Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$

Dividing by 4-3i gives me $$z^2  \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$
which goes to $$z^2 -4z-3zi + 7+i$$

then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after that i can't get the expansion to work.

Can you help me out?

Starting from what you ought to have:

  $(z−(2+\frac{3}{2}i))^2 = -7-i + (2+\frac{3}{2}i)^2$.

we get:

  $(z−(2+\frac{3}{2}i))^2 = -7-i + (4+6i-\frac{9}{4}) = \frac{-21+20i}{4} = (\frac{2+5i}{2})^2$.

The last step I obtained by guessing. If you want a systematic way to find the square root of a complex number if it is a perfect square (the square of a rational complex number) you can use the following technique.

Given $(a+bi)^2 = c+di$ where $a,b,c,d \in \mathbb{R}$:

  $a^2-b^2 = c$ and $2ab = d$ [by comparing real and imaginary parts].

  Thus $4 a^2 - 4 a^2 b^2 = 4 a^2 c^2$ and $4 a^2 b^2 = d^2$ [multiply the first by $2a^2$ and square the second].

  Thus $4 a^4 - 4c a^2 - d^2 = 0$ [Add them together to get a quadratic in $a$].

  Thus $(2a^2-c)^2 = c^2+d^2$ [Complete the square].

  Thus $2a^2-c = \sqrt{c^2+d^2}$ and hence $a = \pm \sqrt{\frac{\sqrt{c^2+d^2}+c}{2}}$.

  Also $2b^2 = 2a^2 - 2c = \sqrt{c^2+d^2}-c$ and hence $b = \pm \sqrt{\frac{\sqrt{c^2+d^2}-c}{2}}$.

  [Note that the choices of sign are dependent, so it may be better to use $b = \frac{d}{2a}$ instead.]

In the above case we get:

  $\sqrt{-21+20i} = \pm \left( \sqrt{\frac{\sqrt{21^2+20^2}-21}{2}} + \sqrt{\frac{\sqrt{21^2+20^2}+21}{2}} i \right) = \pm (2+5i)$.

  [Here I have chosen the signs correctly.]

If $z_2−(4+3i)z+7+i=0$

Then

$\Delta=(4+3i)2−4(7+i)=−21+20i=(2+5i)2$.

From which one can use the quadratic formula to obtain;

$z=3+4i$ and $z=1−i....$


permutation -number of numbers of four different digits formed from the digits of the number 12356 such that it is divisible by 4 The number of numbers of four different digits that can be formed from the digits of the number 12356 such that it is divisible by 4??

Hint: If a number ends in $12$, $32$, $52$, $16$, $36$, $56$, it is multiple of $4$. How many ways can you fill the other two digits?

The answer to this is quite simple. Adding to what @ajotatxe has written, any number is divisible by $4$, if the last $2$ digits are multiples of $4$. From the given list the following are possible in the last $2$ places: $12, 16,32, 36,52, 56$.
Hence $6$ ways.

Now, the remaining $2$ digits can be any of the $3$ digits from the list, as $2$ have been used up in the last $2$ places.

Therefore, the total number of ways is: $3 \times 2 \times 6 = 36$ total ways

Note: I have made the necessary edits to my answer. Apologies as I had mistaken the question... And a bunch of thanks to @Daniel Fischer


Does this recurrence have a closed form limit $x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$? I have a first order nonlinear recurrence relation:

$$x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$$

Here $a,x_0$ are positive constants and $a<x_0$. (Also $x_0=A+B$ and $a=(A−B)^2$, for some $A,B>0$).

For these conditions the recurrence quickly converges to a certain limit, which depends only on $a,x_0$:

$$\lim_{n \to \infty}x_n=X(a,x_0)$$

I don't know if this limit has closed form or not, and there is no general method for dealing with nonlinear recurrence relations.

  Can $X(a,x_0)$ have a closed form and how to obtain it?

I don't need the explicit expression for $x_n$, only the limit.

I tried to turn it into a differential equation, but I don't know if I've done it correctly, and how the solution to the ODE relates to the original problem:

$$x_{n+1}-x_n=-\frac{a}{3^{2n+1}x_n}$$

$$\frac{df(t)}{dt}=-\frac{b}{3^{2t} f(t)}$$

$$\frac{1}{2} f^2=\frac{b}{2\ln 3} 3^{-2t}+C$$

$$f(t)=\sqrt{\frac{a}{3\ln 3} 3^{-2t}+C}$$

If I set:

$$x_n=\sqrt{\frac{a}{3\ln 3} 3^{-2n}+C}$$

I get:

$$C=x_0^2-\frac{a}{3\ln 3}$$

$$\lim_{n \to \infty}x_n=\sqrt{C}=\sqrt{x_0^2-\frac{a}{3\ln 3}}$$

But that's not correct. Does this work only with linear recurrences?

I have also inverted the recurrence:

$$x_n=\frac{1}{2} \left(x_{n+1}+\sqrt{x_{n+1}^2+\frac{4a}{3^{2n+1}}} \right)$$

This works correctly, but I'm not sure how it may help.

I would be astounded if the limit has a nice closed form. Here are some facts I have investigated.

Let $f(z) = 3z - \frac{1}{z}$. Then we easily check that $y_n = \frac{3^n}{\sqrt{a}} x_n$ satisfies $y_{n+1} = f(y_n)$. Thus $x_n$ has the following general form

$$ x_n = \frac{\sqrt{a}}{3^n} f^{\circ n}(y_0) = \sqrt{a} F_n\left(\frac{x_0}{\sqrt{a}}\right), \qquad F_n(z) := z \prod_{k=0}^{n-1} \left(1 - \frac{1}{3f^{\circ k}(z)^2} \right). $$

So it suffices to consider the case where $a = 1$ and then investigate the limit of $F_n(z)$ if exists. To this end, we claim the following:

  Proposition. Let $I = [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$. Then the limit $F(z) = \lim_{n\to\infty} F_n(z)$ converges and defines a holomorphic function on $\Bbb{C}\setminus I$. Moreover,

  There exists a finite Borel measure $\mu$ on $\Bbb{R}$ which is supported on $I$ and satisfies
  \begin{align*}
F(z)
&= z - \int_I \frac{\mu(d\lambda)}{z-\lambda} \\
&= z - \frac{3}{8z} - \frac{9}{640z^3} - \frac{2241}{465920z^5} - \cdots
\quad \text{as } z \to \infty.
\end{align*}
  $F(f(z)) = 3F(z)$.

The proof of this fact is at the end of this answer. Here are some comments:

Identifying $F$ amounts to identifying the measure $\mu$. Although not entirely sure, it seems to me that $\mu$ is discrete and $\operatorname{supp}(\mu)$ is a Cantor-like set. That is, poles aggregate at uncountably many points and they do not average out. I can hardly imagine of an elementary function which exhibits this kind of behavior, so I suspect that $F$ is not an elementary function.
On $(\frac{1}{\sqrt{2}}, \infty)$ the function $f$ has inverse. If we denote by $g = f^{-1}$ the inverse, then for each $\epsilon > 0$ we can prove that

$$ g^{\circ n}\left(\frac{1}{\sqrt{2}} + \epsilon\right) = \frac{1}{\sqrt{2}} + \frac{\ell + o(1)}{5^n} \qquad \text{as } n\to\infty \tag{*}$$

for some constant $\ell = \ell(\epsilon) \geq$ which depends on $\epsilon$. Then in view of the identity

$$ F\left(\frac{1}{\sqrt{2}} + \epsilon\right) = 3^n F\left(g^{\circ n}\left(\frac{1}{\sqrt{2}} + \epsilon\right) \right), $$

if $F(\frac{1}{\sqrt{2}} + \epsilon) \sim c \epsilon^{\alpha}$ for some constants $c > 0$ and $\alpha$, then we must have $\alpha = \frac{\log 3}{\log 5}$. Although I have not formally written down my idea, using a detailed version of $\text{(*)}$ I checked that $F(\frac{1}{\sqrt{2}} + \epsilon) = \Theta(\epsilon^{\alpha})$ for this $\alpha$, where $\Theta$ is a Landau asymptotic notation.

Proof of Proposition. By the extra assumption $a =1$, we have $x_n = F_n(x_0)$ and $y_n = f^{\circ n}(x_0)$. From this, we have the following relations.

$$ F_{n+1}(z) = F_n(z) - \frac{1}{3^{2n+1}F_n(z)}, \qquad F_n(z) = 3^{-n} f^{\circ n}(z). \tag{1}$$

Using the first relation in $\text{(1)}$, we have

$$ \operatorname{Im}(F_{n+1}(z)) = \operatorname{Im}(F_n(z))\left( 1 + \frac{1}{3^{2n+1}|F_n(z)|^2} \right). $$

In particular, $F_n$ is a Nevanlinna function and thus we can write write $F_n(z)$ as

$$F_n(z) = z - \int_{\Bbb{R}} \frac{\mu_n(d\lambda)}{z-\lambda} $$

for some point mass $\mu_n$ on $\Bbb{R}$. (This can also be proved directly by some tedious algebra.) This representation shows that $F_n(z) = z - \mu_n(\Bbb{R})z^{-1} + \mathcal{O}(z^{-2})$ as $z \to \infty$. Plugging this to the recurrence relation in $\text{(1)}$, we find that

$$\mu_{n+1}(\Bbb{R}) = \mu_n(\Bbb{R}) + \frac{1}{3^{2n+1}}. $$

In particular, the total mass of $(\mu_n)$ is bounded. On the other hand, if $x \in \Bbb{R}\setminus I$, then we easily check that $|f(x)| > |x|$. This implies that

$\text{(2)}$ $F_n$ is finite on $\Bbb{R}\setminus I$, and thus $\mu_n$ is supported on $I$.

$\text{(3)}$ $F_n$ converges on $\Bbb{R}\setminus I$, since $F_n(x)$ is monotone and bounded for each $x \in \Bbb{R}\setminus I$.

So it follows from $\text{(2)}$ that $(\mu_n)$ is weakly compact and hence $F_n$ is a normal family on $\Bbb{C}\setminus I$. Then the first claim follows from the identity theorem together with $\text{(3)}$.

Once we know that $F_n$ converges on $\Bbb{C}\setminus I$, then the second claim follows by taking the limit to $F_n(f(z)) = 3F_{n+1}(z)$. Using this, we can compute the Laurent expansion of $F$ near $\infty$. Since $F$ is an odd function, with $a_n = \int_{\Bbb{R}} \lambda^{2n} \, \mu(d\lambda)$ we have

$$ F(z) = z - \sum_{n=0} \frac{a_n}{z^{2n+1}}. $$

Plugging this to the identity $F(f(z)) = 3F(z)$ produces an system of equations for $(a_n)$, from which we can compute $(a_n)$ at least theoretically.

How about asymptotics for a solution?  Something like this:
$$
x_n = 1 + \frac{3a}{8}\;9^{-n} - \frac{81a^2}{640}\;9^{-2n}
+\frac{41553a^3}{465920}\;9^{-3n} + O(9^{-4n})
\qquad\text{as } n \to \infty
$$

addition, suggested by Winther:

Solution has the form
$$
x_n = L \cdot F\left(\frac{a}{L^29^n}\right)
$$
where $F(z)$ is given by a certain power series:
$$
F(z) = 1 + \frac{3z}{8} - \frac{81z^2}{640}
+\frac{41553z^3}{465920} + \dots
$$


Intuition and the fundamental theorem of calculus I think the following example explains the fundamental theorem of calculus quite intuitively. Or more precisely, that's what I thought; now I'm starting to have some doubts.

Suppose $v(t)$ is the velocity of a car driving along the highway. The units for $t$ are in hours and the units for $v(t)$ are in miles per hour. Assume $v(t)$ is continuous and nonnegative. What is the displacement of the car over one hour (ie., $t \in [0,1]$)?

Well, if we subdivide $[0,1]$ into $n$ subintervals of equal length, in each subinterval $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ the velocity doesn't change too much for large $n$ and hence can be approximated by $v(\frac{k}{n})$. Therefore, the displacement in $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ is equal to $\frac{1}{n} v(\frac{k}{n}) + \epsilon(k, n)$ where $\epsilon(k, n)$ is a small error dependent on $k$ and $n$.

Hence $$ \text{Displacement} = x(1) - x(0) = \sum_{k=0}^{n-1}  \frac{1}{n} v\left(\frac{k}{n}\right) + \sum_{k=0}^{n-1} \epsilon(k, n) $$

Note that the above equality holds for all $n$, since we have accounted for the error. If we assume that $ \sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$, then it's easy to see that $$x(1) - x(0) = \lim \sum_{k=0}^{n-1}  \frac{1}{n} v\left(\frac{k}{n}\right) = \int_0^1 v(t) \ dt$$

However, it's not obviously clear to me why $\sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$. Why should this hold intuitively?

