A Basic Limit From Exponentials Reading the proof of exponential derivatives I understand this:
To show that $(2^x)'=\ln 2 \cdot 2^x$ in the proof is used the limit:
$$\lim_{x \to 0} \frac{2^x-1}{x}$$
My question is: ¿How do I prove that this limit exist?
I don't care about its value. If I were going to prove that this limit is equal to
$\ln 2$, I would need the number $e$, and again, this number is defined as the number $a$ such that:
$$\lim_{x \to 0} \frac{a^x-1}{x} = 1$$
In another words if I know that for some constant value $a$ the limit:
$$\lim_{x \to 0} \frac{a^x-1}{x} $$
exists, let's say it's a number $L$ then I could find all the limits like these in function of $L$,
But what makes obvious that this limit exist?

Let $a>0$ and $a\ne1$. We want to prove the existence of  $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.
Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then
$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\
=0 &\text{if }x=1\\
>0 &\text{if }x>1 \end{cases}$$
Therefore, $f$ attains its absolute minimum at $x=1$. So  for all $x>0$, we have
$$f(x)\ge f(1)=0$$
$$x^r\ge rx+1-r$$
So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0
\end{align}
When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0
\end{align}
Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.
When $h<0$,
$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$
As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.
Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.
Source: How does one prove that $e$ exists?


Why is a Zero Vector Space a Vector Space? The first example in my Linear Algebra textbook states this is so by stating the example satisfies axiom $4$, which says there is zero vector, there is an object $u$ in which $0 + u = u + 0 = u$, where $0$ is the zero vector and $u$ is an object or a set of real number.

Is there not already a violation in axiom 1 where the zero space is not closed under addition because the vector space is missing another object v?

The zero set is closed under addition: $0+0=0$. It is closed under scalar multiplication: $\lambda\cdot 0=0$. It is non-empty. Hence it is a vector space, and in fact the only one of dimension $0$ (up to isomorphism and once the field is fixed).

The zero set satisfies the two properties of a subspace.

i) Closed under addition. $0+0=0$

ii) Closed under scalar multiplication $\lambda(0)=0$.

My previous edit was naively wrong.


Finr the first term and the difference of an arithmetic progression, given two relations between its terms In an AP, the 15th term is double the 9th term. If also, the sum of the first 15 and the sum of the first 9 terms added together is 279, find the first term and the common difference.

Alright, so we have to make two equations and solve them simultaneously.

First equation is pretty easy:

$$a + 14d = 2(a+8d)  $$

How do we make the second equation? the formula for sum of an AP is:

$S_n = (\frac n2)[2a + (n-1)d]$ where a is the first term and d is the common difference

We can't use the value of 279 because  that's for the 15th and the 9th term.

The sum of the first $15$ terms is $\frac{15}2 (2a + 14d)$, and the sum of the first $9$ terms is $\frac{9}{2}(2 a + 8d)$.

Their sum together is $279$, so
$$\frac{15}2 (2a + 14d) + \frac{9}{2}(2 a + 8d) = 279$$

is your second equation.

Each new term equals the previous term plus the common difference. So for 9 terms we add the common difference 8 additional times for a total of sum of 1 to 8 with the common difference of 1, which is 36. Hence the second equation:
         9a+36d=279 or a+4d=31
solve with the first equation of
         2(a+8d)=a+14d or a=-2d
The solution is:
  a=-31
  d=15.5


Ideal of $TV$ which trivially intersects $V$ Let $V$ be a vector space over a fied $\mathbb{K}$ and let $TV = \bigoplus_{n \in \mathbb{N}} V^{\otimes n}$ be its tensor algebra. Let $b \colon V \times V \rightarrow \mathbb{K}$ be a bilinear form over V and let $I \subseteq TV$ the ideal of $TV$ generated by the elements of the form $v \otimes v + b(v,v)$.

I need to show that, identifing $V$ with $V^{\otimes 1}$ in $TV$, one has $V \cap I = \{ 0 \}$. It should be quite obvious, but I can't get it.

Edit: I have found a proof in Lawson, Michelsohn: Spin Geometry, but there's a passage which I don't understand:

  Let $\varphi = \sum a_i \otimes(v_i \otimes v_i + b(v_i,v_i)) \otimes b_i \in V \cap I$, where we may assume that the $a_i$'s and $b_i$'s are homogeneous. Let us now consider the sum $\overline\sum a_j\otimes (v_j \otimes v_j) \otimes b_j$, where this sum is taken over those indices with $\operatorname{deg} a_i + \operatorname{deg} b_i$ maximal. Since $\varphi \in V$, $\overline\sum a_j \otimes (v_j \otimes v_j) \otimes b_j = 0$. Then, by contraction with $b$, $\overline\sum b(v_j,v_j) ( a_j \otimes b_j ) = 0$. Proceeding inductively, we prove that $\varphi = 0$

Why do $\overline\sum b(v_j,v_j) ( a_j \otimes b_j ) = 0$? What do they mean by contraction with $b$?

Generators of $I$ are quadratic (not homogeneous because of the additive constant from b(v,v). ) So the ideal generated by them can not intersect lower homegeneous degree elements.


What values of $0^0$ would be consistent with the Laws of Exponents? I am using the following fundamental properties of exponentiation on $N$ as as basis for this discussion:

(1) $0^1 = 0$

(2) $\forall x\in N (x\ne 0 \implies x^0 = 1)$

(3) $\forall x,y\in N (x^{y+1}=x^y\cdot x)$

Missing, of course, is a value for $0^0$. But only $0^0=0$ or $1$ are consistent with the Laws of Exponents:

(4) $\forall x,y,z\in N (x^{y+z}=x^y\cdot x^z)$

(5) $\forall x,y,z\in N (x^{y \space\cdot z}=(x^y)^z)$

EDIT:

From (5), we must have $(0^0)^2=0^{0\times 2}=0^0$. Therefore, $0^0= 0$ or $1$. Is this correct?

Is there any way to eliminate $0$ (or $1$) as a possible value, with reference to the fundamental properties or the laws of exponents?

If $c,d$ are cardinal numbers, then $c^d$ is the cardinal of the set of maps $d \to c$. This works for all cardinal numbers and the usual arithmetic laws hold. For $d=0$ we get $c^0=1$ since there is a unique map $\emptyset \to c$. This holds for all $c$, in particular $0^0=1$. So there is actually no debate what $0^0$ is or not, it is $0^0=1$ by the general definitions. No case distinctions are necessary. Forget about $0^0=0$, this is nonsense.

Follow-up

It can be shown that there exists not just 1 or 2 "exponent-like" functions on N as I suggested here, but an infinite number of them. And from any one of them, we can derive the usual Laws of Exponents. For formal proofs, see Oh, the ambiguity! at my blog.


What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.

  What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?

I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.

For $n < 5$:

$$\begin{align*}
1! &\equiv 1 \pmod{5} \\
2! &\equiv 2 \pmod{5} \\
3! &\equiv 1 \pmod{5} \\
4! &\equiv 4 \pmod{5} \\
\end{align*}$$

So $1! + 2! + \cdots + 100! \equiv 8 \equiv 3 \pmod{5}$.
Therefore the remainder is $3$.
Am I thinking properly?

