary form, one can use repeated divisions by $2$:
\begin{align*}
35\div2&=17+1\\
17\div2&=8+1\\
8\div2&=4+0\\
4\div2&=2+0\\
2\div2&=1+0\\
1\div2&=0+1
\end{align*} The binary expansion is then the remainders of $1$s and $0$s in reverse order: $35=100011_2$.
The computer program combines these two in one. As it halves $b$ and doubles $a$, it adds the $a$s that correspond to odd $b$s, equivalently the $c_i(2^iA)$ when $c_i=1$. An odd $b_i$, of course, gives a remainder of $c_i=1$.
The written algorithm (which I think is the form actually used by the ancient Egyptians, though they used other short-cuts as well) builds the number $b$ in a binary form visually, for example $5=1+4=c_0+c_12^2$, and transfers those $c_i$ as ticks to the doublings of $a$, i.e. $c_07+c_1(2^27)$.

In the written version, they're just picking out the powers of $2$ in the binary representation of the multiplier: $1+4=5$.  I'm not sure why the last line is included.  Once you get as high as $4$, you're done.  Perhaps their method is to keep doubling until you get a number bigger than the multiplier, but then why bother to double the second column?
After all this is how the algorithm works.  In the computer version, the rows where you have odd numbers in the first column correspond to the places where you have $1$s in the binary representation of the multiplier.
To carry out the algorithm, you check off the largest power of $2$, lees than or equal to the multiplier, then subtract it from the multiplier.  The you check off the largest power of $2$ less than or equal to the difference, subtract it, and continue until you get a difference of $0$.  This is neat.  No division at all, just doubling and subtraction.
Where did you find this called "Egyptian multiplication?"  I've always known it as "Russian peasant multiplication."


Need countereample : If a sequence $(a_n) \in l^2 $ , then the sequence $(1/a_n) \notin l^2 $ I want to know the counterexample for the following statement : Given a sequence $(a_n)$ such that $a_i\ne 0 $ for any $i$ :  If the sequence $(a_n) \in l^2 $ , then the sequence $(1/a_n) \notin l^2$.

If $a\in\ell^2$, then $a_n\to0$. Thus $1/a_n\not\to0$, which prevents it from being in $\ell^2$.

Put $a_n=\frac{1}{n}$.
Then $ \sum \frac{1}{n^2}=\frac{\pi^2}{6}$,finite.

But $\sum n^2$ is not in $l^2$.


What does the sum notation $\sum\limits_{j=n}^\infty a_j$ mean? What does the sum notation $\sum\limits_{j=n}^\infty a_j$, $n\in\mathbb{N}$, mean?

Is this the sequence of the sums, i.e:

$$ \sum_{j=n}^\infty a_j=\left( \sum_{j=1}^\infty a_j,\;
  \sum_{j=2}^\infty a_j,\;
  \sum_{j=3}^\infty a_j,\;
  \sum_{j=4}^\infty a_j,\;
  \sum_{j=5}^\infty a_j,\;
   \ldots\right)\ ?
$$

It means the limit of the partial sums when $k\rightarrow \infty$ and $n\in\mathbb N$ is fixed, i.e $$\lim\limits_{k\rightarrow\infty}\sum\limits_{j=n}^{k}{a_j}$$
If it was $$\left\{\sum\limits_{j=n}^{\infty}{a_j}\right\}_{n=1}^{\infty}$$ then it is the sequence that you wrote.

Basically the notation
$$\sum\limits_{j=n}^\infty a_j$$ is the infinite sum of the values of $a_j$ when $j$ changes so:
$$
\sum\limits_{j=n}^\infty a_j=a_n+a_{n+1}++a_{n+2}+a_{n+3}+a_{n+4}+ \cdots
$$ up until $n$ reaches infinity.


If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? Suppose $f\in L^1(\mathbb{R})\cap L^\infty(\mathbb{R})$ is bounded, non-negative and integrable (w.r.t. Lebesgue measure) : does there exist $g$ continuous (non-negative) and integrable such that
$$ f\leq g$$
almost everywhere? If so, can we choose $g$ bounded on top of that?

In general, there does not exist such a $g$. To see this, let $(q_n)_{n \in \Bbb{N}}$ be an enumeration of $\Bbb{Q}$. Using $\sigma$-subadditivity of the Lebesgue measure, it is easy to see that the open set
$$
U := \bigcup_{n \in \Bbb{N}} (q_n - 2^{-n}, q_n + 2^{-n})
$$
has finite measure; in fact, $\lambda(U) \leq 2$. Hence, the indicator function/characteristic function $f := 1_U$ satisfies all of your assumptions.

Now, assume that $f \leq g$ almost everywhere with $g$ continuous, say $f(x) \leq g(x)$ for all $x \in \Bbb{R} \setminus N$, with $\lambda(N) = 0$.

We will show $g \geq 1$, so that $g$ is not integrable. By density of $\Bbb{Q}$, it suffices to show $g(x) \geq 1$ for all $x \in \Bbb{Q}$. Let $x \in \Bbb{Q}$ be arbitrary. It suffices to show that for each $\varepsilon > 0$, there is $x_\epsilon$ with $|x - x_\varepsilon| < \varepsilon$ and $g(x_\varepsilon) \geq 1$.

But $x = x_n$ for some $n \in \Bbb{N}$ and the interval
$$
I := (x - \varepsilon, x + \varepsilon) \cap (x - 2^{-n}, x + 2^{-n})
$$
has positive measure, so that there is some $y \in I \setminus N$. But $I \subset U$, so that $1 = f(y) \leq g(y)$. All in all, $x_\varepsilon := y$ does what we want.

Yes, I think so. Here's an idea of a proof, but there are some issues that I've not resolved:

Let $E_n = \{x : n-1 < f(x)<n\}$ which are all measurable with finite measure, we also have that for large enough $E_n$ that $\mu(E_n) = 0$. We also have that $f \le \sum \chi_j$ a.e. where $\chi_j$ is the characteristic function of $E_j$.

So we have to prove that there's a continuous function $g_j \ge \chi_j$ that is integrable. We want to do that by setting $g_j = 1$ on $E_j$ and that it's support is measurable with limited measure (ie less than $\mu(E_j)+\epsilon_j$ for an $\epsilon_j$ of our choice). Then we would have that $\int g dx < \sum \int \chi_j + \sum \epsilon_j < \infty$  if we select $\epsilon_j$ wisely.

Now since $E_j$ is measurable we have that there's an open set $U_j\supset E_j$ such that $\mu(U_j)<\mu(E_j)+\epsilon_j/2$. Now we use that we can write $U_j$ as an union of contable non-intersecting (open) intervals, $U_j = \bigcup I_j(k)$.

Now we form functions that are $1$ on $I_j(k)$ and drops of fast enough so it's integral is no larger than $\mu(I_j(k))+\epsilon 2^{-k}$. This construct should hopefully be enough to make sure that their sum is continuous and it's integral would be small enough.

Then adding all these functions up would give us $g$.


prove that $\frac{d^2y}{dx^2} = -\frac{a}{y^2}$ for $ x = a\arccos\left(\frac{y-a}{a}\right) - \sqrt{2ay-y^2}$ What is the easiest way to prove that $$\frac{d^2y}{dx^2} = -\frac{a}{y^2}$$ for
$$ x = a\arccos\left(\frac{y-a}{a}\right) - \sqrt{2ay-y^2}$$ where $a$ is a constant.

I tried solving it by differentiation on the right side and isolating the derivative, I did that twice to get the second derivative, what I wrong with my approach?

Make the variable substitution $v = \frac1a(y-1)$ to simplify the expression to

$$\frac xa = \arccos v - \sqrt{1-v^2}$$

Then, use $(\arccos v )'=-\frac1{\sqrt{1-v^2}}$ and take the derivatives of both sides to get

$$ \frac1a = -\sqrt{\frac{1-v}{1+v}}\cdot\frac{dv}{dx}$$

Proceed by differentiate again ...


In $\mathbb{R}^n$, is the dot product the only inner product? Just what the title says. I'm reading, from various resources, that the inner product is a generalization of the dot product. However, the only example of inner product I can find, is still the dot product.

Does another "inner product" exist on the $\mathbb{R}^n$ space?

Yes and are all in the form
$$
\langle x, y\rangle=x^T\cdot A\cdot y
$$
where $x^T$ is the transpose vector of $x$ and $A$ is a $n\times n$ symmetric definite positive matrix.

In fact let $\langle x, y\rangle$ a generic inner product on $\mathbb R^n$ then for every $y$ the function
$$
f_y:x\rightarrow \langle x, y\rangle
$$
is linear from $\mathbb R^n$ to $\mathbb R$ then exists a vector $\alpha(y)\in\mathbb R^n$ such that
$$
\langle x, y\rangle = \alpha(y)^T\cdot x
$$

Observe that
$$
\langle x, ay+by'\rangle = a\langle x, y\rangle + b\langle x, y'\rangle\Rightarrow \alpha(ay+by')=a\alpha(y)+b\alpha(y')
$$
then $\alpha$ is a linear operator from $\mathbb R^n$ in itself then exists an $n\times n$ matrix $A$ such that
$$
\alpha(y)=A\cdot y
$$
so
$$
\alpha(y)^T\cdot x=y^T\cdot A^T\cdot x
$$

Now remember that $\langle x, y\rangle=\langle y, x\rangle$ then you can easly prove that $A^T=A$ and $A$ must be symmetric.

Why now $A$ must be definite positive? Because $\langle x, x\rangle\geq 0$ and holds equality if and only if $x=0$. Applying it to the initial formula we obtain the definition of a definite positive matrix.

The tensor metric allows for the vector norm to remain constant under change of basis vectors, and is an example of inner product (page 15). In the simple setting of basis vectors constant in direction and magnitude from point to point in $\mathbb R^2,$ here is an example:

Changing basis vectors from Euclidean orthonormal basis $\small\left\{\vec e_1=\begin{bmatrix}1\\0 \end{bmatrix},\vec e_2=\begin{bmatrix}0\\1 \end{bmatrix}\right\}$ to

$$ \left\{\vec u_1=\color{red}2\vec e_1 + \color{red}1\vec e_2,\quad \vec u_2=\color{blue}{-\frac 1 2 }\vec e_2+\color{blue}{\frac 1 4} \vec e_2\right\}\tag 1$$ would result in a different norm of the vector $\vec v=\begin{bmatrix} 1\\1\end{bmatrix}_\text{e basis},$ i.e $\Vert \vec v \Vert^2=v_1^2 + v_2^2 = 2,$ when expressed with respect to the new basis vectors.

In this regard, since vector components transform contravariantly, the change to the new coordinate system would be given by the backward transformation matrix:

$$B=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}$$

i.e. the inverse of the forward transformation for the basis vectors as defined in $(1),$ which in matrix form corresponds to

$$F=\begin{bmatrix} \color{red} 2  & \color{blue}{-\frac 1 2}\\\color{red}1& \color{blue}{\frac 1 4}\end{bmatrix}.$$

Hence, the same vector $\vec v$ expressed in the new basis vectors is

$$\vec v_{\text{u basis}}=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix} \frac 3 4\\1\end{bmatrix}.$$

And the norm would entail an inner product with the new metric tensor. In the orthonormal Euclidean basis this is simply the identity matrix. Now the matrix tensor is a $(0,2)-\text{tensor},$ and transforms covariantly:

$$g_{\text{ in u basis}}=(F^\top F) I= \begin{bmatrix}  2  & 1\\-\frac 1 2& \frac 1 4\end{bmatrix}\begin{bmatrix}  2  & -\frac 1 2\\1& \frac 1 4\end{bmatrix}\begin{bmatrix}  1  & 0\\0& 1\end{bmatrix}=\begin{bmatrix}  5  & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}$$

The actual multiplication of basis vectors to obtain the metric tensor is explained in this presentation by @eigenchris. This metric tensor indeed renders the right norm of $\vec v:$

$$\begin{bmatrix}\frac 3 4 & 1\end{bmatrix}\begin{bmatrix}  5  & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}\begin{bmatrix}\frac 3 4 \\ 1\end{bmatrix}=2$$

following the operations here.


Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is

$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}\;,$ bcz Largest side has largest angle.

Now Using $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab}.$ So we get area of $\displaystyle \triangle ABC = \frac{1}{2}ab\cdot \sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$

Now How can i find Maximum area.