The function $v$ is uniformly continuous on $[0,1]$. Given any $\epsilon > 0$, there is an $N$ such that whenever $n \geq N$, we have, for all $k$ and all $x \in [k/n, (k+1)/n]$, the inequality $|v(x) - v(k/n)| \leq \epsilon$. Thus your error $\epsilon(k,n)$ is bounded in absolute value by $\epsilon/n$. Summing over $k$, the total error in the displacement is bounded in absolute value by $\epsilon$, so long as $n \geq N$.

In broad strokes:

Suppose $F(x)$ is the cumulative area under the curve of $f(x)$ from $0$ to $x$ i.e. the green region below?

What is the derivative of $F(x)$?  How fast is $F(x)$ changing as $x$ changes?

$F(x+h)-F(x)$ is the red region.

$F'(x) = f(x)$

Update:

The Riemann Sum.

We can partition the domain

$a=x_0<x_1<x_2<\cdots<x_i<\cdots<x_n =b$

$\int_a^b f(x) dx = \sum_{i=1}^{n} f(x_i^*)(x_i - x_{i-1})$

Where $x_i^*\in [x_{i-1},x_i]$

I say, that if the partition is fine enough, it does matter if we take the biggest possible value of $f(x_i^*)$ the smallest or something in between.

Suppose we choose $x_i^*$ to always produce the smallest $f(x_i)$.  We would call this the lower sum, and we would get the green area.

And if we choose $x_i^*$ to produce the largest $f(x_i)$ we get the upper sum.  The green + red.

And the difference is just the red boxes.

As the partition gets to be increasingly fine, the errors (the red boxes) only get smaller, and eventually their net area goes to 0.


Approximating Sobolev functions in $W^{1,p}(\mathbb{R}_+^n)$ Let $p \geq 1$. I know that there exists a continuous and linear extension operator $$ E : W^{1,p}(\mathbb{R}_+^n) \to W^{1,p}(\mathbb{R}^n) .$$

I read that from the existence of such an extension one can deduce that $C_c^{\infty}(\bar{\mathbb{R}^n_+})$ is dense in  $W^{1,p}(\mathbb{R}_+^n)$, where for $C_c^{\infty}(\bar{\mathbb{R}^n_+})$ I mean smooth functions on $\mathbb{R}_+^n$ whose support is contained in $\bar{\mathbb{R}_+^n}$(I don't know whether this is a standard notation or not).

I tried to use convolution and use that $C_c^{\infty}(\mathbb{R}^n)$ is dense in $W^{1,p}(\mathbb{R}^n)$ but I wasn't able to get anything .

$C_c^{\infty}(\bar{\mathbb{R}^n_+})$ is dense in  $W^{1,p}(\mathbb{R}_+^n)$,

This is not correct. We choose $n=1,p=1$.
Lets assume that we can approximate the constant function $1\in W^{1,p}(\mathbb R^n_+)$ with functions from $C_c^\infty (\overline{\mathbb R_+^n})$.

Let $\phi\in C_c^\infty (\overline{\mathbb R_+^n})$.
Then $\phi(0)=0$.

If $\phi(x) \leq \frac12$ for all $x\in (0,1)$, then we have
$$ \| 1- \phi |_{W^{1,1}} \geq \|1-\phi|_{L^1} \geq \frac12.$$
So we can assume that $\phi(x)>\frac12$ for some $x\in (0,1)$.
Then
$$
\frac12 = \int_0^x \phi'(y) \mathrm dy \leq \int_0^1 | \phi'(y) |\mathrm dy
\leq \|\phi'\|_{L^1} \leq \|1-\phi\|_{W^{1,1}}.
$$
So in any case, we cannot approximate $1$ with $\phi$.


Convergence implies boundness of a sequence but an example confuses me! given the following sequence
$$
p_{n} = \frac{1}{n-1}
$$
I am tempted to say that for n=1 the sequence assumes an infinite (and thus unbounded) value but still the sequence converges to zero as n increases.
I am of course wrong because this contradicts theorem 3.2c of Baby Rudin which states that "if a sequence is convergent in a metric space then it's bounded". What am I missing? Maybe 1/0 is not a mathematical defined value?
I would be greatly thankful for an answer.
Erma

Single values for which $p_n$ is not defined are not relevant when we take the limit as $n\to \infty$ what matters is that $p_n$ is eventually a well defined expression.
For example
$$\lim_{n\to \infty} \frac1{\sqrt{n^2-1\,000\,n-1\,000\,000}}$$
is not well defined for a finite number of values but the limit is equal to $0$ since eventually the expression inside the square root becomes positive and tends to $\infty$.


Is university math all about proofs? Do mathematicians do anything else beside writing proofs? It seems like all the "upper-division" math here are about proving something rather than solving for something i.e. instead solving for $x^2 = -1$, prove that solution to $x^2$ is something adic, on some sort of space, satisfy some vexity, generalizable to some blah...

Yes, mathematicians do a lot more besides writing proofs!  While some already pointed out that “they also have lunch” I assume that you are more interested into the intellectual processes involved in the mathematical activity.

When I¹ examine my intellectual processes, I can see that some exercise my intuition and some exercise my rationality.  My intuition tells me “there is something there” and with my rationality I can organise intuitions, see causal relations between facts and all the like. I am not, by far, versed in that area of philosophy, and I am confident that many people investigated this and wrote beautiful books on that topic. But for the purpose of this answer, I kindly bid the right of being superficial and approximative.

Mathematics is all about the interaction of intuition and rationality. As I study mathematics, my intuition leads me to genuine astonishment and makes me wonder about mathematical phenomenons I experience:

“could the same obstruction be at work in these two seemingly unrelated problems?”
“what is the geometric meaning of that algorithm?”
“where do the solutions of that problem live?”
“why do this topologist and this algebraist run into the same problem?”
“is there a formalism turning this gibberish into a straight computation?”

Producing all these questions is the work of my intuition – and this is a pretty common experience for mathematicians. Sometimes, my intuition is on a big day and can even give me a glimpse at the answer of the question!  This is where my rationality comes into play, and helps me to reconnect the new intuitive idea to the body of knowledge which I am already familiar with.  And this reconnection is what we call a proof, and this is how I can tell other mathematicians how they can approach and tame the intuition I had.

Think of mathematics as a journey in a place where you cannot take pictures of what you see (your intuitions) but you still can draw a “treasure map” to teach others how they can go the same place where you had been (rationality at work).

As a conclusion, proofs are the most visible aspects of mathematics, but as in many other activities, there is much more that one can experience than one can see. If mathematics were all about proofs, then mathematicians would be like computers which randomly explore the space of proved or provable statements by combining known facts together, but this is not the case, as this method would hardly produce anything interesting.

¹ I write this at the first person to provide a vivid and entertaining picture, with the implicit assumption that the case I describe is the common case.

Einstein, Newton, Euler & Hawkins to name but a few of the greats all solved problems, found solutions and created new way's of seeing the world, looking at problems and adding an explanation to all things. Not just maths, but the universe and many other wonders.

Most of it was born in their head, an idea, a theory.  The theory's all have to be proved, and sadly approved, before they are accepted.

Mathematical proofs are the answer to scientific experimentation.  If it can be repeatedly shown to have a consistent outcome, you have proved what started as a theoretical solution.


question about the basis of order topology recently I read the order topology and the basis element,I have a question that if (a,b)$\cap$(c,d)=$\emptyset $,but $\emptyset $is not belongs to any basis elements(ie. (a,b) for a<b),why it can be basis?
thanks

You are misunderstanding the condition:
$$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2 : \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$$
If $B_1 \cap B_2$ are empty, the statement is already true because we have a quantor $\forall$ over the empty set! See vacuous truth.
You can formulate this condition equivalently as: the intersection of any two basic elements is a union of a family of basic elements, and then it works too as the empty set is the result of the empty union (the union of the empty subfamily of the base).

You have to throw the null set in, to make it a basis.  Any topology includes the null set.


Can we have a countable region in a perfect set? I am reading a proof in Baby Rudin that perfect sets in $\mathbb R^k$ are uncountable. The proof seems to assume it is first countable and show that it contradicts the fact that if a collection of compact sets have non empty finite intersections then the entire intersection is nonempty.

However, is it possible for an open neighborhood of a point in an uncountable set, say $x_0,$ to intersect only a countably infinite number of points? Because then it seems as if the same proof can be applied to this subset only.

Consider the set $\{0\}\cup\{1/n : n \in \mathbb{N}\} \cup\{x :1 \leq x\leq 2, x \in \mathbb{R}\}$ which is uncountable and any ball of radius less than one around zero will only have countably many points and is an open neighborhood. Notice that it has isolated points so it is not perfect.

Also consider $\{x : 0 \leq x \leq 1, x \in \mathbb{Q}\} \cup\{x : 1 \leq x \leq 2, x \in \mathbb{R}\}$ which is uncountable, any of the rational numbers on the interval $[0,1]$ can have a ball of sufficiently small radius to contain only a countable number of points and it has no isolated points, but it is not closed so it is not perfect.


How to construct center of homothety for two circles which overlap In general any two circles have two centers of homothety. They have only one center when the circles have the same radius or when the circles have the same center.

Given two circles of different radii with disjoint interiors, their external center of homothety is at the intersection of the external tangents and the internal center is at the intersection of the internal tangents. This is the case because homothety preserves tangency. So the centers of homothety are easy to construct in this case.

If the circles partially overlap, there is still an external homothetyic center at the intersection of external tangents, and an internal homothety center lying somewhere along the line joining the centers of the circles. If the circles overlap completlety (one inside the other), there are two internal homothety centers on the line joining the centers of the circles.

My question is how do you construct (with compass and straightedge) the centers of homothety in the cases where the circle interiors are not disjoint.

In the case of two circles that are not disjoint, I think that you only have one homothety center, which is the one found by the intersecction of the external tangents.
The reason for that is that when you have two circles that are not disjoint the internal tangent no longer exists.
In case you don't know how to find the external tangent of two cercles (it doesn't matter if they are disjoint or not) with compass and straightedge, here you have images and an explanation that shows you how to do it.

http://jwilson.coe.uga.edu/emt669/Student.Folders/Kertscher.Jeff/Essay.3/Tangents.html

Basically, you draw a line between the two centers, you find the medium point of this line and draw a circle centered there that touches both cirle centers using the compass.
After that, you draw a circle centered in the center of the biggest circle with a radious equal to R= R(big circle)-R(small circle).
You draw a line from the center of the biggest circle to the point that you find on the intersection of the two new circles that you've drawn, and follow this line until it intersects with the biggest circle. This point is the first tangent point. If you now draw a parallel line to this one that passes through the center of the small circle, you'll have the second tangent point in the intersection of this line with the small circle. If you draw a line joining both tangent points and make it intersect with the line that joins both circle centers, there you'll have your homothetic center.

For more information, you may want to take a look at this:

https://commons.wikimedia.org/wiki/Category:Homothety_in_circles

Maybe you are looking not for homothety centers but antihomothety centers.
I hope I've been able to help you, and don't hesitate to tell me if not, i'll try to do my best!


Limit of a sum of two variables Recently at my calculus course we are doing derivatives and integrals.

I've stumbled upon a sum that seems to have nothing in common with our current objectives, though I'm sure it does have, but still I have no idea how to solve it.

$$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $$

I tried to represent it as a $$ f_{n}(x) = \sum_{k=1}^{n} \frac{x}{4n - \frac{k^2}{n}}$$ and to resolve it at a point $x = 1$, but I failed because of the two variables instead of one.

EDIT:

Of course, I would like to get some CLUES how to solve it, not the whole solution.

Observe that we have
$$
\int_0^1\frac1{4-x^2}dx=\lim_{n\to\infty} \frac1n\sum_{k=1}^n\frac1{4-\frac{k^2}{n^2}}
$$
by definition of the integral.

