How many degrees of freedom does a line have on a plane? Using $y=ax+b$ I can get a line for every point $(a,b)$ but there are still some lines left $(x=c)$. So it seems that lines are more than points so they have more degrees of freedom than $2$. How many?

You have indeed two degrees of freedom, but the problem is that the correspondence is not "perfect" with points of the plane $\mathbb{R}^2$. It is nicer with points in $\mathbb{R}^{+}\times S^1$, where $S^1$ is the circle and $\mathbb{R}^+$. is the ray of non-negative real numbers.

Let us see what this means, but before let me point out that even though the correspondence in your post doesn't cover one of the lines in the plane, it is ok, to determine degrees of freedom to allow considering correspondences that skip a set of states of the system that is small (has smaller dimension). So, your analysis also shows that there are two degrees of freedom, even though the single line $x=c$ is skipped.

By $S^1$ above we are denoting the circle. For each line on the plane we have its angle with respect to the positive part of the $X$-axis. We only need to consider angles between $0$ and $\pi$ and consider $0$ the same angle as $\pi$. Take into account that when you rotate more than and angle larger than $\pi$ you are getting an slope that can also be obtained with an angle smaller than $\pi$. Each of these angles corresponds to a point in the circle.

Having a line we also have the distance to the line from the origin. This is a non-negative real number.

The distance to the origin and the angle determines the line and conversely each line determines the distance to the origin and the angle.

We can put all of these in formulas. A general line in the plane can be written as
$$ax+by=c$$

The vertical lines fall in the case that $b=0$. This equation is unique up to multiplication by a non-zero constant. For example, multiplying by $r\neq0$ we get an equation $$rax+rby=rc$$ that defines the same line.

There is a unique angle $0\leq \theta<\pi$ such that $b\cdot\tan(\theta)=-a$, where we assume $\theta=\pi/2$ when $b=0$. This is the angle we discussed above.

The distance from the origin to this line is $$d=\frac{|c|}{\sqrt{a^2+b^2}}$$

We can go the other way around. If we have $\theta$ then we can find a pair $a,b$ such that $b\tan(\theta)=-a$. This pair is not unique, but it is up to multiplication by a constant $r\neq0$, which is all we need. One you have an $a,b$ pair you can use the given $d$ and solve for $$c=d\cdot\sqrt{a^2+b^2}$$

We removed the absolute value from $c$ because we are allowed to multiply the whole equation $ax+by=c$ by $-1$ if we want, and still get the same line. We can always assume we want $c\geq0$.

A segment of any line can be scaled, translated and rotated on a plane so there are three DOF. The last two are Euclidean motions that are constants of integration of differential equations of motion in the plane.

EDIT1:

Taking a view that whatever is added on in integration can be viewed as a degree of freedom (DOF).

In the plane, $ \dfrac{dy}{dx}=m $ gives one DOF. Integrating $ y = mx + C $ adds one more DOF, now $ (m,C) $. It is same if we use polar coordinates $ x \cos \alpha + y \sin \alpha = p $, two constants or  parameters $( p, \alpha) $ .

For a straight line segment when limits are included it may be viewed as a DOF in the sense of dilation.


Local minimum and maximum of the function Can anyone help me to solve the following question?
maximize and minimize the function $(10-x)(10-\sqrt{9^2-x^2})$ over $x\in[0,10]$
This is a high school question, so is there any simple trick help solve it? Thanks!

We give a short argument for both maximum and minimum.

To bring out the symmetry, let $y=\sqrt{81-x^2}$.
We study the behaviour of $(10-x)(10-y)$, that is, of
$$xy-10(x+y)+100.\qquad\qquad\text{(Expression $1$)}$$

We will find the maximum and minimum values of Expression $1$, given that
$x^2+y^2=81$ and $x \ge 0$, $y \ge 0$.

Let $w=x+y$ and $xy=v$.
We are interested in the behaviour of
$$v-10w+100. \qquad\qquad\text{(Expression $2$)}$$

But since $(x+y)^2-2xy=x^2+y^2$, we have $w^2-2v=81$.
So we are interested in the behaviour of
$$\frac{w^2-81}{2} -10w +100, \quad\text{that is, of}$$
$$\frac{1}{2}\left(w^2-20w+119\right).  $$

Complete the square.  We get
$$\frac{1}{2}\left((w-10)^2+19\right).\qquad\qquad\text{(Expression $3$)}$$

Now it's over. There is a local minimum at $w=10$.
The minimum value is $19/2$.

The maximum is reached where $w$, that is, $x+y$, reaches a maximum subject to $x^2+y^2=81$.  Since $(x+y)^2+(x-y)^2=2(x^2+y^2)=162$, the maximum value of $(x+y)^2$, and hence of $x+y$, occurs where $x=y$.

We can if we wish find the values of $x$ at which the minimum is reached.  We need to solve the system $x+y=10$, $x^2+y^2=81$. That gives $2xy=10^2-81=19$, so $(x-y)^2=81-19=62$, and therefore $x-y =\pm \sqrt{62}$, and now we can find $x$ and $y$.

Note on Symmetry:  All the way through, we have preserved symmetry between $x$ and $y$. We introduced symmetry in the initial setup, and every step involved only symmetric functions of $x$ and $y$.  Symmetry allowed the easy identification of $x+y$ as a key parameter.

The formal algebraic symmetry comes, in this case, from the underlying geometry. For the problem posed by the OP is fundamentally geometric. It has to do with the interaction between a circle and a rectangular hyperbola.

Generally, one can note the following:

There are single, double and triple roots.
For a simple root of the graph intersects the x-axis Example x1=0
With a double root touch it and  indicating an extreme point. Example x1=0,x2=0
Tripple a root point is a saddle point Example x1=0,x2=0,x3=0.

if your equatiion:

$f(x)=(10-x)\cdot (10-\sqrt{81-x^{2}})$

$f(x)=0$
$(10-x)=0 $

if $ x=10$

$(10-\sqrt{81-x^{2}})=0$

$10 = \sqrt{81-x^{2}}$ | ^2

$100 = 81-x^{2}$

$-19 = x^{2}$

$\sqrt{-19}=x$ ERROR ! you will get complex roots... not useable for Maxima and Minima

root = 10

derivation of $f(x) = y$ is

$y' = - \frac{(-10 + x) x)}{ \sqrt{81 - x^2}} + \sqrt{81 - x^2} -10$

To get the extrema you had to $y'=0$
Xe = root(s) of $y'=0$

and put the roots (Xe) if any there in $y''(Xe)$
$y''(Xe) < 0 $ it's a Maximum
$y''(Xe) > 0 $ it's a Minimum
$y''(Xe) = 0 $ Xe is not an extrema


Graph Theory Cycles in Nonseparable graphs Let $k$ be the maximum length of a cycle in a nonseparable graph $G$. Prove that if $C$ and $C'$ are any two $k$-cycles in $G$, then $C$ and $C'$ have at least two vertices in common.

Nonseparable meaning biconnected. Does anyone have a hint?

Theorem. If the graph G has a vertex v that is connected to a vertex of the component G1 of G, then v is also a vertex of G1.

Suppose there exists one and only edge connecting C to C'. Removing this edge would yield two connected components.

Suppose there exists two edges connecting C to C'. Removing one edge would yield one connected component. C and C' have at least two vertices in common.


How do I simplify $p \lor ((p \lor ¬q) \land (q \lor\neg r)) \lor \neg q \lor r$ using logical rules? I'm trying to use the De Morgan and other rules of simplification but I'm going nowhere. Is this supposed to be hard or am I missing something?

First, use commutivity and association:

$${\quad p\vee\big((p\vee \neg q)\wedge(q\vee \neg r)\big)\vee\neg q\vee r\\=(p\vee\neg q\vee r)\vee\big((p\vee\neg q)\wedge(q\vee\neg r)\big)\\=}$$

Next use distribution of $\vee$ over $\wedge$.

Then it shall become clear what to do to polish it off.

Using Boolean algebra,

$$\begin{align} p + (p + \bar q) \cdot (q + \bar r) + \bar q + r &= p + p q + p \bar r + \overbrace{\bar q q}^{=0}  + \bar q \bar r + \bar q + r\\\\ &=p \underbrace{(1 + q + \bar r)}_{=1} + 0 + \bar q \underbrace{(\bar r + 1)}_{=1} + r = \color{blue}{p + \bar q + r}\end{align}$$


Is there a notation for defining if a function is recursively defined? Background: I'm working on a paper on high level synthesis, a method which combines the use of programming with digital circuit design. One of the methods I want to explain is called loop unrolling which makes use of concurrency to speed up execution, by duplicating the processing elements.

Unrolling can only be done on any function iff there is no internal relation between iterations, an example of a function with this property would be vector multiplication. My best guess would be something along the lines of $ g \triangleq  \{f \in ?? | \forall n \in \mathbb{N}: f[N] does\ not\ contain\ a\ term\ for\ f[N-n]\}$.

Is there a standard or good notation to describe the set of functions which do not contain a recursive term?

Edit: I originally accepted the answer as I could see that there was a slight mismatch between the question as a whole and the formulation of the question in the final paragraph. The previously accepted answer did answer the question as written, although not as intended. I'll clear it up a bit more, apologies for the mess.

As far as I know, there is no specific notation.

You might write something like

$$f[N]:=\alpha(\{a,b,c,\cdots\})=\alpha(V)$$

where $\alpha$ denotes the expression (or more generally the algorithm) that evaluates the value to be assigned to $f[N]$, and $V=\{a,b,c,\cdots\}$ is the set of variables that appear explicitly in $\alpha$.

Then the absence of a dependency is expressed by

$$f[N-n]\notin V.$$

On second thoughts, though your notation seems to imply an array, this is unlikely as you say $\forall n$. But then, saying that the value of $f_N$ (at iteration $N$) depends on the value $f_{N-n}$ (at iteration $N-n$) only makes sense if the process has enough memory (variables) to realize such a dependency. This is a much more complicated situation.

An example would be a running sum on, say, three elements,

i= 0
a= b= c= s= 0
while True:
    a= b; b= c; c= K[i]
    i= i + 1
    s= s - a + c

Edit

Since I am unable to delete this answer because it has been accepted, I'm going to say that there appears to be no standard way to write what you want, as all functions of the natural numbers (and likely other sets of numbers) can be written using a recursive defintion. You might as well just write it out in English or define the class of functions that can be calculated without depending on previous terms in the sequence in the paper.


How many elements of order 7 in a simple group of order 168? I'm not sure my proof is correct because I don't use the fact that the group is simple anywhere...

Since $168=2^3 \cdot 3 \cdot 7$, there exists a subgroup of order 7. Let $n_7$ be the number of Sylow 7-subgroups. Since $n_7 | 24$ and $n_7 \equiv 1$ mod $7$, $n_7 = 8$. Since every group of prime order is cyclic, there exist 8 elements of order 7.

I think that I still need to prove that there are no other elements of order 7 and that's where the fact that the group is simple would come in.

The $8$ Sylow subgroups of order $7$ each contain $6$ elements of order $7$ together with the identity. Since an element of order $7$ generates a group of order $7$ these elements are all distinct. There are $8 \times 6=48$ elements of order $7$.

You use the fact that the group is simple in asserting that there are $8$ subgroups of order $7$. It is always possible to have $1$ subgroup of a given order (so long as it is a factor of the group order e.g. the cyclic group of order $168$) - but a single subgroup of order $7$ would be normal, and the group would not be simple.

168=8⋅3⋅7
the elements of sylow 7-groups of grop of order 168
the order of 7 subgroups is 7. number of elements between 0-168 which is 1 mod 7 are 2,8,15,22,29...
since 1 and 8 are the only positive integers congruent to 1 mod 7 and divides 168, the number of sylow 7-subgroups is 8.
then there are 8x6=48 elements of order 7 since identity is of order 1 and common in all the 8 sylow 7-subgroups


All of the dihedral groups are factor groups of the infinite dihedral group. Show that $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1 \}$ and write $\alpha : \mathbb Z_2 \rightarrow \operatorname{Aut}(\Bbb{Z})$ for the nontrivial homomorphism. The semidirect product $\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2}$ is called the infinite dihedral group. Show that the dihedral groups $D_{2n}$ are all factor groups of the infinite dihedral group.

We know that $\Bbb{Z}$ only has two generators, $1$ and $-1$. So there are only two possibilities: sending $1$ to $1$ and sending $1$ to $-1$. So $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1\}$. There is an example in our textbook that shows that $\Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2 \cong D_{2n}$. So I only need to show that $$(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z}_2)/\langle n \rangle \cong \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2$$ right?