Thanks a lot.

To complement the other answer in less theory-heavy terms, you can view it as:

$$1! + 2! + 3! + 4! + 5! (1 + 6 + [6 \times 7] + \dots + [6 \times \dots \times 100])$$

The final term is clearly divisible by $5$; so we just need to look at $1! + 2! + 3! + 4!$ and its remainder, as you did.

You are right.

Notice that the unit digit of the sum is $3$, because all the terms in the sum end up with a zero from $5!$ and hence to find the unit digit, we just need to find the unit digit of the sum $1!+2!+3!+4!=33$ which is $3$.

So, when the sum $1! + 2! + 3! + \cdots + 100!$ is divided by $5$, it leaves a remainder $3$.


Time shifted function in differential equation from a control systems problem Suppose we had the following differential equation with zero initial conditions:
 $$\ddot{x}\left(t\right)+2\dot{x}\left(t\right)+x\left(t\right)=0.5u\left(t\right)$$
where $u\left(t\right)$ is given via the following graph:

Analytically  $$u\left(t\right)=2t(h(t)-h(t-1))$$
where $h(t)$ is the Heaviside unit step function.
This leaves us with the following ODE which I intend to solve using the superposition principle along with some other tricks:
$$\ddot{x}\left(t\right)+2\dot{x}\left(t\right)+x\left(t\right)=t(h(t)-h(t-1))$$

Firstly, I considered the differential equation:
$$\ddot{x}\left(t\right)+2\dot{x}\left(t\right)+x\left(t\right)=t(h(t))$$ with zero initial conditions.
It has a solution in the form of:
$$ x_1(t)=(2e^{-t}+te^{-t}+t-2)h(t)$$
The next step is to solve the differential equation$$\ddot{x}\left(t\right)+2\dot{x}\left(t\right)+x\left(t\right)=-t(h(t-1))$$ The initial conditions here ought to be $x(1^-)=x_1(1^-)=0.1036$ and $\dot{x}(1^-)=\dot{x_1}(1^-)=0.264$
Since "we are  starting at $t=1$", the solution for the homogeneous part should be in the form of:
$$x_{2h}=Ae^{-(t-1)}+B(t-1)e^{-(t-1)}$$ and we have one more solution in the form of $$x_{2p}=Ct+D$$
Using the initial conditions for $t=1^-$, I arrive to the solution:
$$x_2(t)=h(t-1)(-0.8964e^{-(t-1)}+0.3676(t-1)e^{-(t-1)}-t+2)$$
With the final solution being $x(t)=x_1(t)+x_2(t)$.
Here it is graphically shown, alongside with the system input as $0.5u(t)$:

I am sure there shouldn't be a discontinuity at $t=1$, from an engineering standpoint, the input suddenly vanishes, but the system gets an additional kick of energy. Makes no sense. I hope someone can point out where I made a mistake, assuming no numerical errors were made.

Either you compute the solution of
$$
L(u)=A+B
$$
as superposition (sum) $u=u_A+u_B$ of the solution of $L(u_A)=A$ and $L(u_B)=B$

or

you split the domain into subintervals with "standard" solutions and adapt the constants so that you get a differentiable solution.

Mixing both will result in problems. Since you largely follow the first approach, and the Heaviside functions "switch on" parts of the right side, your initial conditions for the second function should be, to enable a smooth sum, $x_2(1)=0=\dot x_2(1)$ and your full solution is then $x(t)=x_1(t)+x_2(t)$.

To demonstrate how the domain splitting works: First, for $t<0$ the zero initial conditions imply for $L(u)=(\frac d{dt}+1)^2u=0$ the zero solution.

On $0\le0\le 1$ you get $L(u)(t)=t$ which with the trial solution $a+bt$ gives $(2b+a)+bt=2t$ or $b=1$, $a=-2$, thus
$$
u(t)=(A+Bt)e^{-t}+t-2=(A+Bt)(1-t+O(t^2))+t-2=A-2+(-A+B+1)t+O(t^2)
$$
resulting in $A=2$, $B=1$ for differentiable continuation of the zero solution at $t\le0$.

On the third segment $t\ge 1$ one has again $L(u)=0$ thus
$$
u(t)=(A+Bt)e^{-t},\ u'(t)=(B-A-Bt)e^{-t}
$$
so that at $t=1$
$$
(A+B)e^{-1}=3e^{-1}-1, \ -Ae^{-1}=-2e^{-1}+1
$$
which implies $A=2-e$, $B=1$.

Thus
\begin{align}
u(t)&=\begin{cases}
0&t<0\\
(2+t)e^{-t}+t-2&0\le t\le 1\\
(2-e+t)e^{-t}&1<t
\end{cases}
\\
&=[(2+t)e^{-t}+t-2]h(t)-(e^{1-t}+t-2)h(t-1)
\\
&=[(2+t)e^{-t}h(t)-e^{1-t}h(t-1)+(t-2)(h(t)-h(t-1))
\end{align}

$x_1(t)$ must equal zero for $t\geq1$, which it currently does not. The correct solution could read something like this:
\begin{equation}
x_1(t)=(2e^{-t}+te^{-t}+t-2)\big(h(t)-h(t-1)\big).
\end{equation}


Change the bounds of a summation formula I'm currently reading my notes and this appeared and I tried making the left side change into the right side but it's not working. I know you're supposed to use a change in lower bound and use the summation identity but I still can't get n(n-1)/2 appear. Isn't the identity of summation j = n(n+1)/2?

Please check your result once again.

$$\begin{align}3 \sum_{j=2}^n j - (n-1) & = 3 \left(\sum_{j=1}^n j - 1\right) - n + 1 \\ & = \frac{3n(n+1)}{2} - 3 - n + 1 \\ & = \frac{3n(n+1)}{2} - n - 2\end{align}$$

Use something like $$\sum_{j=27}^\infty a_j\;=\;\sum_{j=1}^\infty a_{j+26}$$.


Trouble with showing continuity of an integral. Let $f\in L^1(0,\infty)$. Define $g(t)=\int_0^\infty e^{-tx}f(x)dx$. Prove that $g$ is bounded and continuous on $[0,\infty)$ and $\lim_{t\to\infty}g(t)=0$.

$\lim_{t\to \infty} g(t) = \lim_{t\to \infty} \int_0^\infty e^{-tx}f(x)dx$. Can we say this is the same as $\int_0^\infty \lim_{t\to \infty} e^{-tx}f(x)dx = \int_0^\infty 0f(x)dx=0$?
Note that $\{e^{-tx}\}$ is a positive decreasing sequence so for all $x\in (0,\infty)$.  it obtains it maximum when $t=1$. So $|e^{-nx}f(x)| < e^{-x}f(x)$
furthermore for each $t$ $e^{-tx}$ is a positive decreasing function so its supremum is on the interval $(0,\infty)$ is $1$ which implies  $|e^{-nx}f(x)| < e^{-x}f(x)<f(x) \in L^1$. In otherwords for $h_t=e^{-tx}f(x)$, $|h_t|\leq f(x)\in L^1$. Is $\{h_t\}\in L^1$? Yes, of course because it is bounded from above by a $L^1$ function. Thus we know that we CAN pass the limit through the integral and so $\lim_{t\to\infty}g(t)=0$.