Let $f(t)=\frac{1}{4-t^2} $ then $$\frac{1}{4n -\frac{k^2}{n}} =\frac{1}{4-\left(\frac{k}{n}\right)^2} \cdot \left(\frac{k}{n} -\frac{k-1}{n}\right) =f\left(\frac{k}{n} \right)\cdot \left(\frac{k}{n} -\frac{k-1}{n}\right)$$
Hence
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{4n -\frac{k^2}{n}} =\lim_{n\to\infty} \sum_{k=1}^n f\left(\frac{k}{n} \right)\cdot \left(\frac{k}{n} -\frac{k-1}{n}\right)= \int_0^1 f(v)dv =\ln \sqrt[8]{\frac{5}{3}}$$


Compare largest root of two polynomials For $1\leq a, a+1<b$ with $a,b\in\mathbb{N}$ let

$f:=f(a,b):=\sum_{i=2}^{b-a}i,$
$g:=g(a,b):=f(a,b)+a+b+1$
$h:=h(a,b):=f(a,b)-(b-a-1)$

and consider the polynomial
$$
p(t):=t^g-\left(\sum_{i=h}^f t^i\right).
$$

I would like to show that the largest (in absolute value) root of $p(t)$ is smaller than the largest (in absolute value) root of the polynomial
$$
q(t):=t^{a+b+1}-t^{a+b}-1.
$$

We can factor out $t^h$, i.e.
$$
p(t)=t^h\cdot\left(t^{g-h}-\sum_{i=1}^{b-a-1}t^i -1\right),
$$
so
$$
p(t)=0\Leftrightarrow t^h=0\quad\text{ or }\quad t^{g-h}-\sum_{i=1}^{b-a-1}t^i-1=0
$$

In this proof, we only need that $a,b\in\mathbb{Z}_{\geq 0}$ with $a<b$.  We have
$$p(t)=t^{\frac{(b-a-1)(b-a-2)}{2}}\left(t^{2b}-\sum_{k=0}^{b-a-1}\,t^k\right)\,.$$
Observe that each of the two polynomials $p(t)$ and $q(t)$ has a unique positive real root (using Descartes's Rule of Signs, or something similar---there are a multitude of ways to show this simple result).

Write $\alpha$ and $\beta$ for the positive real roots of $p(t)$ and $q(t)$, respectively.  Using Rouché's Theorem, $\alpha$ and $\beta$ are roots of $p(t)$ and $q(t)$, respectively, with the largest moduli amongst all (real or complex) roots. (Rouché's Theorem is not necessary here; this result can be proven via the Triangle Inequality for complex numbers.)

   Let $B(r,z) \subseteq \mathbb{C}$ denote the open ball of radius $r>0$ centered at $z\in\mathbb{C}$.  To use Rouché's Theorem for $p$, consider the disc $B(\alpha+\epsilon,0)$, where $\epsilon>0$ is arbitrary.  At the boundary, $$\big|z^{2b}\big|>\left|-\sum_{k=0}^{b-a-1}\,z^k\right|$$ for all $z\in \partial B(\alpha+\epsilon,0)$, whence the polynomial $p(t)$ has the same number of roots in $B(\alpha+\epsilon,0)$ as the polynomial $t^{2b}$, which is $2b$.  That is, all roots of $p(t)$ are in $\bigcap\limits_{\epsilon>0}\,B(\alpha+\epsilon,0)$, which is the same as the closed ball $\bar{B}(\alpha,0)$.  For $q$, you can do something similar, but I am using the Triangle Inequality to deal with it now.  Note that $\big|q(z)\big|\geq q\big(|z|\big)$ for all $z\in\mathbb{C}$ by the Triangle Inequality.  If $z$ is a complex number such that $|z|>\beta$, we get $$\big|q(z)\big|\geq q\big(|z|\big)>0\,.$$  Thus, all roots of $q$ are in the closed ball $\bar{B}(\beta,0)$.

If $b=a+1$, then clearly $\alpha=1$ and $\beta>1$ (because $q(1)<0$).  From now on, suppose that $b>a+1$, which means $\alpha>1$ and $\beta>1$ (noting that $p(1)<0$ and $q(1)<0$).

Suppose contrary that $\alpha\geq \beta$.  Since $\beta^{a+b}(\beta-1)=q(\beta)+1=1$, we obtain
$$\alpha^{a+b}\left(\alpha-1\right)\geq \beta^{a+b}\left(\beta-1\right)=1\,.$$
Now, since $p(\alpha)=0$ and $\alpha\neq 0$, we get that
$$\alpha^{2b}=\sum_{k=0}^{b-a-1}\,\alpha^k=\frac{\alpha^{b-a}-1}{\alpha-1}\,.$$
That is,
$$1\leq \alpha^{a+b}(\alpha-1)=\frac{\alpha^{b-a}-1}{\alpha^{b-a}}<1\,,$$
which is a contradiction.

Part of the claim (namely about the maximal positive root of $p$) follows by straightforward computation:

Note that
$$\tag1 q(t)=(t-1)t^{a+b}-1$$
As $q(1)=-1$ and $q(t)\to+\infty$ as $t\to+\infty$, there is clearly a root of $q$ outside $[-1,1]$, i.e., we need only consider $|t|>1$.

Let $t_\min$, $t_\max$ be the smallest and largest root of $q$, respectively. As  seen above, $t_\max>1$ and $q(t)>0$ for $t>t_\max$.
Using the formula for geometric series, we verify
$$\tag2\frac{ p(t)}{t^{f+1}}=\frac{q(t)}{t-1}+\frac{t^{b-1}}{t-1}$$
Then using $q(t)\ge 0$ for $t\ge t_\max>1$, it follows from $(2)$ that $p(t)>0$ for all $t\ge t_\max$.
In particular,

  The maximal root of $p$ is strictly smaller than the maximal root of $q$.


Finding a function $\phi$ so that $\phi(X)$ is a random variable In my notes, Chebyshev's inequality is formulated as follows:
$$
\forall a \geq 0: P(X \geq a) \leq \dfrac{1}{\phi(a)}E[\phi(X)]
$$
Which is true when $X$ is a random variable and  $\phi\geq 0$ and non-decreasing for $-\infty < x < \infty$.

$\phi(X)$ is simply said to be a function $\phi: \mathbb{R} \rightarrow \mathbb{R}$, so that X is a random variable.

Which conditions apply so that $\phi(X)$ is still a random variable?

Edited answer after realizing my mistake:

The function $\phi$ only needs to be measurable for $\phi(X)$ to be a random variable. As @Did says, a non-decreasing function is trivially measurable.

More formally, if $X$ is a random variable defined on  $(\Omega,\mathscr A)$, then for any Borel (measurable) function $\phi$ on $\mathbb R$, $\phi(X)$ is also a random variable.

This is because for all $x\in\mathbb R$, we have $$\{\omega\in\Omega:\phi(X(\omega))\le x\}=\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}$$

As $\phi$ is measurable, $\phi^{-1}((-\infty,x])$ is a Borel (measurable) set and as $X$ is a random variable, $$\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}\in \mathscr A\quad,\forall\,x\in\mathbb R$$


If I flip a coin 1000 times in a row and it lands on heads all 1000 times, what is the probability that it's an unfair coin? Consider a two-sided coin. If I flip it $1000$ times and it lands heads up for each flip, what is the probability that the coin is unfair, and how do we quantify that if it is unfair?

Furthermore, would it still be considered unfair for $50$ straight heads? $20$? $7$?

First of all, you must understand that there is no such thing as a perfectly fair coin, because there is nothing in the real world that conforms perfectly to some theoretical model.  So a useful definition of "fair coin" is one, that for practical purposes behaves like fair. In other words, no human flipping it for even a very long time, would be able to tell the difference.  That means, one can assume, that the probability of heads or tails on that coin, is $1/2$.

Whether your particular coin is fair (according to above definition) or not, cannot be assigned a "probability". Instead, statistical methods must be used.

Here, you make a so called "null-hypothesis": "the coin is fair".  You then proceed to calculate the probability of the event you observed (to be precise: the event, or something at least as "strange"), assuming the null-hypothesis were true.  In your case, the probability of your event, 1000 heads, or something at least as strange, is $2\times1/2^{1000}$ (that is because you also count 1000 tails).

Now, with statistics, you can never say anything for sure.  You need to define, what you consider your "confidence level".  It's like saying in court "beyond a reasonable doubt".  Let's say you are willing to assume confidence level of 0.999  .  That means, if something that had supposedly less than 0.001 chance of happening, actually happened, then you are going to say, "I am confident enough that my assumptions must be wrong".

In your case, if you assume the confidence level of 0.999, and you have 1000 heads in 1000 throws, then you can say, "the assumption of the null hypothesis must be wrong, and the coin must be unfair".
Same with 50 heads in 50 throws, or 20 heads in 20 throws.  But not with 7, not at this confidence level.  With 7 heads (or tails), the probability is $2 \times 1/2 ^ {7}$ , which is more than 0.001.

But if you assume confidence level at 95% (which is commonly done in less strict disciplines of science), then even 7 heads means "unfair".

Notice that you can never actually "prove" the null hypothesis.  You can only reject it, based on what you observe is happening, and your "standard of confidence".  This is in fact what most scientists do - they reject hypotheses based on evidence and the accepted standards of confidence.

If your events do not disprove your hypothesis, that does not necessarily mean, it must be true!  It just means, it withstood the scrutiny so far.  You can also say "the results are consistent with the hypothesis being true" (scientists frequently use this phrase).  If a hypothesis is standing for a long time without anybody being able to produce results that disprove it, it becomes generally accepted.  However, sometimes even after hundreds of years, some new results might come up which disprove it.   Such was the case of General Relativity "disproving" Newton's classical theory.

Given the chance for a head-up is 0.5:

$(1/2)^{1000} = 9.332636 \times 10^{-302}$

50 times?

$(1/2)^ {50} = 8.8817842 \times 10^{-16}$

Depends on your standard of unfairness, I would say it is pretty unfair for cases 50 and 1000 of straight heads.


Solution check for counting in a list This problem involves lists made from the letters T,H,E,O,R,Y, with repetition allowed.

How many 4-letter lists are there that don’t begin with T, or don’t end in Y ?

Just want to make sure my solution is right and my logic isn't flawed.

My solution:

A = 4 letter lists that don't begin with T

B = 4 letter lists that dont end in Y

$|A| = 5 * 6 * 6 * 6 = 1080$

$|B| = 6 * 6 * 6 * 5 = 1080$

$|A \cup B | = 1080 + 1080 = 2160$

We may simplify the problem by using De Morgan's laws:

\begin{align*}
  |A\cup B| &= \overline{\overline{|A\cup B|}} \\
  &= \overline{|\overline{A}\cap \overline{B}|} \\
  &= |U|-|\overline{A}\cap \overline{B}| \\
  &= 6^4-6^2 \\
  &= 1260
\end{align*}

Hence, for a $n$-letter list, the number of such lists is $6^n-6^{n-2}$.

We can also calculate it like that:

We calculate the number of 4-letter lists without restrictions and from this number we subtract these one,which begin with T and  these,which end in Y.But subtracting the number of lists that begin with T and the number of lists that end in Y,we subtract twice the number of lists that begin with T AND end in Y,so wehave to add this number once.

So,it is:

$$6^4-1 \cdot 6 \cdot 6 \cdot 6 -6 \cdot 6 \cdot 6 \cdot 1+1 \cdot 6 \cdot 6 \cdot 1$$

If you have questions,feel free to ask..


Find $A^x$ where A is a $2\times 2$ matrix So, I need to calculate $A^x$ where

$A=\pmatrix{4&-3\\3&-2}$

First of all,  $A^x = e^{x\ln A}$, so I get that

$\ln A = \pmatrix{2&0\\-1&1}$

Then I use that

$f(A)= T^{-1}f(J)T$

Here $J$ is the Jordan Canonical form. I got that

$J=\pmatrix{1&1\\0&1}$

and

$T=\pmatrix{3&1\\2&0}$, $T^{-1}=\pmatrix{0&\frac{1}{2}\\1&\frac{-3}{2}}$

What should I do next? Any help would be greatly appreciated.

The result is this one
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
3x+1 & -3x\\
3x & -3x+1\\
\end{array}\end{array}\right)
\end{equation}
Now I have to exit, but as soon as I'm back I will explain it if needed...

Edit>>:
There are many ways to reach the same answer. The more plain one is that:
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
ax+b & cx+d\\
ex+f & gx+h\\
\end{array}\end{array}\right)
\end{equation}
since $A^0=I$ we have $d,f=0$ and $b,h=1$, and since $A^1=A$ we have $c=-3$ and $e=3$.
Then imposing
$$
a+b=4, g+h=-2,
$$
we obtain $a=3$ and $g=-3$.

So for example
\begin{equation}
A^{10}=\left(\begin{array}{c}
\begin{array}{ccccc}
31 & -30\\
30 & -29\\
\end{array}\end{array}\right)
\end{equation}

Hint: using the Mercator series  $$\ln(1+x) = x - \frac{x^2}2 + \frac{x^3}3 + \ldots$$

You can write $$\ln(J)=\ln(I+X) = X$$ where $X=\pmatrix{0&1\\0&0}$ so $X^2$ and higher terms are zero.


Topological entropy of circle homeomorphism is zero. True or false? may I know if it is true that $\ f: S^1 \to S^1$ a homeomorphism, then $h_{top}(f) = 0$, where $h_{top}$ stands for topological entropy.