Suppose that $\Bbb{Z} = \langle b \rangle$ and $\Bbb{Z}_2 = \langle a \rangle$. We can see that $(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2})/\langle n \rangle = \{b^ia^j \text{mod n} : b^i \in \Bbb{Z}, a^j \in \Bbb{Z}_2\} = \{b^ia^j : b^i \in \Bbb{Z}_n, a^j \in \Bbb{Z}_2\} = \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z_2}$.

Is my answer correct? I'm just curious because it seems too short.

Thank you in advance

Your answer has a couple of issues as discussed in the comments:

You are using $n$ to denote the pair $(n,0)$ (recall that the elements of $\mathbb Z\rtimes \mathbb Z_2$ are defined as pairs).
More importantly, you are basically arguing that $\mathbb Z\rtimes \mathbb Z_2/\langle n\rangle$ and $\mathbb Z_n\rtimes \mathbb Z_2$ are equal as sets, which isn't really true or relevant; you want to show that they are isomorphic as groups.

You are correct however in your argument that $\mathbb Z$ has only two automorphisms, and your intuition that $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ is also right. To prove this, we consider the map $f:\mathbb Z\rtimes \mathbb Z_2\to \mathbb Z_n\rtimes \mathbb Z_2$ defined by $f((a,b))=([a]_n,b)$. Clearly this is a surjective map with kernel $\langle (n,0)\rangle$. We have
$$\begin{align}
f((a,b)(c,d)) &=f((a+(-1)^bc,b+d))\\
&=([a+(-1)^bc]_n,b+d)\\
&=([a]_n+(-1)^b[c]_n,b+d)\\
&=([a]_n,b)+([c]_n,d)\\
&=f((a,b))f((c,d))
\end{align}$$
thus $f$ is a homomorphism, so by the First Isomorphism Theorem we have $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ as desired.

The way to show this, is to note that any of the dihedral groups have a cyclic condition.  When this condition is removed, the elements are infinite.

In essence, the cyclic groups can be written with a $mod(Z)$ condition, while the infinite group is just $z$.  Thus, every cyclic group becomes the infinite group with a cyclic remainder.


Finite abelian group generated by elements of maximal order I'm trying to prove that a finite abelian group $G$ is generated by elements of maximal order.

I can sort of see why that happens in a vague instinctive kind of way but no real hard logic. So far, I have tried to use this lemma:

Let $G$ be a finite abelian group and a be an element of maximal order in $G$. Then any element $b$ is such that $|b|$ divides $|a|$.

Then each cyclic group generated by the maximal order element contains exactly 1 element of each possible order for an element in the group. I have tried expanding from this idea in a few directions but I don't think it's the correct way to go since it's not going anywhere.

Any help appreciated. Thanks guys.

Let $G$ be a finite abelian group, then $G = \mathbb{Z}_{n_1}\oplus\cdots\oplus \mathbb{Z}_{n_k}$ for some $n_1|\cdots|n_k$. Then clearly these elements
$$\{(1,\cdots,1),(0,1,\cdots,1),(0,0,1,\cdots,1),\cdots,(0,\cdots,0,1)\}$$
are of maximal order $n_k$ and they generate $G$.

let a be an element of a prime order such that is not belong to the subgroup of G generated by all maximal elements then ab is not belong to the set of the generators of maximal elemens for some b of maximal elements.


Is $\frac {\sin^2(x)}{x^2}$ Lebesgue integrable on $[0,\infty)$? How would someone show $$x\mapsto\frac{\sin^2(x)}{x^2}$$ is Lebesgue integrable on $[0, \infty)$? Also that

$$\int_{[0,\infty)} \frac {\sin^2(x)}{x^2} d\lambda = \int_{[0,\infty)} \frac {\sin(x)}{x} d \lambda$$

My thoughts so far:

$$\frac{\sin^2(x)}{x^2} \leq \frac{1}{x^2}$$
Since $1/x^2$ is Lebesgue integrable, then $\sin^2x/x^2$ must be as well.

But I don't know how to proceed with showing the other part.

Since
$$
\frac{\sin x}{x} \to 1\quad\text{as $x\to 0$},
$$
there is some $\delta > 0$ and some $M > 0$ such that
$$
\bigg|\frac{\sin^2 x}{x^2}\bigg|< M\quad \text{if $0\le x < \delta$}.
$$
Since everything in sight is non-negative, by additivity,
$$
\int_0^\infty\frac{\sin^2 x}{x^2}\,dx = \int_0^\delta\frac{\sin^2 x}{x^2}\,dx + \int_\delta^\infty\frac{\sin^2 x}{x^2}\,dx,
$$
and the first integral is convergent because it is bounded above by $M\delta$. The second integrand is bounded above by $x\mapsto \frac{1}{x^2}$, which belongs to $L^1\big([\delta,\infty)\big)$, by the $p$-test if you like. Hence $x\mapsto \frac{\sin^2 x}{x^2}$ is Lebesgue integrable on $[0,\infty)$.

Show that $f(x)=\frac{\sin^2\!x}{x^2}$ is continuous in $[0,\infty)$ and you are done.


How to find $\lim_{(x,y)\rightarrow(0,0)} \frac{x^2\sin^2(y)}{x^2+3y^2}$ Find the following $$\lim_{(x,y)\rightarrow(0,0)}  \frac{x^2\sin^2(y)}{x^2+3y^2}$$

How to solve this limit? I have tried to use polar system, like sub everything with $x=rcos\theta$ and $y=\sin\theta$, but r cancel each other.

The way it is written, $\frac{x^2 sin(y)}{x^2+ 3x^2}= \frac{x^2 sin(y)}{4x^2}$ the $x^2$ terms cancel, leaving $\frac{sin(y)}{4}$ which goes to 0 as y goes to 0.
  Did you mean $\frac{x^2 sin(y)}{x^2+ 3y^2}$?  In that case, replacing x by $r cos(\theta)$ and y by $r sin(\theta)$ gives $\frac{r^2 cos^2(\theta)sin(r sin(\theta)}{r^2 cos^2(\theta)+ 3r^2 sin^2(\theta)}$.  You are correct that the $r^2$ terms cancel, leaving $\frac{cos^2(\theta)sin(r sin(\theta)}{cos^2(\theta)+ 3sin^2(\theta)}$.  Since that depends on "$\theta$", there is no limit.


number of terms in the expansion containing powers of $x$ How do i find the number of terms containing powers of $x$ in the expansion of: $$(1+x)^{100}(1+x^2-x)^{101}$$

I tried using $(1+x)((1+x(1+(x)^2-x))^{100})$ which simplified into : $$(1+x)(1+x^3)^{100}$$
but i'm not sure if this is the correct approach and also what do i do to simplify it further to get the answer?

It's a multiple choice question with options A:202, B:302, C:301 and D:101
please explain the method to solve questions of this type :)

There are 101 terms in the first factor, of which 100 have a power of $x$ greater than 0. The second factor can be written:

$$
\sum_{k_1 + k_2 + k_3 = 101} 1^{k_1} \cdot x^{2 k_2} \cdot x^{k_3}
$$

(just imagine the product multiplied out) so you are asking how many sets of (k_1, k_2, k_3), all integers at least 0, are such that $k_1 + k_2 + k_3 = 101$ with $k_1 < 101$ (that one gives the term 1, there is just one of those).

Let's find out how many solutions $k_1 + k_2 + k_3 = 101$ has, this is like chopping a line of 101 $*$ into three pieces, say by separating with $|$ (this is called a stars and bars argument, for obvious reasons). But then you have a total of $101 + 2$ positions to be filled with 101 stars and 2 bars, that can be done in $\binom{101 + 2}{2} = 5253$ ways, of which you subtract 1 for $k_1 = 101$.

Combining your factors, you have 101 terms in the first factor, $5252$ with $x$ from the second, and the terms with $x$ in the product are the result of multiplying any term from the first factor with a term containing $x$ from the second, i.e., $101 \cdot 5252 = 53042$ (as long as no simpolifications happen).

A lower bound is that the result is a polynomial of degree $100 + 2 \cdot 101 = 302$, so there are at most $301$ terms with powers of $x$. But there are negative terms, so cancellation can/will happen, and you get less.

there are no 301 terms present. The 1st person and 2nd gave the same answer. But your analogies are wrong anyway. Aand your simplication is perfect Mr.Krazkat. Actually you have got (1+x)(1+x^3)^100 right. The binomial expansion gives 101 terms for (1+x^3)^100 and these 101 terms are multiplied by 1 and x and you get a result like this- (1+X)('101 terms'). So, after multiplying 101 terms with '1', you get 101 terms and after multiplying with x you will get another 101 terms like this- (1+x)('101 terms')=1('101 terms')+x('101 terms'). So there will be no cancellations because all the powers of x will be different. So there are total 101+101=202 terms.


Why does simplifying a function give it another limit I'm asked:

$$\lim_{x\to 1} \frac{x^3 - 1}{x^2 + 2x -3}$$

This does obviously not evaluate since the denominator equals $0$. The solution is to:

$$\lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+3)}$$
$$\lim_{x\to 1} \frac{x^2 + x + 1}{x + 3}$$
$$\frac{1+1+1}{1+3} = \frac{3}{4}$$

My question: what is actually happening? How can simplifying a function give it another limit? Is it a complete other function and if so why would it be relevant to our original question?

Perhaps you (and I think many beginners in calculus) have the notion that limit of a function is evaluated by plugging the value of the variable. Thus for the function $$f(x) = \frac{x^{3} - 1}{x^{2} + 2x - 3}\tag{1}$$ you say that its limit as $x \to 1$ can not be evaluated by plugging $x = 1$ because denominator vanishes.

It is very important to understand that limit of a function at a point is something completely different from the value of the function at that point. A limit of a function at a point has nothing to do with the value of the function at that point, but it has everything to do with the values of the function near that point.

However, there is a catch!! There are many functions seen commonly in calculus for which the limit at a point turns out to be the same as their value at that point and hence for such functions it is possible to evaluate the limit just by plugging. I think it is this behavior of some functions combined with the fact I described in last paragraph which creates utter confusion in the mind of a beginner. At one stage I mentioned that limit is not same as value of a function and next I mention that for some functions the limit is same as its value. Really confusing!

The only way to sort out this confusion is to learn to identify at least some basic types of functions which have this nice property that their limit at a point is same as their value at that point. Such functions are called continuous functions. Using a series of theorems on limits it can be proved in a step by step manner that any function made up of algebraic (which includes polynomials and rational functions), trigonometric (direct and inverse), logarithmic and exponential functions using a finite number of arithmetic operations and compositions is continuous wherever it is defined. The type of function described in last sentence is called an elementary function.

The function $f(x)$ given by equation $(1)$ is an elementary function and we need to calculate its limit as $x \to 1$. From what we have mentioned in last paragraph it is clear that we can evaluate its limit simply by plugging provided it is defined at the point under consideration. The trouble is that $f(x)$ is not defined at $x = 1$. Then we use the fact mentioned in the beginning that $\lim_{x \to 1}f(x)$ has nothing to do with its value at $x = 1$. The limit operation $\lim_{x \to 1}$ ensures that $x \neq 1$ and we can now use any sort of transformation on $f(x)$ under the assumption that $x \neq 1$ and try to simplify it in the form of an elementary function which is perhaps defined at $x = 1$.

Here we are lucky and by cancelling the factor $x - 1$ from numerator and denominator we reach another function $$g(x) = \frac{x^{2} + x + 1}{x + 3}\tag{2}$$ Note that both $f$ and $g$ are different functions ($f$ is not defined at $x = 1$ whereas $g$ is defined there), but $f(x) = g(x)$ as long as $x \neq 1$. Hence as far the limit operation $\lim_{x \to 1}$ is concerned both $f(x), g(x)$ have same behavior. And now we see that $g(x)$ is also an elementary function and it is defined at $x = 1$ and hence its limit as $x \to 1$ is same as its value $g(1) = 3/4$. Thus limit of $f(x)$ as $x \to 1$ is also $3/4$.

The denominator is not equal to 0, it tends to 0. And it gets closer to 0 at diferent speed than the numerator.

By simplifying you are getting the relation beween both so you can tell where the division tends.


How to calculate probability of winning bowl pool? I currently have a spreadsheet that keeps track of players and their respective picks as to who they think will win each college bowl game. At the beginning, before any game is played, each player has equal probability of winning the whole thing. What I'm looking for is an expression that will give me their updated probability after each bowl game is played.