Claim: $g(t)$ is bounded. Since $\{e^{-tx}\}\leq 1$ for all $t$ and all $x \in (0,\infty)$ we have $\int_0^\infty e^{-tx}f(x)dx\leq \int_0^\infty f(x)dx <\infty$ since $f(x) \in L^1$.

Claim: $g(t)$ is continuous. I attempt to show this using an $\epsilon$, $\delta$ proof but run into problems.

I wonder, how can I find $\delta$, is there another more convenient way to show that $g(t)$ is continuous, and lastly are there any other concerns with the validity of this proof? Thanks in advance!

This is what I have tried so far...

Let $\epsilon>0$. We want to find a $\delta$ s.t. if $|t_2-t_1|<\delta \implies |g(t_2)-g(t_1)|<\epsilon$. Consider $|g(t_2)-g(t_1)|=|\int_0^\infty e^{-t_2x}f(x)dx-\int_0^\infty e^{-t_1x}f(x)dx|=|(e^{t_2}-e^{t_1})\int_0^\infty e^{-x}f(x)dx|\leq |(e^{t_2}-e^{t_1})||\int_0^\infty e^{-x}f(x)dx|$ and we want this quantity less than $\epsilon$. But we know that $e^{-x}f(x) \in L^1$ so we can let $|\int_0^\infty e^{-x}f(x)dx|=||h_n||_1=M$ some number in the reals. So we want $|e^{t_2}-e^{t_1}|<\epsilon /M$.

$g(t)$ is continuous. This follows from the Dominated Convergence Theorem and the fact that $g(t_1)=\lim_{t\to t_1} g(t)= \int_0^\infty \lim_{t\to t_1} e^{-tx}f(x)dx= \int_0^\infty  e^{-t_1 x}f(x)dx$.


Calculating an improper integral involving $\log$ Could somebody help me to solve the integral below?
$$\int\limits_0^1\frac{\ln x}{x^{a}(1-x)^{b}}\,\mathrm{d}x,$$
where $a,b \in (0,1)$.
It may be helpful to know the definition of the Beta function:
$$\textbf{B}(r,s):=\int\limits_0^1x^{r-1}(1-x)^{s-1}\mathrm{d}x, \quad r,s>0.$$

Thanks!

It is straightforward to see by differentiation under the integral that we have

\begin{align}I(a,b)&=\int_0^1\frac{\ln(x)}{x^a(1-x)^b}~\mathrm dx\\&=\frac\partial{\partial a}\int_0^1\frac1{x^a(1-x)^b}~\mathrm dx\\&=\frac\partial{\partial a}\mathbf B(1-a,1-b)\end{align}

which can further be written out using the Gamma function and it's derivatives given in terms of the digamma function.

This sounds like homework, and you did not include what you have done.

Anyway, another way to write this in terms of the beta function and its derivative is to integrate by parts.


Prove Existence of a Circle There are two circles with radius $1$, $c_{A}$ and ${c}_{B}$. They intersect at two points $U$ and $V$. $A$ and $B$ are two regular $n$-gons such that $n >  3$, which are inscribed into $c_{A}$ and ${c}_{B}$ so that $U$ and $V$ are vertices of $A$ and $B$.

Then suppose a third circle, $c$, with a radius of $1$ is to be placed so that it intersects $A$ at two of its vertices $W$ and $X$ and intersects $B$ at two of its vertices $Y$ and $Z$.

Details and Assumptions:

Assume that $U,V,W,X,Y,Z$ are all distinct points.
$U$ lies outside of $c$.
$V$ lies inside of $c$.

Given all of these details, prove that there exists a regular $2n$-gon which comprises of $W,X,Y,Z$ as its 4 vertices.

This answer focuses on identifying families of solutions to the problem described in the question.

I've made two provisional conjectures in order to make progress with the problem:

The result can be stated for three $2n$-gons rather than two $n$-gons and one $2n$-gon.
Solutions have mirror symmetry. Or equivalently, in any solution there are two pairs of $2n$-gons which have the same degree of overlap. [This turns out to be false - see 'Solution family 5' below. However, this condition is assumed in Solution families 1-4.]

[Continuation 6: in an overhaul of the notation I've halved $\phi$ and doubled $m$ so that $m$ is always an integer.]

If we define the degree of overlap, $j$, between two $2n$-gons $(n>3)$ as the number of edges of one that lie wholly inside the other, then $1 < j < n$.

If
$$
\phi = \frac{\pi}{2n}
$$
is half the angle subtended at the centre of the $2n$-gon by one of its edges, then  the distance between the centres of two overlapping $2n$-gons is
$$
D_{jn} = 2\cos{j\phi}
$$
Consider a $2n$-gon P which overlaps a $2n$-gon O with degree $j$. Now bring in a third $2n$-gon, Q, which also overlaps O with degree $j$ but is rotated about the centre of O by an angle $m\phi$ with respect to P, where $m$ is an integer.

The distance between the centres of P and Q, which I'll denote by $D_{kn}$ for a reason that will become apparent, is
$$
D_{kn} = 2D_{jn}\sin{\tfrac{m}{2}\phi} = 4\cos{j\phi} \, \sin{\tfrac{m}{2}\phi}
$$

We now demand that P and Q should overlap by an integer degree, $k$, so that
$$
D_{kn}  = 2\cos{k\phi}
$$
This will ensure that all points of intersection coincide with vertices of the intersecting polygons, and thus provide a configuration satisfying the requirements of the question (with the proviso that the condition does not guarantee that there is a common area of overlap shared by all three polygons).

We have omitted mention of the orientation of the polygons, but it is easily shown that this is always such as to achieve the desired overlap.

Combining the two expressions for $D_{kn}$ gives the condition

$$
2\cos{j\phi}\, \sin{\tfrac{m}{2}\phi} = \cos{k\phi}
$$
or (since $n\phi=\pi/2$)
$$
2\cos{j\phi}\, \cos{(n-\tfrac{m}{2})\phi} = \cos{k\phi} \tag{1}
$$

The configurations we seek are solutions of this equation for integer $n$, $j$, $k$ and $m$.

In the first example in the question $n = 12, j = 8, k = 6, m = 12$.

In the second example $n = 15, j = 6, k = 10, m = 6$.

[Continuation 6: for solutions under the constraint of conjecture 2, $m$ is always even, but in the more general case $m$ may be odd.]

I'll now throw this open to see if anyone can provide a general solution. It seems likely that $j$, $k$ and $m/2$ must be divisors of $2n$ [this turns out to be incorrect], and I have a hunch that the solution will involve cyclotomic polynomials [this turns out to be correct].

Continuation (1)

I've now identified 3 families of solutions consistent with conjecture 2 (mirror symmetry), all involving angles of 60 degrees. There may be others.