You would be better off using Heron's formula. But there is an even better way: what is the maximum value that $\frac12ab\sin \theta$ can attain, if $a \le 1$, $b \le 2$, and $\theta$ can be any angle? Is this value attainable given the additional constraint $2 \le c \le 3$?

As the above user said a better approach would be to look at the area of a triangle as $\frac12 ab \sin{x}$, where $a$ and $b$ are the lengths of two adjacent sides and $x$ is the measure of the angle between these two consecutive sides. The maximum value of $\sin{x}$ occurs when $x=\pi/2$ which gives a value of 1. We now try to maximize the length of the legs and get that the maximum area is $2\cdot 1/2=1$. We make sure that we can have 2 and 1 as leg lengths by checking if $c$ fits in the inequality that was given in the problem, and indeed $\sqrt{5}$ is between 2 and 3.


Is arg(f(z)) continuous when f is analytic? My question is same with the title.

When f(z) is analytic(or just continuous), arg(f) is differentiable(or just continuous)?

If not, what restrictions are needed for arg(f) to be continuous?

Thanks!

$\operatorname{Arg}(z)$, if defined in any sensible way on all of $\Bbb C$, isn't continuous, no matter how you slice it.

If $f(z)$ is continuous and never zero, $\operatorname{Arg}(f(z))$ can be continuous. There are two ways to make this work:

If there is some continuous, non-self-intersecting curve from the origin to $\infty$ that $f(z)$ never hits, you can use that as the branch cut for $\operatorname{Arg}$, and everything works nicely.
Accept the multivalued nature of $\operatorname{Arg}$, and let $\operatorname{Arg}(f(z))$ be multivalued as well. Then it's not strictly speaking a function, but locally it behaves like a function.

In either case, it is continuous. An $\varepsilon$-ball around $\operatorname{Arg}(f(z))$ corresponds to a sector around $f(z)$ in $\Bbb C$. And assuming $f$ is continuous and $f(z)\neq0$, you can always find a $\delta$-ball around $z$ which, when mapped with $f$, lands inside this sector.


What should be the intuition when working with compactness? I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow. The point is that I think I'm not really getting the idea on compactness. I mean, in $\mathbb{R}^n$ the compact sets are those that are closed and bounded, however the guy who answered this question and had his answer accepted says that compactness is some analogue of finiteness.
That's the first problem: In my intuitive view of finiteness, only boundedness would suffice to say that a certain subset of $\mathbb{R}^n$ is in some sense "finite". On the other hand there's the other definition of compactness (in terms of covers) which is the one I really need to work with and I cannot see how that definition implies this intuition on finiteness.
Also, I feel it's pretty strange the covers people use when they want to deal with compact sets. To prove a set is compact I know they must show that for every open cover there's a finite subcover; the problem is that I can't see intuitively how one could show this for every cover. Also when trying to disprove compactness the books I've read start presenting strange covers that I would have never thought about. I think my real problem is that I didn't yet get the intuition on compactness.
So, what intuition should we have about compact sets in general and how should we really put this definition to use?
Can someone provide some reference that shows how to understand the process of proving (and disproving) compactness?

The following story may or may not be helpful. Suppose you live in a world where there are two types of animals: Foos, which are red and short, and Bars, which are blue and tall. Naturally, in your language, the word for Foo over time has come to refer to things which are red and short, and the word for Bar over time has come to refer to things which are blue and tall. (Your language doesn't have separate words for red, short, blue, and tall.)

One day a friend of yours tells you excitedly that he has discovered a new animal. "What is it like?" you ask him. He says, "well, it's sort of Foo, but..."

The reason he says it's sort of Foo is that it's short. However, it's not red. But your language doesn't yet have a word for "short," so he has to introduce a new word - maybe "compact"...

The situation with compactness is sort of like the above. It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.

But in some sense you've never encountered the notion of compactness by itself before, isolated from the notion of discreteness (in the same way that above you've never encountered the notion of shortness by itself before, isolated from the notion of redness). This is just a new concept and you will to some extent just have to deal with it on its own terms until you get comfortable with it.

In my opinion, I think it's okay that you treat compactness as a clever trick as you said. But as it is an abstract concept, I think it's recommended to take the concept as itself, not comparing to real examples.
Maybe the concept of compactness is motivated from Heine-Borel property, but even without a motivation, I think the definition of compactness itself is very natural.
There are many branches of mathematics and each branch has its main vocabularies. For example, in number theory, one's main vocabulary is an "integer". Just like this, "openness" is a main vocabulary in general topology. (The question of asking why openness is a main vocabulary is a completely different question and this is well-answered in wikipedia.) So when you do general topology, you play with open sets. And definitely it's hard to make an argument on infinite open objects, but compactness ease this so that when you play with compactness you can only consider finite objects (for each compact set).
If you take concepts in this way, definitions of Lindelöf space and paracompact space and etc are very natural just like compactness. I'm quite sure no one has a clear "real life example" of paracompactness but almost all of people here think the definition is very natural.


Show that a sequence of reals is cauchy So I right away notice that this $\large\frac{1}{2^n}$ will go to zero, and this can just be replaced by an epsilon.

However, it is not apparent to me how to transition about the statement given about the two sequential elements to any arbitrary element. (if that language is bad, I mean i'm given that $|x_{n+1} - x_n| < \large\frac{1}{2^n}$ $\forall n > N$, and I want to transition that to a statement of $|x_{m} - x_n| < \large\frac{1}{2^n}$ $\forall n > m$. Then the right side just becomes an epsilon as n goes infinity.

Observe that by hypothesis, $$\sum_{k=n}^{\infty} |x_{n+1} - x_{n} | < \frac{1}{2^{n-1}}.$$ Therefore, you can show that $\{x_n\}$ is Cauchy, since given any $\epsilon > 0, \exists N$ so that $\frac{1}{2^{N-1}} < \epsilon$. Therefore, $\forall n,m > N, |x_n - x_m| \le \sum_{n=N}^{\infty} |x_{n+1} - x_n| < \epsilon$. There is a hidden triangle inequality in this last step, but I think you've got it from here.

Let $m=n+1$.
Since $n>N$, then $m>N$. So it becomes $|x_m-x_n|<\epsilon$ (letting $\epsilon=\frac{1}{2^n}$) for $m,n>N$ which is the definition of Cauchy sequence.


Simplifying a complex number into $a + bi$ form I am trying to write $3^i$ in $a + bi $ form. I know

$$ 3^i = e^{i \log 3 } = e^{ i ( \log 3 + 2 \pi k ) }= e^{i \log 3} e^{2 \pi ki} = e^{i \log 3} = \cos ( \log 3 ) + i \sin( \log 3) $$

is this correct?

Assuming you want all possible values,
$$3^i=e^{i\log3}=e^{i(\ln3+2\pi k i)}$$
and so on (I am using $\ln$ to denote the natural logarithm of a real number).  Note that you have a couple of small errors in your working.


Integration of $\int \tan^2 x \ dx $ without the use of trig identities I am trying to integrate $\int \tan^2 x \ dx $  without the use of trig identities, although I know that using $ 1 + \tan^2 x \equiv \sec^2 x $ makes the question trivial.

Using the exponential form of tan(x) I arrived at

$$ -\int \dfrac{(e^{2ix}-1)^2}{(e^{2ix}+1)^2} \ dx $$

I have not studied complex integration, so I am wary of using substitutions of such as $ u = e^{2ix} + 1 $ because I do not know if they are valid.

Taking a look here would be great.

Note that $$\tan^2 x=\sum_{n=1}^{\infty}\frac{t_{n+1}}{(2n)!}x^{2n},$$ with $t_n$ being tangent numbers and defined as $\displaystyle t_n=\frac{2^{2n}(2^{2n}-1)|B_{2n}|}{2n}, n=1,2,...$. Therefore
\begin{align}
\int \tan^2x dx&=\sum_{n=1}^{\infty}\frac{t_{n+1}}{(2n)!}\int x^{2n}dx\\
&=\sum_{n=1}^{\infty}\frac{t_{n+1}}{(2n)!}\frac{x^{2n+1}}{2n+1}dx\\
&=\sum_{n=1}^{\infty}\frac{t_{n}}{(2n-1)!}x^{2n-1}dx-t_1 x\\
&=\sum_{n=1}^{\infty}\frac{t_{n}}{(2n-1)!}x^{2n-1}dx- x\\
&=\tan x -x
\end{align}

$$\int\tan^2(x)\,dx$$
With IBP
$$\int \tan^2(x)\,dx=x \tan^2(x)-\int 2x \sec^2 x \tan x\,dx\tag{1}$$
$u=\tan(x)$
$$\int 2x \sec^2 x \tan x\,dx=\int 2 u \tan^{-1} u\,du\tag{2}$$
With IBP again
$$\int 2 u \tan^{-1} u\,du=\frac{u^2 \tan^{-1} u}{2}-\frac 12 \int \frac{u^2}{1+u^2}\,du\tag{3}$$
With PFD
$$\int \frac{u^2}{1+u^2}\,du=\int 1- \frac{1}{1+u^2}\,du=u-\tan^{-1}u\tag{4}$$
Plugging into $(3)$
$$\int 2 u \tan^{-1} u\,du={u^2 \tan^{-1} u}- {u+\tan^{-1} u} \tag{5}$$
Back substituting from $(2)$
$$\int 2x \sec^2 x \,dx\tan x={x\tan^2 x}-{\tan(x)+x}$$
Plugging into $(1)$
$$\int \tan^2(x)\,dx=x \tan^2(x)-{x\tan^2 x}+{\tan(x)+x}$$
Plugging into the original integral:
$$\int \tan^2(x)\,dx= \tan(x) + x$$


How to write absolute value as a "true" function Here's the basic absolute value ... what?

\begin{align*}
 |x| = \left\{ \begin{array}{r@{\quad \mathrm{if} \quad}l}
 x  &  x > 0, \\
 0  &  x = 0, \\
 \!\! -x  &  x < 0. \end{array} \right.
 \end{align*}

Is this a function, an "expression," or just some short-hand statement? How would you categorize this? Could it be a function if I wrote it like this?

\begin{align*}
 f(x) = \left\{ \begin{array}{r@{\quad \mathrm{if} \quad}l}
 x  &  x > 0, \\
 0  &  x = 0, \\
 \!\! -x  &  x < 0. \end{array} \right.
 \end{align*}

or is this not really a proper algebra function format?

That is correct; it is a piecewise function. One comment: you may simply define $f(x) = x$ when $x \geq 0$ and $f(x) = -x$ when $x < 0$. There is no need to break it into three pieces by separating out the $x = 0$ case.

The second equation is valid. $f (x) $ is a shorthand notation for some function that takes in some value $x $ on a specified domain and returns some value on a range. So, given the definition of $|x|$ in your question, if you label it as $f (x) $ then you have $f (x)=|x|$ which is essentially what you wrote for your second equation.


What is the basic difference between differentiable, analytic and holomorphic function? The function f(z) is said to be analytic at z∘ if its derivative exists at each point z in some neighborhood of z∘, and the function is said to be differentiable if its derivative exist at each point in its domain.
So whats the difference?

Holomorphic functions are complex functions, analytic functions are not necessarily complex.

Both, holomorphic and analytic functions, are infinitely continuous differentiable. But a differentiable functions is not necessarily infinitely differentiable, moreover: an infinitely differentiable function is not necessarily analytic or holomorphic.

Take $f(z) = z^2 \cdot 1_{\mathbb Q}$. This function is not holomorphic but it is differentiable at $0$.


Finding expected number coin flips to get 2 consecutive heads First, I know what the right answer is, and I know how to solve it. What I'm trying to figure out is why I can't get the following process to work.

The probability that we get 2 consecutive heads with one flip is 0. The probability that we get 2 consecutive heads with 2 flips = 1/4. The probability of getting 2 consecutive heads with 3 flips = 1/6. The probability of getting 2 consecutive heads with 4 flips = 2/10 = 1/5. And the probability of getting 2 consecutive heads with 5 flips = 3/16.

Am I doing something wrong. I don't see any easy way to use these numbers to solve the original problem of finding the expected number of coin flips to get 2 consecutive heads.

The problem is that your experiment is not well-defined. To solve the mentioned expectation problem, you would define the experiment as follows:

You toss a coin until you get two heads in a row and then you would stop. So there are no two consecutive heads before that.

each of the probabilities, that you have calculated in your process, is related to the following experiment:

you toss a coin n times. what is the probability of having 2 heads in a row.