I believe this statement is true, but I cannot prove it.

Here, we use the spanning set definition of topological entropy. Since $f$ is uniformly continuous on $S^1$, for arbitrary $\epsilon>0$, there exists a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ if $|x-y|\leq\delta$. Let $S_n$ be the cardinality of the minimal $(n,\epsilon)-spanning$ set. Then we study the relations between $S_n$ and $S_{n+1}$.

Denote by $\{I_i,i=1,...,S_n\}$ the open interval cover of $S^1$ corresponding to the minimal $(n,\epsilon)-spanning$ set. We consider an open interval $I_i=(x,y)$. Then we have $d(f^i(x),f^i(y))<\epsilon, i=0,1,...,n.$ Therefore, it's critical to determine $d(f^{n+1}(x),f^{n+1}(y))<\epsilon$. In order to ensure $d(f^{n+1}(x),f^{n+1}(y))<\epsilon$, we devide the interval $(f^{n}(x),f^{n}(y))$ or $(f^{n}(y),f^{n}(x))$ into $\epsilon/\delta$ subintervals. Then each of such subintervals has a preimage interval of $f^n$ in $I_i=(x,y)$ for $f$ is a homeomorphism. Thus we obtain $S_n\cdot \epsilon/\delta$ intervals cover $S^1$ such that $d(f^i(x),f^i(y))<\epsilon, i=0,1,...,n+1,$ for $x,y$ in the same interval. By the minimality of spanning set, so we have $S_{n+1}\leq S_n\cdot \epsilon/\delta.$ By the definiton of topological entropy $h(f)$, $$0\leq h(f)=\lim_{\epsilon\rightarrow 0}\overline{\lim_{n\rightarrow\infty}}\frac{\log{S_n}}{n}\leq\lim_{\epsilon\rightarrow 0}\epsilon/\delta=0.$$ Hence $h(f)=0.$


Showing the irreducibility of $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ in $\mathbb{Q}[x]$ I would like to show the irreducibility of  $x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$ and $x^8 - 120 x^6 + 4360 x^4 - 45600 x^2 + 15376$ in $\mathbb{Q}[x]$. In both cases Eisenstein criterion fails. I also attempted some linear changes of variables but nothing seems to work. Any help?

Irreducibility of the first polynomial
$$f(x) = x^8 - 60 x^6 + 1160 x^4 - 7800 x^2 + 8836$$
can also be deduced as follows.

Recall the usual business with Gauss's lemma. If $f(x)$ factors in $\Bbb{Q}[x]$, it also factors in $\Bbb{Z}[x]$. Let's assume contrariwise that
a non-trivial factorization $f(x)=g(x)h(x),g(x),h(x)\in\Bbb{Z}[x]$ exists.
Without loss of generality the leading coefficients of both $g$ and $h$ are equal to one.

A potentially useful feature of $f(x)$ is that modulo five it becomes very sparse. More precisely
$$
f(x)\equiv x^8+1\pmod 5.
$$
In $\Bbb{F}_5[x]$ we have the factorization
$$
x^8+1=x^8-4=(x^4-2)(x^4+2).
$$
These quartic polynomials are actually irreducible in $\Bbb{F}_5[x]$. We have
$$x^8+1\mid x^{16}-1.$$ Therefore any zero of either factor (in some extension field of $\Bbb{F}_5$) must be a root of unity of order sixteen. But $16\nmid 5^\ell-1$ for $\ell=1,2,3$ meaning that the field $\Bbb{F}_{5^4}$ is the smallest extension field containing such roots of unity. Therefore their minimal polynomials over $\Bbb{F}_5$ have degree four.

At this point we can conclude that the only remaining way $f(x)$ can factor in $\Bbb{Z}[x]$ is as a product of two irreducible factors of degree four, and
$$
g(x)\equiv x^4+2\pmod 5,\qquad h(x)\equiv x^4-2\pmod 5.
$$

Another feature of $f(x)$ is that it has even degree terms only. In other words, $f(x)=f(-x)$. Therefore $f(x)=g(-x)h(-x)$ is another factorization. But, factorization of polynomials is unique, so we can deduce that either $h(x)=g(-x)$ (when also $h(-x)=g(x)$),
or we have both $g(x)=g(-x), h(x)=h(-x)$.

Claim. It is impossible that $h(x)=g(-x)$.

Proof. Assume contrariwise that $h(x)=g(-x)$. If $g(x)=x^4+Ax^3+Bx^2+Cx+D$, then $h(x)=x^4-Ax^3+Bx^2-Cx+D$. Expanding $g(x)h(x)$ we see that the constant term is $D^2=8836=94^2$. Therefore we must have $D=\pm94$. But, earlier we saw that the constant terms of $g,h$ must be congruent to $\pm2\pmod5$. This is a contradiction.

Ok, so we are left with the possibility $g(x)=g(-x)$, $h(x)=h(-x)$. In other words, both $g(x)$ and $h(x)$ share with $f(x)$ the property that they have even degree terms only. Let's define $F(x),G(x),H(x)$ by the formulas
$$f(x)=F(x^2),\quad g(x)=G(x^2),\quad h(x)=H(x^2).$$
The above considerations can be summarized as follows. If
$$
F(x)=x^4-60x^3+1160x^2-7800x+8836
$$
is irreducible, then so is $f(x)=F(x^2)$. Furthermore, the putative factors must satisfy the congruences
$$G(x)\equiv x^2+2\pmod5,\quad H(x)\equiv x^2-2\pmod5.$$

A miracle is that depressing $F(x)$ produces a surprise:
$$
R(x)=F(x+15)=x^4-190x^2+961.
$$
The substitution $x\mapsto x+15$ does not change anything modulo five, so the only possible factors of $R(x)$ must still be congruent to $x^2\pm2\pmod5$.

Irreducibility of $R(x)$ follows from this. The constant term of $R(x)$
is
$$
R(0)=961=31^2,
$$
and this has no factors $\equiv\pm2\pmod5$.

The other octic surrenders to similar tricks:
$$f(x)=x^8 - 120 x^6 + 4360 x^4 - 45600 x^2 + 15376.$$
Again, $f(x)\equiv(x^4-2)(x^4+2)\pmod 5$. The constant term
$15376=(2^2\cdot31)^2$ is a square of an integer $\equiv\pm1\pmod5$,
ruling out a factorization of the form $g(x)g(-x)$. Again, we are reduced
to proving that
$$
F(x)=x^4-120x^3+4360x^2-45600x+15376
$$
is irreducible. Depressing this gives
$$
R(x)=F(x+30)=x^4-1040x^2+141376\equiv(x^2-2)(x^2+2)\pmod5.
$$
This time the constant term $R(0)=2^6\cdot47^2=376^2$ has more factors, so we need a different argument. However, we can repeat the dose! $\pm 376\equiv\pm1\pmod5$, ruling out the possibility of a factorization of the form $R(x)=G(x)G(-x)$ as above. So the remaining possibility is a factorization of the form
$$
R(x)=(x^2-A)(x^2-B)
$$
with integers $A$ and $B$. But, the equation
$$
x^2-1040x+141376=0
$$
has no integer roots. Irreducibility follows.

Both these polynomials have irreducible remainders in $\mathcal{F}_2[x]$, field of integers modulo 2, when we try to divide each by $x^2+x+1$.

The reference is to Artin's Algebra, Proposition 12.4.3

  Let $f(x) = a_n x^n + \dots + a_0$ be an integer polynomial, and let $p$ be a prime integer that does not divide the leading coefficient $a_n$. If the residue $\bar{f}$ of $f$ modulo $p$ is an irreducible element of $\mathcal{F}_p[x]$, then $f$ is an irreducible element of $\mathcal{Q}[x]$.

This means if we divide our polynomials in field $\mathcal{F}_2[x]$, and we are lucky (= this often works) to get a remainder which is an irreducible polynomial in this field, then the original polynomial is irreducible.

In field $\mathcal{F}_2[x]$ the following polynomials are irreducible, $x^2+x+1$ and $x+1$.

If we divide the original polynomials by $x^2+x+1$, the remainders are

$$16575 + 8959 x \equiv 1+x \mod 2$$

$$60855 + 49959 x \equiv 1+x \mod 2$$

Note

If any of these two polynomials had proper divisors in $\mathcal{Q}[x]$, then it had proper divisors in $\mathcal{Z}[x]$ as well, Artin 12.3.6.

This might suggest that if any of these polynomials had proper divisors in $\mathcal{Z}[x]$, then divisors would be monic polynomials with limited choice of the last term, ie $8836=2^2 \times 47^2$. But I do not have a deep insight to pursue it further.


$\square+\square+\square=30$, with boxes filled using $1, 3, 5, 7, 9, 11, 13, 15$, possibly repeated. How? From the days I started to learn Maths, I've have been taught that

  Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
  $$3 + 5 + 1 = 9$$

OK, but look at this question

This question was solved and the answer was 30, how it was possible? Need a valid explanation please.

You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.

You should not believe everything you read in a photo on the internet.


Self-avoiding walk on $\mathbb{Z}$ How many sequences $a_1,a_2,a_3,\dotsc$, satisfy:
i) $a_1=0$
ii) ($a_{n+1}=a_n-n$ or $a_{n+1}=a_n+n$)
iii) $a_i\neq a_j$ for $i\neq j$
iiii) $\mathbb{Z}=\{a_i\}_{i>0}$

Are the two alternating sequences the only solutions?

$a_1,a_2,a_3,..=0,1,-1,2,-2,3,-3,4,...$
or
$a_1,a_2,a_3,..=0,-1,1,-2,2,-3,...$

Is there a sequence satsifying i),iii),iiii) and ii) ($a_{n+1}=a_n-n^2$ or $a_{n+1}=a_n+n^2$) ?

Only uncountably many. See this link to MO.

Two obvious solutions are :-

0, 1, 3, 6, 10, 15, ...  an = Sigma(n-1)
0, -1, -3, -6, -10, -15, ... an = -Sigma(n-1)

Here, I am considering only one of the Formulae at a time. Either an = a(n-1) - n OR an = a(n-1) + n for all the numbers in the series. And of course, here ai != aj (for i != j)

Now, considering the solutions that you gave :

$a_1,a_2,a_3,..=0,1,−1,2,−2,3,−3,4,...$

AND

$a_1,a_2,a_3,..=0,−1,1,−2,2,−3,...$

Here, you have used both the formulae in alternate fashion, starting from either one of them and thereby getting two series.

Now, consider an = a(n-1) - n as "Going Backwards" AND

an = a(n-1) + n as "Going Forwards".

Observe one solution below :-

$0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, ...$

Here I go forward 3 times, till I reach $6$ and then toggle between going backwards and going forwards from then on. Since my first few (say r) steps were forward, it could be possible that going backwards from some ak could lead to a number which already exists in the series and thereby violating the condition ai !+ aj (For i != j). Whenever such violation occurs, I would forward once again and then try to go backwards and the process continues.

Similarly, we can also go backwards a few steps (more specifically more than 1 step), and then keep toggling. Whenever there is a violation, we go back once again and try to come forward and the process continues.

When our initial step (before toggling) is only 1 time, then we get the two solutions that you gave. And there is no conflict in any step in that case. This can be generalized to going in one direction for r times, and then toggling. This would lead to INFINITE SUCH SERIES, since r E N (Natural Numbers) which are infinite.

However, this is only one such set of infinte series. We could go in any different fashion as follows :

$0, 1, 3, 6, 2, -3, -9, ... $(Going forward till 6 and then going back)

Or it can be any other random order too.

So, the possible number of sequences that satisfy all 4 conditions would be INFINITE.

And for the answer to the second question, series satisfying an+1=an−n2 or an+1=an+n2,

Following two are again the obvious solutions :

$1. 0, 1, 5, 14, ...  $(Going forward)
$2. 0, -1, -5, -14, ...$ (Going backward)


The transpose of a linear injection is surjective. Let
$$T:V\longrightarrow W$$
be a linear map (of vector spaces), and let
\begin{eqnarray}
T^*:W^* &\longrightarrow& V^* \\
f\ &\longmapsto& f\circ T
\end{eqnarray}
be its transpose (or dual map).

How to show that if $T$ is injective then $T^*$ must be surjective, and that if $T$ is surjective then $T^*$ must be injective?

If $T$ is injective then it has a inverse $T^{-1}:T(V)\rightarrow V$. Take $v\in V^\star$ and define $f\in W^\star$ by $f(x)=v(T^{-1}(x))$. Then $T^\star(f)=f\circ T=v\circ T^{-1}\circ T=v$. Hence,$T^{*}$ is surjective.