Solution family 1

This family is defined by setting $j=2n/3$. This means that half the angle subtended at the centre of O by its overlapping edges is $\tfrac{\pi}{3}$ radians or 60 degrees. Since $\cos{\tfrac{\pi}{3}} = \tfrac{1}{2}$ it reduces equation 1 to
$$
\cos{(n-\tfrac{m}{2})\phi} = \cos{k\phi}
$$
so there are solutions with
$$
n-\tfrac{m}{2} = k
$$
(where $\tfrac{m}{2}$ is an integer) subject to $2 \le k \le n-1\,\,$, $1 \le \tfrac{m}{2} \le n-2\,\,$ and $3|n$.

The first example in the question belongs to this family. The complete set of solutions for $n=12$ combine to make this pleasing diagram:

Solution family 2

This family has $m=2n/3$. This makes $\cos{(n-\tfrac{m}{2})\phi}=\cos{(\pi/3)} = \tfrac{1}{2}$, which reduces equation 1 to
$$
\cos{j\phi} = \cos{k\phi}
$$
so (given that $j<n$ and $k<n$)
$$
j = k
$$
These solutions have threefold rotational symmetry. The only restriction is that $n$ must be divisible by 3. Example ($n=9, j=k=4, m=6$):

Solution family 3

This family is the most interesting of the three, but yields only one solution. It is defined by setting $k=2n/3$ so that $\cos{k\phi}=\cos{\tfrac{\pi}{3}} = \tfrac{1}{2}$. Equation 1 then becomes

$$
2\cos{j\phi}\,\cos{(n-\tfrac{m}{2})\phi} = \tfrac{1}{2}
$$
which may be written in the following equivalent forms:
$$
\cos{(n+\tfrac{m}{2}-j)\phi} + \cos{(n+\tfrac{m}{2}+j)\phi} = -\tfrac{1}{2} \tag{2}
$$
$$
\cos{(n-\tfrac{m}{2}-j)\phi} + \cos{(n-\tfrac{m}{2}+j)\phi} = \tfrac{1}{2} \tag{3}
$$
Solutions to these equations can be found using the following theorem relating the roots $z_i(N)$ of the $N$th cyclotomic polynomial to the Möbius function $\mu(N)$:

$$
\sum_{i=1}^{\varphi(N)} {z_i(N)} = \mu(N)
$$
where $\varphi(N)$ is the Euler totient function (the number of positive integers less than $N$ that are relatively prime to $N$) and $z_i(N)$ are a subset of the $N$th roots of unity.
Taking the real part of both sides and using symmetry this becomes:
$$
\sum_{i=1}^{\varphi(N)/2} {  \cos{(p_i(N) \frac{2\pi}{N})}  } = \tfrac{1}{2} \mu(N) \tag{4}
$$
where $p_i(N)$ is the $i$th integer which is coprime with $N$.

The Möbius function $\mu(N)$ takes values as follows:

$\mu(N) = 1$ if $N$ is a square-free positive integer with an even number of prime factors.

$\mu(N) = −1$ if $N$ is a square-free positive integer with an odd number of prime factors.

$\mu(N) = 0$ if $N$ has a squared prime factor.

Equation 4 thus provides solutions to equations 2 and 3 if $\varphi(N) = 4$, $\mu(N)$ has the appropriate sign and the cosine arguments are matched.

The first two conditions are true for only two integers:

$N=5$, with $\mu(5)=-1$, $p_1(5) = 1, p_2(5) = 2$

$N=10$, with $\mu(10)=1$, $p_1(10) = 1, p_2(10) = 3$.

We first set $N=5$ and look for solutions to equation 2.

Matching the cosine arguments requires firstly that
$$
2j \frac{\pi}{2n} = (p_2(5)-p_1(5))\frac{2\pi}{5}
$$
from which it follows that
$$
5j = 2n
$$

$n$ must be divisible by 3 to satisfy $k=2n/3$, so the smallest value of $n$ for which solutions are possible is $n=15$, with $k=10$ and $j=6$. All other solutions will be multiples of this one.
Matching the cosine arguments also requires that
$$
(n+\tfrac{m}{2}-j) \frac{\pi}{2n} = p_1(5) \frac{2\pi}{5}
$$
which implies $m=6$.

This is the solution illustrated by the second example in the question.

Setting $N=10$ and looking for solutions to equation 3 yields the same solution.

Continuation (2)

Solution family 4

A fourth family of solutions can be obtained by writing equation 1 as

$$
\cos{(n+\tfrac{m}{2}-j)\phi} + \cos{(n+\tfrac{m}{2}+j)\phi} + \cos{k\phi} = 0 \tag{5}
$$

and viewing this as an instance of equation 4 with $\varphi(N)/2 = 3$ and $\mu(N) = 0$. There are two values of N which satisfy these conditions, $N = 9$ and $N = 18$, which lead to three solutions:

For $N = 9$:
$$
n=9, j=6, k=8, m=2
\\n=9, j=4, k=4, m=6
$$

For $N=18$:
$$
n=9, j=2, k=2, m=6
$$

However, these are not new solutions. The first is a member of family 1 and the last two are members of family 2.

Continuation (3)

Solution family 5

Rotating a $2n$-gon about a vertex by an angle $m\phi$ moves its centre by a distance
$$
2\sin{ \tfrac{m}{2}\phi} = 2\cos{(n-\tfrac{m}{2})\phi} = D_{n-m/2,n}.
$$
If $m$ is even the rotated $2n$-gon thus overlaps the original $2n$-gon with integer degree $n-\tfrac{m}{2}$, and a third $2n$-gon with a different $m$ may overlap both of these, providing another type of solution to the problem.

Solutions of this kind may be constructed for all $n \ge 3$. The diagram below includes the complete set of such solutions for $n=5$. A similar diagram with $n=12$ (but with a centrally placed $2n$-gon of the same size which can only be added when $3|n$) is shown above under Solution family 1.

This family of solutions provides exceptions to conjecture 2: not all groups of three $2n$-gons overlapping in this way show mirror symmetry.

Continuation (4)

If we relax the condition set by conjecture 2, allowing solutions without mirror symmetry, we need an additional parameter, $l$, to specify the degree of overlap between O and P (which is now no longer $j$).

The distances between the centres of the three $2n$-gons are now related by the cosine rule:

$$
D_{nk}^2 = D_{nj}^2 + D_{nl}^2 - 2 D_{nj}D_{nl}\cos{m_k\phi},
$$
where a subscript $k$ has been added to $m$ to acknowledge the fact that $j$, $l$ and $k$ can be cycled to generate three equations of this form. These can be written
$$
\\ \cos^2{J} + \cos^2{L} -  2 \cos{J} \cos{L} \cos{M_k} = \cos^2{K}
\\ \cos^2{K} + \cos^2{J} -  2 \cos{K} \cos{J} \cos{M_l} = \cos^2{L}
\\ \cos^2{L} + \cos^2{K} -  2 \cos{L} \cos{K} \cos{M_j} = \cos^2{J}
$$
where
$$
J = j\phi,\, L = l\phi,\, K = k\phi,
\\M_j = m_j\phi,\, M_l = m_l\phi,\, M_k = m_k\phi
$$

The same result in a slightly different form is derived in the answer provided by @marco trevi.