By the way, according to the last experiment, you have miscalculated the probabilities in your process.

The definition of expected value in this case is:

$$E[x]=\sum_{x=2}^{\infty}xP[x]$$

and

$$P[x]=\frac{x-1}{2^x}$$

Combining these yields:

$$E[x]=\sum_{x=2}^{\infty}\frac{x^2-x}{2^x}$$

Therefore $E[x]=4$.


probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?

Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.

There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478

I'm not sure how to get this without the help of a computer though.

Approach via inclusion-exclusion and De Morgan's laws.

Let $A$ be the event that you roll at least two $1$'s.  Let $B$ be the event that you roll at least one $2$.

We are attempting to calculate then $Pr(A\cap B)$

$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$

We calculate each term on the right now that they are in simpler form.

$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring.  Either exactly one $1$ occurs or no $1$'s occur.  In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities.  In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities.  Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.

$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s.  This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$

$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s.  For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$

This gives a final probability of:

$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$

$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$

As I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).

Well, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$.

Thus the answer is $P = \frac{(3^3)/3!}{(6^6)/6!} = \frac{5}{72}$.


When to simply plug in infinity when evaluating limits to infinity. Please don't destroy me as I am teaching myself calculus.
Anyway, my question pertains to this problem:
The problem with solution.
This is the solution to the limit of $x-$ $\sqrt{x^2-7}$ as $x$ approaches -$\infty$.
My question is, when is it okay to just plug in -$\infty$ like they did in the solution?
Also, when I did the problem, I got an answer of 0. Here is my step by step solution:
My solution
Feel free to critique it and give me and tips, tricks, and advice.
Thank you!
Edit: I made an error. The denominator in my solution should be a 0 as -1 +1 = 0.My final result based on the work I did should be 0/0.  Nevertheless, my answer is still incorrect.

Your basic building blocks for infinite limits are these:
$\lim_{x \to \infty} c = c$
$\lim_{x \to \infty} x = \infty$ and $\lim_{x \to -\infty} x = -\infty$
$\lim_{x \to \infty} 1/x = 0$ and so $\lim_{x \to \infty} 1/x^n = 0$ for $n > 0$
The way I would have rearranged it is
$$x-\sqrt{x^2-7} = x - |x|\sqrt{1 - 7/x^2}$$ and noting that we are headed into the negative numbers, $|x| = -x$, we have
$$x+x\sqrt{1-7/x^2} = x\big(1+ \sqrt{1-7/x^2}\big) $$
and using the product rule for limits,
$$\lim_{x \to -\infty} x\big(1+ \sqrt{1-7/x^2}\big) = \lim_{x \to -\infty} x\cdot \lim_{x \to -\infty}\big(1+ \sqrt{1-7/x^2}\big)$$
I gathered all the finite parts so the last limit in the expression is $2$, while the first limit gives $-\infty$ so the final answer is $-\infty$.  So I don't plug in $\infty$ until the last possible moment, hoping that they will all cancel out beforehand without my having to guess what indeterminate forms such as $\infty - \infty$ might equal in this particular problem.

The limit of a sum is the sum of the limits (if it exists):
$$\lim_{x \to a} f(x) = b \quad\text{and}\quad \lim_{x \to a} g(x) = c \quad\text{implies}\quad \lim_{x \to a} f(x) + g(x) = b + c$$
Let $a = -\infty$, $f(x) = x$, $g(x) = -\sqrt{x^2-7}$. Then $b = c = -\infty$. Thus the answer is $-\infty$.
It is not always ok to simply plug in $\infty$. For example,
$$\lim_{x \to \infty} (x + a - x) = \lim_{x \to \infty} a = a$$
but
$$\lim_{x \to \infty} (x + a) - \lim_{x \to \infty} x = \infty - \infty$$
where $\infty - \infty$ is an indeterminate form. For a formalization of the idea of when you can "plug in" $\infty$, see the extended real number line:

By adjoining the elements $+\infty$ and $-\infty$ to $\mathbb{R}$, it enables a formulation of a "limit at infinity", with topological properties similar to those for $\mathbb{R}$.
To make things completely formal, the Cauchy sequences definition of $\mathbb{R}$ allows defining $+\infty$ as the set of all sequences $\left\{ a_n \right\}$ of rational numbers, such that every $M \in \mathbb{R}$ is associated with a corresponding $N \in \mathbb{N}$ for which $a_n > M$ for all $n > N$. The definition of $-\infty$ can be constructed similarly.

Note the permitted arithmetic operations.


Different probability space in the same sample space (probability) Can someone give me examples to the following problem: Exist 2 different probability space on the same sample space?
a probability space is a triple (Ω, σ-algebra , P) P - probability function, Ω - sample space

Thank you very much :) I try to understand.

Given we have a $\Omega$ we can construct various $\sigma$-algebras on $\Omega$. Let's say we construct two different $\sigma$-algebras: $\mathcal{F}_1, \mathcal{F}_2$ on the same $\Omega$. For each $\sigma$-algebra we then define a probability measure $P_1, P_2$ (one for each $\sigma$-algebra). Then, we have two different probability spaces on the same $\Omega$, namely: $(\Omega, \mathcal{F}_1,P_1)$ and $(\Omega, \mathcal{F}_2,P_2)$.

If we were to use the same probability measure $P$ for both $\sigma$-algebras. Then, we have the two different probability spaces: $(\Omega, \mathcal{F}_1,P)$ and $(\Omega, \mathcal{F}_2,P)$.

Background:

1)

Yes, two different $\sigma$-algebras can have the same probability measure. Remember: a $\sigma$-algebra on $\Omega$ must (among other things) include $\Omega$ and $\varnothing$, and then it may or may not include more sets all of which must be subsets of $\Omega$. A probability measure gives probability to each set in a $\sigma$-algebra so for any $\sigma$-algebra defined on $\Omega$ the measure 1 will be assigned to $\Omega$ and the measure 0 to $\varnothing$, then if the $\sigma$-algebra contains more sets those will be assigned probabilities some where in between 0 and 1.

2)

A probability space (a.k.a. probability triple/probability model) is represented as: $(\Omega,\mathcal{F},P)$, where

$\Omega$ is a sample space (a set of objects),

$\mathcal{F}$ is a $\sigma$-algebra on $\Omega$ (a set of subsets of $\Omega$),

$P$ is a probability measure on $\mathcal{F}$ (a function assigning probabilities to each subset of $\mathcal{F}$).


Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:

If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.

Initial attempts at AM-GM are not very helpful for me, I don't think.

$x^8+1≥2x^4,$ and $x^4+1≥2x^2$, so $(x^8+1)(x^4+1)≥4x^2(2x^2-1)$. It is easy to show $(x^8+1)(x^4+1)≥4x^2(2x^2-1)≥(x^2+1)^2$ if $x≥\sqrt{\frac{3}{7}},$ but that's insufficient.

I doubt that this problem turns into a Vieta bash due to the nature of having $x^8$ on the LHS. Initial attempts at Newton's sums were giving what looked like annoying numbers and something more bashy than what I would guess is a more elegant solution.

We begin with the following observation $$(x^8+1)(x^4+1)=((x^4)^2+1^2)((x^2)^2+1^2)\geq(x^4\cdot x^2+1\cdot1)^2=(x^6+1)^2.$$
By the Cauchy-Schwarz Inequality $$\prod_{cyc}(x^8+1)(x^4+1)\geq\prod_{cyc}(x^6+1)^2.$$
Thus, it's enough to prove that
$$\prod_{cyc}(x^4-x^2+1)\geq1$$ and since $$x^4-x^2+1\geq\left(\frac{x^2+1}{2}\right)^2,$$ it's enough to prove that
$$\prod_{cyc}(x^2+1)\geq8$$ and since by the Cauchy-Schwarz Inequality again $$\prod_{cyc}(x^2+1)(1+y^2)\geq\prod_{cyc}(x+y)^2,$$ it's enough to prove that
$$(x+y)(x+z)(y+z)\geq8,$$ which is true because
$$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)=\frac{8}{3}(x+y+z)=$$
$$=\frac{8}{3}\sqrt{(x+y+z)^2}\geq\frac{8}{3}\sqrt{3(xy+xz+yz)}=8$$
Finally, equality is achieved when $$x=y=z=1$$

COMMENT.-I find this inequality quite difficult to prove by ordinary means. However it seems to me we could do something valuable as follows: The maximum of $f(x)=\dfrac{(x^2+1)^2}{(x^8+1)(x^4+1)}$ is equal to $1.695$ and it is taken at $x=0.717$ Putting $f(x)$ and $f(y)$ maximum we must have $z=\dfrac{3-xy}{x+y}=1.7335502$ so $f(z)=0.01936916$. This way we verify that in fact
$$1\ge f(0.717)f(0.717)f(1.73355)=0.0556392$$ what makes quite plausible that $$1\ge f(x)f(y)f(z)$$ which is the proposed inequality.


Prove that a norm that satisfies the parallelogram inequality defines an inner product In the opening chapter of a functional analysis book I had this question:

Prove that for a norm $||\cdot||$ that if for all vectors $u$ and $v$ it is true that $2||u||^2 + 2||v||^2 = ||u+v||^2 + ||u-v||^2$ then there is an inner product that results in this norm. (That is $\left<u,u\right> = ||u||^2$ ).

So, I figured that if there is an inner product it ought to be definable by $\left<u,v\right> = \frac{||u+v||^2 - ||u-v||^2}{4}$ and I would proceed to show that this satisfies the conditions of inner products.

However, when I go to prove that $\left<\lambda u, v\right> = \lambda \left<u,v\right>$ I'm completely stuck. I can't see any way of demonstrating this.

Does anyone know how I can complete this proof or of an alternate method of proving this?

Here are some hints.

The inner product is definable by means of the polarization identity,
$$
\langle x,y \rangle = \frac{1}{4}\left(||x+y||^2 - ||x-y||^2\right) + \frac{\rm i}{4}\left(||x+{\rm i}y||^2 - ||x-{\rm i}y||^2\right) .
$$
Showing strict positivity & symmetry is easy. Showing linearity under addition is half a page, but very doable (use the parallelogram law).

Showing linearity under scalar multiplication calls for a far deeper insight - at least the way I did it; maybe there are easier ones. Essentially, you have to build your way up from showing such linearity over $\mathbb{Q}_+$ to showing it for $\mathbb{R}_+$ (closure - use continuity), then to $\mathbb{R}$ (enough to see what happens under multiplication by $-1$) & then to $\mathbb{C}$ (split into real & imaginary parts; see what happens under multiplication by $\rm i$). So why is linearity over $\mathbb{Q}_+$ doable? Well, because it boils down to using additive linearity which you showed in half a page above :).

NB: I'd be very interested to know whether someone knows of a simpler proof.

Hint: Consider the case that $\langle u, u \rangle = ||u||^{2}$ and expand out $||u+v||^{2} + ||u-v||^{2}$ to see what you get


When a prime ideal is a maximal ideal In a commutative ring with unit every maximal ideal is prime. Under what conditions does the converse occur?

When the ring has Krull dimension equal to zero.  If we're talking about integral domains then every prime ideal of $R$ is maximal if and only if $R$ is a field (since $0$ is a prime ideal in any integral domain).

A ring $R$ with identity is artinian if and only if it is noetherian and every prime ideal is a maximal ideal by Hopkins theorem.


How do I compute the specific map between two isomorphic finite C* algebras? Starting with a finite C* algebra $\mathcal{A} \subset M_{n}\left({\mathbb C}\right)$ (complex $n\times n$ matrices), $\mathcal{A}$ is known to be isomorphic to a canonical algebra of the form $$\mathcal{A} \sim \bigoplus_k {\mathcal{A}_{k}}$$where each C* subalgebra $\mathcal{A}_k \sim M_{n_{k}}\left({\mathbb C}\right)$ is determined by the set of minimal nonzero self-adjoint central projections of $\mathcal{A}$. More specifically for my problem involving quantum noiseless subspaces, $$\mathcal{A} \sim \bigoplus_k {M_{p_{k}}\left({\mathbb C}\right)\otimes {\mathbb 1}_{q_{k}}}$$ where ${\mathbb 1}_{q_{k}}$ is the $q_{k}\times q_{k}$ identity matrix.

I know how to compute the central projections of $\mathcal{A}$ and determine the block structure (i.e. the sets $\left\{ p_k\right\}$ and $\left\{ q_k\right\}$) of this canonical algebra, which I'll denote $\mathcal{A}'$.