On the other hand, suppose that  $T$ is surjective. If $T^\star f=T^\star g$, then $f(T(x))=g(T(x))$ for all $x\in V$. Hence $(f-g)(T(x))=0$ for all $x\in V$. Because $T$ is surjective you can conclude that $f-g=0$, or $f=g$.

Partial Solution:

$T$ is injective $\implies T^*$ is surjective: Choose $f\in V^*.$ We need to find out (using the injectivity of $T$) a linear transformation $f':W\to F$ such that $f'T=f.$  Let's try to reduce the proposed problem a bit more using the fact that two linear transformations agree everywhere if and only if they agree on a basis of the  vector space considered as the domain. Thus our reduced task is to find out a linear transformation $f':W\to F$ such that $Tf'=f$ on a chosen basis of $V.$  The problem can a little more be reduced to define $f'$ using the fact that any mapping defined on a basis of a vector space extends to a unique linear transformation. Thus our aim is to define $f'$ on a basis of $W$ such that the effect of the composed map $f'T$ on the basis is identical with that of $f.$ How can it be done? Naturally by considering the effect of the pre-image when it exists (and here the injectivity comes into play) and nullify the effect otherwise. Can you construct $f'$ now?


Valid AM-GM inequality proof? We have to prove that $$\frac{(x_1+x_2+x_3+...+x_n)}{n} \geq (x_1\cdot x_2\cdot x_3\cdots x_n)^{1/n}$$

Attempt: Raising both sides to the nth power gives $\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$
This is equivalent to proving that if a sum of random numbers is a fixed number, then their maximum product occurs when all the number are equal. So we have to find which combination of numbers gives the maximum product if their sum is fixed.

Obviously $$\left({x_1+x_2+x_3+...x_n}\right)^{n} > x_1x_2x_3...x_n$$

then, a maximizer must exist.

Now suppose that we have this combination of products: $x_1x_2x_3...x_n$
If i take the arithmetic mean of any 2 numbers and replace them by their arithmetic mean, I will get a bigger product (By the simple 2 case AM-GM inequality which is easy to proof), obviously the arthimetic mean of 2 numbers will not change the sum of the numbers, we can write that:

$$\frac{x_1+x_2}{2}\cdot\frac{x_1+x_2}{2}\cdot x_3 \cdots x_n\ \geq x_1x_2x_3\cdots x_n$$

Which means if we have any combination of product of numbers, I could increase that product by making any 2 numbers equal (arthimetic mean) without changing the sum of the numbers.

Now supposing that the maximum product does not have all of the numbers equal, then I can increase its product by having 2 numbers equal, so this mean that it's not the maximum product, then, by contradiction, the maximum product must occur when all the numbers are equal (Sum divided by the number of numbers)

Which means that $$\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$$

Is this a valid proof? It's my first time here and I don't know how to do the math notations thingy on this website so forgive me, thanks.

That's pretty good.

One aspect I'd want to fiddle with is the existence of the maximizer.  You argued that it's bounded (which as a commenter noted could be made more explicit) and so a maximizer exists; this requires some kind of compactness argument, and the continuity of $\mathbb R^n\to\mathbb R$, $(x_1,\dotsc,x_n)\to\prod_{k=1}^n x_k$.  That's all fine, but it would be better to make it explicit.

If you don't want to use topological concepts like that, then the argument can be adjusted not to assume the existence of the maximizer, but to prove it.  The idea is to repeatedly replace pairs of $x_i$ until they're all the same, and argue (as you did) that we can do this keeping the AM constant but increasing the GM.  The only tricky bit is that, if you replace pairs of numbers with their average, this process might not ever stop.  (Example: $(1,2,2)\to (\frac32,\frac32,2)\to (\frac32,\frac74,\frac74) \to\dotsm$.)  So the replacement step has to be tweaked.  The resulting proof can be found in a note by Dijkstra, EWD1140.

Hint: Your idea is correct. Here is a north to let things rigorous.

Consider $f, \psi : U \to \mathbb R$ defined as $$ f(x) = x_1 \cdot x_2  \cdots x_n \,\,\, \text{and} \,\,\, \psi (x) = x_1 + x_2 + \ldots + x_n$$

for all $x = (x_1, x_2, \ldots , x_n) \in U$. Fix $s > 0$ and search for the critical points of $f|_M$ where $M = \psi^{-1}(s)$.

Observe that $\mathrm{grad} \, \psi (x) = (1,1,\ldots, 1)$ for any $x \in U$ and $\mathrm {grad}\, f (x) = (\alpha_1 , \ldots , \alpha_n)$, with $a_i = \prod_{j\neq i} x_j$. Show that the maximum is necessarily in $M$ and is the only critical point of $f|_M$.


Why $9$ & $11$ are special in divisibility tests using decimal digit sums? (casting out nines & elevens) I don't know if this is a well-known fact, but I have observed that every number, no matter how large, that is equally divided by $9$, will equal $9$ if you add all the numbers it is made from until there is $1$ digit.
A quick example of what I mean:

$9*99 = 891$
$8+9+1 = 18$
$1+8 = 9$

This works even with really long numbers like $4376331$
Why is that? This doesn't work with any other number. Similarly for $11$ and alternating digits sums.

Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.

The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means:
$$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$
The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.

But since every multiple of $9$ is a multiple of $9$, you will always get $9$.

Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.

If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.

This is a special case of casting out nines, which in turn is a special case of modular arithmetic. It is what is behind many divisibility tests (e.g., for $2$, $3$, $5$, $9$, and $11$).

Coda. This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.

$9$ is also special for accounts in finding book keeper errors of transposed digits.

$124+256+345=725$

$124+265+345=734$

$734-725=9$


Find $a_{ij}$ in $v=\sum\limits_{1\le i<j\le n} a_{ij}e_i\wedge e_j$ If $v=\sum_{i=1}^n\lambda_{1i}e_i$ and $w=\sum_{i=1}^n\lambda_{2i}e_i$ are $2$ vectors in $\mathbb R^n$, then one can write $v\wedge w$ as a linear combination of $\{e_i\wedge e_j:1\le i<j\le n\}$, what are then coefficients in that case, i.e. $v=\sum\limits_{1\le i<j\le n} a_{ij}e_i\wedge e_j$, how can I express $a_{ij}$ with $\lambda_{1i}$ and $\lambda_{2i}$ ?

Well there are $\binom{n}{2}$ possibilities for $e_i\wedge e_j$ so at least the combination of  $\lambda_{1i}$ and $\lambda_{2i}$ cannot be just the product

Why don't we try a little example and see if we can figure this out?

Suppose $v = a e_1 + b e_2$ and $w = c e_1 + d e_2$ (so either $n = 2$ or we decide all the other coefficients are zero). Then
\begin{align*}
v \wedge w
&= (a e_1 + b e_2) \wedge (c e_1 + d e_2)
\\
&= ac e_1 \wedge e_1
+ ad e_1 \wedge e_2
+ bc e_2 \wedge e_1
+ db e_2 \wedge e_2
\\
&= (ad - bc) e_1 \wedge e_2
\end{align*}
because $e_2 \wedge e_1 = - e_1 \wedge e_2$ (and $e_i \wedge e_i = 0$ for all $i$).

Can you now work out the general case? If not try another example with only three nonzero coefficients. It'll come.

You're right that there are $\binom{n}{2}$ possibilities. However, remember that $e_k$ are all basis vectors (This is an assumption I'm making based on notation. If it were not for this, then I doubt there is a neat expression). Therefore, all of the $$a_{ij} = \lambda_{1i}\lambda_{2j}e_i \wedge e_j$$
Will be $0$ when $i \neq j$ since $e_i \wedge e_j = 0$

However, when $i = j$, then $e_i \wedge e_j = 1$

Therefore,

$$
a_{ij} = \lambda_{1i} \lambda_{2j} \ \ \ \text{when} \ \ \ i = j \\
a_{ij} = 0 \ \ \ \text{when} \ \ \ i \neq j
$$

However, since your index runs in a way that $i < j$, you can never have $i = j$. Therefore, the entire product collapses to $0$


Domain of a random variable - sample space or probability space? In most probability theory texts you'll see a random variable definition that goes something like this:

A probability space is a triple $(\Omega, \mathcal{F}, P)$, a real valued random variable is a function $X:\Omega \rightarrow \mathbb{R}$

The computer scientist in my head can't get around what seems to be a type-error.  Is a random variable dependent on $P$ or not.  My intuition says it is, which is why we say things like 'uniformly distributed random variable', but in this case shouldn't a random variable be a function from the probability space, i.e. $X:(\Omega, \mathcal{F}, P) \rightarrow \mathbb{R}$.  In fact this is how it is defined in Wolframs mathworld.

Alternatively, one could imagine in a functional programming language a function which counts the number of heads from a set of coin tosses.  This function could be independent of whether or not the coin is fair or biased, etc, and hence in some way independent of $P$.

I understand some of this may be just for convenience when speaking, but this is how a random variable is rigorously defined, so I believe I must be missing something.  In other words, my question is, is a random variable associated with probabilities or independent of it.

Your observation is reasonable, but your suggested cure for the problem, making $(\Omega,\mathcal F,P)$ the domain of $X$, won't work because the domain of a function needs to be a set (or a type or something like that).  My impression is that, when people refer to a function $X:\Omega\to\mathbb R$ as a random variable, they always do so in the context of a probability-space structure ($\mathcal F$ and $P$) on $\Omega$.  If no such structure is given, then I wouldn't call $X$ a random variable. And if there is uncertainty about which structure is intended, then, as you said, notions like "distribution" of $X$ will not be well-defined.

If I had to formalize the notion of random variable, in Bourbaki style, I would probably say that a random variable is a pair consisting of a probability space $(\Omega,\mathcal F,P)$ together with a function $X:\Omega\to\mathbb R$.  As with many mathematical concepts, one often omits mentioning part of an entity (in this case the probability space) when it is understood from the context.

A random variable $X$ is a measurable map from the probability space $(\Omega, \mathcal{F}, P)$ to another measure space $(S, \mathcal{S})$ (usually $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. It is not just a function between $\Omega$ and $S$ - the measurability of the map (which intrinsically depends on the measure spaces $(\Omega, \mathcal{F}, P)$ and $(S, \mathcal{S})$) is a key part of the definition of the random variable $X$.

It is just cumbersome to write out the entire measure space every time, so we just write $X: \Omega \to S$ and keep the measures and $\sigma$-fields implicit.


Using 6 Sided Dice to Get an Even Distribution of Numbers $1$ to $50$ Is this possible?
I have a $6$ sided dice (call it a $d6$).
I want a way of getting an even distribution of the numbers $1$ to $50$ using that dice with no or minimal rerolls.
A way with a lot of rerolls.
A. Roll d6. $1 - 3 $means use table $A$, $4 - 6$ means use table $B$.
B. Table $A$ needs you to roll $2$ dice. First Dice gives you the column and second dice gives you the row. The numbers $1$ to $25$ are in the table, with the other $11$ spaces telling you to reroll. (Table B is similar, but with numbers $26$ to $50$).
The only problem with this is there are $22$ reroll spaces.
What I want is a way to get an even distribution, but with no rerolls or very few.

Rolling three dice gives you a number from $1$ to $216$.  For a start, you can use the numbers $1-200$ and reroll the last $16$ but remember what the number is.  Now one more roll gives you a number $1-96$ by multiplying your remembered number by $6$, subtracting the new roll, and adding $1$.  You can use $1-50$ and reroll $46$ values.  Again remember the remainder, roll again, and you get a number $1-6\cdot 46=1-276$.  You can use $1-250$ and keep $1-26$.  Keep going.  This has to be optimal because you are using all the information you get.  I haven't computed the average number of throws.

Probably the best you can do is to convert any $3$ rolls to a number $1-50$ as follows : If the rolls give $a,b,c$ , then the number is $$36(a-1)+6(b-1)+c$$ If $a>2$ , then you can stop since the number must then exceed $50$, so you can ignore all trials where the first roll is more than $2$. If we have $a=2$ , the second roll is not allowed to exceed $3$, again you can save a roll if $3$ is exceeded.


Is $ds$ and curvature has the max value at the same point? If there is a plane curve, define as a $y=f(x)$, and my guess is if curvature of this curve $\frac{y^{\prime\prime}}{(1+(y\prime^2))^\frac{3}{2}}$ has a max value at ${x=x_0}$ and $\sqrt{dx^2+dy^2}$ has the max value at same point. that two is necessity, or sufficiency or both of them, or not anything? I try to derive them and both have $0$at same point, but this method not worked well.