$M_j$, $M_l$ and $M_k$ are the angles of the triangle formed by the centres of the three polygons. Since these sum to $\pi$ we have
$$
m_j + m_l + m_k = 2n
$$

The sine rule gives another set of relations:
$$
\frac{\cos{J}}{\sin{M_j}}  = \frac{\cos{L}} {\sin{M_l}} = \frac{\cos{K}}{\sin{M_k}}
$$

In general the $m$ parameters are limited to integer values (as can be seen by considering the symmetry of the overlap between a $2n$-gon and each of its two neighbours). But they are now not necessarily even.

Looks like false to me...

For $n=4$, which seems to satisfy $n>3$ requirement, the $A$ and $B$ squares can not intersect at opposite vertices (because they would be the same, $A=B$), so they'll have to share one side. Then there's no $X$ or $Y$ vertex between $U$ and $V$ where $C$ octagon might cross $A$ and $B$...


Is $\lambda < \kappa \leq 2^\lambda$ special? It is known that if cardinal $\kappa$ is such that there is a cardinal $\lambda$ with $\lambda < \kappa \le 2^\lambda$ then any $\kappa$-complete ultrafilter on $\kappa$ is principal.

Do such $\kappa$ have a special name and is this property equivalent to some other well known properties? Is this equivalent to being measurable?

It seems pretty special to me and relates to GCH and critical points of elementary embeddings.

$\mathfrak{c}$ has the property but $\omega$ doesn't.

A consequence is that any ultrafilter on $\mathbb{R}$ closed under countable intersections is principle (assuming CH).

As Asaf points out the name of such $\kappa$s is not a strong limit

$\kappa\text{ is not a strong limit}$

$\iff\neg\forall\lambda\left(\left(\lambda<\kappa\right)\to\left(2^{\lambda}<\kappa\right)\right)$

$\iff\exists\lambda\neg\left(\neg\left(\lambda<\kappa\right)\vee\left(2^{\lambda}<\kappa\right)\right)$

$\iff\exists\lambda\left(\left(\lambda<\kappa\right)\wedge\neg\left(2^{\lambda}<\kappa\right)\right)$

$\iff\exists\lambda\left(\lambda<\kappa\le2^{\lambda}\right)$.

When $\kappa$ is not a strong limit then any $\kappa$-complete ultrafilter
$U$ on $\kappa$ is principal (ie.has the form $\left\{ A\subseteq\kappa\mid a\in A\right\} $). This is proved here.

Therefore in particular no infinite cardinal of the form $2^{\lambda}$
can have a $2^{\lambda}$-complete free ultrafilter, and that includes
$\mathbb{R}$ which has cardinality $2^{\omega}$.

Note that if $U$ is an ultrafilter on $\mathbb{R}$ and CH is false
then there is a $\omega<\kappa<2^{\omega}$ so the above theorem would
enable us to conclude that if $U$ is $\kappa$-complete then it is
principal.

A measurable cardinal is an uncountable cardinal $\kappa$ with a
non-principle $\kappa$-complete ultrafilter and so is a strong limit.
If the GCH is true then $\lambda^{+}=2^{\lambda}$ for every infinite
cardinal so if measurable cardinals exist they are not successor cardinals.


Why are There No "Triernions" (3-dimensional analogue of complex numbers / quaternions)? Since there are  complex numbers (2 dimensions) and quaternions (4 dimensions), it follows intuitively that there ought to be something in between for 3 dimensions ("triernions").

Yet no one uses these. Why is this?

It's because there isn't one! (Indeed, Hamilton was originally searching for such a thing, and found the quaternions instead; it was only later that people understood why he hadn't been successful, initially.)

The quaternions - along with the real numbers and the complex numbers - have a number of nice properties: specifically, they form a real division algebra. This is a mouthful, but basically amounts to:

Addition/multiplication of quaternions satisfy the ring axioms.
We can divide by quaternions.
We can multiply a quaternion by a real (and this "scalar multiplication" satisfies the basic properties it should).

It turns out the only finite-dimensional real division algebras are $\mathbb{R}$, $\mathbb{C}$, and the quaternions; see this. (I include associativity in the definition of algebra: if we allow non-associative algebras, then the octonions also count.)

By the way, there is a way to (sort of) keep going past the quaternions: the Cayley-Dickson construction. This produces things like the octonions and the sedenions, and other delightfully weird algebraic structures. However, it has a couple drawbacks:

Each time you apply Cayley-Dickson, the dimension of the starting algebra doubles. So this won't help us get to $3$.
Also, you keep losing nice properties. Passing from the reals to the complex numbers, we lose order; going from the complexes to the quaternions, we lose commutativity of multiplication. If we keep going, we lose associativity of multiplication, in increasing degrees: the sedenions are even less associative than the octonions, etc.

All of these other answers go way over my head, but this is the way I think about it.

The complex numbers can be represented on a 2-dimensional plane, but they are an extension of the one-dimensional real number line. They are not really 2-dimensional, it's just a convenient way for us to represent them.  When we expand the reals from 1 dimension into 2 dimensions, the corresponding complex numbers must double their dimension as well, going from 2 to 4. That is why they appear to "skip" 3.

Perhaps I am wrong, or this argument was folded into the other explanations, and I just didn't see it.


Proof $f$ is Differentiable. I have been looking through the stack and can not find a direct answer to this question. I recently read that we say $f$ is defferentiable on $I$, if $f$ is differentiable at every point on an open interval $I$. It almost reminds me of the rigorous deffinition of a limt, $\forall\epsilon>0\hspace{0.2cm}\exists\hspace{0.2cm}\delta=\delta(\epsilon)>0$ such that $0<|x-a|<\delta\implies|f(x)-L|<\epsilon$, and how we can use said definition to write all kinds of $\epsilon-\delta$ proofs regarding limits. Proving continuity, proving uniqueness of limits, so on and so forth.

My question is: What is the standard proof writing format (such as $\epsilon-\delta$ proofs for limits) for proving differentiability based of the rigorous definition of a derivative? Assuming there is a rigorous definition of a derivative similar to that of limits, please provide an example proof.

The derivative is a limit. You can just apply the definition of the limit to prove differentiability rigorously. Namely, a function $f$ defined on an open $U \subset \mathbb{R}$ is differentiable at a point $x_0 \in U$ if there is a real number $L$ such that for all $\epsilon>0$ there is a $\delta>0$ such that $0<|x-x_0| < \delta$ implies $\left|\frac{f(x) - f(x_0)}{x - x_0} - L\right| < \epsilon $.

You might wonder, why don't we ever use $\epsilon-\delta$ arguments with derivatives? Well, it's because we don't need to. The proofs of the differentiation rules (e.g., product rule, chain rule) require an $\epsilon-\delta$ argument, but after we've proven these theorems, we never need to use $\epsilon-\delta$ arguments again.

My proof of derivative implies continuity:
Think about this: For all ϵ>0 there exists $δ>0$ such that $0<|x-c|<δ$ implies $|\frac{f(x)−f(c)}{x−c}|<ϵ$.

Start my proof: Since f is derivative at $c ∈ R$,
There exists δ>0 such that $0<|x-c|<δ$ implies $|\frac{f(x)−f(c)}{x−c}|=|f(x)-f(c)|\times \frac{1}{|x-c|}$.