By isomorphism, I mean there exists some unitary matrix $U$ such that $\mathcal{A} = U^\dagger \mathcal{A}' U$. My question is: how do I compute $U$, or where can I look to find out how to compute it?

Any assistance or advice would be greatly appreciated.

The most direct way is to decompose the tautological representation of $\mathcal A$ on ${\mathbb C}^n$ to irreducible components. Each of them corresponds to $M_{p_k}$ for certain $k$ and has dimension $p_k$. I guess that projectors to these subspaces are the thing the original poster calls “central projections”, since they belong to the center of $\mathcal A$.

I cannot say how one can compute these things because I do not know how exactly the matrix subalgebra is specified.


Proof of ASA , SAS , RHS , SSS congruency theorem I have tried searching in many places for some good proofs of these theorems but couldn't find them anywhere . Even my math teacher cannot explain it to me and says that these theorems just work.

I am a ninth grader so please try to explain in simple terms .

With "Angle-Side-Angle" we know the third angle and use the sine rule $\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}(=2R)$ to determine the unknown sides.

With "Side-Angle-Side" we use the cosine rule to determine the third side opposite the given angle $a^2=b^2+c^2-2bc \cos A$, and then the other angles from the same formula.

With "Side-Side-Side" we use the cosine rule to determine the angle opposite each side.

This always puzzled me until I realised that the sine and cosine rules effectively encoded the uniqueness of these constructions. No-one ever pointed out the connection, which is obvious when you see it.

In a right-angled triangle if we know the hypotenuse and one side, we find the other side using Pythagoras (a stripped down cosine rule), and then we have three sides and an angle and have reduced to one of the other cases.

In the general case SSA the cosine rule gives a quadratic for the third side, and there are two solutions. With a right-angled triangle the two solutions coincide.


Most general linear transformation which leaves the origin fixed and preserves all distances? I'm working on the following problem from Ahlfors (complex analysis):

Prove that the most general (linear) transformation which leaves the origin fixed and preserves all distances is either a rotation or a rotation followed by reflexion in the real axis.

I have the part about it being a rotation. Roughly,

Let $Sz = \frac{az+b}{cz+d}$. $S(0)=0$ implies that $b=0$. $|Sz|=|z|$ implies $|cz+d|=|a|=constant$ for all $z$, so that $c=0$ and $|a|=|d|$. Together they imply $Sz = \frac{a}{d} z$ where $|a/d| = 1$.

I'm not really sure where the reflexion in the real axis part of the question comes in. Is "reflexion in the real axis" the mapping $x+iy \mapsto -x+iy$ ? Would that not mean that it is no longer a linear transformation?

My suggestion,

S(z)=az/d. Take T(z)=-az/d and R(z)=-z.
 Then T is a rotation, R is a reflexion in the real axis and RT(z)=S(z).
 So S is a rotation followed by reflexion in the real axis.


Is there a rule for prime numbers? I've passed by this article:
http://gauravtiwari.org/2011/12/11/claim-for-a-prime-number-formula/

and this paper:
http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/kaddouraAMS73-76-2012.pdf

They say that there is a formula such that when you give it (n) then it returns the n-th prime number. Where other articles states that no formula discovered so far that does such thing.

If the formula exists indeed, then why from time to time they discover a new largest prime number known ever. It would be very simple using the formula to find a larger one.

I just want to ensure whether such formula exists or not.

It depends on what you mean by "formula". Certainly no formula is known such that using it to find a new very large prime would be "very simple".

Well there have been a # of functions developed all of which-I believe-have started at 1 & gone up. The problem is: they are sloooooow! If it takes l000 steps to obtain all primes up to #7 ... well! The thing about them though is they show weird relationships to other things: One uses pentagonal #s; one only pi & e; & I found a relationship between primes & the Fibonacci #s.So they are fascinating to mathematicians.

Now, I just took a look at Kaddoura's formula in the website you gave. I'm impressed that he appears to be giving exactly what you are asking for. You can now see what the key Q is: how fast is it, assuming it works. But he gives us a way to check: the Mathematica info to run it. Most people do not have it; so connect up with a local University & befriend a grad student!

We might at least imagine K's formula to be faster than E's sieve, but we can ck it. Possibly it's faster for a lower n, but not for higher; but who knows?

Another big Q for you is what range of #s you are interested in. Oh, say you are around 10^6=1,000,000 & your function-with a certain computer-takes 3 secs to solve f(n). Then we estimate it will take 1,000,000 times as long to find f(n) for 10^12. Modern problems related to cryptography were in the range of 10^150 ten yrs ago(for the product of 2 primes) & have steadily increased. To give you an idea, there are 10^80th protons in the universe. And 10^150 is 10^70 times larger!

Note that the commentator, on the other site mentioned, claimed it a threat wrt cryptography. Totally not, unless it can deal with very, very large #s.

OK! lets' just suppose K's formula gives f(n) quickly for any #, no matter how large. In that case it could bust certain old ones because of their structure. RSA, of course, would immediately change such in defense. It would probably also necessitate an increase in n. And that would increase times for coding & decoding. Offhand I would not expect a serious increase.

Hamzeh, I hope this has helped.


If $f'$ tends to a positive limit as $x$ approaches infinity, then $f$ approaches infinity Some time ago, I asked this here. A restricted form of the second question could be this:

  If $f$ is a function with continuous first derivative in $\mathbb{R}$ and such that $$\lim_{x\to \infty} f'(x) =a,$$ with $a\gt 0$, then $$\lim_{x\to\infty}f(x)=\infty.$$

To prove it, I tried this:

There exist $x_0\in \mathbb{R}$ such that for $x\geq x_0$,
$$f'(x)\gt \frac{a}{2}.$$
There exist $\delta_0\gt 0$ such that for $x_0\lt x\leq x_0+ \delta_0$
$$\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&\gt -\frac{a}{4}\\
\frac{f(x)-f(x_0)}{x-x_0}&\gt f'(x_0)-\frac{a}{4}\\
&\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\
f(x)-f(x_0)&\gt \frac{a}{4}(x-x_0)\end{align*}.$$
We can assume that $\delta_0\geq 1$. If $\delta_0 \lt 1$, then $x_0+2-\delta_0\gt x_0$ and then $$f'(x_0+2-\delta_0)\gt \frac{a}{2}.$$
Now, there exist $\delta\gt 0$ such that for $x_0+2-\delta_0\lt x\leq x_0+2-\delta_0+\delta$ $$f(x)-f(x_0+2-\delta_0)\gt \frac{a}{4}(x-(x_0+2-\delta_0))= \frac{a}{4}(x-x_0-(2-\delta_0))\gt \frac{a}{4}(x-x_0).$$ It is clear that $x\in (x_0,x_0+2-\delta_0+\delta]$ and $2-\delta_0+\delta\geq 1$.

Therefore, we can take $x_1=x_0+1$. Then $f'(x_1)\gt a/2$ and then there exist $\delta_1\geq 1$ such that for $x_1\lt x\leq x_1+\delta_1$ $$f(x)-f(x_1)\gt \frac{a}{4}(x-x_1).$$
Take $x_2=x_1+1$ and so on. If $f$ is bounded, $(f(x_n))_{n\in \mathbb{N}}$ is a increasing bounded sequence and therefore it has a convergent subsequence. Thus, this implies that the sequence $(x_n)$:
$$x_{n+1}=x_n+1,$$ have a Cauchy's subsequence and that is a contradiction. Therefore $\lim_{x\to \infty} f(x)=\infty$.

I want to know if this is correct, and if there is a simpler way to prove this. Thanks.

Since you assume that $f$ has first derivative continuous on $\mathbb{R}$, there is an alternative to the Mean-value theorem, namely the Fundamental theorem of calculus (the former is very useful for cases when $f'$ is not assumed continuous). So, from $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. Since $f'$ is assumed continuous, we can apply the the Fundamental theorem of calculus to obtain
$$
f(x) - f(M) = \int_M^x {f'(u)\,du}  \ge \int_M^x {\frac{a}{2}\,du}  = \frac{a}{2}(x - M),
$$
for any $x > M$.
Since the right-hand side goes to $\infty$ as $x \to \infty$ (note that $M$ is fixed), so does $f(x) - f(M)$. Hence $
\lim _{x \to \infty } f(x) = \infty $.

EDIT: The above was just to give an alternative to the the Mean-value theorem (used in Amitesh Datta's answer); its  analog for the Mean-value theorem is as follows.

From $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. For any $x > M$, since $f$ is continuous on the closed interval $[M,x]$ and differentiable on the open interval $(M,x)$, by the Mean-value theorem there exists $c = c(M,x) \in (M,x)$ such that
$$
f(x)-f(M)=f'(c)(x-M).
$$
Since $c \geq M$, $f'(c) > a/2 $, and hence
$$
f(x)-f(M) \geq \frac{a}{2}(x - M).
$$
The conclusion is as before.

Note: As already indicated above, the proof using the Mean-value theorem has the advantage that it works for any differentiable function $f: \mathbb{R} \to \mathbb{R}$.

I will try to prove is in a different way which can be much simpler - using visualization.

Imagine how will a function look if it has a constant, positive slope -
A straight line, with a positive angle with the positive x axis.
Although this can be imagined, I am attaching a simple pic -

(Plot of our imaginative function - $f(x)$ vs $x$)

As per the situation in the question, for $f(x)$ the slope exists (and is finite) at all points, so it means that the function is continuous. Since the slope is also constant at $\infty$, $f$ has to be linear at $\infty$. Thus, the graph of the function should be similar to the above graph.

(assume the value of x to be as large as you can imagine.)

Hence, putting the above situation mathematically, we have,

If $\lim_{x\to \infty}\ f'(x) =a\qquad(with\ a>0)$

then $\lim_{x\to\infty}\ f(x)=\infty.$


Proving $\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $ $$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx=0 $$

I consider if it can be solved by other methods without Riemann lemma.
I try my best to do it as follow:

  Let $t=nx$
  $$\lim_{n\to \infty}\int_0^\pi\frac{\sin\left(nx\right)}{1+x^2}dx$$
  \begin{eqnarray}
&=&\lim_{n\to \infty}\int_0^{n\pi}\frac{\sin\left(t\right)}{1+(\frac{t}{n})^2}d\frac{t}{n}\\
&=&\lim_{n\to \infty}\int_0^{n\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\
&=&\lim_{n\to \infty}\sum_{k=0}^{k=n-1}\int_{k\pi}^{(k+1)\pi}\frac{n\sin\left(t\right)}{n^2+t^2}dt\\
\end{eqnarray}

Then I can not go on. Who can tell me how to prove it? Thank you.

Integration by parts is enough:

$$\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx =\left.\frac{\frac{1}{n}(1-\cos(nx))}{1+x^2}\right|_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\frac{2x\,(1-\cos(nx))}{(1+x^2)^2}\,dx,\tag{1}$$
so:
$$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\frac{2\pi^2}{\pi^2+1}\right)=\frac{2}{n}.\tag{2}$$
Using the Cauchy-Schwarz inequality to bound the second integral in $(1)$, it is possible to improve the last inequality up to:
$$\left|\int_{0}^{\pi}\frac{\sin(nx)}{1+x^2}\,dx\right|\leq\frac{1}{n}\left(\frac{2}{\pi^2+1}+\sqrt{\frac{3\pi}{2}\int_{0}^{\pi}\frac{4x^2}{(1+x^2)^4}\,dx}\right)<\frac{14}{9n}.\tag{3}$$

For each integral of the sum $|I_k| = \left|\displaystyle \int_{k\pi}^{(k+1)\pi} \dfrac{n\sin t}{n^2+t^2}dt\right| \leq \displaystyle \int_{k\pi}^{(k+1)\pi} \dfrac{1}{n}dt = \dfrac{\pi}{n}$. Taking the limit $n \to \infty$ we have: $I_k \to 0$ , $\forall k$. You're done.


Laplace transform of a signal multiplied to a time function Laplace transform of multiplication of two time functions can be calculated easily using standard formulae. But I have a problem in which signal $x(t)$ is multiplied to a time function. I have to do control analysis of the problem.
Feedback loop diagram

Here, $z(t)=cos(\Omega t) x(t)$.

For control analysis, I need to get relationship between $Z(s)$ and $X(s)$.

As per defination of Laplace transform,  $X(s)=\int_{0}^{\infty} x(t) e^{-s t} dt$, it can be derived

$Z(s) = \int_{0}^{\infty} x(t) e^{j \Omega t} e^{-s t} dt = X(s-j \Omega)$
But, this relationship cannot be used for control analysis like root locus and many other things.