In the example below, what is the probably (in percentage) of Braden winning the whole thing, given his score (5 of 8), everyone else's scores, and games played vs total number of games?

I wouldn't try to do this analytically, since such an analysis would probably have to make assumptions about the games that don't hold water (e.g., that the result of each game is $50$-$50$).

Instead, I would approach it with a Monte Carlo simulation.  As best you can, estimate the probability of each game's result.  One possibility is to use the Vegas odds.  This will take some care, since the odds will include the vigorish (the house percentage), so this has to be scaled down.

For instance, if the odds are given as money lines: Say the bet on Team A is $+170$ and the bet on Team B is $-200$.  That means that a bet of $\$100$ on Team A wins $\$170$ if they win, while a bet of $\$200$ on Team B wins $\$100$ if they win.

Converted into odds, those bets are $17$ to $10$ and $10$ to $20$ against, respectively.  Converted, in turn, into probabilities, that's $10/27$ and $2/3$.  Obviously, those add up to more than $1$, and the excess of $1/27$ reflects the vigorish.  These can be corrected in different ways, but the easiest is to scale them down to $10/28$ and $18/28$.

Once you've done that for each of the $33$ remaining games, run a Monte Carlo on them, and tally each person's score for every run.  A million or so runs should be enough to figure out winning probabilities to three significant digits.

I think a coin flip approach could be used. What are the odds of any given player to guess the most coin flips (game winners) correctly? Variables would be 1) number of guesses they currently have correct, 2) number of guesses other players have correct, 3) number of differences in the remaining picks they have with every other player, then 4) If every possible permutation could be ran, what is the probability that this player to have more correct picks.

Can anyone set up the equation?

To edit... for THIS scenario, players do not care about odds of winning based on who are the favorites of any given game, but odds of winning if EITHER team can win (i.e., coin flip). The favorites of the game do not matter. Rather, what is the mathematical probability that any one player can possibly win over the opponents.  Can anyone set up THAT equation?


Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:

Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.

Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.

Multiplying the given expressions together:

\begin{align}
\frac{47}{10} &= (a+b+c)\bigg(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\bigg)   \\ \\
&= \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a} \\ \\
&=3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \\ \\
\end{align}

$$\Longrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} = \frac{17}{10}$$

The only possible numbers a, b, c that add up to 6 are 1, 2, 3.

Substituting these in the first equation gives you 47/60 so they are the correct values and there is no repeated number for example a and b both being 1 and c being 4.

So simply substituting in the second equation gives you 17/10 immediately.

The three fractions in each equation are all the available permutations.

There is no need for any other manipulations.


Prove that $f$ has the maximum value on $[-a, a]$. Let $a > 0$.
Let $f(x) := 2 x \sin \frac{1}{x} - \cos \frac{1}{x}$ for $x \in [-a, 0) \cup (0, a]$ and $f(0) := 0$.
Prove that $f$ has the maximum value on $[-a, a]$.
My attempt is the following:
Let $b \in (0, a)$.
Then, $f$ has the maximum value $\max f_{[b, a]}$on $[b, a]$.
Let $-c \in (-a, 0)$.
Then, $f$ has the maximum value $\max f_{[-a, -c]}$ on $[-a, -c]$.
If we can choose $b$ and $-c$ such that if $x \in (-c, b)$, then $f(x) \leq \max\{\max f_{[b, a]}, \max f_{[-a, -c]}\}$, then the proof is done.

Here's an idea for you. In a very small neighborhood of $\;x=0\;$ the function is bounded, since
$\;2x\sin\frac1x\xrightarrow[x\to0]{}0\;$ and $\;\left|\cos\frac1x\right|\le1\;$ , so say $\;|f(x)|\le M_1\;$ and $\;f(x_1)=M_1\;$ for $\;x,\,x_1\in(0,\,\epsilon)\;$ ,  with $\;\epsilon>0\;$ .
Now, at $\;[\epsilon,\,a]\;$, the function $\;f(x)\;$ is continuous and thus, by Weierstrass theorems, it both
is bounded there and attains in that interval its maximum/minimum value, say
$$\;f(x_2)=M_2\ge f(x)\,,\,\,\forall\,x\in[\epsilon,\,a],\,x_2\in[\epsilon,a]\;$$
Finally, just put the above together...and get your proof.

Here is a rough proof.
Fix $a$.
As stated in the comments, we need only consider the interval $[0,a]$.
Let $\alpha = \min \{ \frac{a}{2}, \frac{1}{4\pi} \}.$
We break up the interval into two sets $A=[0,\alpha]$ and $B=[\alpha,a]$.  The set $B$ is compact, and $f(x)$ is continuous and bounded on $B$.  Therefore, $f(x)$ attains a maximum on $B$, call it $f^\star$.  $\forall x  \in A, f^\star \ge f(x),$ so $f^\star$ is the maximum on $[0,a]$ and thus on $[-a,a]$.


Question from Indian olympiad If four real numbers a, b, c, x satisfy $abc\neq 0$ and
$$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}$$
then either a = b = c, or a + b + c = 0

Here is my attempt
------- is the part which can be filled by symmetry

The question is why am I getting the condition which is supposed was not true, also initially I divided by $a-b$ so $a=b$ doesn't make sense now???
I don't want a new solution to this question, I want to know why my solution is giving me weird results

You assumed that $a,$ $b$ and $c$ are different numbers and you got a contradiction.

It ends the proof!

I like the following way:

Let $$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=k.$$
Thus, $$bx-cx=ka-c$$
$$cx-ax=kb-a$$ and
$$ax-bx=kc-b,$$ which gives
$$0=k(a+b+c)-(a+b+c),$$
which gives $$a+b+c=0$$ or $$k=1.$$

For $k=1$ we obtain:
$$(b-c)x=a-c,$$
$$(c-a)x=b-a$$ and
$$(a-b)x=c-b.$$
Now, let $a\neq b$.

Thus, $c\neq a$ and from here $b\neq c$.

Thus, $$\frac{a-c}{b-c}=\frac{b-a}{c-a}=\frac{c-b}{a-b},$$ which gives
$$(a-b)^2=(a-c)(c-b),$$ $$(a-c)^2=(a-b)(b-c)$$ and
$$(b-c)^2=(a-b)(c-a),$$ which gives that $c$ is placed between $a$ and $b$, $b$ is placed between $a$ and $c$ and $a$ is placed between $b$ and $c$, which is impossible and we got a contradiction.

Id est, $a=b$ and we obtain: $$a=b=c.$$


complex numbers equation with 3rd power I have problem with this equation. Can someone help mi with it?
\begin{align*} z^3+3z^2+3z &= i - 1 \end{align*}
That's what I have:
\begin{align*} (z+1)^3 = i \end{align*}

If ${(z+1)^{3}=i}$, what we essentially want to do is find the cube roots of ${i}$. This is much easier than it sounds. Firstly, we represent ${i}$ in polar form. This is simply
$${e^{i\left(\frac{\pi}{2} + 2n\pi\right)}}$$
This can be seen with an argand diagram (it's just an application of Euler's formula):
$${=cos\left(\frac{\pi}{2}+2n\pi\right)+i\sin\left(\frac{\pi}{2}+2n\pi\right)=i}$$
Now raising this to the one third power gives
$${\Rightarrow z+1=e^{\frac{i}{3}\left(\frac{\pi}{2}+2n\pi\right)}}$$
Simply plugging in ${n=\{0,1,2\}}$ will give the three roots we need,
${z+1=e^{\frac{i\pi}{6}}, e^{\frac{5i\pi}{6}},e^{\frac{3\pi i}{2}}}$
(plugging these again into Euler's formula and subtracting $1$ will give you your answers for $z$).

Yes, that is correct. And, since $i=e^{\pi i/2}$, $z+1$ is one of these three numbers:

$e^{\pi i/6}\left(=\frac{\sqrt3}2+\frac i2\right)$,
$e^{5\pi i/6}\left(=-\frac{\sqrt3}2+\frac i2\right)$,
$e^{9\pi i/6}(=e^{3\pi i/2}=-i)$.


The meaning of differentiation of $x$ with respect to $y$ The physical meaning of the differentiation of $x$ with respect to $y$ is the rate of change of $x$ with respect to $y$. But, I am finding it difficult in understanding the geometrical interpretation of the expression $\dfrac{\mathrm{d}x}{\mathrm{d}y}$. On a given curve on any given points, what does the expression actually tells us?

I currently know that differentiating the equation of a curve gives the tangent of that curve. But how can we understand that tangent line in terms of $y$ and $x$?

Differentiating $x$ with respect to $y$ gives you the gradient of the tangent at the point $\left(x, y(x)\right)$ on the curve $y(x)$. The gradient of the tangent at the point $\left(x, y(x)\right)$ indicates the rate of change of $y(x)$ at that specific point. Because on a curve, the gradient varies, differentiating is basically taking the limit of the $\dfrac{\Delta y}{\Delta x}$ as $\Delta x \to 0$ and hence finding the gradient or rate of change of a function at a specific point on a curve.

Remember $y$ is just a shorthand notation for $y(x)$ as in $y$ is a function that takes a number as an input and outputs a number. Think of $\frac{d}{dx}$ as a function that takes a function as input and outputs a function. Don't forget the definition $ \frac{d}{dx}y(x) = \lim_{h->0} \frac{y(x+h)-y(x)}{h}$. So the outputted function gives us information about how the function acts on points close to $x$.


Truth and Lie Problem in Probability A fair die is rolled once and the outcome is observed by person $A$. $A$ then reports the outcome to person $B$ and $B$ writes it on a paper slip which shows that $6$ appeared on the die. Given that $A$ speaks the truth $3$ out of $4$ times and $B$ writes the truth (i.e, what A told him) $7$ out of $8$ times, what is the probability that $6$ actually appeared on the die? Given that when $A$ and $B$ are lying they tell or write a random number from $1$ to $6$ other than the one which they have observed (in case of $A$) or been told (in case of $B$).
Since I'm not that good at math so I made a code that imitated the given scenario and iterated it a billion times. The probability, thus obtained, approached $0.1107$. But I want to solve this problem mathematically.

Let $B$ be the event that $B$ writes $6$ and $R$ be the event that a $6$ was rolled.  We need to calculate $$P(R|B)=\frac{P(R\cap B)}{P(B)}$$

Let $p$ be the probability that $A$ tells the truth. The probability that $B$ is told that a $6$ was rolled is $$\frac16p+\frac56(1-p)\left(\frac15\right)=\frac16p+\frac16(1-p)=\frac16$$ since either a $6$ is rolled and $A$ tells the truth or a $6$ is not rolled, $A$ lies, and $A$ chooses to report a $6$ among all possible false values.  Therefore, the probability that $A$ reports $6$ to $B$ is $\frac16$ -- independent of the probability that $A$ tells the truth.

Obviously, the same calculation applies to the second half.  There is a $\frac16$ probability that $B$ is told to write a 6 and that is the probability that he will write a $6$ independently of how trustworthy he is.  Thus, $P(B)=\frac16$.

Now we need to find $P(R\cap B)$, the probability that both a $6$ was rolled and written.  For this, we do need the trustworthiness probabilities.  To have a 6 rolled and written, we need a six rolled AND (either A and B both tell the truth or A and B both lie and B chooses to write 6), which is $$\frac16(\frac34\cdot\frac78+\frac14\cdot\frac18\cdot\frac15)=\frac{53}{480}$$

Thus $$P(R|B)=\frac{{53}/{480}}{1/6}=\frac{53}{80}$$

Nefore you look at the paper the probability that a $6$ was rolled is $1/6$.

If you want to see how that probability changes when you know  B wrote $6$, draw a tree showing all the scenarios, with probabilities

roll 6 (probability 1/6)
   A reports 6 (probability 3/4)
       B writes 6
       B writes 1-5
   A reports 1-5 (probability 1/4)
       B writes 6
       B writes 1-5
roll 1-5 (probability 5/6)
   A reports 1-5 truthfully (probability 3/4)
       B writes 6
       B writes 1-5
   A reports 1-5 when lying (probability ??)
       B writes 6
       B writes 1-5
   A reports 6 when lying (probability ??)
       B writes 6
       B writes 1-5

Finish the tree, then multiply the probabilities down each branch  and sum the ones where B wrote $6$. Find the fraction of that when you sum the paths that start with a roll of $6. You will have to decide what happens when A or B lies.