We assign $ϵ=\frac{ϵ_1}{|x-c|}>0$, then $|f(x)-f(c)|*\frac{1}{|x-c|}<ϵ$ implies $|f(x)-f(c)|<ϵ=\frac{ϵ_1}{|x-c|}$.

Hence for all $ϵ>0$, There exists $δ>0$ such that $0<|x-c|<δ$ implies $|f(x)−f(c)|<ϵ$, i.e. we have $\lim_{x→c} f(x)=f(c)$.


Phase-type distribution - moments I was doing my best to calculate moments of phase-type distribution.
Density of phase-type distribution is
$$f(x)=\alpha e^{Sx}S_{0}$$
($\alpha$ is $1\times m$ vector; $S_{0}$ is $m\times 1$ vector; $S$ is $m\times m$ matrix) for all $x>0$, where $e^{Sx}$ is matrix expotential.
So first of all I checked if integral of density over $\mathbb{R}$ is equal to $1$:
$$\int_{0}^{\infty}\alpha e^{Sx}S_{0} \ dx = [\alpha S^{-1}e^{Sx}S_{0}]_{x=0}^{\infty} = -\alpha S^{-1}S_{0}$$
So, the following equality must hold: $-\alpha S^{-1}S_{0}=1$
Now I tried to compute expected value:
$$\int_{0}^{\infty}x\alpha e^{Sx}S_{0} \ dx = [x\alpha S^{-1}e^{Sx}S_{0}]_{x=0}^{\infty} - \int_{0}^{\infty}\alpha S^{-1}e^{Sx}S_{0} \ dx = [-\alpha S^{-2}e^{Sx}S_{0}]_{x=0}^{\infty} = \alpha S^{-2}S_{0}$$
But according to wikipedia it should be equal to:
$$-\alpha S^{-1}\textbf{1}$$

Can anyone tell me where is the mistake?

Note that $S^{-1}S_0 = -1$, your answer is actually the same as that given by Wiki.


Find the sum of the infinite series $\sum_{n=3}^\infty [3/(n(n+3))]$ I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ \frac{1}{n} - \frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.

$$ \sum_{i=3}^\infty\frac{3}{n(n+3)} $$

The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.

You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:

$\frac 13 - \frac 16 + \frac 14 - \frac 17 + \frac 15 - \frac 18 + \frac 16 - \frac 19 + \frac 17 - \frac 1{10} + ...$

Note how, starting from $\frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $\frac 15$. So your entire series "telescopes" to become the really short $\frac 13 + \frac 14 + \frac 15  = \frac{47}{60}$.

This is a special case
of the following theorem:

Given $k$ reals
$(a_k)_{j=1}^k$,
the sum
$\sum_{n=1}^{\infty} \sum_{j=1}^k \dfrac{a_j}{(n-1)k+j}
$
converges
if and only if
$\sum_{j=1}^k a_j= 0$.

This problem is the case
$k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
Therefore the sum converges.


Showing $\operatorname{Aut}(\mathbb{R})$ is abelian I have an exercise (not for a class) that asks whether $\operatorname{Aut}(\mathbb{R})$ (field automorphisms) is abelian.  I know that $\operatorname{Aut}(\mathbb{R})$ is just the trivial group, but is there a nice way to see that it is abelian without knowing the group?

This is Part (c) of a question and Parts (a) and (b) were that $\operatorname{Aut}(\overline{\mathbb{Q}})$ is infinite and nonabelian.

$\mathbb R$ is a $\mathbb Q$-vector space, so every $\mathbb Q$-linear function is in $Aut(\mathbb R)$ and $End_{\mathbb Q}(\mathbb R)$ isn't abelian, so $Aut(\mathbb R)$ isn't abelian.

If you mean $Aut(\mathbb R)$ like the group of ring automorphism of $\mathbb R$, the only one is the identity.


Probability question (Birthday problem) I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.

My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:

$$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$

The basic idea was right, and a small modification is enough.

Line up the people in some arbitrary order. There are, under the usual simplifying assumption that the year has $365$ days, $365^{23}$ possible birthday sequences. Under the usual assumptions of independence, and that all birthdays are equally likely, all these sequences are equally likely. The assumption "equally likely" is not correct, though it is more correct for people than for eagles.

Now we count how many ways we can have precisely $2$ people have the same birthday, with everybody else having a different birthday, meaning different from each other and also different from the birthday of our birthday couple.

The couple can be chosen in $\binom{23}{2}$ ways. For each of these ways, the couple's birthday can be chosen in $365$ ways. And the birthdays of the others can be chosen in what is sometimes called $P(364,21)$ ways. (I have always avoided giving it a name.)  So the number of birthday assignments that satisfy our condition is
$$\binom{23}{2}(365)P(364,21), \quad\text{that is,}\quad \binom{23}{2}(365)(364)(363)\cdots (344).$$
For the probability, divide by $(365)^{23}$.

Remark: Or else argue this way, which may be closer in spirit to your thinking. Pick two people, $i$ and $j$. What is the probability these two have the same birthday, and all other birthdays are different from this one, and different from each other? Whatever $i$'s birthday is, the probability that $j$'s matches it is $\frac{1}{365}$. By the usual birthday argument, the probability that the birthdays of the other people are different, and different from the birthdays couple's, is
$$\frac{364}{365}\frac{363}{365}\frac{363}{365}\cdots\frac{344}{365}.$$
Multiply the above expression by $\frac{1}{365}$.
Finally, sum the result over the $\binom{23}{2}$ ways to choose $i$ and $j$, that is, multiply by $\binom{23}{2}$.

This question has been asked in more ways than there are Buddhas.  Try looking here https://math.stackexchange.com/search?q=birthday+problem

birthday problem - expected number of collisions <-how about this one, you should be able to find an answer from one of the answers of the linked problem.

I am not trying to be an arse; however, with one of the most widely known probability question of the existence of time, you have to check this website before posting.

trying to save you from being harshly downvoted.


Syllogism proof using set notation $$B\subseteq C \wedge B\nsubseteq A^c \Rightarrow A\nsubseteq C^c$$

So I need to prove that $$\exists x \in A\,\ x\notin C^c$$

so far I'm up to the point where I can say that $$\exists x \in C , x \in A$$

I've noticed that if I swap the A and C around then I have my answer but what steps do I take to be able to swap it around or is the ending statement simply false?