  Can someone help me for getting relationship between $X(s)$ and $Z(s)$? How can convolution theorem be used in this regard?

Multiplication in the time-domain corresponds to convolution in the Laplace domain.


checking the diffeomorphism between 2 surfaces I tried to show that a surface $x^4+y^2+z^2=1$ and the unit sphere are diffeomorphic.

Since a diffeomorphism between them is not chosen easily, I would apply a theorem using invertibility of Jacabian.

For this process, I must know a surface patch of $x^4+y^2+z^2=1$, but I don't know that.

Q1. How do I choose the surface patch?

Q2. Is my plan is correct to show that they are diffeomorphic?

Let $M=\{(x,y,z): x^4+y^2+z^2=1\}$. The (radial projection) map $F\colon \Bbb R^3-\{0\}\to S^2$ given by $F(x) = x/\|x\|$ is smooth, hence restricts to a smooth map $f\colon M\to S^2$. It is easy to check that $f$ is a bijection. You can then prove that it is a diffeomorphism by noting that for any $p\in M$, $$T_pM\cap \{tp: t\in\Bbb R\}=\{0\} \tag{$\star$}$$ (i.e., $T_pM$ is transverse to the line spanned by $p$), and so $\ker df_p = \ker dF_p\cap T_pM = \{0\}$. Thus, by the inverse function theorem, $f$ is a local diffeomorphism, and we're done.

EDIT: I guess I should elaborate a bit on the claim ($\star$). If $p=(a,b,c)$, the normal vector to $T_pM$ is $(2a^3,b,c)$. If the line spanned by $p$ were contained in $T_pM$, we'd have $(2a^3,b,c)\cdot (a,b,c) = 0$, which leads us to the system
\begin{align*}
2a^4+b^2+c^2&=0 \\ a^4+b^2+c^2&=1
\end{align*}
This means $a^4=-1$, which has no solutions.

It is my suggestion (whenever possible) to avoid explicit chart-checking and to work with the implicit representation of the manifold.

It is obvious that the map sending $x$ to $\pm x^2$ defines diffeomorphisms between the upper and the lower hemispheres of your surfaces. You just need to check that things are OK on the equator. Can you?

A different way is to check that both surfaces are strictly convex, then they are diffeomrophic.


Proving isomorphism in lattices The question is as follows:

Let f be a monomorphism from a lattice $L$ to a lattice $M$.Show that $L$ is isomorphic to a sublattice of $M$.

My attempt:

Since $f$ is a monomorphism from a lattice $L$ to a lattice $M$ therefore it will be a lattice homomorphism which is injective.
Since, $f$ is injective therefore every element of $L$ will be mapped into a unique element in $f(L)$ and hence $M$,and also every element of $f(L)$ will have a unique preimage.

Thus $f:L\to f(L)$ is bijective and hence a lattice homomorphism.
Further $f(L)$ is a sublattice of $M$ and hence $L$ is isomorphic to a sublattice of $M$.

Am I correct ?

One thing that can be done is
First show that f(L) is sublattice of M
After that
Show that L is isomorphic to f(L) which is nothing but the sublattice of M


Hitting time of Brownian Motion with a drift Let $X_t =x+bt+\sqrt{2}W_t$, where $W_t$ is a standard Brownian motion.
Let $T=\inf\{t: |X_t|=1\}$. I am trying to find $\mathbb{E}[T]$ for the case $b\neq0$.

Firstly, I am going to apply Girsanov to change the measure and the drift:
$$M_t = e^{-\frac{b}{\sqrt{2}}W_t-\frac{b^2}{4}t},$$
If $\frac{d\mathbb{P}}{d\mathbb{Q}}|\mathcal{F}_t=M_t$, then
$\mathbb{E}[T|\mathcal{F_t}]=\mathbb{E}^\mathbb{Q}[TM_t|\mathcal{F_t}]$. And under $\mathbb{Q}$, $X_t$ is driftless BM.

So $\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{\sqrt{2}}W_t-\frac{b^2}{4}t}|\mathcal{F_t}]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{2}X_t-\frac{b^2}{4}t+\frac{b}{2}x}|\mathcal{F}_t]$

If I managed to show $T<\infty$ a.s., or otherwise, I could arrive at:
$$\mathbb{E}[T]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{2}X_T-\frac{b^2}{4}T+\frac{b}{2}x}]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b^2}{4}T}]\mathbb{E}^\mathbb{Q}[e^{-\frac{b}{2}X_T}]e^{\frac{b}{2}x}$$
Now
$$\mathbb{E}^\mathbb{Q}[e^{-\frac{b}{2}X_T}]=e^{-\frac{b}{2}}\mathbb{P}^\mathbb{Q}(X_T=1)+e^{\frac{b}{2}}\mathbb{P}^\mathbb{Q}(X_T=-1)=e^{-\frac{b}{2}}\frac{1-x}{2}+e^{\frac{b}{2}}\frac{x+1}{2}$$

I have two questions:

1) How to compute $\mathbb{E}^\mathbb{Q}[Te^{-\frac{b^2}{4}T}]$

2) How to get around the problem that $T$ might not be finite?

Edit: 2) it's clear that $T<\infty$ under $\mathbb{Q}$, and so $T$ is $\mathbb{P}$-a.s. finite.

If I understand Tom's problem correctly, we are looking for the
two-sided first passage time ${\bf T}$ in a Wiener process $\{X(t), t \geq 0\}$
starting at $X(0)=x, \, -1 < x < +1$, with drift $b$, variance $2$,
and absorbing barriers at $-1$ and at $+1$.
This is a standard problem treated in many sources such as
Darling and Siegert (1953) who in their Theorem 6.1 and eq. (6.6) derive
the basic differential equation for $f(x) = {\sf E}[{\bf T}|X(0)=x]$,
which (in the present notation) is
$f^{''}(x) + b \, f^{'}(x) \; = \; -1$,
with $f(-1)=f(+1)=0$.
The solution is easily seen to be
$$
f(x) \; = \; {\sf E}[{\bf T}|X(0)=x] \; = \;
\frac{1}{b} \, \left\{ 2 \, \frac{1 - \exp[-b(1+x)]}{1 - \exp(-2b)} \, - \, (1+x) \, \right\}
$$

I think $e^{-bX_t}$ is a martingale, something like that is right, anyway.  Using it you have $e^{bx} = e^{-ba}\mathbb P(X_T = a) + e^{ba}\mathbb P(X_T = -a)$, and $ 1 = \mathbb P(X_T = a) +\mathbb P(X_T = -a)$, from which you determine $\mathbb P(X_T = a)$ and $\mathbb E(X_T)$. Then Wald says $\mathbb E (X_T) = b \mathbb E (T)$


$L^p$ subspaces of $L^1$ Suppose I have a closed subspace $S$ of $L^1([0,1])$ such that each $f\in S$ lies in some $L^p([0,1])$ space for $p>1$. How can I prove that $S$ itself lies in some $L^p([0,1])$ for $p>1$? I would like to apply some sort of Banach-Steinhaus argument but can't see how to proceed.

Let's prove this by contradiction. Assume that there exists some $\{p_n\}\subset \mathbb{R}$ such that $p_n$ decreases strictly to $1$ and such that there is an $f_n\in S$ such that $p_n = \sup\{p > 1 : f_n\in L^p([0, 1])\}$. Without loss of generality, let $\|f_n\|_{L^1([0, 1])} = 1$ and $f_n(x)\geq 0$ for all $x\in [0, 1]$ and $n\in \mathbb{N}$. Note that as $[0, 1]$ is a finite measure space, $L^q([0, 1])\subset L^p([0, 1])$ for all $1\leq p < q$. Then, consider $f(x) := \sum_{n=1}^{\infty} 2^{-n}f_n(x)$ ($f(x)$ may be infinite at some points $x\in [0, 1]$). We have by Lebesgue's monotone convergence theorem that $$\|f\|_{L^1([0, 1])} = \lim_{N\to \infty} \left\|\sum_{n=1}^N 2^{-n}f_n\right\|_{L^1([0, 1])} = \lim_{N\to \infty} \sum_{n=1}^N 2^{-n} = 1$$ Furthermore, $$f(x)\geq \sum_{n=1}^N 2^{-n}f_n(x)\geq 2^{-N}f_N(x)$$ for all $N$, so given any $p > 1$, we let $N$ be such that $p_N < p$. We then have that $$\|f\|_{L^p([0, 1])}\geq 2^{-N}\|f_N\|_{L^p([0, 1])} = \infty$$ Therefore, $f\notin L^p([0, 1])$ for any $p > 1$, which contradicts our precondition on $S$. Thus, there is no sequence $\{p_n\}\subset \mathbb{R}$ such that $p_n$ decreases strictly to $1$ and such that there is an $f_n\in S$ such that $p_n = \sup\{p > 1 : f_n\in L^p([0, 1])\}$. This implies that there is some $p > 1$ such that for all $f\in S$, $f\in L^p([0, 1])$ (as otherwise, we could construct a sequence $\{p_n\}$ as above). Therefore, $S\subset L^p([0, 1])$.

Consider the subspace generated by $f_p(x)=\frac{1}{x^{2/p}}$ for $p=2,3,4,...$.

This space cannot be contained in a single $L^p$.


Dividing an equilateral triangle into N equal (possibly non-connected) parts It’s easy to divide an equilateral triangle into $n^2$, $2n^2$, $3n^2$ or $6n^2$ equal triangles.
But can you divide an equilateral triangle into 5 congruent parts? Recently M. Patrakeev found an awesome way to do it — see the picture below (note that the parts are non-connected — but indeed are congruent, not merely having the same area). So an equilateral triangle can also be divided into $5n^2$ and $10n^2$ congruent parts.

Question. Are there any other ways to divide an equilateral triangle into congruent parts? (For example, can it be divided into 7 congruent parts?) Or in the opposite direction: can you prove that an equilateral triangle can’t be divided into $N$ congruent parts for some $N$?

(Naturally, I’ve tried to find something in the spirit of the example above for some time — but to no avail. Maybe someone can find an example using computer search?..)
I’d prefer to use finite unions of polygons as ‘parts’ and different parts are allowed to have common boundary points. But if you have an example with more general ‘parts’ — that also would be interesting.

Recently Pavel Guzenko found a way to divide an equilateral triangle into 15 congruent parts (and also into 30 congruent parts).

As the equilateral triangle has three axes of symmetry, congruent equilateral triangles can be split into a square number of shapes, multiplied by $1, 2,$ or $3$, so $6$ also works.

Any equilateral triangle can be split into $n^2$ sub-equilateral triangles,
First, we need to find a square number that is divisible by $n$. As seen in the picture you posted above, each row of the triangle adds the next odd number of triangles: $1,3,5,7,9...$ This is the pattern used by square numbers:
$1=1||1+3=4||4+5=9||9+7=16||16+9=25...$ For this reason, any equilateral triangle can be split into $n^2$ equal shapes.

Due to the three axes of symmetry, any equilateral triangle can be divided into $2,3,$ or $6$ shapes, as seen in the picture below.

If we combine these two possibilities, we can get an equilateral triangle into $n^2,2n^2,3n^2,$ or $6n^2$ shapes. No other shapes are possible because of these two laws.

So, we can only split an equilateral triangle into $n=2,3,4,6,8,9,12,16,18,20,24,25,etc...$ congruent shapes. However, like in the example you gave, splitting a triangle into separate congruent parts is indeed possible.