Find the rate of change of the frequency when D, L, σ and T are varied singly. I'm reading Calculus made easy to learn the notation (I know derivatives with the limit/prime style) and also some integral calculus which I haven't seen at school yet. You can check it here: https://www.gutenberg.org/files/33283/33283-pdf.pdf

$\color{red}{\texttt{page 32}}$

  (9) The frequency n of vibration of a string of diameter D, length L
  and specific gravity σ, stretched with a force T, is given by

    $$n = \frac{1}{DL} * \left( \frac{gT }{ πσ} \right)^{1/2}$$

  Find the rate of change of the frequency when D, L, σ and T are varied
  singly

The answers are on $\color{red}{\texttt{page 252}}$. I know you have to treat everything as a constant except for what you are differentiating at the moment. For dn/dD you only work with D and every other letter is constant, for example.

My answers for dn/dD and dn/dL match the ones in the book... however, when I have to 'differentiate singly' stuff inside the square root I just can't get the algebra right (dn/dT and dn/dσ).

I post the derivative w.r.t T. You can first seperate the variable T from the others:

$n=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \sqrt{T}$

Then you write $\sqrt T$ as $T^{1/2}$

$n=\underbrace{\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}}_{\large{\begin{matrix}   \texttt{constant in case of} \\  \texttt{ differentiation w.r.t. T}   \end{matrix}}}\cdot T^{1/2}$

Therefore you need the derivative of $T^{1/2}$

$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{1/2-1}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{-1/2}$

And $T^{-1/2}=\frac{1}{\sqrt{T}}=\sqrt{\frac{1}{T}}$

Now you can put all the terms under the radical together.

$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}\cdot \frac{1}{2}=\frac{dn}{dT}=\frac{1}{2DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}$

You can try the other derivative for yourself.

Recall $\sqrt{gT} = \sqrt{g} * \sqrt{T}$.
As the book indicates, when calculating partial derivatives, we let only one variable change while the others are constant.

Say you're looking for $\frac{dn}{dT}$, then think of $n$ as

$n = constant * \sqrt{T}$.

Now calculating $\frac{dn}{dT}$ should be more clear. $\frac{dn}{d\sigma}$ follows the same method.


Interior point in convex set on n.v.s I'm reading Peter Lax's Functional analysis, and I have a question about a definition:

  Definition. $X$ is a linear space over the reals, $S$ a subset of $X$.  A point $x_0$ is called an interior point of $S$ if for any $y$ in $X$ there is an $\epsilon$, depending on $y$, such that
  $$x_0 + ty \in S \qquad\text{for all real $t$, $|t|<\epsilon$.}$$

I'm wondering if this definition of interior point is equivalent to the classical one (topological interior) in the case of a convex set. I was looking for a counterexample but I couldn't find it.

It  is not the same thing. It is what is called the algebraic interior. The standard counterexample is below where the origin is not in the topological interior.

EDIT (again) : what follows is 100% wrong and it is a good example of how NOT to write an answer.

EDIT: What's below is not true; it refers to the definition of "algebraic interior of a convex set", which is another thing. The definition in the original question is exactly the same as the usual one of interior point in the topological sense.

They're not equivalent. Consider a line segment in the plane. Its midpoint, or any point not in the extremes, is in the "algebraic interior" but not in the topological interior, since the latter is empty.


$f(x) = ax \text{ mod } m$, given that $\text{gcd}(a,m) = d$ what values does $f(x)$ take on? What values are in the range of $f(x) = ax \: (\text{mod } m)$ for $x \in \{0,...,m-1\}$.

I know that the answer should be all multiples of d however I'm not sure how one would reach that conclusion

Hint: $xa+ym=c$ has a solution (in integers $x$ and $y$) iff $d\mid c$.
For this is a linear diophantine equation.


Is There an Analytical Method for Solving the Given Equation? I recently encountered this problem on a problem set:

  Given that
  $$\log_2(2 + \sqrt{3}) < \log(3\pi)$$
  find the number of roots of the equation
  $$4\cos(e^x) = 2^x + 2^{-x}$$

I consulted my teacher for the solution and he checked the solution he had and told me that it was not possible to solve this manually as the solution he had was via graphing the functions and it was not possible without a graphing calculator or an app.

While using a graphing calculator the initial inequality is not used but it must have been given for a reason.

Graphically 4 solutions are obtained.

I want to know if there is really no way to solve this question using only calculations which are doable by hand.

Call the LHS and RHS of the equation

$L(x) = 4\cos(e^x)$

and

$R(x) = 2^x+2^{-x}$

respectively.

On the interval $[0,\ln{\pi}]$, $L(x)$ strictly decreases from $4\cos{1}\approx2.16$ to $-4$. On the same interval, $R(x)$ strictly increases from $2$ to $2^{\ln{\pi}}+2^{-\ln{\pi}}\approx2.66$. We then deduce that there is exactly one root in $[0,\ln{\pi}]$.

On the interval $[\ln{\pi},\ln{\frac{3}{2}\pi}]$, $L(x)\le0$ while $R(x)\gt0$ so there are zero roots in this interval.

On the interval $[\ln{\frac{3}{2}\pi},\ln{\frac{5}{2}\pi}]$, $L(x)$ is concave down ($L''\lt0$). On the same interval, $R(x)$ is concave up ($R''\gt0$). So there are either 0 or 2 roots in this interval. At $x=\ln{2\pi}$, $L(x)=4$ is greater than $R(x)=2^{\ln{2\pi}}+2^{-2\ln{\pi}}\approx3.85$. Yet on the ends of the interval $L(x) \lt R(x)$. We deduce that there are exactly 2 roots in this interval.

On the interval $[\ln{\frac{5}{2}\pi},\infty)$ there are no more roots because $R(x)>4$ and $L(x)\le4$.

For $x\lt0$ as $x\rightarrow-\infty$, $L(x)$ asymptotically approaches $4$ with continuously decreasing (negative) slope. In the same interval $R(x)$ is increasing with continuously increasing (negative) slope. We deduce that there is exactly one root in this interval.

For a total of four roots.


Disproving convergence of an integeal I remember that there is a calculus statement that says the following:
Assume you have an integral of a function (analytical, 1D) with infinite upper limit - if the limit of the function is not zero - the integral diverge.

More formally:
Assume f(x) is analytical function over the interval $[1,\infty]$
If
$\lim_{x\to \infty} f(x)$ $\neq$ 0
Then
$\int_{1}^\infty f(x)dx$
Diverge

Does this kind of statement exists? If so what is it called?

Would I be able to use the above to prove that
$\int_{1}^\infty x^4 \cos(x^3)dx$
diverge?

Thanks in advance

The way you can proceed is by evaluating this integral for a finite limit i.e.

$$I(l) = \int_{1}^{l}x^{4}cos(x^3)dx$$

Then you can analyse
$$L = \lim_{l\rightarrow \infty} I(l)$$

You need to distinguish between the limit existing and being nonzero, and the function not having a limit. The former is what you are talking about in the theorem, the latter is what occurs in the example.

The statement you give is true. It is akin to the result about convergent series' terms converging to zero: suppose that $ f(x) \to a \neq 0 $. Then given $\varepsilon>0$, there is an $N$ so that $ \lvert f(y) - a \rvert < \varepsilon $ for $y>N$. In particular this is true with $\varepsilon=a/2$ But then if $y,z>N$, $ \int_y^z f $ lies between $ (z-y)a/2 $ and $ 3(z-y)a/2 $. But this is not bounded as $z \to \infty$, so the improper integral does not satisfy the Cauchy criterion and so does not converge.
For the example, the function does not converge to a limit. This does not tell you anything in itself: for example, $ \int_0^{\infty} \sin(x^2) \, dx $ converges, but the integrand does not tend to a limit. You have to compute manually what is going on. What you can do is change variables to $ t = x^{1/3} $, turning the finite integral into a multiple of $ \int_1^{b^{3}} t^{2/3} \cos t \, dt $, which can be integrated by parts to produce an integral that we can recognise tends to a limit, plus a term that diverges.


Square root of permutation I have been thinking about square roots of permutations. I realized that the square of cycle of odd length is a cycle with the same length and the square of cycle with even length is multiplication of two cycles of half length.

Can the number of roots of some permutation be found in explicit way? What other facts about these roots are known?

You can find a general formula in the following reference: Jesús Leanos, Rutilo Moreno, and Luis Manuel Rivera-Martınez."On the number of mth roots of permutations." AUSTRALASIAN JOURNAL OF COMBINATORICS 52 (2012): 41-54.


Volume of triangular pyramid given 4 vertices I want to calculate the colume of the polyhedra with vertices $(a_1,0,0),(0,a_2,0),(0,0,a_3),(a_1,a_2,a_3)$. My solution is:
Using translation and a reflection I am considering the polyhedra  $(a_1,a_2,0),(0,a_2,a_3),(a_1,0,a_3),(0,0,0)$. I calculated the area of the triangle with vertices $(a_1,a_2,0),(0,a_2,a_3),(a_1,0,a_3)$ and found $A=|a_1a_2|/2$ and the distance from $(0,0,0)$ to the plane passing from the other 3 vertices and found $H=\frac{2|a_1a_2a_3|}{\sqrt{a_1^2a_2^2+a_2^2a_3^2+a_1^2a_3^2}}$. Finally the volume of the pyramid is $V=1/3 A*H=\frac{|a_1a_2||a_1a_2a_3|}{3\sqrt{a_1^2a_2^2+a_2^2a_3^2+a_1^2a_3^2}}$.

Is this correct and also is there a more simple solution using integration maybe?

A perhaps simpler approach is to start with the unit cube. Let
$$A=(1,0,0),\ B=(0,1,0), C=(0,0,1),\ D=(1,1,1)$$
and note that if the volume of this tetrahedron is found, then your coordinate version can be found by applying three successive transforms, one carrying $(x,y,z)$ to $(a_1x,y,z),$ the next taking $(x,y,z)$ to $(x,a_2,z),$ and the third carrying $(x,y,z)$ to $(x,y,a_3z).$ These three multiply volumes respectively by $a_1,a_2,a_3$ and so your desired volume is the unit cube tetrahedron's volume multiplied by the product $a_1a_2a_3.$

For the unit cube version, note that the plane through $A,B,C$ has equation $x+y+z-1=0.$ Thus the vector $(1,1,1)$ is normal to that plane, and we can find where that normal hits the plane by putting $x=y=z=t$ into $x+y+z-1=0,$ giving $t=1/3$ so that the normalfrom $D$ to the plane of the triangle $ABC$ hits the triangle at the point $(1/3,1/3,1/3)$ and the length of that normal is $\sqrt{(2/3)^2+(2/3)^2+(2/3)^2}=2/\sqrt{3}.$ The area of the triangle $ABC$ is, since it is equilateral of side length $\sqrt{2},$ conveniently $\sqrt{3}/2,$ so applying the cone volume formula we get
$$\frac{1}{3}\cdot \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}=\frac{1}{3}$$
for the volume of the tetrahedron $ABCD$ inside the unit cube.

Thus in your coordinate choice the volume is $a_1a_2a_3/3$ after stretching this as outlined above.

A simpler proof of unit cube tetrahedron volume. The complement of the tetrahedron $ABCD$ consists of four smaller tetrahedra, each the hull of one of the omitted vertices of the unit cube and the three of $A,B,C,D$ which are adjacent to it. These four tetrahedra are all congruent, and each may be viewed as a cone of altitude one over a triangle of area $1/2$, so has volume $(1/3)\cdot(1/2) \cdot 1=1/6.$ Thus the complement of tetrahedron $ABCD$ has volume $4\cdot(1/6)=2/3,$ leaving $1/3$ for the volume of $ABCD.$

You can also find the volume of the pyramid, or any continuous shape for that matter, by setting up a triple integral. It is best to first draw a picture of the figure to help you set up your bounds of integration. Your method is much simpler to compute for this example; however, the integral method is much more general.


Read-only Turing machine recognizes only regular languages? Show that the Turing machines, which have a read only input tape and constant size work tape, recognize precisely the class of regular languages.

According to wiki : A read-only Turing machine or Two-way deterministic finite-state automaton (2DFA) is class of models of computability that behave like a standard Turing machine and can move in both directions across input, except cannot write to its input tape. The machine in its bare form is equivalent to a Deterministic finite automaton in computational power, and therefore can only parse a regular language.

  Can you explain in for formal way, please?

If your machine can not modify the input tape and only has a working tape of fixed finite size, the amount of information that can be stored by the machine is finite (cardinality of the work tape's alphabet to the power of cells on the tape), hence you could do without the working tape by simply increasing the number of internal states of the machine, so this is then just a finite state automaton (with a large set of states).