$B \subseteq C$ means: $\forall x \ (x \in B \to x \in C)$.
And $B \nsubseteq A^c$ means: $\exists x \ (x \in B \land x \notin A^c)$, i.e: $\exists x \ (x \in B \land x \in A)$.
For $(\exists x∈A)x∉C^c$, see Restricted quantifiers; it is an abbreviation for: $\exists x \ ( x∈A \land x∉C^c)$ that in turn means: $\exists x \ (x∈A \land x∈C)$.
Now we can complete the syllogism:

$\exists x \ (x \in A \land x \in B)$,
$\forall x \ (x \in B \to x \in C)$;

therefore:

$\exists x \ (x∈A \land x∈C)$.

$$B\subseteq C \wedge B\nsubseteq A^c$$

by the second part we have $\exists x\in B,x\notin A^c$, we can change $\notin A^c$ into $\in A$ to get $$\exists x\in B,x\in A$$from the first part of the original statement we can change $B$ to $C$ to get$$\exists x\in C,x\in A$$change the order we are looking at it to get $$\exists x\in A,x\in C$$
changing the order is okay since those statements are simply properties of $x$, there is no order to those properties, they all are just... exists you can say. it is like saying: i am a human and i am standing VS i am standing and i am a human.

we can now change $\in C$ into $\notin C^c$ to get $$\exists x\in A,x\notin C^c$$hence we get $$A\nsubseteq C^c$$


Prove that if $f''(x)+25f(x)=0$ then $f(x)=Acos(5x)+Bsin(5x)$ for some constants $A,B$ Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable.

Prove that if $f''(x)+25f(x)=0$ then $f(x)=Acos(5x)+Bsin(5x)$ for some constants $A,B$

Consider $g(x):= f(x)-Acos(5x)-Bsin(5x)$ where $A$ and $B$ are chosen so that $g(0)=g'(0)=0$ [this portion is given, so can someone explain why it is important?]

For reference, we note that
$g(x):= f(x)-Acos(5x)-Bsin(5x)$
$g'(x):= f'(x)+5Asin(5x)-5Bcos(5x)$
$g''(x):= f''(x)+25Acos(5x)+25Bsin(5x)$

Consider the derivative of $\frac{25}{2}g(x)^2+\frac{1}{2}g'(x)^2$

$25g(x)g'(x)+g''(x)g'(x)$ if and only if $g'(x)[25g(x)+g''(x)]$

Following substitution, we conclude that

$g'(x)[25g(x)+g''(x)]=g'(x)(0)=0$

Therefore, by the Constancy Theorem, $\frac{25}{2}g(x)^2+\frac{1}{2}g'(x)^2=C_0$, where $C_0$ is a constant

$25g(x)^2+g'(x)^2=C$, where $C$ is a constant

Thus,

$25(f(x)-Acos(5x)-Bsin(5x))^2 + (f'(x)+5Asin(5x)-5Bcos(5x))^2=C$

Where do I go from here?

By steps: $g(0)=g'(0)=0$ is important because you have a homogenous differential equation, therefore zero initial data implies that the solution is zero everywhere.

After obtaining that $25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2=C$, use initial data $g(0)=g'(0)=0$.

We chose $A$ and $B$ to get $f(x)-A\cos(5x)-B\sin(5x)=0$  and $f'(x)+5A\sin(5x)-5B\cos(5x)=0$ at $x=0$. Therefore we can find $C$ by putting $x=0$:
$$25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2=C = 25(f(0)-A\cos(0)-B\sin(0))^2 + (f'(0)+5A\sin(0)-5B\cos(0))^2=0.$$

Hence  $$\forall x\quad 25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2 =0.$$
You have the sum of squares equal to zero, therefore each term is zero itself. We can conclude that $\forall x\, (f(x)-A\cos(5x)-B\sin(5x))^2=0$ and, therefore,
$$\forall x\quad  f(x)=A\cos(5x)+B\sin(5x).$$

We conclude that all solutions of the initial differential equation has the above form.

Is it easier just to substitute?
$$f(x)=Acos(5x)+Bsin(5x)$$
then
$$\dot f(x)=-5Asin(5x)+5Bcos(5x)$$
$$\ddot f(x)=-25Acos(5x)-25Bsin(5x)$$
finally:
$$-25Acos(5x)-25Bsin(5x)+25( Acos(5x)+Bsin(5x)) =0 $$
?


How is the second derivitive derived? As everyone knows that the derivitive of a function is notated as $\frac{dy}{dx}$

The question is:
How is the second derivitive $\left(\frac{d^2y}{dx^2}\right)$ notation derived?

This is just notation. Let $y=f(x)$. Then $\frac d{dx}(y)=f'(x)$.  Now
$$f''(x)=\frac d{dx}\left(\frac d{dx}(y)\right)=\left(\frac d{dx}\right)^2(y)=\frac {d^2}{dx^2}(y).$$

Edit: I misread the OP question and thought that practical examples were asked for.

The second derivative arises from considering the first derivative as an independent function, and is obtained by derivation.
Notation (like the other answerer mentioned), is mainly of two types:
Leibniz notation or:
$ {d\over dx} (f(x))$ is the first order derivative
${d\over dx} ({d\over dx}(f(x)) = {d^2\over dx}f(x)$

Lagrange Notation instead uses "prime" marks:
Derivative of $f(x) = f'(x)$, derivative of $f'(x) = f''(x)$ and so on.

Practical example:
$$
f(x) = x^3
$$
$$
f'(x) = {d\over dx}f(x) = 3x^2
$$
$$
f''(x) = {d\over dx}f'(x) = 6x
$$
A pratical example is however the change in kinetic energy of a moving body.


Find the maximum of $x$ satisfies the cubic equation Let $a,b,c,d$ be integers with $1\leq a,b,c,d\leq10$, and $x$ be real such that

$ax^3-bx^2-cx-d=0$. How to find the maximum of $x$ ?

Thanks in advance.

From Descartes' rule of signs, it is evident the cubic $f(x) = ax^3-bx^2-cx-d$ has exactly one positive root for all allowable $a, b, c, d$, so we just need to find its maximum value.

Let $r$ be the positive root of $p(r)=r^3-10(r^2+r+1)$. It is clear $p(x)> 0$ for $x> r$, as $r$ is the only one positive root, where the sign turns from negative to positive.

Now if $x>r$, we have $f(x)\ge p(x)>p(r)=0$.  Thus $f(x)$ cannot have any root $> r$.

Numerically $r \approx 10.993$, but if an exact formula is needed, you need to use a method like Cardano's formula to get $r = \frac13 \left(10+\sqrt[3]{5 (317+3 \sqrt{1401})}+\sqrt[3]{5(317-3 \sqrt{1401})}\right)$.

discriminant of this equation is
$c^2b^2-4(db^3+c^3a)+18abcd-27d^2b^2$ thus when the  discriminant will have max value that equation will have max value for $x$.or use vieta's formula


Velocity Measurement Error Estimate I have 2 position estimates (along with their measurement error) and a difference in time between estimates.  I estimate velocity using

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this.  I assume it has to use Sigma_PosA and Sigma_PosB.  I would also assume it's relative to DeltaT and/or abs(PosA - PosB).  What is the velocity measurement variance/standard deviation?

sigmav = sqrt((sigmaA)2 + (sigmaB)2)  / (DeltaT)

EDIT:

(Corrected an error above-- DeltaT should not be squared.)