Is $\sqrt[3]{-3} = -\sqrt[3]{3}$ true? Well, Wolfram Alpha says no. Why?

$$\sqrt[3]{-3} = -\sqrt[3]{3} \iff (\sqrt[3]{-3})^3 = (-1)^3(\sqrt[3]{3})^3 \iff -3=(-1)3 \iff -3=-3$$
So I would say it is true if we are talking about real numbers.
In fact in "Analisi Matematica" of Paolo Maurizio Sardi (CIttàStudi edizioni), we define roots in the following manner.
Theorem 1.7.1 Given an integer $n \ge 1$, and a real number $\alpha >0$, there exists one, and only one, real number $\beta>0$ such that
$$\beta^n=\alpha \mbox.$$
Definition 1.7.3 The number $\beta >0$  found in the Theorem above is called $n$-th root of $\alpha$ and is denoted by
$$\beta = \sqrt[n]\alpha \mbox.$$
Observation: $(\sqrt[n]\alpha )^m =\sqrt[n]{\alpha^m}$
Definition 1.7.4 Let $m\in \mathbb{Z}$, let $n\ge 1$ be an integer and let $\alpha >0$. We set
$$\alpha^{m/n}=(\sqrt[n]\alpha )^m $$
Observation: we now want to define all of this for negative real numbers. Let $-a<0$ and let $n \ge 1$ be an integer:
(1) if $n=2k$ is even, it is clear that there does not exists $\beta\in \mathbb{R}$ such that
$$\beta^{2k}=(\beta^2)^k=-\alpha \mbox,$$
so we cant' do this in the reals;
(2)if $n=2k+1$ is odd and let $\beta >0$ be such that $\beta^{2k+1}=\alpha$ ($>0$), then
$$(-\beta)^{2k+1}=-\alpha$$
so $-\beta$ is called the $n$-th root of $-\alpha$ an is denoted by
$$-\beta=\sqrt[n]{-\alpha} \mbox.$$
You could then continue with defining roots with real exponent (going back to $a>0$), showing that the following definition is well-defined:
Definition 1.7.5:
(1) if $\alpha \ge 1$ and $\beta >0$ we set
$$\alpha^{\beta}:= \sup\biggl{\{} \alpha^{m/n}:\frac{m}{n} \le \beta \biggr{\}}$$
(2) if $0<\alpha < 1$ and $\beta >0$ we set
$$\alpha^{\beta}:=\frac{1}{(1/\alpha)^{\beta}}$$
(3) if $\alpha >0$ and $\beta >0$ we set
$$\alpha^{-\beta}=\frac{1}{\alpha^{\beta}$$
(4) $\alpha^0=1$
(defined for every $\alpha>0$ and $\beta$ reals).


What restrictions lay on truth sets of predicates A(x) and B(x) if the given predicate is true Suppose we are given 2 predicates
$A(x)$ and $B(x)$ with domain $M$.
Suppose next we are given the following predicate
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$
which we know is true, so
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x))) = 1$$
The question is how does it restrict the truth sets of $A(x)$ and $B(x)?$
It is obvious that we have
$$\neg (A(x) \land B(x)) = 1 \\
A(x) \land B(x) = 0\\
A(x) = 0 \lor B(x) = 0$$
So from that we get that either truth set for $A(x)$ is $E_A \neq M$ or truth set for $B(x)$ is $E_B \neq M$.
But knowing that
$$\forall x(A(x) \rightarrow B(x)) = 1\\
\forall x(\neg A(x) \lor B(x)) = 1$$
I have no idea how to link it to useful information on truth sets of $A(x)$ and $B(x)$, any suggestions?

You are getting hung up on truth sets when there is only one variable in the problem.  Focus on one element $x$ of $M$ and ask whether $A(x)$ and $B(x)$ can be true because all the elements are equivalent.  You have found that both $A(x)$ and $B(x)$ cannot both be true but $A(x) \implies B(x)$.  You should be able to derive that $A(x)$ is false and $B(x)$ can be anything.  Check that in the original axiom and it works.

1
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$
2
$$\neg (A(x) \land B(x)) \equiv (\neg A(x)) \lor (\neg B(x))$$
$$(\forall x(A(x) \rightarrow B(x))) \equiv (\neg A(x)) \lor B(x)$$
3
$$(\neg A(x) \lor \neg B(x)) \land (\neg A(x) \lor B(x)) \equiv (\neg A(x))$$


Normal distribution with dice I'm wondering how to control the normal distribution that comes from summing dice rolls only using different numbers of dice, different combination of types of dice (d4, d6, d8, d10, d12, d20) and simple math (+,-,•,/ and perhaps ^)?

I know that if you only use one type of die the mean value and the standard deviation increases linearly with the number of dice with a constant that depends on the number of sides on the dice.

I also realize that I can shift the distribution with plus and minus. But I don't know if there is a equally simple way to control the standard deviation.

In summary: how do I control the standard deviation of a normal probability distribution only using different combination of dice and simple math?

Here is a blogpost that gives you an overview of the distributions of summed dice as the number of dice increases. In short, as the number increases, it becomes increasingly well modelled by the normal distribution.
However, there is a small gap between the analytic solution that we get for the probability distribution of dice and the normal distribution. My intuition is that this will not be important - because rolling the vast number of dice that would allow statistical tests to detect this gap is probably not something you will do in this game.
If you are happy using d6s then a roll of $n$d6 is relatively well approximated by
$ \mu    = \frac{6n+1}{2}$,
$ \sigma = 0.175\sqrt n $
when $n$ gets sufficiently large. I imagine I would use $n=4$ or 5.

I think you are using words that you don't understand.  The sum of multiple dice does not follow a normal distribution.  See my explanation for the sum of two $n$-sided dice case in my answer here.
Supposing you remove the word "normal," you can see how the mean and standard deviation interact by the linearity of expected value.

The expected value of a discrete random variable, $X$, is given as $E[X] = \sum\limits_{x\in\mathcal{M}} x\cdot Pr(X=x)$ (where $\mathcal{M}$ is the sample space)
The expected value operator is linear.  I.e., $E[\alpha_1 X_1 + \alpha_2 X_2] = \alpha_1 E[X_1] + \alpha_2 E[X_2]$

The mean of a random variable $X$ is defined to be $E[X]$

Let $X$ be a random variable described by the result of a die throw.  If $X$ is a fair $n$-sided die with sides labeled $1,2,\dots,n$, then the mean, $\mu = E[X]$ is equal to $\frac{n+1}{2}$.
The variance of $X$ in this case is $\frac{n^2-1}{12}$.

Let $X$ be a random variable described by the result of a die throw.  If $X$ is a fair $n$-sided die with sides labeled $1,2,\dots,n$, then the expected value of $\frac{1}{X}$ is $H_n\cdot \frac{1}{n}$ where $H_n$ is the $n^{th}$ Harmonic Number

Suppose our random variable, $Y$, is a linear combination of random variables.  Let $X_1$ be the random variable described by the result of an $a_1$-sided die, $X_2$ be the random variable described by the result of an $a_2$-sided die, ..., $X_k$ be the random variable described by the result of an $a_k$-sided die.
Let $Y=\alpha_1 X_1 + \alpha_2 X_2 + \dots \alpha_k X_k$
Then $E[Y] = E[\alpha_1 X_1 + \dots + \alpha_k X_k] = \alpha_1 E[X_1] + \dots + \alpha_k E[X_k] = \alpha_1\frac{a_1+1}{2} + \dots + \alpha_k\frac{a_k+1}{2}$
Note, that for the case where $X_1, X_2,\dots, X_k$ all describe the result of rolling an $n$-sided die, you get $E[X_1+X_2+\dots+X_k] = E[X_1]+\dots+E[X_k] = \frac{n+1}{2}+\dots + \frac{n+1}{2} = k\cdot \frac{n+1}{2}$, matching your intuition.
When you use multiplication, $E[XY] = E[X]E[Y]-Cov(X,Y)$, however noting that when $X$ and $Y$ are independent (which they will be when $X$ and $Y$ are for different dice [not necessarily different types of dice]) it implies that $Cov(X,Y)=0$ and that $E[X,Y] = E[X]E[Y]$
This gives us a way to calculate the mean of any combination of results of dice and $+,-,\times$.  Noting that division is just multiplication by the multiplicative inverse of the other, we get that $E[X/Y] = E[X\cdot \frac{1}{Y}] = E[X]E[\frac{1}{Y}]$ and the term on the right will rely on the $n^{th}$ harmonic number.
I see no easy way of giving a general rule to use when using exponentiation of one random variable to the power of another apart from referring you to calculate it directly using the definition of expected value.  For finding the expected value of a random variable to an integer power, I refer you to the definition of Moments and to Generalized Harmonic Numbers as well as for the specific case of $E[X^2]$ being the $n^{th}$ Square Pyramidal Number divided by $n$.

For standard deviation, $\sigma$, of a random variable $X$ note that $\sigma^2=Var[X] = E[(X-E[X])^2] = E[X^2]- E[X]^2$
With all of this information, you should have the tools to calculate whatever you are looking for.

Example
You have three fair dice, a d4, a d6, and a d8.  Let $x,y,z$ represent the results of throwing the d4, d6, and d8 respectively.  Suppose that you are playing a game where you get to attack and deal $z+xy-2$ damage.  What is the expected amount of damage and what is the standard deviation?
The average amount of damage is: $E[Z+XY-2] = E[Z]+E[X]E[Y] - 2 = \frac{8+1}{2} + \frac{4+1}{2}\cdot\frac{6+1}{2} - 2 =4.5 + 3.5\cdot 2.5 - 2 = 11.25$
The variance is: $Var[Z+XY-2] = E[(Z+XY-2)^2] - (E[Z+XY-2])^2 = E[Z^2 + X^2Y^2 + 4 + 2XYZ - 4Z -4XY] - 11.25^2$
$= E[Z^2] +E[X^2]E[Y^2]+4+2E[X]E[Y]E[Z] - 4E[Z] - 4E[X]E[Y] - 11.25^2$
$=\frac{(8+1)(16+1)}{6} + \frac{(4+1)(8+1)}{6}\frac{(6+1)(12+1)}{6}+4+2\cdot 2.5\cdot 3.5\cdot 4.5 - 4\cdot 4.5 - 4\cdot 2.5\cdot 3.5 - 11.25^2$
$=42.4375$
The standard deviation then is $\sqrt{42.4375}\approx 6.51$.


Prove that a theory $\Gamma$ is consistent if and only if there is a structure $M$ so that $M$ $\models$ $\Gamma$. $\Gamma$ is consistent, then either $\Gamma \models \varphi$ or $\Gamma \models \neg \varphi$ (but not both) for all sentences $\varphi$.

Can we assume $M$ is a structure such that $M \models \varphi$ for all $\varphi \in \Gamma$? I think this is incorrect because if so, $M \models \Gamma$ will be true trivially. We are assuming what we want to prove. If this is wrong, what else can I try?

On the other side, from $M \models \Gamma$, we have $M \models \varphi$ for every $\varphi \in \Gamma$.

Then can we say $M \not\models \neg\varphi$? And how can we finish the prove with $\neg (M \models \varphi \land \neg\varphi)$?

From right to left is the easy one: if $M \vDash \Gamma$ then $ M \vDash \varphi$ for all $\varphi\in \Gamma$ (by definition of $M\vDashφ$). Ans since M is a structure, you have that whenever $M\vDash \varphi $ then $M\not \vDash\neg \varphi$ (By definition of it being a structure), and finally you have $M⊨ \varphi \land \psi$ iff $M\vDash\varphi$ and $M\vDash \psi$ by formal semantics of the $\land$, so you can't have $\Gamma \vDash\varphi \land \neg \varphi$. So: if $\Gamma$ were inconsistent, we would have $\Gamma\vDash\varphi \land \neg \varphi$, and thus $M \vDash\varphi \land \neg \varphi$, and thus both $M\vDash \varphi $ and $M \vDash \neg \varphi$. But as we just saw, that is impossible. So, $M$ is consistent.

From left to right is a good bit harder! As pointed out, we can't assume $\Gamma$ is complete ... but we can extend $\Gamma$ into $\Gamma$' such that $\Gamma$' is complete: Roughly, you consider any of the sentences $\varphi$ that can be generated from your language for which $\Gamma \not \vDash \varphi$ and $\Gamma \not \vDash \neg \varphi$, and add them one by one to $\Gamma$ (one by one, so that once you have added $\varphi$, you don't end up adding $\neg \varphi$ as well ... it is a good thing that for any enumerable language all sentences from that language are enumerable as well!).

And, now you can show that any consistent and complete set of sentences $\Gamma$ has a model by focusing on the atomic statements of $\Gamma$ and translating those into a structure $M$ so that $M$ indeed becomes a model for $\Gamma$.

So: given that $\Gamma$' has a model, $\Gamma$ has a model as well (the same one!)

But there are lots of technical details to this proof!

If $\Gamma$ is consistent (i.e. $\Gamma\models\varphi\longrightarrow\Gamma\not\models\neg\varphi$) then clearly there exists a formula $\varphi$ such that $\Gamma\not\models\varphi$. So $\Gamma$ has a model $M\models\Gamma$, since otherwise every model of $\Gamma$ (since there are none) would satisfy $\varphi$. This may seem a bit tricky, to use a void hypothesis, but if you think a moment you will realise that it is not.