Calculate the third partial derivative If $u = vt^{\frac{1}{3}}$ and $v=h(y), y=it^{\frac{1}{3}}\sin(iwx)$ I need to determine $ u_{xxx}$ and $u_t$
I have :
\begin{align}
u_x &= u_v \cdot v_y \cdot y_x\\
&= t^{\frac{1}{3}}v_y(-wt^{\frac{1}{3}}\cos(iwx))\\
&=-wt^{\frac{2}{3}}\cos(iwx)v_y
\end{align}

You have $y=it^{1/2}\sin iwx$. So $\frac{dy}{dx}=-wt^{1/2}\cos iwx$. Then
$$\begin{split}u_x&=-wt^{2/3}h'(y)\cos iwx\end{split}$$
To get from $u_x$ to $u_{xx}$ you need to use the product rule. $u_{xx}=-wt^{2/3}\left[\left(\frac{d}{dx}h'(y)\right)\cos iwx+h'(y)\frac{d}{dx}\cos iwx\right]$. Now you can apply the chain rule now that you've applied the product rule.
$$\begin{split}u_{xx}&=-wt^{2/3}\left[h''(y)(-wt^{1/2}\cos^2 iwx)+h'(y)(-iw\sin iwx)\right]\\
&=w^2t^{2/3}\left[h''(y)t^{1/2}\cos^2 iwx+h'(y)i\sin iwx\right]\\
u_{xxx}&=w^2t^{2/3}\left[h'''(y)t^{1/2}(-wt^{1/2}\cos^3 iwx)+h''(y)t^{1/2}2\cos iwx (-iw\sin iwx)+h''(y)i\sin iwx (-wt^{1/2}\cos iwx)+h'(y)i^2w\cos iwx\right]\\
&=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-2iwt^{1/2}h''(y)\cos iwx \sin iwx-it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-3it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&=-w^3t^{2/3}\cos iwx\left[t^{2/3}h'''(y)\cos^2 iwx+3it^{1/2}h''(y)\sin iwx +h'(y)\right]\end{split}$$


Evaluating $\int_a^b (x-\alpha)(x-\beta) \ dx$ without expanding the integrand Is there a property for definite integrals of the form:

$$\int_{a}^{b} (x-\alpha)(x-\beta) \ dx$$

that would allow one to evaluate it without expanding the integrand?

Thanks!

While expanding seems the most straightforward solution, we can also integrate by parts

$\begin{align}
I&=\int_{a}^{b} (x-\alpha)(x-\beta) \, dx
\\&=\frac{1}{2}(x-\alpha)^2(x-\beta)\bigg|_{x=a}^{x=b}-\frac{1}{2}\int_{a}^{b} (x-\alpha)^2 \, dx
\\&=\frac{1}{2}(x-\alpha)^2(x-\beta)-\frac{1}{6}(x-\alpha)^3\bigg|_{x=a}^{x=b}
\end{align}$

$$\int (x-\alpha)(x-\beta)dx = \int (x^2-(\alpha+\beta)x+\alpha\beta )dx = \frac{x^3}{3}-(\alpha+\beta)\frac{x^2}{2}+\alpha\beta x+k$$
where $k$ is constant.


Same perimeter and area for a circle and an ellipse For a given circle, is there exist an ellipse with same perimeter and area as to that circle?

If not, that is my suspicion, is in three-dimension parallel question: For a given sphere, is there exist an ellipsoid with same surface area and volume as to that sphere?

No, because the isoperimetric quotient $A/P^2$ is smaller for a non-circular ellipse than for any circle.

I cannot give you a deep rigourous argument, but surface tension is the reason a soap-bell is spherical so...

A circle has a specific eccentricity, and you choose another eccentricity while the area is fixed, there are no more parameters left to adjust for the perimeter.


Bézout's theorem intuition I am not taking algebraic geometry or trying to prove anything. I'm just looking for a simple intuitive understanding of Bézout's theorem for the case when one of the curves is a constant function. From https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Examples
we have the example: "Two distinct non-parallel lines (in the same plane) always meet in exactly one point. Two parallel lines intersect at a unique point that lies at infinity."

But take y = 1 and y = x. These two lines meet once at (1,1), but the product of the degrees of these curves is 0*1 = 0.

I feel like I'm missing something very simple or don't understand the hypotheses of the theorem... can someone please help explain?
Thanks!

The line described by $y=1$ is the zero set of a certain polynomial in the two variables $x$ and $y$, namely $y-1$. This polynomial has degree $1$, which is the exponent of $y$ in the first term. So, by definition, the degree of that curve is $1$.

On the other hand, the line described by $y=1$ is also the graph of a certain polynomial in the single variable $x$, namely $1$. This polynomial has degree $0$. But that isn't the polynomial relevant to Bézout's theorem!

The equation $y=1$ is linear, as is any equation $$ax+by+c=0,$$ where $a,b,c$ are constants. These are all of first degree since the operations on the variables involve nothing more than scalar multiplication and addition.


Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even

My Attempt, Case by case analysis:

Case 1: a is odd, b is odd. From the first equation,

$odd^2 + odd^2 = c^2$

$odd + odd = c^2 \implies c^2 = even$

Squaring a number does not change its congruence mod 2.

Therefore c is even

$ a + b + c = odd + odd + even = even$

Case 2: a is even, b is even. Similar to above

$even^2 + even^2 = c^2 \implies c$ is even

$a + b + c = even + even + even = even$

Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even

$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$

Therefore c is odd

$a + b + c = odd + even + odd = even$

We have exhausted every possible case, and each shows $a + b + c$ is even. QED

Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$

Consider $(a+b+c)^2$

Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$

Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.

and since squares of odd is odd and evens is even $a+b+c$ has to be even.


Proof that a Combination is an integer From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.

This is clearly an integer, however I was curious as to why the expression

$$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.

So far I figured:

$n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.

Well, one noncombinatorial way is to induct on $n$ using Pascal's triangle; that is, using the fact that ${n \choose k} = {n-1 \choose k - 1} + {n-1 \choose k}$ (easy to verify directly) and that each ${n - 1 \choose 0}$ is just $1$.

One way to get there: any product of $k$ consecutive positive integers $d := \frac{n!}{(n-k)!} = \prod^k (n-i+1)$ is divisible by $k!$.

You can assure yourself of this by considering divisibility by the primes and prime powers up to $k$. For example, $k!$ has $\lfloor k/p\rfloor$ multiples of $p$, and $d$ will have either the same number of multiples of $p$ or one more, since the "first" multiple of $p$ in $k!$ will be at $p$, whereas it may be earlier in $d$.

The quantity I have denoted by $d$ is the $k$-descending factorial of $n$, sometimes written $n^{\underline k}$ or $(n)_k$.


"All true theorems are logically equivalent" I've seen the phrase "all true theorems are logically equivalent" thrown around here, when people ask if a theorem X and a theorem Y are logically equivalent.

What is meant by this? Are they just referring to the fact that an implication with a true consequent is always a true statement, therefore the biconditional $P \Longleftrightarrow Q$ is true for any true $P$, $Q$, or do they mean something more meaningful?

Yes, that must be what people who throw around that phrase mean: If $P$ and $Q$ are both true (or provable in some particular theory), then $P\Leftrightarrow Q$ is also true (or provable in that theory).

However, this is not actually what "logically equivalent" means in logic. The usual meaning of that is that $P$ and $Q$ are logically equivalent if and only if $P$ has the same truth value as $Q$ in every interpretation. Or, equivalently, $P$ and $Q$ are logically equivalent if $P\Leftrightarrow Q$ is provable without using any non-logical axioms.

One can speak about being equivalent relative to some theory -- for example the Axiom of Choice and Zorn's Lemma are equivalent relative to ZF (or"given ZF"), which simply means that ZF proves AC$\Leftrightarrow$Zorn. People often just say that "AC and Zorn are equivalent", in which case they are leaving which theory they are talking about implicit. Usually it is clear from the context what the underlying theory is.

The word "logically" should not be used in the latter case, though.

The only logical component of a statement is whether it is true or false. That is logic. So saying they are logically equivalent means that they are either both true or both false.

We may not know whether they are true or false, but we may still be able to show that there truth values coincide. In other situations, the truth or false of the statement "depends" on context (variables). In this case, logical equivalence means they have the same truthiness in every scenario.


Uniform convergence of $ U_n(x) = \sum_{n=0}^{+ \infty} (-1)^n \ln ( 1 + \frac{x}{1+ nx} ) $. We consider the series of functions:
$$U_n(x) = \sum_{k=0}^{n} (-1)^k \ln \left( 1 + \frac{x}{1+ kx} \right)  ,~ x \geq0.$$

Prove that $U_n$ is convergent.

Study the uniform convergence of $U_n$.

Study the normal convergence.

We consider $U(x) = \sum_{n=0}^{+ \infty} (-1)^n \ln ( 1 + \frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.

Compute $U'(x)$.

I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $\lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ \sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?

This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = \sum_{k=0}^m \ln \left( 1+ \frac{x}{1+2kx} \right) - \ln \left( 1+ \frac{x}{1+(2k+1)x} \right)$$ which (after some calculations) turns out to be
$$\sum_{k=0}^m \ln \left( 1+ \frac{x^2}{(1+2kx)(1+2(k+1)x)} \right)$$

As $m \to + \infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $\Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.

Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).

As for differentiability of $U(x)$, you have to apply this limit .


When is the sheaf corresponding to a vector bundle on a smooth manifold coherent? In algebraic and analytic geometry, vector bundles are usually interpreted as locally free sheaves of modules (over the structure sheaves). They are in particular examples of quasi-coherent sheaves. If the bundle is of finite rank, then the sheaf is actually coherent and this is good for certain cohomology groups to have finite rank for example.

I think the equivalence of vector bundles and locally free sheaves holds as well for the categories of topological spaces and smooth manifolds, and locally free sheaves are in particular quasi-coherent. The question is, when are they coherent? maybe it is best to ask first about the structure sheaf itself. In algebraic geometry the structure sheaf is coherent for neotherian schemes and in analytic geometry the structure sheaf of a complex manifold is coherent. What about smooth manifolds for example?

Since coherence is a local property and is preserved by finite direct sums, the general question reduces to the structure sheaf. But this is coherent only in trivial cases.

  If $M$ is a smooth manifold whose connected components have positive dimension, then $C^{\infty}_M$ is not coherent.

For a proof, see Prop. 7.3.8 in the course notes by Andrew Lewis on sheaf theory. By a projection argument, it suffices to deal with $M=\mathbb{R}$. If $f$ is a smooth function, which is $>0$ on $\mathbb{R}_{>0}$ and $=0$ on $\mathbb{R}_{\leq 0}$, then the kernel of the multiplication map $f : C^{\infty}_{\mathbb{R}} \to C^{\infty}_{\mathbb{R}}$ is not of finite type: Loot at the stalk at $0$. The kernel $I$ is given by those germs of smooth functions vanishing on $\mathbb{R}_{\geq 0}$. If $\mathfrak{m}$ is the maximal ideal of functions vanishing at $0$, then we have $I = \mathfrak{m} I$ by Taylor expansion. Since $I \neq 0$, Nakayama's Lemma shows that $I$ is not finitely generated.

If you're using the [sheaf of continuous sections] and [etale space] functors they do not give an equivalence of these categories. The problem is that locally constant sheaves are covering spaces, and vector bundles aren't covering spaces. They're not even (topologically) etale in general over the base space.

Calling a sheaf coherent over a space that isn't a scheme is silly, since [the definition of coherence] implicitly uses the structure sheaf of the space. A manifold doesn't really have a structure sheaf(it does have a natural sheaf of smooth functions, but I don't know that it makes much sense to talk about coherent modules over such a sheaf of rings, as I don't think there is a good, canonical, functorial way to take a module of the global section and 'tilde'(to use Hartshorne notation) it into a sheaf.)


Find the maximum value of $f(x,y)=x^2+y^2$ in the region bounded by $y=\frac x2,y=-\frac x2$ and $x=y^2+1$,including the boundary lines The figure of the problem  where $y=\frac x2,y=-\frac x2$ and $x=y^2+1$ enclose the region is this:

The shaded region is the enclosed area. Now the max of $x^2+y^2$ in this area is exactly what in this region?Is it the point where $y=\frac{x}{2}$ intersects $x^2+y^2$?