I assume you are using the parametrization $r(t)=(t,f(t))$

Then the maximum of the speed is at

$$\frac{d^2 s}{dt^2}=\frac{f'(t)f''(t)}{\sqrt{1+f'(t)^2}}=0$$

Which would imply that $f'(t)=0$ or $f''(t)=0$, but we reject the case for $f'(t)=0$ because curvature cannot be defined on non regular curve.

So that means the speed is at maximum when the second derivative of the curve is $0$.

$$\kappa '(t)=\frac{f'''(t)(1+f'(t)^2)^{\frac 32} - \frac 32 (1+f'(t)^2)^{\frac 12} (2f'(t)f''(t))f''(t)}{(1+f'(t)^2)^3}$$

It is equal to zero when the numerator is $0$ :

$$f'''(t)(1+f'(t)^2)^2 -3f'(t)f''(t)^2=0$$

If $f''(t)=0$ , we must have $f'''(t)=0$ also in order for the curvature to be at maximum, so we cannot conclude that.

In general, the curvature of a plane curve is an intrinsic property in the sense that it is independent of the parametrization and rigid transformation. So usually there would be no relation between the curvature and the derivative of the curve.


Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :

$2^{x+2}.3^{\frac{3x}{x-1}} =9$

My working :

Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2  \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$

Now how to proceed further in this problem... please suggest thanks....

As $\log_aa=1$ and $\log_a(a^m)=m\log_aa=m$

taking logarithm wrt $3,$

$$(x+2)\log_32+\frac{3x}{x-1}=2$$

$$\implies(x+2)\log_32=2-\frac{3x}{x-1}=-\frac{x+2}{x-1}$$

$$\implies (x+2)\left(\log_32+\frac1{x+1}\right)=0$$

We know if $a\cdot b\cdot c\cdots=0;$  at least one of $a,b,c,\cdots$ is zero

Think of something to multiply the entire equation by that will remove any $x$'s from the denominator.


Riesz representation theorem: an application Let $L_n\colon C_c(\mathbb{R})\rightarrow \mathbb{R}$ with
$$ L_nf = \frac{1}{n}\sum_{k=0}^{n}  e^{k/n} f\left(\frac{k}{n}\right).  $$
Prove that $\lim_{n\rightarrow \infty}L_nf=:Lf$ exists and then apply the Riesz representation theorem to $L$.

Until now I proved the existence of the limit.

  Do you have any idea of how to continue the exercise?

Using the Riemann sum, one readly observes that
$$ \lim_{n\to\infty}L_nf = \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n}  e^{k/n} f\left(\frac{k}{n}\right) =\int_0^1e^xf(x)dx $$
for each $f\in C_c(\Bbb R). $ whence set $$Lf= \int_0^1e^xf(x)dx $$

Since $x\mapsto e^x$ is in $L^2(0,1)$ there is not problem if we extend $L$ by taking $f\in L^2(0,1)$

that is
$$Lf= \int_0^1e^xf(x)dx=\langle f,e^x\rangle_{L^2} ~~~~~ f\in L^2(0,1).$$
this is directly the Riesz representation theorem.


Is there a compact normal injective operator on a infinite dimensional Hilbert space? Is there a compact, normal and injective operator $T$ in an infinite-dimensional Hilbert space $H$ such that $sp(T)$ is finite?

Thanks.

Related: Is there a injective compact operator TT such that #sp(T)<∞?

This is not possible.  One way to show this is the following:

If $T$ is normal, and has finite spectrum, then each element of the spectrum is an eigenvalue (see theorem 12.29 of Rudin's Functional Analysis). If $T$ is also injective, then $0$ is not in the spectrum, because if it were it would be an eigenvalue.  But then $T$ cannot be compact, because $H$ is infinite-dimensional.

No, the spectrum of a compact operator in an infinite dimensional Banach space contains $0$, so it is not injective since every element of the spectrum is an eigenvalue.

https://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators#Statement


Verify Mean Value Theorem for $\,f\left(x\right)=x^3\,$ I am trying to verify the Mean-Value Theorem for $\,f\left(x\right)=x^3.\,$ So far I have:

$$f'\left(c\right)=\frac{b^3-a^3}{b-a}=b^2+ab+a^2=3c^2$$

Question: since we know that $b>a$, how can we show that at least one solutions of $c$ lies between $a$ and $b$?

I could not find a simpler solution than the following one.

Mean Value Theorem: There exists $c\in (a,b)$ such that $3c^2=a^2+ab+b^2$.

Denote $c_1 = \sqrt{\frac{a^2+ab+b^2}{3}} > 0$ and $c_2 = -\sqrt{\frac{a^2+ab+b^2}{3}} < 0$. We need to prove that: at least one of the following two inequalities holds:
$$(1) \quad a<c_1<b$$
$$(2)\quad a<c_2<b$$

Consider 3 cases:

$0\le a <b$: In this case $(1)$ is true because $c_1^2-a^2 = \frac{1}{3}(b-a)(b+2a) >0$ and $b^2-c_1^2 = \frac{1}{3}(b-a)(2b+a) >0$.
$a < b \le 0$: Similarly, in this case $(2)$ is true because $a^2-c_2^2 >0$ and $c_2^2-b^2 >0$.
$a\le 0 <b$: in this case $a<c_1$ and $c_2<b$ are already true, we only need to show that at least one of the following two are true: $c_1<b$ or $c_2>a$, or equivalently: $c_1^2<b^2$ or $c_2^2 < a^2$. This is easy by contradiction: if $c_1^2 \ge b^2$ AND $c_2^2 \ge a^2$ then $c_1^2+c_2^2 \ge a^2+b^2 \Leftrightarrow 2(a^2+ab+b^2)\ge 3(a^2+b^2) \Leftrightarrow 0 \ge (a-b)^2$, obviously false.

We are done :)

P/s: a remark for those who are trying to apply the intermediate value theorem:
If we denote $g(x) = 3x^2 - (a^2+ab+b^2)$ then we have $g(a)g(b) = -(a-b)^2(2a+b)(2b+a)$, not always negative.


Proving that $\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$ $f:[a,b]\rightarrow R$ that is integrable on [a,b]

So we need to prove:

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

1.) So we'll use a property of definite integrals: (homogeny I think it's called?)

$$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$

2.) Great, now using the fundamental theorem of calculus:

$$-1\int_{-b}^{-a}f(x)dx=(-1)^2\int_{-a}^{-b}f(x)dx=\int_{-a}^{-b}f(x)dx$$

This is where I'm stuck. For some reason I think it might be smarter to skip step 2, to leave it asL

$$-1\int_{-b}^{-a}f(x)dx$$

because graphically, we've "flipped" the graph about the x-axis, but we're still calculating the same area. Proving that using properties seems to have stumped me.

I prefer hints over solutions, thanks.

Your first step is mistaken: it seems that you mistake $\int _a ^b (-f) (x) \ \Bbb d x$ for $\int _a ^b f (-x) \ \Bbb d x$; these two are completely different, and the homogeneity property applies only to the first formula, not to the second.

Just use the substitution $y = -x$, this will solve the problem in no time.

There is a small error in your first step, you didn't change the sign of the limits. Always keep in mind that after making any substitution, don't forget to change the limits. Since you have made a substitution $x=-t$, so change the limits accordingly. So corrected step 1 is $$-\int_{b}^af(x)dx$$
Now you can reverse the limits to get the correct answer i.e.
$$\int_{a}^bf(x)dx$$


Rules of inference help? I have a question here I need to solve:

For each of the arguments below, formalize them in propositional logic. If the argument
is valid identify which inference rule was used, and formulate the tautology underlying the
rule. If the argument is invalid, state whether the inverse or converse error was made.

(a) All cheaters sit in the back row.
George sits in the back row.
∴ George is a cheater.

For this one I came up with the following:

$C$ = is a cheater

$B$ = sits in back row

$G$ = George

$x$ = students

$\forall$$x$($C(x)$ $\rightarrow$ $B(x)$)

$B(G)$

∴ $C(G)$

From there, I wrote that the converse error was made, since it didn't state that just because you sit in the back row you're a cheater.

The following one is harder for me.

(b) For all students x, if x studies discrete math, then x is good at logic.
Dawn studies discrete math.
∴ Dawn is good at logic.

$x$ = students

$M$ = studies discrete math

$L$ = good at logic

$D$ = Dawn

$\forall$$x(M(x)$$\rightarrow$$L(x))$

$M(D)$

∴$L(D)$

I know this makes sense, but im not sure which rule of inference is made here. how do i determine that? If I got any of the statements wrong please let me know. Thanks for the help.

Your answer to the first question is correct.

The second argument is of the form:
$$\begin{array}{} P \to Q \\ P \\ \hline \therefore Q \end{array}$$

This rule of inference is known as modus ponens. This shows up a lot in logic, and so you should memorize it.

There are many other rules of inference. Here are a few:

Modus Tollens
$$\begin{array}{} P \to Q \\ \lnot Q \\ \hline \therefore \lnot P \end{array}$$

Conjunction
$$\begin{array}{} P \\ Q \\ \hline \therefore P \wedge Q \end{array}$$

Here are a few websites with much more on them:

http://www.philosophypages.com/lg/e11a.htm
https://en.wikipedia.org/wiki/List_of_rules_of_inference


What are the isomorphisms in the category of vector spaces, $Vec_K$? Here is an interesting question from "The Rising Sea" book of Algebraic Geometry: http://math.stanford.edu/~vakil/216blog/FOAGapr2915public.pdf
1.2.3 Example. Consider the category $Vec_K$ of vector spaces over a field $K$. The objects are $K$-vector spaces, and the morphisms are linear transformations. What are the isomorphisms?
I would say that the isomorphisms are the invertible linear transformations, is that correct?

Question: "I would say that the isomorphisms are the invertible linear transformations, is that correct?"
Answer: If $C$ is any category and $U,V$ objects in $C$, we define $U$ to be isomorphic to $V$ (written $U \cong V$) iff there are morphisms (in $C$) $f:U \rightarrow V$ and $g: V\rightarrow U$ such that $f\circ g=Id_V$ and $g \circ f=Id_U$. In your case the maps $f,g$ are invertible linear maps/linear transformations between vector spaces.


How should I approach solving non-linear equations? I need help creating a method for a program I'm making. I've worked on this countless hours and I can not seem to figure it out. what I need: A method that returns $x$. My variables ( initialized to a given number ):

double initial, initial2, initial3,
    Kab, coefficient1, coefficient2, coefficient3

Equation:

kab = ((initial2+coefficient2*X)^coefficient2 *
       (initial3+coefficient3*X)^coefficient3)  /
      (initial-coefficient1*X)^coefficient1

I have tried binary search in order to solve it and that didn't work.

Please be patient with me. I am a high school student and I haven't had much advanced math in order to solve these kinds of problems.

Thank you so much!

Knowing nothing about your parameters, except that they are real numbers I would suggest to use
$$
F(x) =\frac{(i_2 + c_2 x)^{c_2}(i_3 + c_3 x)^{c_3}}{(i - c_1 x)^{c_1}}
$$
and solve numerically for zeros of
$$
f(x) = F(x) - k_{ab}
$$
This is possible e.g. by bisection, or by using Newton-Raphson iteration.


Pattern in the list of square-triangular numbers. I was just reading about  Square Triangular numbers on its Wikipedia page (https://en.wikipedia.org/wiki/Square_triangular_number) and noticed a peculiar pattern in them. Given below is the list of the first $11$ such numbers-

$$0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881,$$ $$55420693056, 1882672131025, 63995431761796, 2172602007770041$$

The last digits of these numbers seem to repeat in the following order - $0,1,6,5,6,1{\dots}$

Can anyone give a mathematical explanation of why this is happening. Or is it just a coincidence?

A square number has only one of the property:

$$n \equiv 0,1,4,5,6,9, \pmod {10}$$

Also a triangular number is of the form :

$$K = \frac{a(a+1)}{2}$$

Hence for triangular numbers :

$$K \equiv 0,1,3,5,6,8 \pmod{10} \space\space\space\space\space\space\space\space\space \text  {(Why? )}$$

Hence for a number to be both square and triangular ,

$$\text{Number} \equiv 0,1,5,6 \mod 10$$

Only possible ending digits are  $ \in \{0,1,5,6\} $


Markov chain simulation I'm wondering whether there is an algorithm to simulate a discrete Markov chain with a specific number of occurrences of state knowing the transition matrix way.

For example, how to simulate in R a Markov chain of length $n$ with $p$ occurrences ($p < n$)

TransitionMatrix<- matrix(c(0.7, 0.3, 0.4, 0.6),byrow=TRUE, nrow=2)
colnames(TransitionMatrix) <- c('0','1') row.names(TransitionMatrix) <- c('0','1')

First a reformulation of the question, then a pseudo-algorithm to solve it.