The converse is clear, right? If $\Gamma$ has a model, the inconsistence of $\Gamma$ would imply that some $\varphi$ is true and false in that model.


Proof: A convergent Sequence is bounded A part of the proof says that if $n\le N$, then, the sequence
$x_n \le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.

I'm not capturing the intuition of the above. Even more perplexing is, if it the case where the sequence is monotonic decreasing, then for every $n\le N$, it is obvious that $x_n$ will not be $\le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.

I greatly appreciate some input.

Well, $|x_n| \le \max\{|x_1|,\ldots, |x_{N-1}|\}$ for all $n \color{red}{<} N$ because if $n < N$, $|x_n|$ is one of the numbers $|x_1|,\ldots, |x_{N-1}|$, and some number in this list is the largest, i.e., $\max\{|x_1|,\ldots, |x_{N-1}|\}$.

If $\{a_n\}$ be a convergent then for any $\epsilon>0$: then there exist a positive integer $n_0$ such that
$$
|a_n-l|\le \epsilon\:\text{  for all }\:n> n_0
$$
and this implies that $|a_n|\le\epsilon +l$.
Now let $M=\{|a_1|,|a_2|,|a_3|,\ldots,|a_n| ,|\epsilon+l|\}$: then $|a_n|\le M$ is bounded for all $n\in\Bbb N$ and hence sequence $\{a_n\}$ is bounded.


Application of the conservation of mass. I am having trouble doing this problem that has multiple parts to it and has applications to physics so I'm not sure if this belongs in here, but it is for a real analysis class in mathematics.

We are to assume that a tracer is being transported by a moving fluid in a one dimensional medium. Let $u(x,t)$ be the density of the tracer at position $x$ at time $t$. Also let $c(x,t)$ be the velocity of the fluid at $(x,t)$. The following conservation principle can be applied to any interval $a\le x\le b$. $d\over dt$ $($Total mass in $[a,b])$ $= ($Rate of flow past $x=a )$ - $($Rate of flow past $x=b )$

(a.) For any interval $[a,b]$, find an expression for the total mass of the tracer in $[a,b]$ at time $t$

(b.) Find an expression for the left side of the conservation principle stated above. It is an integral on $[a,b]$. Assume that the function $u$ satisfies the hypothesis of the theorem on differentiating under the integral.

There are other parts but I will not post them for now. So for (a.) I understand that. I am having a lot of problems just getting this question started on all parts, so any help to get started would be very helpful. I am familiar more so with math than physics so I am having trouble with the latter. Thank you for any help with this problem.

You can think of the interval as a vessel.  The amount of something that is in the interval after a bit of time is the amount that was in there at the start plus what came in minus what went out.  In your equation everything is flowing in the positive direction, so what comes in is what passes $a$ and what goes out is what passes $b$.  That is the conservation principle you cite.  It depends on the fact that there is no production or destruction in the interval.

You are expected to use the functions $u(x,t)$ and $c(x,t)$ to determine the amount of tracer coming in and leaving and write an equation for the amount of tracer in the interval based on this.


At what coordinates should the fourth vertex be located? A parallelogram is drawn on a coordinate grid so that three vertices are located at $(3, 4)$, $(-2, 4)$ and $(-4, 1)$.  At what coordinates should the fourth vertex be located?

Hint:  Have you drawn a picture?  There are three answers, one seeming more obvious to me than the others.  You can connect the points with two line segments to make two sides of the parallelogram  in three ways.  Then through the two points that have only one side drawn so far, draw a line parallel to the other side.  Where the two new lines intersect is the point you want.

Added:  another way to think of it is that a parallelogram can be divided into two congruent triangles by either diagonal.  One of the triangles is upside down relative to the other.  Your three points form a triangle, which you can reflect in any of the three sides.

Keep the points as A,B,C. Find mid point(A1)of B and C, Find A2 as A.A2:A2.A1 = -2:1 (external). A2 will be the 4th point opposite to A. IIIy find B2, C2.


How do I solve $e^{iz} = 3i$? I am stuck on the question that asks me to solve $e^{iz} = 3i$.
I rewrote $e^{iz}$ as $e^{ix-y}$ and thus as $e^{ix}\cdot e^{-y}$ and thus as $e^{-y}\cdot (\cos x + i\sin x)$.
I then wrote $e^{-y}\cos x = 0$ and $e^{-y}\sin x = 3$. When I try to solve this, however, I find $x = \pi/2 + 2k\pi$ as solution for the first expression, but I can't find a solution for $e^{-y}\sin x = 3$, because there are no solutions for $\sin x = 3$.
I then tried to find y through solving $e^{-y}\cdot \cos(\pi/2 + 2k\pi)=0$, this however gives me $y = \log 0$, which has no solutions. What am I doing wrong? Thanks!

You're doing good: if $z=x+iy$, then $e^{iz}=e^{-y}(\cos x+i\sin x)$, so you have
$$
\begin{cases}
e^{-y}\cos x=0 \\[4px]
e^{-y}\sin x=3
\end{cases}
$$
From the first equation you get $\cos x=0$, so $x=\pi/2+2k\pi$ or $x=-\pi/2+2k\pi$. Substituting in the second equation, the former gives
$$
e^{-y}=3
$$
so $y=-\log 3$. The latter instead gives $-e^{-y}=3$, which has no solution.

Use $z=x+iy$ to obtain

$$\exp(iz)=\exp(ix-y)=\exp(-y)\exp(ix)=\exp(-y)\cos(x)+i\exp(-y)\sin(x)=3i$$

By comparing the real and imaginary part we obtain two simultaneous equations:

$$\exp(-y)\cos x = 0 \implies \cos x = 0 \text{ as }\exp(-y)\neq 0$$
$$\exp(-y)\sin x = 3.$$


Explicit formula for $(I+E)^{-1}$ for E is Idempotent Matrix I wanted to find formula for  $(I+E)^{-1}$
I know that this $I+E$ is invertible .
Because this is equivalent to show that -1 is not eigenvalue of E ,Idempotent matrix.

Main Problem occured when I had write power expansion of term $(I+E)^{-1}$=$(I-E+E^2-E^3.....)$ Actually which I this not converging .
How to find Inverse of I+E?
Any Help will be appreciated

$I-\frac 1 2 E$ is the inverse of $I+E$.

Here is a perhaps more direct way to arrive at the answer. You were trying to use $\frac1{1+E}=1-E+E^2-\dots$. The problem is that this is only valid when $|E|<1$. However, since $E$ is idempotent, its eigenvalues may be either $0$ or $1$, so $|E|<1$ may not hold.

However, $-\frac12+E$ will have eigenvalues in $\{-\frac12,\frac12\}$, so $|-\frac12+E|=\frac12<1$. Therefore, we can instead do
$$
\frac1{1+E}=\frac{2/3}{1-(\frac13-\frac23E)}=\frac23\Big(1+\big(\frac13-\frac23E\big)+\big(\frac13-\frac23E\big)^2+\dots\Big)\tag{*}
$$
You still have to work out what that nasty infinite summation is. The best method I could think of to evaluate the terms $\big(\frac13-\frac23E\big)^n$ was to work out small cases to find a pattern and prove it by induction. Fortunately, you can show
$$
\big(\frac13-\frac23E\big)^2=\frac19
$$
which quickly implies that
$$
\big(\frac13-\frac23E\big)^n=
\begin{cases}
\frac1{3^n} & n\text{ is even}\\
\frac1{3^n}-\frac2{3^n}E & n\text{ is odd}
\end{cases}
$$
which allows you to collect the coefficients of $E$ in $(*)$ and arrive at a simple formula for $(1+E)^{-1}$.


Assume $f(x)$ has continuous derivative on $[0,1]$ ,$|f'(x)| \leq M$ Assume $f(x)$ has continuous derivative on $[0,1]$ with $|f'(x)| \leq M$ and
$f(0)=f(1)=0$.

Prove that
$$\left| \int_0^1 f(x) \ dx \right| \leq \frac{M}{4}$$

A hint: Try to imagine how fast $f$ can grow from the ends towards the center of the interval.

By the way, you can replace $\leq$ by $<$ in the claim.

Hint: Let $g(x)=x-a$ where $a=\frac{3 + \sqrt{6}}{6}$. Using the Cauchy Inequality
$$ \left(\int_0^1f'(x)g(x)dx\right)^2\le\int_0^1(f'(x))^2dx\int_0^1(g(x))^2dx $$
I think you will get the desired inequality. Note that
$$ \int_0^1(x-a)^2dx=\frac{1}{4}, \int_0^1(x-a)f'(x)dx=-\int_0^1f(x)dx. $$


Four color theorem disproof? My brother in law and I were discussing the four color theorem; neither of us are huge math geeks, but we both like a challenge, and tonight we were discussing the four color theorem and if there were a way to disprove it.

After some time scribbling on the back of an envelope and about an hour of trial-and-error attempts in Sumopaint, I can't seem to come up with a pattern that only uses four colors for this "map". Can anyone find a way (algorithmically or via trial and error) to color it so it fits the four color theorem?

Starting at the top, going clockwise:

center: 1, 2, 3
middle: 2,4,3,4,2,4
outside: 1

I hope this helps $\ddot\smile$

To answer the "algorithmically" question, this map has some regions that only border four others. There is a relatively short, algorithmic proof that if you can 4-colour all but one of the regions of a map, and the last region, R, only borders four others (call them R_1, R_2, R_3, R_4 in clockwise ordering about R), then you can colour the whole map. To do this, work as follows. It is easy unless R_1 to R_4 already use all the colours. Say they are blue, green, red, yellow in that order around region R. Now we try to recolour R_1 to red, in the hope that this will allow us to colour R blue. If we do that, we have to recolour any red region which borders R_1 blue. Then we have to recolour any blue region which borders one of these red, and so on. We keep doing this until we run out of things to recolour. Now one  of two things happens: either we can colour R blue or we had to recolour R_3 blue. In the latter case, there was a chain of red and blue regions stretching from R_1 to R_3, but there can't also be a chain of green and yellow regions stretching from R_2 to R_4 (they have to cross somewhere, but the fact that they use different colours means that they can't). So now if we try the same trick recolouring R_2 yellow, any yellow regions next to R_2 green, and so on, this time we won't have to recolour R_4, and we will be able to colour R green.

We've now shown that if we can colour everything apart from a region that meets four (or fewer) others, we can do this recolouring trick and then colour the last region. Similarly, if we had a colouring of some of the regions, and one of the uncoloured regions only bordered four coloured ones, we can recolour in this way and then colour that region as well. So if we can find an ordering of regions such that each one borders at most four previous ones, we can progressively recolour in this way. In this map we can -- basically progressively remove regions that only have four neighbours in what's left, then reverse that ordering -- so this algorithm will work.

This method was the basis of Kempe's incorrect proof of the 4-colour theorem, and was used by Heawood to prove the 5-colour theorem (using five colours we are ok so long as there is always a region we can remove which borders at most five others, but that is true for any plane map). It can be used to easily find a 4-colouring of Martin Gardner's "April Fools" map, which would be very difficult to find by trial and error.


Limit at infinity of two root functions What is the limit of $f(x) = \sqrt{|x|} - \sqrt{|x-1|}$ as $x\to\infty$? I can plot this and see that $f(x)$ is going to zero but I would like to solve this algebraically. I feel like I need to put both terms under the same root.

Since, as $x \to \infty$, $x$ eventually becomes greater than $0$, we can drop the absolute values without a problem.

Consider multiplying and dividing by $\sqrt{x} + \sqrt{x-1}$. Then on the top you can utilize the difference of squares formula, i.e. $(a-b)(a+b) = a^2 - b^2$, and see that

$$\begin{align}
\lim_{x \to \infty} \sqrt{x} - \sqrt{x-1} &= \lim_{x \to \infty} (\sqrt{x}- \sqrt{x-1}) \cdot \frac{\sqrt{x}+ \sqrt{x-1}}{\sqrt{x}+ \sqrt{x-1}} \\
&= \lim_{x \to \infty}\frac{1}{\sqrt{x}+\sqrt{x-1}}
\end{align}$$

The limit of this final expression is clearly $0$, and thus

$$\lim_{x \to \infty} \sqrt{x} - \sqrt{x-1} = 0$$


Compute angle between two points in a sphere. Exist an equation that provide the angle between two points in a sphere.