If the feasible region is the light blue region then the maximum for $x^2+y^2$ is $\infty$ otherwise if the feasible region is the white one the the maximum is attained at $(x_0,y_0)=(0,1)$ with value $x_0^2+y_0^2 = 1$

$x^2+y^2$ is is square of distance from origin thus the point furtherest from the origin i.e $(2,1)$ and $(2,-1)$ lying on intersection of the parabola and the line are the optimal points with $f_{max}=5$


Evaluating $\lim\limits_{x\to \infty}\frac{x^2}{2^x-1}$ I would like to evaluate
  $$\lim_{x\to \infty}\frac{x^2}{2^x-1}$$
  Without using L'HOSPITAL's rule nor series.

I tried more than one
technique such the sub
$$x=\frac{1}{y}$$
But i could not get the solution ?

Using the hint provided by Jyrki Lahtonen in the comments, we may sneakily apply the binomial theorem in order to obtain
$$ 2^n - 1 = (1+1)^n - 1 = -1 + \sum_{k=0}^{n} \binom{n}{k} \ge \binom{n}{k} $$
for any choice of natural number $k$ less than $n$.  This implies that
$$ \frac{n^2}{2^n - 1} \le \frac{n^2}{\binom{n}{k}}. \tag{1}$$
By choosing $n$ large enough, it is possible to find a value of $k$ such that $\binom{n}{k} \gg n^2$.  For example, if $k = 3 < n$, then we have
$$ \binom{n}{3} = \frac{n!}{(n-3)!3!} = \frac{n(n-1)(n-2)}{3\cdot 2}. $$
Plugging this into (1), we get
$$\frac{n^2}{2^n - 1}
\le \frac{n^2}{\binom{n}{3}}
= 6 \frac{n^2}{n(n-1)(n-2)}
= \frac{6n^2}{n^3 - 3n^2 + 2n}.
$$
The expression that we want to limit is positive for any $n > 0$, thus applying the squeeze theorem we can take limits to get
$$ 0 \le
\lim_{n\to\infty} \frac{n^2}{2^n - 1}
\le \lim_{n\to\infty} \frac{6n^2}{n^3 - 3n^2 + 2n}
= \lim_{n\to \infty} \frac{\frac{6}{n}}{1 - {\frac{3}{n}} + \frac{2}{n^2}}
= 0.
$$
(The last expression, obtained via multiplication by $\frac{1/n^3}{1/n^3}$ is perhaps not strictly necessary, but I think that it aids in understanding—more generally, we might simply recall that the limit (at infinity) of a rational expression is 0 if the degree of the numerator is smaller than the degree of the denominator; $\infty$ if the numerator has greater degree, and some nonzero constant if the numerator and denominator have the same degree.)

EDIT:  I left off a step of the computation that seemed obvious to me, but for the sake of completeness, all that has been shown above is that if
$$ \lim_{x\to \infty} \frac{x^2}{2^x - 1} $$
exists, then it must be equal to 0.  To finish the argument, we might note that
$$ \frac{x^2}{2^x - 1}
\le \frac{\lceil x \rceil^2}{2^{\lfloor x \rfloor} - 1}
= \frac{(\lfloor x \rfloor + 1)^2}{2^{\lfloor x \rfloor} - 1}
= \frac{\lfloor x \rfloor^2}{2^{\lfloor x \rfloor} - 1} + \frac{2\lfloor x \rfloor + 1}{2^{\lfloor x \rfloor} - 1}. $$
The first term is exactly the one analyzed with $n = \lfloor x \rfloor$.  The second term can by analyzed similarly.

Alternatively, with
$$ f(x) = \frac{x^2}{2^x - 1}, $$
look at $f'(x)$, and note that $f'(x) < 0$ for $x$ sufficiently large (using the same kinds of estimates as above; this requires a little work, but is not unreasonably difficult).  Since the derivative is negative, the function is decreasing.  Moreover, the function is nonnegative for all $x > 1$, and so bounded below by zero.  As $f(x)$ is monotonically decreasing and bounded, it must have a limit.  By the above argument, the limit is zero.

Note that

$$\frac{x^2}{2^x-1}=\frac{2^x}{2^x-1}\cdot \frac{x^2}{2^x}\to1\cdot0=0$$

indeed

$$\frac{2^x}{2^x-1}=\frac{1}{1-\frac{1}{2^x}}\to 1$$

and since eventually, notably for $x>10$, we have that $2^x>x^3$

$$\frac{x^2}{2^x}<\frac{x^2}{x^3}=\frac1x\to0$$

EDIT

Proof by induction of $2^n>n^3$.

Base case

$n=10\implies2^{10}=1024>10^3=1000$

Induction step

We need to prove that

$2^n>n^3\implies 2^{n+1}>(n+1)^3$

Let observe that

$$2^{n+1}=2\cdot 2^n\stackrel{\text{inductive hypothesis}}\ge 2\cdot n^3 \stackrel{\text{?}}\ge (n+1)^3$$

and the last inequality is true since

$$2\cdot n^3 \ge (n+1)^3=n^3+3n^2+3n+1\iff n^3 \ge 3n^2+3n+1 \quad \square$$


Weight of watermelons after percentage of water is evaporated. A stock of watermelons of the initial weight of $500 \space\text{kg}$
  has been put in a store for a week.

  Initially the percentage of water in the watermelons makes up $99 \% $
  of the weight,at the end of the week,due to evaporation, such
  percentage dropped to $98 \%$.

  How much do the watermelons weight at the end of the week ?

$A)250 \space\text{kg}$,$B) 400 \text{kg} $,$C)480 \text{kg} $,$D)495 \text{kg}$

My effort

I have that  water makes up  $99 \%$ of the weight of the watermelons,so it weights $495 \space\text{kg}$ .

After the week I have that $2 \% $ of the water evaporated  so we loose $\cfrac{2}{100}\cdot 495 =9,9 \text{kg}$  so the watermelons should weight $490,1 \text{kg}$ which is none of the answers given ...

The solution is given as option $A)250  \text{kg}$ ,but I don't understand why.

As you said, initially water weights $495 \text{kg}$ and dry mass weights $5\text{kg}$. After the week, dry mass weights still $5\text{kg}$ and makes up $2$% of the watermelon. Then watermelons should weight $5\cdot 50 = 250(\text{kg})$.

At the beginning you count with 1% equivalent to 5 kg of solid mass, this mass shouldn't change for obvious reasons (it can't evaporate), nevertheless the problem is suggesting a drop of 1%, 99% to 98%.

=> Initial equivalence:

.01 => 5 kg

From the water compound now represent 98% of the total mass, where is the other 2%? well it stills being 5 kg nevertheless now represent a 2%.
to know what is the total weight, the 100% we use the following operation.

$\frac{5 kg}{0.02} = 250$.


Find $f(0),f'(0),f''(0)$ Suppose $f(x)$ has second derivative on some neighbourhood around $x=0$, and $$\lim_{x\to 0}\left(1+x+\frac{f(x)}{x}\right)^{1/x}=e^3$$ Find $f(0),f'(0),f''(0)$

I know $f(0)$ and $f'(0)$ must be $0$ or the limit wouldn't exist. Then after some rough calculation, I guess $f''(0)$ should be $2$. My problem is I have trouble to write the proof with strict words.

By the given condition we get via taking logarithm (and the fact that $\log$ function is continuous) $$\lim_{x\to 0}\frac{1}{x}\log\left(1+x+\frac{f(x)}{x}\right)=3\tag{1}$$ and by multiplying with $x$ we can see that $$\lim_{x\to 0}\log\left(1+x+\frac {f(x)} {x} \right) =0$$ Next the continuity of $\exp$ function gives us $$\lim_{x\to 0}\left(1+x+\frac{f(x)}{x}\right)=1$$ so that $$\lim_{x\to 0}\frac{f(x)}{x}=0\tag{2}$$ Multiplication by $x$ gives us $$f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}x\cdot \frac {f(x)} {x} =0\cdot 0=0$$ and then $$f'(0)=\lim_{x\to 0}\frac {f(x) - f(0)}{x}=\lim_{x\to 0}\frac{f(x)}{x}=0\text{ (via (2))}$$ To evaluate $f''(0)$ we need more effort. It is known that the second derivative $f''(0)$ exists and hence the limit $$f''(0)=\lim_{x\to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x\to 0}\frac{f'(x)}{x}$$ exists. By L'Hospital's Rule we have $$f''(0)=2\lim_{x\to 0}\frac{f(x)}{x^2}\tag{3}$$ It is now time to revisit equation $(1)$. Using the fundamental limit $$\lim_{t\to 0}\frac {\log(1+t)} {t} =1$$ the equation $(1)$ can be rewritten as $$\lim_{x\to 0}\dfrac{x+\dfrac{f(x)} {x} }{x}=3$$ or $$\lim_{x\to 0}\frac{f(x)}{x^2}=2$$ and by equation $(3)$ we see that $f''(0)=4$.

To sum up $$f(0)=f'(0)=0,f''(0)=4$$ As the above solution shows we don't really need the existence of second derivative $f''$ in some neighborhood of $0$ but only its existence at the point $0$.

We can use  it  this to solve this problem.
first we can use this limit representation of $e^k$

$$
{
e^k=\lim_{n\to\infty} \left( 1+\frac{k}{n} \right)^n
\qquad (1)
}
$$
 and changing n to 1 / x and substituting k = 3

$$
{
e^3=\lim_{x\to0} \left( 1+
3x \right)^{\frac{1}{x}}
\qquad (2)
}
$$

Then using (2) we can see that

$$\lim_{x\to 0}(1+x+\frac{f(x)}{x})^\frac{1}{x}=\lim_{x\to0} \left( 1+
3x \right)^{\frac{1}{x}} $$
And afther that we see that
$$x+\frac{f(x)}{x}=3x$$
$$\frac{f(x)}{x}=2x$$
$$f(x)=2x^2$$

Although this form of equating terms can not be used habitually, since the limit of two funcuons are equal does not mean that two functions are the same. But in this case it can be deduced from the statement that all possible functions f have the same derivative in 0, since otherwise the statement would contradict itself and have multiple solutions. Then with analyzing the derivatives of the simplest function that can be f is sufficient to extrapolate it to all the f.

So we can easily solve the problem by replacing  the zeros and  the derivatives
$$f(0)=2\cdot 0^2 =0$$
$$f'(0)=4\cdot 0 =0 $$
$$f''(0)=4$$
Thanks for reading and goodbye.


How to pick the "best" line equation that fits data and passes through the origin? I have some data shown in the image. Each symbol corresponds to a different dataset. As it can be seen the data initially behaves like a line and after some points it follows other behaviour.

The idea is to fit a line that passes through the origin and that the line "is good enough" approximating the behaviour of the data where it is linear.

What I tried is to do the following: I pick a minimum of data (e.g. $m_i$) for each data set $D_i = \{d_{ij}=(x_j,y_j)_i, j\in\{1,...,N_i\}\}$, where $N_i = |D_i|$ is the number of data of each set, and $x_1=0, x_p < x_q$ if $p < q$ . Now, for $D_i$ we consider subsets $S_{ik}=\{d_{ij}=(x_j,y_j)_i, j\in\{1,...,k_i\}\}$, where $k_i \in \{m_i,...,N_i\}$. Each of these subsets has associated with it a correlation coefficient $p_{ik}$. My idea was to pick the maximum of these coefficients and to fit a line $y=mx$ through the data in $S_{ik}$ associated with $\max_k p_{ik}$.

The problem with this is that the correlation coefficient tells how the $y$ data behaves with respect to the $x$ data, i.e. how good a line would fit the data. This line intercepts the $y$ axis at a point that in general is not $0$.

My next try was to instead fit a line $y=mx$ for each $S_{ik}$ and then compute the mean squared error MSE. Then pick the subset that is associated with the minimum value of MSE. This also fails because of course the minimum value is the one associated with the subset with less number of data (MSE penalizes the variation very hard, I look for something "softer").

Is there a way of knowing where the data behaves like a line passing through the origin? I was thinking of taking a small subset of data S_{ik}, fitting a line $y_i=mx$, computing the MSQ. Then adding one data point to the subset and computing its error, then compare it with the MSQ and decide whether to continue the process of adding data, or stopping.

Any help is very much appreciated.

I worked this out
a number of years ago.

If you go the the process
of minimizing
$\sum (y_i-ax_i)^2$
as a function of $a$,
you get
$a
=\frac{\sum x_iy_i}{\sum x_i^2}
$.