Reformulation: Let $n\geqslant1$ and assume that $X=(X_t)_{1\leqslant t\leqslant n}$ is a Markov chain of length $n$ on the state space $\{0,1\}$ with transition probability matrix $\begin{pmatrix}1-a & a\\ b& 1-b\end{pmatrix}$ for some $a$ and $b$ in $(0,1)$.

Fix $k\leqslant n$ and consider the set $\Gamma_k^n$ of paths $\gamma=(x_t)_{1\leqslant t\leqslant n}$ of length $n$ visiting $k$ times the state $1$. For every $x$ and $y$ in $\{0,1\}$ and every path $\gamma$ in $\Gamma_k^n$, let $N_\gamma(xy)$ denote the number of times $t\leqslant n$ such that $(x_t,x_{t+1})=(x,y)$. Finally, let $M(\gamma)=N_\gamma(01)$.

Then, forgetting a possible discrepancy of $\pm1$ here and there, one sees that $N_\gamma(10)=M(\gamma)$, $N_\gamma(00)=n-k-M(\gamma)$ and $N_\gamma(11)=k-M(\gamma)$, hence, for every $\gamma$ in $\Gamma_k^n$,
$$
P(X=\gamma)=
(1-a)^{N_\gamma(00)}a^{N_\gamma(01)}b^{N_\gamma(10)}(1-b)^{N_\gamma(11)},
$$
is also, since $n$ and $k$ are fixed,
$$
P(X=\gamma)= (1-a)^{n-k-M(\gamma)}a^{M(\gamma)}b^{M(\gamma)}(1-b)^{k-M(\gamma)}\propto c^{M(\gamma)},
$$
where
$$
c=\frac{ab}{(1-a)(1-b)}.
$$
Thus, there exists some $A_k^n$ independent of $\gamma$ such that, for every $\gamma$ in $\Gamma_k^n$,
$$
P(X=\gamma\mid X\ \text{visits}\ 1\ \text{exactly}\ k\ \text{times})=A_k^n\cdot c^{M(\gamma)},
$$
and the question is to simulate a random path in $\Gamma_k^n$ following this distribution.

Pseudo-algorithm: This assumes one is able to generate uniformly subsets $T$ of $\{1,2,\ldots,n\}$ of size $k$.

Choose a path $\gamma$ uniformly in $\Gamma_k^n$, that is, choose uniformly a subset $T\subseteq\{1,2,\ldots,n\}$ of size $k$ and define $\gamma=(x_t)$ by $x_t=1$ if $t$ is in $T$ and $x_t=0$ otherwise.
Compute $M(\gamma)$.
Accept $\gamma$ with probability proportional to $c^{M(\gamma)}$.
Repeat until a path $\gamma$ is accepted. Return $\gamma$.

To generate $T$, the following procedure might prove useful:

Let $T=\varnothing$.
Choose $x$ uniformly in $\{1,2,\ldots,n\}$. If $x$ is not already in $T$, add $x$ to $T$.
Repeat until $T$ has size $k$. Return $T$.

Suppose you have transition probabilities $p_1 \leq p_2 \dots \leq p_k$, $\sum p_k = 1$. Define the distribution function in pseudo code:

function cdf(state) { a = math.random() % randomly generated number in U(0,1)
if 0 <= a < p_1, return 1
if p_1 <= a < p_1 + p_2, return 2
...
if p_1 + ... + p_{k-1} <= a < 1, return k

In the above, I have assumed that the transition probabilities are all the same, i.e. independent of the state. Otherwise, the $p_i$ will have to depend on state.

The above code is just saying that if I have numbers $p_1$, ..., $p_n$ such that $\sum p_i = 1$, then I can partition the unit interval $[0,1]$ with these numbers. The area of the rectangle [$p_1 + ... + p_{k-1}, p_1 + ... + p_k$] with unit height is $p_k$, and that's the probability we want for the $k^{th}$ transition. This is the standard way of generating probability distributions, since if $X$ has cdf $F$, then $F(X) \sim U(0,1)$.

Now pick an initial state (say $s = 1$) and a state you want to measure, say $r = 4$.

count = 0
for i = 1 : n
   s = cdf(s) // transition with the appropriate probabilities
   if s == 4, count++
   end
end

I should mention this algorithm is the worst.


find the limits : $\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?$ Find the limits (without Stirling's approximation ) :
$$\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?$$

My Try :
$$\frac{1\times 3\times 5\cdots \times(2n-1)}{2\times 4\times 6\cdots \times(2n)}=\frac{(2n)!}{4^n(n!)^2}$$
now what ?

This is
$$\prod_{n=1}^\infty\left(1-\frac1{2n}\right).$$
Its logarithm is
$$\sum_{n=1}^\infty\left(-\frac1{2n}+O(n^{-2})\right).$$
Does that sum converge/diverge/what?

maybe it will help you. the answer in the image is a partial result

Since the root of k is necessary for it to converge to a value, I must assume that the limit of the production tends to zero, and that its rate of growth is equal to that of the square root of k


Is this series convergent or divergent? $\frac{n}{1+e^{-n}}$ I don't know what to do. I was thinking of root test but I get stuck whenever I do it. Any help will be appreciated

Hint: apply the divergence/trivial test.

Just apply the limit test$$\lim_{n \to \infty} \frac{n}{1 + e^{-n}} = \lim_{n \to \infty} \frac{n}{1 + \frac{1}{e^n}}= \frac{\infty}{1 + \frac{1}{e^\infty}} = \frac{\infty}{1 + 0} = \infty$$

and so since the limit is not equal zero then the series is said to diverge

check this link also

  http://mathworld.wolfram.com/LimitTest.html


Does there exist a function whose graph contains any finite collection of points in the plane, no two with the same x coordinates, up to translation? Given any finite set of points in $\mathbb{R}^{2}$, no two with the same $x$-coordinates, it is easy to find a function $f:\mathbb{R}\rightarrow\mathbb{R}$ whose graph passes through all of them. In fact, Lagrange interpolation tells us functions as nice as polynomials will do the job. Can we do it with just one function?
The obvious answer is no, since if $f(x)=y$ then any set of points containing $(x,y)$, $z\neq y$, will never be covered by $f$. But what if we are allowed to translate this function not only horizontally but also vertically? Does there exist a function $f$ such that no matter what finite set of points we start with, some set of points in $f$ contains the "shape" of those finite points?
My first thought was that of the pathological solutions to the Cauchy functional equation $f(x+y)=f(x)+f(y)$. But I don't know how to prove this does/doesn't work, and as a sort of bonus I'd prefer something which is more axiomatically efficient (this certainly doesn't look like a problem requiring even all of $\sf{ZF}$).
I have a suspicion the answer may be no, however. My reasoning, intuitively, is that there are uncountably infinitely many sets of finite points, even up to translation. Each collection of (at least $2$) finite points has nonzero width. From here I wish to use an argument similar to that which states every collection of pair-wise disjoint intervals in $\mathbb{R}$ is countable, where we note that even if we were to try and go "finite set by finite set" and fit the function to one finite set, then to the next, and so on, we would only cover countably many finite sets (since $\mathbb{Q}\subset\mathbb{R}$ is dense and countable). Of course the issue here is that the function may satisfy many finite sets in the same interval, and in fact we can have uncountably infinitely many finite sets satisfied in any given interval.
I would not be surprised if this were possible. I would be somewhat surprised, however, if it were still true should we add certain smoothness conditions on the function. Perhaps differentiability, or infinite differentiability. Or maybe just continuity. To prevent the question from becoming too broad, I'll simply ask: Is it possible with no restrictions on $f$? Is it still possible if $f$ is continuous?

For real $(a,b,c)$ the simple example of a parabola with symmetry axis parallel to $x-$ axis:
$$y= a+\sqrt{ b x +c}$$


Compute the number of positive integer solutions to the equation $x_{1} + x_{2} + x_{3} = 25$. Where 1 $\leq x_{1}, x_{2}, x_{3}$ $\leq$ 15. The book I am using for my Combinatorics course is Combinatorics:Topics, Techniques, and Algorithms.

Compute the number of positive integer solutions to the equation $x_{1} + x_{2} + x_{3} = 25$. Where 1 $\leq x_{1}, x_{2}, x_{3}$ $\leq$ 15.

This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin:

Proof:

One way to do this is to define $w=x+x^2+x^3+ \cdots +x^{15}$ and then cube $w$ and determine the coefficient of $x^{25}$ in the expanded version of $w^3.$ That coefficient comes out $168$ when I used maple. I don't know any "clever" way to get this without the aid of something like maple, and this approach would be quite a mess by hand. Each term of $w^3$ which contributes to the $x^{25}$ term represents one way to get a sum of 25 using three summands from 1 to 15.

The number of unrestricted positive solutions to $x_{1} + x_{2} + x_{3} = 25$ is given by stars and bars and is

$$N_u=\binom{24}{2}$$

To count the number of "negated" solutions that violate $1\le x_{1}, x_{2}, x_{3}\le 15$, note by symmetry that this will be three times the number of solutions for which $x_1>15$ (and $x_2+x_3=25-x_1$). For each such $x_1$, $x_2$ can range from $1$ to $25-x_1-1$, so we get

$$N_n=3\sum_{x_1=16}^{23}{(24-x_1)}=3\sum_{k=1}^{8}{k}=3\frac{8\times9}{2}=108$$

So the number of solutions subject to the restriction is

$$N=N_u-N_n=\binom{24}{2}-108=276-108=168$$


Integral with $e$ , $ \int e^{x^3}\;dx$ $ \int e^{x^3}\;dx$
  so, i'm searching for answer this question, i think that this not is too easily, but i think this integral not exist solution undefined, this integral would be easy if  had the derivative out but i'snt happen

There is no elementary function, say $\mathrm{f}$, with the property that
$$\frac{\mathrm{df}}{\mathrm{d}x}=\mathrm{e}^{x^3}$$

the function of $\int e^{x^2} dx$ is undefined.
most likely the function of $e^{x^3}$ is also undefined.
there exists a error function ($erf$) which approximates the integral
you can read more over that here http://en.wikipedia.org/wiki/Error_function


integral with several answers Could you help me please with a question about integrals? Can an integral have more than one answer? For example with this integral:
$$\int\sqrt{1+\sqrt{1-x^2}}dx$$
Doing by replacing u=$\sqrt{1-x^2}$, I have this solution:
$$2\sqrt{1-\sqrt{1-x^2}}-\frac23 \left(1-\sqrt{1-x^2} \right)^{3/2},$$
and another solution doing by replacing $\sin \theta =x/1$, I get:
$$2\sqrt{2} \sin(\arcsin(x/2))-\frac43 \sqrt{2} \sin^3(\arcsin(x/2))).$$
The graph of each one is in this picture:

Thanks in advance.

The error in one of your solutions lies in the fact that you have to consider the sign of the $x$ values used when performing the integration. In particular, using $u=\sqrt{1-x^2}$, we cannot say that $x=\sqrt{1-u^2}$ because $x$ may be negative whereas the square root term cannot. Also I believe you forgot to use the fact that $\sqrt{x^2}=|x|$ in your working. The trigonometric substitution has given you the correct answer for the integral in the interval $x\in[-1,1]$. We can get the correct algebraic solution by using the identity
$$\sin{\left(\frac{\arcsin{(x)}}2\right)}=\text{sgn}(x)\sqrt{\frac{1-\sqrt{1-x^2}}2}$$
for $x\in[-1,1]$. Here $\text{sgn}(x)$ denotes the Sign function. This gives the value of the integral as
$$2\text{sgn}(x)\sqrt{1-\sqrt{1-x^2}}-\frac23\text{sgn}(x) (1-\sqrt{1-x^2})^{3/2}+C$$
instead of your solution. The two answers are then the exact same function.

Yes, an integral can be have more than one "answer", hence the importance of the constant that is added when finished integrating. More correctly, if two functions $f$ and $g$ they have the same derivative, so we can say that these differ by a constant :
$$f'=g' \, \Rightarrow \, f=g+c \, \textrm{ for some } c$$
Think of a simpler example, let
$$\int 2 \sin x \cos x \ dx$$
then, if we take $u=\sin x$, this integral is equal to $\sin^2 x$ and if we take $u=\cos x$ the integral is $-\cos^2 x$.

As you can see, the integral has two (or more) answers, and that clearly $\sin^2 x \neq -\cos^2 x$.

In fact, this two functions differ by a constant, by $1$ :
$$\sin^2 x = -\cos^2 x \color{red}{+1}$$


Axiomatic Metacategories In Mac Lane's Categories for the Working Mathematician, he starts off with the notion of a metacategory, where we have a "collection" of objects a, b, c...; a collection of arrows f, g, h..., and then some extra structure (axioms ensuring things like cod $f$, id$_a$, etc., are well-defined). He then says that a category is "any interpretation of these axioms within set theory."