What I'm looking is not easy to explain, anyway assume to have two points in a sphere than connect this tho points with a line (A and B in the figure) and than how can I compute the angle between the line and a fix reference line or a kind of orientation of reference (as if the two points were in a normal plane)? Is there any way to do something like that?

I'm going to give two answers. The first I'll derive, but it uses some linear algebra, which may not be familiar to you, and converts to "xyz" coordinates to do its work. The second uses just angles, and I'll just give you the formula and a references. Ideas needed:

I'll represent an arrow from the origin $(0,0,0)$ to a point $(a, b, c)$ by writing $[a, b, c]$, and call it a "vector". I call this "the vector corresponding to the point $(a,b,c)$", just to have a name for it.
"cross product of vectors": $[x, y, z] \times [a,b, c] = [yc - zb, za - xc, xb-ya]$. If the two vectors are nonzero and not parallel, then the cross product is a vector perpendicular to both.
"dot product of vectors": $[x, y, z] \cdot [a, b, c] = ax + yb + zc$.
"length of a vector $v = [x, y, z]$: we define $\| v \| = \sqrt{x^2 + y^2 = z^2} = \sqrt{ v \cdot v }.$
If $v$ and $w$ are nonzero vectors, then the angle $\theta$ between then satisfies $ \cos \theta = \frac{v \cdot w} {\|v \| ~ \|w\|}$.
There's a function, $atan2$, in most software packages, that takes a non-origin point point $(x, y)$ of the plane and produces $atan2(x, y)$, which is the angle, measured counterclockwise from the positive $x$-axis, to the ray from $(0,0)$ through $(x, y)$. (It's a lot like "arctan", but gives correct answers for points in the 3rd and fourth quadrant by checking the sign of $x$.)
A vector $[x, y, z]$ represents a point with spherical coordinates, where the "latitude coordinate" $\theta$ is $atan2(x, y)$, and the "longitude coordinate" is $\phi = \cos^{-1} z$. This assumes that the coordinates are set up so that the $xy$ plane is the equatorial plane, and the two "poles" lie at $(0,0,1)$ and $(0,0, -1)$, and that we are measuring in units that make the radius of the sphere "1". Note that the $\phi$ coordinate of a point on the equator is 0, with the phi coordinate being $\pi/2$ (or 90 degrees) a the north pole. If you are used to measuring from $0$ (at the north pole) to $\pi$ (or 180 degrees) at the south pole, you'll need a conversion.
The inverse transformation, from $(\phi, \theta)$ coordinates to $xyz$ is $(x, y, z) = (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$.
Points $(x,y,z)$ on the plane $Ax + By + Cz = 0$ (which intersects the unit sphere in a great-circle, because the plane passes through the origin, which is the center of the sphere) can also be described by saying that each point $(x, y, z)$ corresponds to a vector $v = [x, y, z]$ with the property that $n \cdot v = 0$, where $n$ is the vector $[A, B, C]$. We call $n$ "a normal vector for the plane."

I think that's it. :)

Now let's look at your diagram. I'm going to call the center of the sphere $U = (0,0,0)$, just to have a name for it.

The great circle containing $B$ and $C$ lies in a plane that's perpendicular to the vectors $b$ and $c$ for $B$ and $C$. One such vector is $n_1 = b \times c$. The same goes for the great circle through $A$ and $B$. If we say that $a$ is the vector corresponding to $A$, then one normal to the plane of that second great circle is $n_2 = a \times b$.

The cool observation here is that the angle $\theta$ that you're looking for at $B$ is exactly the angle between these two planes...which is also exactly the angle between their normal vectors! And from item 5 above, you can compute this by
$$
\theta = \cos^{-1} \left( \frac{n_1\cdot n_2}{\|n_1\|~ \|n_2 \|}\right).
$$

So your complete "formula" for computing the angle is this:

Convert each point, $A$, $B$, $C$ in polar coordinates, to rectangular, using item 8, to get coordinate triples that we treat as vectors $a$, $b$, $c$.
Compute $n_1 = b \times c$, $n_2 = a \times b$, using item 2.
Compute $\| n_1 \| $ and $\| n_2 \|$ using item 4, and $n_1 \cdot n_2$ using item 3.
Compute $\theta = \cos^{-1} \left( \frac{n_1\cdot n_2}{\|n_1\|~ \|n_2 \|}\right)$.

That wasn't so bad, was it? :)

In general, the formula for the cosine of the angle at $B$ is

$$
\cos B = \frac{\cos b - \cos a \cos c}{\sin a \sin c}
$$
where in this case, I'm using "B" to denote the angle you've drawn at point $B$, but "$b$" to indicate the angle subtended by the points $A$ and $C$ from the origin $U$, and similarly for $a$ and $c$. (You've used these lower-case letters to label three arcs, and I'm just saying that they denote not the length of the arc in miles, but rather the measure of the arc (a number between $0$ and $180$ degrees, or between $0$ and $\pi$ radians).

Computing those subtended angles? I'd be inclined to use rectangular coordinates and dot products as I did before. :)

Also: the angle at "B" is ambiguous: if you pick the angle $\pi + B$, you'll still eventually get to point $A$. If you're writing a program, you may find that the inverse cosine messes you up sometimes because of this.

Most books on celestial navigation derive these formulas, as will any book on spherical trigonometry.

An equation that provides the angle between two points in a sphere:

Let the angle between two points A and B be $ω$, then

$$cos(ω)=cos(θ_1-θ_2)*cos(φ_1-φ_2)$$

where, $A=[L_1,θ_1,φ_1]=(x_1,y_1,h_1)$, $B=[L_2,θ_2,φ_2]=(x_2,y_2,h_2)$.

$L=\sqrt{r^2+h^2}$ , $r=\sqrt{x^2+y^2} , θ=tan^{-1}(\frac{y}{x}),  φ=tan^{-1}(\frac{h}{r})$.


Where is the lost dollar? Somebody explained me this problem, but I am not sure to understand what is wrong.

Three people go to a bar to have a drink.
To pay the bill everyone gives $10.
The barman gives back $5 coins.
Every people take $1 back and leave the $2 other dollar for the barman.

If everybody give \$10 and get \$1 back, they paid \$9 right?

3 * $9 + $2 = $29

Where is the missing dollar?

The math is simples:

10*3 = 30 and 30-25 = 5
(1*3)+2= 5

∴

There is no loss.


What are some good math study methods for a high school sophomore? First, I'm not trying to make this sound like a "poor-me" story. I understand fully that every decision I've made leading to this is my fault. I am genuinely looking for advice.

So, I am a high school student who is a sophomore and feel I have developed an interest in mathematics a bit late. As an elementary and even middle school student, I was always not paying attention in most math classes because they didn't grab my interest. I think the reason for this is because most of my teachers were "old-fashioned" and didn't really bother to explain the awesome things you can do with math. So on my end of things, I only saw manipulating numbers which didn't seem productive to me. I have always had a deep interest in computers and how they work (you can see where this is going most likely). So, upon entering my freshman year of high school I developed a passion for computer programming. Now realizing that large portions of math are required to major in computer science, I am beginning to worry a bit. A large amount of people I meet automatically assume I am good at math, and I feel sort of guilty for wasting those years of my life barely paying attention. Currently I am beginning to enjoy math, though.

I feel that most math teachers don't capture the interest of the students because they don't show the engineering side of things when it comes to math. I love creating original things and have been told I am very creative. I would like to use that creativity and produce creations using math, but I think I'm a bit behind.

My math final is tomorrow, so I'm rather stressed out. The actual questions in this post are:

What are some good math study methods?
What should I do this summer to strengthen my math skills?

You might want to read this post on how to study math: it's about good study methods /strategies for math. There are some great contributions/answers to that question.

Second, it is never too late to begin anew. Though you may feel like you've wasted "years" of your life, you've still got lots of time, and it seems you've got the motivation to build your math skills and explore your interests further and prepare for college.  While it will take discipline to get up to speed and continue in math, and lots of persistence and patience, be sure to dabble in what excites you about math. Explore and have fun with math!.

Wish you the best, and welcome to the wonderful world of mathematics!

Khan Academy

Khan Academy is a non-profit organization that has a lot of instructional videos on you-tube and practice problems on their site. It helped me get through physics and calculus.


$2^x - a$ touches $\log_2(x)$ I was playing around with the functions $2^x$ and $\log_2(x)$. As they are the inversions of each other, I thought there was a simple number $a$ for which $2^x - a$ touches $\log_2(x)$.

Using trial-and-error, for $a = 2$ looked like the functions just touched each other, but on closer inspection they did cross each other: at $x = 1$ and at approximately $x = 1.04759$.

Zoomed in:

Is there an exact solution to $a$ for which $2^x - a$ touches $\log_2(x)$? With "touches" I mean that they just touch each other, like a tangent does to a function.

If the graphs "just touch", then the tangents are parallel. So you want
$$(2^x - a)' = (\log_2 x)'$$
This leads to
$$(\ln 2)2^x = \frac{1}{(\ln 2)x}$$
or $x2^x = (\ln 2)^{-2}$.

Since $y= x2^x$ is strictly increasing on $(0,\infty)$ (the derivative is $2^x(1+x\ln 2)$), has $\lim\limits_{x\to 0^+}x2^x = 0$ and $\lim\limits_{x\to \infty}x2^x = \infty$, there is one and only one value of $x$ where the equality holds.

So there is one and only one value of $x$ where the tangents of $y=2^x-a$ and $y = \log_2 x$ are parallel. Picking $a$ to equal the value of $\log_2(x)$ at the point $x$ in which this holds, which is the unique solution to $x2^x = (\ln 2)^{-2}$, gives the unique solution.

The solution to $x2^x = (\ln 2)^{-2}$ can be "obtained" by using the Lambert $W$ function.

Since
$x2^x = xe^{x\ln 2}$, we can rewrite the equation as $(x\ln 2)e^{x\ln 2} = \frac{1}{\ln 2}$. Since $W(1/\ln  2)$ is the value $k$ such that $ke^k = \ln 2$, we conclude that $x = \frac{1}{\ln 2}W\left(\frac{1}{\ln 2}\right)$ is where $x2^x = (\ln 2)^{-2}$.

So this is the value of $x$ at which $y = 2^x - a$ (for any $a$) and $y=\log_2(x)$ have parallel tangents. We want to pick $a$ so that the two functions have the same value as well, so we want $2^x - a = \log_2(x)$; that is
$$a = 2^{\frac{1}{\ln 2}W(\frac{1}{\ln 2})} - \log_2\left(\frac{1}{\ln 2}W\left(\frac{1}{\ln 2}\right)\right) = 2^{\frac{W(1/\ln 2)}{\ln 2}} - \log_2W\left(\frac{1}{\ln 2}\right) + \log_2(\ln 2).$$

The values of $a$ and $x$ are
$$\begin{align*}
a&\approx 1.9993335366784208872185237594934841819876651500164506943525\ldots\\\
x&\approx 1.0237165016039817739129111076311195719147116014095574420368\ldots.
\end{align*}$$

...and it's about  2.00971; so, you were pretty close, pimvdb


Ring homomorphisms form an abelian group Let ${\rm Hom}(\mathbb{Z}_n,\mathbb{Z}_m)$ be the set of all ring homomorphisms. I am not assuming $f(1) = 1$. My question is,

  Is it possible to give an abelian group structure in this set?

I know this is true for group homomorphism as the co-domain is an abelian group. But, in the case of ring homomorphisms, $f+g$ need not satisfy the ring homomorphism conditions. What is the natural operation that makes ${\rm Hom}(\mathbb{Z}_n,\mathbb{Z}_m)$ into an abelian group?

I am trying to prove the following equation on pg no. 259 in this paper: https://fada.birzeit.edu/bitstream/20.500.11889/4045/1/homo-monthly.pdf

Kindly share your thoughts. Thank you.

To give a (not necessarily unital) ring homomorphism from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is the same as giving an idempotent in $\mathbb{Z}_m$ whose order divides $n$. Now, it is well known that the idempotents in any unital commutative ring form a Boolean ring, with the same multiplication and the "sum" of $e$ and $f$ being $e+f-2ef$. Also, it is easily seen that if $e$ and $f$ have order diving $n$, then so does $e+f-2ef$. Finally, with this addition, the identity element is still zero and every element is its own inverse. Hence, there is an isomorphic abelian group structure on the set of ring homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ with $(f +' g)(k)=k[f(1)+g(1)-2f(1)g(1)]$ (using the prime mark to avoid confusion with the pointwise sum).