Here's how the math works out:

Let
$D
=\sum_{i=1}^n (y_i-ax_i)^2
$.
Then

$\begin{array}\\
\frac{\partial D}{\partial a}
&=\sum_{i=1}^n \frac{\partial (y_i-ax_i)^2}{\partial a}\\
&=\sum_{i=1}^n -2x_i(y_i-ax_i)\\
&=-2(\sum_{i=1}^nx_iy_i-a\sum_{i=1}^nx^2_i)\\
&=0
\qquad\text{when }a = \dfrac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx^2_i}\\
\end{array}
$

For this value of $a$,

$\begin{array}\\
D
&=\sum_{i=1}^n \left(y_i-ax_i\right)^2\\
&=\sum_{i=1}^n \left(y^2_i-2ax_iy_i+a^2x^2_i\right)\\
&=\sum_{i=1}^n y^2_i-2a\sum_{i=1}^nx_iy_i+a^2\sum_{i=1}^nx^2_i\\
&=\sum_{i=1}^n y^2_i-2\dfrac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx^2_i}\sum_{i=1}^nx_iy_i+\left(\dfrac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx^2_i}\right)^2\sum_{i=1}^nx^2_i\\
&=\sum_{i=1}^n y^2_i-\dfrac{\left(\sum_{i=1}^nx_iy_i\right)^2}{\sum_{i=1}^nx^2_i}\\
\end{array}
$

Note that both the value of $a$
and the value of $D$
can be easily updated
when a data point
is either added or removed.
That way,
you can examine how $D$ changes
(though you probably want to compute
$\frac1{n}D$)
and stop when it gets too large.

Are you required to use a straight line?  It looks to me like a quadratic would fit better and require only a little more trouble.


How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of variables... but it became more complicated . Is there an idea to solve this system of equation ?
  Thanks in advance

So here are the 3 equations:

$$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$

As suggested by transcenmental,

$\sqrt{x-1}-\sqrt{y-1}=4\sqrt 3 - 2 \sqrt{y-1}$ and multiplying with the first eq. gives

$$
x - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})
$$

For the second eq., use

$$-\sqrt{y-4}+\sqrt{z-4}=4\sqrt 3 - 2 \sqrt{y-4}$$ and multiplying with the second eq.

$$
z - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})
$$

Plugging into the last one gives an eq. in y:

$$
\sqrt{y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})-9}+\sqrt{ y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})-9}=4\sqrt3
$$

This is pretty akward, but $y = 28/3$ is a solution (by computer). From here the others follow, namely

$$
x = 52/3$$

and

$$
z = 76/3$$

EDIT:

with a little bit of hindsight and a little bit of psychology, you could argue
as follows (with a twinkling of an eye):

suppose the person asking the question prefers a reasonably nicely looking solution (psychology 1). Then  all variables  should either be multiples of 3 or of $1/3$ to get rid of the $\sqrt 3$ on the RHS. Let's try $1/3$ (hindsight 1). So let $x = x' / 3$ etc. Now assume further that also the numerator of the variables should be nice, e.g. no roots etc. (psychology 2). Then we should have that $x'-3\cdot 1$ and  $x'-3\cdot 9$ are "nice" squares (likewise with the other variables). If we want it even nicer, they should be squares of integers (hindsight 2).

So $x' = 3 + n^2$ and $x' = 27 + m^2$. Now start playing. "Nice" integers n and m will be reasonably small (psychology 3). $x' = 52$ does it nicely, with $n=7$ and $m=5$. A small number of trials, to match all three variables, will then give the solution....

Solving $1$:-

$\sqrt{x-1}+\sqrt{y-1}=4\sqrt3\\$

$\text{Squaring}$

$\begin{align}x-1+y-1+2\sqrt{(x-1)(y-1)}&=48\\to\\50-x-y&=2\sqrt{(x-1)(y-1)}\\\end{align}$

$\text{Squaring}$

$\begin{align}x^2  + y^2+ 2 x y- 100 x  - 100 y + 2500&=4 x y - 4 x - 4 y + 4\\to\\x^2  + y^2 + 2496 &=   96 x+ 96 y +2 x y \end{align}$

Continue for the other equations, then solve.


Let $\xi$ random vector and $\zeta^{x}:=x^{T}\xi$ random variable. Is it correct to say that $\int x^{T}y f_{\xi}(y)dy=\int tf_{\zeta^{x}}(t)dt$? Let $\xi\in\mathbb{R}^{m}$ be a random vector with  joint density function $f_{\xi}:\mathbb{R}^{m}\rightarrow \mathbb{R}$. Let $x\in \mathbb{R}^{x}$ be a nonzero vector. We consider the random variable $\zeta^{x}:=\langle x,\xi\rangle=x^{T}\xi$, we called $f_{\zeta^{x}}:\mathbb{R}\rightarrow\mathbb{R}$   the density function of $\zeta^{x}$.

The question: Is it correct to say that $$\int_{\mathbb{R}}\cdots\int_{\mathbb{R}} \langle x,y \rangle f_{\xi}(y)dy_{1}\cdots dy_{m}=\int_{\mathbb{R}}tf_{\zeta^{x}}(t)dt  .?$$

Yes, you can write $E[g(\xi)]$ as either $\int_{\mathbb{R}^m} g(y) f_\xi(y) \, dy$ or $\int_{\mathbb{R}} t f_{g(\xi)}(t) \, dt$. The first is the "Law of the unconscious statistician"
while the second is one definition of the expectation of a random variable.


Some doubts in 2's complement I am doing one question in which we have to find 2's complement of 43. I know to find 2's complement we have to find 1's complement and then add 1 into it to get 2's complement.

So here is how I am doing it -

Binary value of 43 is 00101011. (We have to use 8-bits to represent)

1's complement is 11010100.

2's complement is 11010101.

But the original answer is 00101011.

I used some online converter and they are giving same answer as the original one. But i don't find any mistake in my method. So please help me to understand my mistake and how to get the original answer.

Edit -

This is the link to online converter : Click Here.

Update -

As said by one user that my question is wrong or it is wrong in textbook. It is not. Same wording of question above in my question. Here's the link from which I am solving this question before.

Click here to see.

And I wrote answer later from this link. As you can see in my answer below.

You were correct and the textbook is still wrong. Not surprising given what a slapdash job they did of putting it together. I'm quoting from the second page you linked:

Explanation: In$2$’s complement representation, positive numbers are represented as their representation and negative numbers are represented by first doing $1$’s complement, then adding $1$ to the result.

Notice the lack of a space between "in" and "$2$’s"? Just that little detail gives me serious doubts as to whether they even know what the hell they're talking about.
Well, to be fair, they do know what they're talking about, they just don't know how to explain it simply and correctly. "Represented as their representation"? What does that even mean? "And negative numbers are represented by first doing $1$’s complement" on what? On their absolute value or on some random number?

So $43$ is represented as $00101011$.
Note that option represents $-43$.

Are they saying that $00101011$ represents both $43$ and $-43$? Maybe they meant to write something else between those two sentences but just didn't get around to it.
The answer is indeed $11010101$ like you got, which means $-43$ in a signed $8$-bit representation using two's complement, or $213$ in an unsigned $8$-bit representation. If they wanted $00101011$, they should have asked for the the $8$-bit representation of $43$.
And lastly, what sort of material are you supposed to read and understand before taking the quiz? Here's an explanation of two's complement from Cornell University: https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html
Although the author of those Cornell notes says the topic doesn't need a lengthy explanation, I think you will find his explanation far more detailed, correct and easy to undestand than anything on GeeksForGeeks.org.
And if after reading that you still have "doubts," try the following exercises with pencil on paper:

$00000101 + 00000011$
$11010101 + 00000011$
$11011000 - 00000011$
$11010101 \times 11111111$

Your answer is correct. In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value.

Here's Wikipedia explanation you can see.


Subspace of strictly positive entries Can someone show me how to solve this question.

  Is the set of all vectors in $R^3$ with strictly positive entry a subspace?

Thanks

To determine whether something is a linear a subspace you have to verify that it is itself a vector space.

There are usually about two things that go wrong: Either $0$ is not in the space or $x,y$ are in the space but $x+y$ is not.

This later "flaw" is generally referred to as "not being closed" which is short for "not closed with respect to vector addition".

These are the two things that go wrong most often but sometimes it also happens that there are no inverses. After, first and foremost, a vector space is a group and if it does not satisfy the group axioms it cannot be a group.

In your example, two of the three things I mention go wrong. Do you see which?

Is the set of all vectors in $\mathbb{R}^3$ with strictly positive entry a subspace?

Let
$$
\mathbb{R}_{>0}^3 := \left\{ \left. \left( x,y,z \right)^\mathsf{T} \in \mathbb{R}^3 \,\right|\, x > 0 \wedge y > 0 \wedge z > 0 \right\}
$$
denote the set of all vectors with strictly positive entries. It is a subset of $\mathbb{R}^3$.

In order for a subset to be a subspace of a given vector space, it is required that it contains the zero vector of the original vector space.

The zero vector $\left( 0,0,0 \right)^\mathsf{T} \in \mathbb{R}^3$ is, however, not in $\mathbb{R}_{>0}^3$. Therefore, $\mathbb{R}_{>0}^3$ is not a subspace of $\mathbb{R}^3$.


Find the sup of this set with condition $x^2+y^2+z^2=1$ let $x,y,z>0$,and such $x^2+y^2+z^2=1$,show that  $$\min{((x-y)^2,(y-z)^2,(z-x)^2)}< \dfrac{1}{5}$$and show that $\dfrac{1}{5}$ is best constant. If it is not possible to find the maximal value, is it possible to give some reasonable upper bound?

Say $x\le y \le z$. So we have to show that at least one of the differences $y-x$, $z-y$ is  $<\frac{1}{\sqrt{5}}$. Assume not. Then
$x>0$, $y >\frac{1}{\sqrt{5}}$, $z > \frac{2}{\sqrt{5}}$, hence their sum of squares is larger than $1$. One can easily see that the constant $1/5$ cannot be decreased, by using $(x,y,z)$ close to the triplet $(0, \frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$.

The best upper bound is $1/5$, and is achieved at six points on the sphere. These points have coordinates being the six permutations of $2/\sqrt{5},1/\sqrt{5},0$

To solve this, notice that given a point $P(x_0, y_0, z_0)$, $(x_0-y_0)^2$ is the square of the distance of point $P$ to the plane $x-y=0$, and similarly for $(y_0-z_0)^2$, $(z_0-x_0)^2$. So to maximize $\min\{(x_0-y_0)^2 ,(y_0-z_0)^2,(z_0-x_0)^2\}$, you should find a point on the sphere that's simultaneously far away from all these three planes.

If you imagine looking down at the sphere from the point $(1,1,1)$, you'll see the three planes cutting the sphere at $60$ degrees angle between them. Then it's clear from geometry which points should maximize $\min\{(x_0-y_0)^2 ,(y_0-z_0)^2,(z_0-x_0)^2\}$


Probability question: proving independence. Let $X,Y,X_n,Y_n: \Omega \rightarrow \mathbb{R}, n \ge 1$ be random variables.

If for all $n\ge1$, $X_n$ and $Y_n$ are independent and if $(X_n,Y_n) \xrightarrow[]{d}(X,Y)$, then X and Y are independent.

I found that if two random variables are independent then for $(\xi_1,\xi_2) \in\mathbb{R}^2$
$$\mathbb{E}(e^{i\langle
(\xi_1,\xi_2), (X_n,Y_n)\rangle})=\mathbb{E}(e^{i\langle\xi_1, X_n\rangle})\cdot\mathbb{E}(e^{i\langle\xi_2, Y_n\rangle})\\ \implies \lim_{n\rightarrow \infty}\mathbb{E}(e^{i\langle
(\xi_1,\xi_2), (X_n,Y_n)\rangle})=\lim_{n\rightarrow \infty}\mathbb{E}(e^{i\langle\xi_1, X_n\rangle})\cdot\mathbb{E}(e^{i\langle\xi_2, Y_n\rangle})=\mathbb{E}(e^{i\langle
(\xi_1,\xi_2), (X,Y)\rangle})$$
But since it isn't given that $X_n,Y_n$ are weakly convergent then how do I proceed?

It feels like I am missing something very trivial, I am trying to read Brownian Motion
An Introduction to Stochastic Processes by Schilling. It feels like there is a lot that I already need to know before tackling this book. I have had a basic course in Stochastic Processes, can someone recommend a book to bridge the gap.

$(X_n,Y_n)$ converging to $(X,Y)$ in distribution implies that $X_n$ converges to X in distribution and $Y_n$ converges to Y in distribution so your argument using characteristic functions is correct; it is also the best way to prove the result.