It seems we have this so we can talk about things like a "metacategory of sets." But I'm failing to grasp how this definition is strictly axiomatic, or how it escapes from set theory. In other words, what is a collection, if it isn't a set?

Edit: In my effort to figure this out, I came across an explanation that metacategories are models of a particular first-order theory, and categories are a subset of those which conform to set theory. This makes sense, but I'm fuzzy on this area of math and its details. In particular, how does a presentation in terms of symbols and sentences defend a presentation which talks about actual things, like collections and functions?

What MacLane is doing is basically providing a definition of what a category is in any theory of collections.

Here he uses the term collection in a very technical sense as oppesed to set.

Without going to much in the details you can think to collections as to thing for which it makes sense to say that they have elements. Sets are collections which are elements of other collections, these are the collections of theories like ZFC.

As you pointed out this is required to build things like the category of sets. The problem is that if you had only collections that are sets then you wouldn't be able to provide a collection of all sets (which cannot be a set by Russell's paradox), hence you wouldn't be able to provide a category of categories.

If you admit collections that aren't sets (as for instance in NBG) you become able to build a collection of all set (which will be a collection, though not a set).

Hope this helps.

Unfortunately, you are in an area where people contemptuously defend textbook summaries against what is found in original sources with insulting language involving "history" or "philosophy".

What you need to look at is the distinction between intension and extension as the foundations of mathematics evolved in the late nineteenth century and early twentieth century.  I have not personally found a convincing defense of category theory in terms of extensionality.  So, MacLane's discussion of metacategory should be understood in the context of intensions.  The import of extensionality in the foundations of mathematics can be understood by critically considering Frege's distinction between "a concept" and "the extension of a concept".  Concepts are intensions.  The extension of a concept corresponds with the collection of individuals with respect to which a concept can be instantiated as being true.  And, one ought not overlook the significant role of the definite article "the" in the expression "the extension of a concept".

In Frege's writings there is an intense criticism of "intensional logicians" because intensions are not grounded by a semantic conception of truth.  Today, this difference would be characterized by the contrast between a correspondence theory of truth and a coherence theory of truth.  To say that mathematics is extensional is to say that its semantic theory is based upon a correspondence theory.

This is why one speaks of Tarski's semantic theory as "materially adequate" or the classical connective for implication as the "material conditional".  Tarski's theory treats terms as denoting objects, and, truth tables as functions have a domain over the "truth objects" introduced by Frege.  What is "material" here corresponds to the fact that what is presumably denoted are extant individuals.

Generally speaking, functions and collections are not viewed as primitive from an extensional point of view (with regard to "truth") because they are not individuals.  This originates with Aristotle who called individuals "primary substance" and the collections of individuals making up "classes" as "secondary substance".  The Aristotelian system did not treat of collections as objects.  So, one must make an adjustment for what happened when Russell introduced his theory of types.  Individuals continue to ground the theory, but, instead of just one notion of "secondary substance" related to partitions of a universe of objects, one must invoke a hierarchy of types based upon the predicate "is a set of".

The distinction between intension and extension motivating Frege's work arises with Leibniz.  He describes his own view of logic as inverting the order relation of the Aristotelian term logic semantics.  This is the origin of the identity of indiscernibles which has been rejected as a logical principle by the modern philosophy community (in spite of generating contradictions when combined with the substitutivity interpretation of identity in the first-order calclus).  The Aristotelian logic is extensionally grounded.  The Leibnizian logic is intensional.

With regard to your question involving the difference between collections and sets, it is becoming fashionable to invoke mereology as a notion of "aggregate" that may not be the same as a collection.  Michael Potter maintains that the conflation between "membership" and "subset" is a certain sign of mereology and claims that one can find it in Dedekind's work.  On such an account, some of the business of the last 150 years has been to clarify the distinction between "membership" and "subset".  Arguably, the many valued properties of first-order logic (obtaining over individuals comprising set domains) is the result of this hard work.

One of the problems with thinking about "sets" in terms of "collections" is that there are many in the philosophy community who think about "sets" in terms of "extensions" (by which I refer to comprehension axioms).  In my non-standard and uninformed view, Cantor concerned himself with collections and the arithmetization of mathematics.  That is a far cry from extensions of concepts given through comprehension.

I cannot reconcile the arguments of others.  What I can tell you is that if you view sets as collections, the advocates of first-order logic as mathematical logic will reject many conclusions because  collections are  fundamentally intensional and second-order logic does not have the nice properties of first-order logic.  What one has are two views of set theory which are not clearly delineated in the introductory literature.  One has the first-order interpretation corresponding with Skolem's criticisms of Zermelo.  Then one has the iterative conception of set which corresponds more closely with a view presented by Zermelo after those criticisms.

And then, of course, there is the category-theoretic work of Lawvere who attempts to justify his positions on the basis that Cantor's views had never been properly represented in Zermelo's axiomatization.  Since Cantor developed his ideas as a contemporary of Dedekind, a number of his positions may also be tainted with the mereological conflation discussed by Potter.  So, Lawvere's justification may have support.  But, it will very likely lie with intensional mathematics.

I wish you all the best with finding an answer to your question.  I certainly have not yet answered it for myself.  But, before you let anyone convince you of a first-order theory of metacategories, let me suggest that you read the paper on natural equivalences by Eilenberg and MacLane that led to category theory.  It is far better to trust yourself reading original sources than it is to trust textbook authors.  And, you should try to understand what is being represented before accepting that a purported representation is faithful simply because it is "symbolic".


How to prove that $\sum_{n}^{\infty}u_{n}^{2}\left(u_{1}+\cdots+u_{n}\right)^{-1}<\infty$ under these conditions? Suppose that $u_n$ is a decreasing sequence of positive numbers that converges to zero. Suppose moreover that $S_n = \sum_{k=1}^n u_k$ diverges. I would like to prove that the sum $\sum u_k^2 / S_k$ is finite, which I believe is true. Indeed, sum of the type $\sum u_k / S_k^{1+\epsilon}$ are finite, but I have not been able to exploit that. Comparison with an integral of the type $\int (f')^2(x)/f(x) \, dx$ has not revealed particularly helpful to gain any intuition. Have also tried to prove that this is a Cauchy sequence, but I always end up missing something.

Thank you.

Consider a sequence that really slowly tends to $0$. For example

$$u_n = \frac{1}{\sqrt{\log (n+1)}}.$$

Then you have

$$\frac{n}{\sqrt{\log (n+1)}} \leqslant S_n \leqslant \frac{n}{\sqrt{\log 2}},$$

and

$$\frac{\sqrt{\log 2}}{n\log (n+1)} \leqslant \frac{u_n^2}{S_n} \leqslant \frac{1}{n\sqrt{\log (n+1)}},$$

so in that case

$$\sum_{n=1}^\infty \frac{u_n^2}{S_n} = +\infty.$$

Generally, you have $S_n \leqslant n u_1$, and hence

$$\frac{u_n^2}{S_n} \geqslant \frac{u_n^2}{nu_1},$$

so it is a necessary condition that $\sum \frac{u_n^2}{n} < \infty$. That condition is however not sufficient, as $u_n = (\log (n+1))^{-t}$ for $\frac{1}{2} < t \leqslant 1$ shows, in that case $S_n \sim \frac{n}{(\log (n+1))^t}$ and $\frac{u_n^2}{S_n} \sim \frac{1}{n(\log(n+1))^t}$. On the other hand, we have $S_n \geqslant n u_n$, so the condition $\sum \frac{u_n}{n} < \infty$ is sufficient.

Hint: use that $S_n \geq nu_n$ for every $n \in \mathbb{N}$.
We then have $\frac{u_n^2}{S_n} \leq \frac{u_n}{n}$

If we call $U_n \overset{\Delta}{=} \sum_{i=1}^n \frac{u_k^2}{S_k}$, what can you say about the general term of the sum?


What are examples of local rings with a nontrivial prime ideal? I've just learned about the idea of local ring, but the only concrete examples in the book are $\{p/q \in \mathbb{Q} \mid q \mbox{ odd}\}$ and the power series ring $k[[x]]$ for some field $k$. But the only prime ideals they have are just $\{0\}$ and their maximal ideal.

Can anybody provide me with more concrete examples of local rings with "non-trivial" prime ideals?

First consider the polynomial ring $R=F[x,y]$ for a field $F$. One chain of prime ideals in this ring is $(0)\subseteq (x)\subseteq (x,y)$. Now $(x,y)$ is a maximal ideal, but there are other maximal ideals, for example $(x+1,y)$.

The easiest way to eliminate the other maximal ideals is to pass to the localization at the prime $M=(x,y)$ so that the new ring $R_M$ is a local ring. It is a property of localization that the prime ideals contained in $M$ will have prime counterparts in $R_M$, with the same containment relations and everything. Thus the chain $(0)_M\subseteq (x)_M\subseteq (x,y)_M$ will be a properly ascending chain of prime ideals in the new ring $R_M$.

Even if you are not handy with localization now, you probably will need to be soon, so trying to understand this type of example is worthwhile.

You will soon learn about localization and that will provide many many examples.

An example: Let $R$ be the ring of rational functions $f/g\in\mathbb Q(x,y)$ such that, when expressed in irreducible terms, the denominator is not divisible by $x$.


When the roulette has hit 5 reds why shouldn't I bet to black? First some context, I'm not a mathematician, not even close (as you will soon see) I do grasp some things about it but in a need to know basis, so plain english answers are appreciated (too).

I can't figure out this one even though some have tried very hard to explain/convince me.

So let's say I'm on a roulette (let's leave out the zero, or make it a coin for that matter).

The roulette hits 5 (or any amount of) times red on a row.

My question is why should't I bet to black (or better said why it is the same the one I pick).

I DO UNDERSTAND THE BASIC PRINCIPLE, that there is a material reality which makes the ball randomly fall on any of the two colors (which gives the options a 50-50 chance).

But, I also understand that if you "say": I bet red will come out six times on a row, you do have a very low probability of that happening so:

How is it that if you have seen the ball fall 5 times on a row on red, and you bet on red, you are NOT betting on 6 times on a row on red (that does have lower probability).

Thanks in advance!

I hope I was clear enough with my question, if not, please ask for any clarifications needed.

Sorry for the VERY plain English, feel free to modify or suggest a change to anything that may be misleading.

Lots of people have trouble with this.  If you bet before any spins on six in a row, the chance of winning is 1/2^6=1/64.  But having seen the five reds, as you say, the chances are 1/2 each on red and black for the next spin.  It is also true for any specific series of six spins, at the start the chance of that series is 1/64.  If you were to bet on RBBRRB, that is also 1/64.  Now that you have seen five reds, there are only two series of six that are possible:  RRRRRR and RRRRRB.  They each started out 1/64 and are now each 1/2.

Hope this helps

Why wouldn't a series have an increasingly lower probability of extending than a single event has of happening once?

Obviously, each unique spin has a probability of 50% (academically, by excluding the greens) for either color.

However, the overall proportion of total reds vs blacks should be 50%.  So every time another of the same color is added to the group, the probability of the group/sample containing ONLY that one color decreases.  Is there really a 50% probability that a sample size of 100 (e.g.) will contain 100 reds or blacks?  When a person gambles on the roulette spin, he may be betting FOR a single spin, or AGAINST a series.

Isn't saying that the wheel has no memory like saying the wheel hasn't had a haircut in (x) weeks?  A gambler isn't betting on the wheel, he's betting on the statistics.  Past data would appear to be relevant due to the sampling.

The trick is to be in the right place at the right time: at what point will the group or sampling attempt to correct itself to ensure an equal distribution of reds and blacks?  Because obviously it is inevitable.


Basic algebra. Please help. When $1200$ is divided by $N$ and $N+x$, the remainder in both cases is $35$.

What is the maximum value of x?

What can we say about $1200-35=1165$ in relation to $N$ and $N+x$? Particularly with regards to divisibility.

Hint 2: Look at all factors of $1165$, see what you can say.

I will use Haskell to brute-force this problem. Here's an extremely brief GHCi session:

Prelude> [ n | n <- [1..1200], 1200 `mod` n == 35]
[233,1165]

Because $1200 = (233 \times 5) + 35 = 1165 + 35$. Since $1165-233 = 932$, we have $x = 932$. Alternatively, we could have used a sexier list comprehension, as follows:

Prelude> [ x | n <- [1..1200], x <- [1..(1200-n)], 1200 `mod` n == 35, 1200 `mod` (n+x) == 35]
[932]

The interpretation of these "experimental" results is left to the reader.