I guess the idea here is that you go to the definitions and the elementary properties.

The hypothesis is that for each $n\in\mathbb N$ the r.v. $X_n$ and $Y_n$ are independent, and a theorem/property/whatever states that this happens if and only if (important that is an equivalence, not a one way implication!)
$$F_{X_nY_n}(x,y)=F_{X_n}(x)F_{Y_n}(y)$$
(that is, the joint cumulative distribution is the product of both marginal cumulative distributions).

You also know that $(X_n,Y_n)\xrightarrow{\mathcal D}(X,Y)$ which means that $F_{X_nY_n}(x,y)\to F_{XY}(x,y)$ as $n\to \infty$ (except maybe for points $(x,y)$ where $F_{XY}$ is not continuous).

Imagine that you could prove, maybe as a consequence of the aforementioned, that $F_{XY}(x,y)=F_X(x)\cdot F_Y(y)$ (not evident at all!), what would you conclude?

UPDATE: Resolution

Well, the answer is that $F_{XY}(x,y)=F_X(x)\cdot F_Y(y)$ would mean that $X$ and $Y$ are independent (because the property I mentioned above is an equivalence, an if and only if).

So this reduces to prove that
$$F_{X_nY_n}(x,y)=F_{X_n}(x)F_{Y_n}(y)\to F_{X}(x)F_Y(y)\text{ as } n\to \infty.$$

But this will be true if we show that $F_{X_n}(x)\to F_X(x)$ and the same for $Y$.

Now, as $F_{X_n}(x)=\lim_{y\to+\infty}F_{X_nY_n}(x,y)$, we can say that
$$\lim_{n\to\infty}F_{X_n}(x)=\lim_{n\to\infty}\lim_{y\to+\infty}F_{X_nY_n}(x,y)$$
and if limits are to be swaped, then
$$\lim_{n\to\infty}F_{X_n}(x)=\lim_{y\to+\infty}\lim_{n\to\infty}F_{X_nY_n}(x,y)=\lim_{y\to+\infty}F_{XY}(x,y)=F_X(x).$$

In the same way, we can prove that
$$\lim_{n\to\infty}F_{Y_n}(y)=F_Y(y),$$
and so $$F_{XY}(x,y)=\lim_{n\to\infty}F_{X_nY_n}(x,y)=\lim_{n\to\infty}F_{X_n}(x)F_{Y_n}(y)=F_X(x)F_Y(y),$$
which proves $X$ and $Y$ are independent.

Notes:

1) The swapping of the limits is justified by the fact that $\lim_n F_{X_n}(x,y)$ is finite for every $y\in\mathbb R$ (except maybe at discontinuities of $F_{XY}$ which are at most countably infinite) as well as $\lim_{y\to+\infty}F_{X_nY_n}(x,y)$ is finite for every $n\in \mathbb N$ and the convergence is uniform.

2) It's worth noting that incidentally it's also been proven that if $(X_n,Y_n)\xrightarrow{\mathcal{D}}(X,Y)$, then $X_n\xrightarrow{\mathcal{D}}X$ and $Y_n\xrightarrow{\mathcal{D}}Y$.


Big O proof for logarithmic function Taking a class in discrete math and going through one of the textbook examples, I am asked to prove that

$T(n)$ is $O(\log{}n)$

Where $T(n)= 5\log_{2} (2n) +7$.
I understand that this means that I must prove that $5\log_{2} 2n +7 \leq\lambda \log(n)$ for some $\lambda$, for $n$ in a neighborhood of infinity. How exactly would I proceed with this?

You can write inequality as $\lambda \geq 5+\frac{12}{\log_2 n}$ which is satisfied by all $\lambda \geq 17$ and constant $n_0=2$. This gives a tight upper bound and also proves that $T(n)=O(\log_2 n)$.

You can use some simple rules to work with O.

First rewrite $T(n)$ as $5(1+\log n)+7=5\log n+12$. Then combine these rules:

$12=O(1)$, and $f=O(1)\implies f=O(\log n)$
$5\log n=O(\log n)$ by definition
$O(\log n)+O(\log n)=O(\log n)$

Detailed computations:

$12<\log n$ for $n>2^{12}=4096$, hence $T(n)<6\log n$ for $n>4096$.


A rigorous computation of the mean time to empty a stack of given size We are given a set of M objects in a stack S. In each iteration, an object (say O) is popped out of the stack. With probability $p_1$, O gives rise to 2 new objects which are added to S, thus S now contains $M+1$ objects. With probability $p_2$, O gives to rise to 1 new object and added to S, thus S contains M objects . With probability $1-(p_1 + p_2)$, O does not generate any new object, so S contains $M-1$ objects.
How to compute the expected number of iterations before which the stack becomes empty?

An intuitive approach-: If we denote $z$ to be the change in the number of objects in S at each iteration, then $E[z] = p_1\cdot 1 + p_2 \cdot 0 + (1-p_1-p_2)\cdot -1 = −1+2p_1+p_2$. So the expected number of iterations for stack to become empty is $M/(1-2 p_1 - p_2)$. Further assume, $1-2p_1-p_2 > 0$, to prevent unbounded stack growth.
If the intuitive approach is correct, can someone provide me a hopefully simple but rigorous proof as to why it is correct.
More specifically, I am trying to wrap my head around the idea that without computing a probability distribution of the number of items in S at each iteration, simply computing the expectation of the change in number of items is sufficient.

One wants to compute every $E(T_M)$, where $T_M$ denotes the number of moves necessary to empty a stack of size $M$.

Until the stack becomes empty, its size performs a random walk on the set of nonnegative integers $M$, whose steps from every positive $M$ are $M\to M+1$, $M\to M$ and $M\to M-1$, with respective probabilities $p_1$, $p_2$ and $p_3=1-p_1-p_2$. Thus, if $p_1\geqslant p_3$, every $E(T_M)$ is infinite.

If $p_1<p_3$, that is, if $2p_1+p_2<1$, we first make use of the elementary remark that, to empty a stack of size $M$, one must first reach a stack of size $M-1$, then reach a stack of size $M-2$, and so on until a stack of size $0$. By homogeneity, this shows that $$E(T_M)=ME(T_1)$$ for every positive $M$. On the other hand, starting from size $1$, the usual one-step decomposition yields
$$E(T_1)=1+p_1E(T_2)+p_2E(T_1)+p_3\,0$$
thus, $$E(T_1)=\frac1{1-2p_1-p_2}$$ and, finally, for every $M$,

  $$E(T_M)=\frac M{1-2p_1-p_2}$$

Let $X_n$ denote the stack size at step $n$, and let $Z_n$ denote the change in stack size on the $n$th step. Then $X_0=M$, $X_{n}=X_{n-1}+Z_{n-1}=X_{n-2}+Z_{n-2}+Z_{n-1}=\ldots=X_0+(Z_1+\ldots+Z_n)$; that is, $$X_n=M+\sum_{i=1}^nZ_i. $$
Therefore
$$E[X_n]=M+n\,E[Z_1]$$
assuming that the $Z_i$ are iid; consequently, the value of $n$ for which $E[X_n]=0$ is such that  $0=M+n\,E[Z_1]$. Thus, using the result for $E[Z_1]$ that you've already found, $$n=-{M\over E[Z_1]}={M\over 1-2 p_1 - p_2}.$$


Show that this is a stopping time Show that $\sigma=\inf \{ t\ge 0 : |B_t|= \log t \}$  is a stopping time with
respect to $(\mathcal F_t^B)_{t\ge0}$.

I've been trying to put the set $\{\sigma\le t\}$ equal to a countable union and then showing that this union belongs in $\mathcal F_t^B$. I am struggling to derive a countable union to this as I am only familiar with when $B_t$ is equal to a number and not a function of $t$.

definitions
$(\Omega,\mathcal F, \mathbb P)$ is a probability space.  $(\cal F_t)_{t \ge 0}$ is a stochastic basis: that is, for each real $t \ge 0$, $\mathcal F_t$ is a sigma-algebra contained in $\mathcal F$, and if $s \le t$ then $\mathcal F_s \subseteq \mathcal F_t$.  Perhaps we even assume the stochastic basis is right-continuous, namely
$$
\mathcal F_t = \bigcap_{s>t}\mathcal F_s
$$
for all $t$.  $(B_t)$ is a stochastic process adapted to $(\mathcal F_t)$: that is, for each $t$, $B_t : \Omega \to \mathbb R$ is a random variable, measurable with respect to $\mathcal F_t$.  Define random variable $\sigma$ as above, $\sigma :  \Omega \to [0,+\infty]$.  We want to show $\sigma$ is a stopping time for $(\mathcal F_t)$: that is, for every real $t$,
$$
\{\sigma \le t\} := \{\omega \in \Omega : \sigma(\omega) \le t\}
$$
belongs to $\mathcal F_t$.

Hint:
$$[\sigma\leqslant t]=\bigcup_{r\in\mathbb Q,r\leqslant t}[|B_r|=\log r]$$
Of course, to see why this identity holds is part of the fun...

Recall that a random variable $\tau : \Omega \to [0, \infty]$ is a stopping time with respect to the filtration $(\mathscr{F}_t)_{t \geq0 }$ if $\{ \tau \leq t \} \in \mathscr{F}_t$ for all $t \geq 0$. Therefore, to see that $\tau = \inf \{ t \geq 0 : \left| B_t \right| = \log t \}$ is a stopping time, consider that \begin{eqnarray*}
\{ \tau \leq t \} &=& \bigcap_{0 \leq s \leq t} \{ \omega : \left| B_s(\omega) \right| = \log s \} \\
&=& \bigcup_{s \in [0,t] \cap \mathbb{Q}} \{ \omega : \left| B_s(\omega) \right| = \log s \},
\end{eqnarray*} where the last equality follows from the right-continuity of the Brownian motion and the density of $\mathbb{Q}$ in $\mathbb{R}$. Now the sets $\{ \omega : \left| B_s(\omega) \right| = \log s \}$ are all $\mathscr{F}_s$-measurable, for $s \leq t$. Moreover, since the filtration is increasing, these sets are all $\mathscr{F}_t$-measurable. Since we have a countable union of $\mathscr{F}_t$-measurable sets, it follows that $\{ \tau \leq t \} \in \mathscr{F}_t$ and $\tau$ is a stopping time.


Determining whether a given function can be considered as a probability function Recently I was given the following problem.

The sample space $\Omega = \{ 1, 2, 3, 4 \}$ and the function P is defined as follows:

$$P(\emptyset) = 0, P(\Omega) = 1, P(\{1,2\}) = \frac12, P(\{3,4\}) = \frac12, P(\{3\}) = \frac13, P(\{4\}) = \frac16 , P(\{1,2,3\}) = \frac56, P(\{1,2,4\}) = \frac23.$$

The question is whether $P$ can be considered a probability function.

My issue with this problem is that I am not sure whether it is definitely possible to answer the question. On the one hand, if P is a probability function, then for each two mutually exclusive events the sum of their probabilities should be $1$. In this case, $\{1\}$ and $\{2,3,4\}$ are mutually exclusive events, but $P(\{1\}) + P(\{2,3,4\})$ is not known. On the other hand, there is not enough information to claim the opposite. It seems to me that the question itself is incorrect, but I am not quite sure because I do not have enough experience in probability theory.

So, is $P$ a probability function?

When I took probability 1, there was a triple $(\Omega, P, \mathcal F)$ consisting of a sample space, a probability measure, and a collection of events, subsets of $\Omega$ to which probabilities are assigned.  The domain of $P$ is  $\mathcal F$.

So I'd give a different answer to this question.  The set of events $\mathcal F$ has not been specified, so we can choose it.  If we take $\mathcal F$ to be all subsets of $\Omega$, then Jeremy's answer is correct: at best $P$ is incompletely specified.  But if we take $\mathcal F$ to be generated by just the subsets $\{1,2\}, \{3\}, \{4\}$, then $P$ is a probability on that $\mathcal F$.  If asked, in this case, what $P(\{1\})$ was,  we could answer: ha-ha, $\{1\}$ is not measurable, is not an event, is not in $\mathcal F$, and it doesn't have (and doesn't need) a probability.

This seems like a trick question. Much information is provided about a possible probability function, $P$. Since the sample space is $\omega = \{1, 2, 3, 4\}$, the probability function must describe the probability of each element of the sample space. Events $1$ and $2$ are defined in three of equalaties together. Since the probability of events $1$ and $2$ aren't defined individually and they are part of the sample space this probability function $P$ is incomplete and not fully defined.

$P$ is not a probability function.