It sounds as if the measurements are independent, so the errors are uncorrelated. We want the standard deviation of a linear combination of (two) variables:

$V = \frac{(B-A)}{\Delta_t} = \frac{1}{\Delta_t}B - \frac{1}{\Delta_t}A$

$\sigma_V^2= \sum_i^n a_i^2\sigma_i^2 = (\frac{1}{\Delta_t})^2\sigma_B^2 + (\frac{1}{\Delta_t})^2\sigma_A^2 = (\frac{1}{\Delta_t})^2(\sigma_A^2 + \sigma_B^2)$

$\sigma_V = \sqrt {(\frac{1}{\Delta_t})^2(\sigma_A^2 + \sigma_B^2)} = \frac{\sqrt{\sigma_A^2 + \sigma_B^2}}{\Delta_t}$

(I'm new to math.stackexchange-- gotta say I'm lovin' the MathJax.)

You want the propagation of error, or propagation of uncertainty. Since you're subtracting them, the formula is

sqrt(sigma_posA * sigma_posA + sigma_posB * sigma_posB)


Existence of Subspace so direct sum gives the orignal vector space. Let $V=\{f:X \to \mathbb{R}\}$ be the vector space of real valued functions.

Fix $x_0 \in X$

Define W=$\{f\in V|f(x_0)=0\}$

Is there a vector space $Y$-a subspace of V such that $V=W \oplus Y$

Attempt..

I think No.

Consider the set X to be $\mathbb{R}$.Then we are looking into the real valued functions from real numbers.
choose $x_0=1$

Then we know that W is the subspace of $V$ which consists of functions $f:\mathbb{R} \to \mathbb{R}$ which vanish at $x=1$.

Now for the sake of contradiction I will assume the existence of such a subspace Y such that $V=W \oplus Y$.

Then $Y$ must contain all the functions which do not vanish at $x=1$.
choose $f_1(x)=1+2x$ and $f_2(x)=-3x$

however $f_1(x)+f_2(x)=1-x$ which does not belong to $Y$ contradicting the additive closure property.(note-This is because $f_1(x)+f_2(x)=1-x\in W$)

Am I Correct?

If not please help me solving this one.

Thanks

Think of direct sums as partitioning all the information in the vector space into two completely unrelated pieces of data.

So, your original vector space is all functions.  Now, you're given that one part of your description for "any function" is about its behavior at absolutely every point other than $x_0$.

What information is left to be specified about such a function?

No, it is possible to have such a subspace Y. I have constructed such one.
Let
$$
\begin{align*}
Y&=\{f\in V\mid f(x_0)=0\Rightarrow f(x)=0\text{ for all }x\in X\}\\
&=\{f\in V\mid f(x_0)\neq0\}\cup\{f\equiv 0\}.
\end{align*}
$$
Let $f\in V$.  Consider $f_1\in W$ and $f_2\in Y$ defined by
$$
\begin{align*}
f_1(x)&=\begin{cases}0 & x=x_0\\ f(x) & x\neq x_0\end{cases}\\
f_2(x)&=\begin{cases}f(x) & x=x_0\\ 0 & x\neq x_0\end{cases}
\end{align*}
$$
Then $f(x)=f_1(x)+f_2(x)$ for all $x\in X$.
Therefore $V=W+Y$.
Now, it is easy to check that $W\cap Y=\{0\}$. So, $V=W\oplus Y$.


Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$. A function $f : (a, b) \rightarrow \mathbb{R}$ is said to be increasing on $(a, b)$ if $f(x)\leq f(y)$ for all $x < y \in (a, b)$. Suppose that $f$ is differentiable on $(a, b)$. Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$.

For the backward direction, I try to use Mean Value Theorem to show but I don't know how to show that $f$ is continuous on $[a,b]$. Can anyone guide me ?

You don't need continuity on $[a,b]$, you only need it on $[x,y]$. And you have that, since $x$ and $y$ are in $(a,b)$, and our function is differentiable and therefore continuous on $(a,b)$. So it is continuous on $[x,y]$.

Now apply the MVT to conclude that $\frac{f(y)-f(x)}{y-x}=f'(c)$ for some $c$ in the interval $(x,y)$, and use the fact that $f'(c)\ge 0$.

You did not ask about the other direction, so I assume that is under control.

if a function is differentiable on (a, b), then it is continuous on (a,b). so the condition's given, you don't have to prove it.
and for the converse part, you have to use the definition, not the mean value theorem cause mvt saying about ONE C in between x,y. the converse part you need to prove f'(x) >0 for ALL c in(a,b).


On which subspace $W\subset C^{\infty}[0,1]$ is $D(f)=xf'(x)$ a bounded operator provided all functions in $W$ are flat functions? First we introduce the concept "Flat function":
Definition: A smooth function $f:\mathbb{R}\to \mathbb{R}$ is called a flat function at origin if $f^{(k)}(0)=0$ for all $k=0,1,2,\ldots $.
let $V=\{f\in C^{\infty}[0,1]\mid \text{f is flat at origin}\}$. We equip $V$ with $|\quad|_{\infty}$.
Is there an infinit dimensional subspace $W\subseteq V$ which is invariant under the differential operator
$D(f)=xf'(x)$ and $D$ is a bounded operator on $W$?
The motivation comes from page 43 remark 2 of the following paper:
http://mcs.qut.ac.ir/article_243944.html

To me, it seems unlikely that such a subspace $W$ can exist. If you define $Q:V \to V$ by $(Qf)(x) = \log(x) f(x)$, then you can check that $$QD-DQ=\mathrm{id}_V$$ which is a version of the canonical commutation relation. I don't know off the top of my head whether realizing the canonical commutation relation while having one of operators bounded is out of the question in the setup you have given, but it at least seems rather far fetched.
Added: Actually, what I wrote above may not be an issue after all. What if $Q$ doesn't map  $W$ (or a dense subspace of $W$) back into $W$? I'll leave this answer up anyway just in case it is somehow helpful.

Other thoughts:
Let me also throw in some disorganized  notes which might be helpful (and which are anyway the way I arrived at the operator $Q$ above).

$D=x \frac{d}{dx}$ is the generator of the scaling action $(\phi_t f)(x)=f(e^tx)$. Actually, if your domain is in $C^\infty([0,1])$, it is more natural to reverse the flow because we only have a sensible scaling map $\phi_t : C^\infty([0,1]) \to C^\infty([0,1])$ when $t \leq 0$ and you would probably prefer a 1-parameter semigroup defined for $t \geq 0$. Boundedness of $x \frac{d}{dx}$ would correspond to norm-continuity (as opposed to strong operator continuity) of this semigroup on the subspace $W$.
You can check that the map $\iota:C^\infty((0,1]) \to C^\infty([1,\infty))$ defined by $(\iota f)(x)=f(1/x)$ maps your space $V \subseteq C^\infty((0,1])$ exactly onto the subspace $V' \subseteq C^\infty([1,\infty))$ consisting of functions all of whose derivatives decay rapidly at $+\infty$. That is the Schwartz functions, restricted to $[1,\infty)$.
If you conjugate the scaling action through the inversion map $\iota$ above, it is still the same scaling action (though it gets time reversed).
Now you can use another identification  $j : C^\infty([1,\infty) \to C^\infty([0,\infty))$ defined by $(j f)(x)=f(e^x)$. This carries $V'$ onto a subspace $V''$ which is contained in the subspace of functions in $C^\infty([0,\infty))$ with Schwartz behaviour at infinity. You don't hit them all though, for example $x \mapsto e^{-x}$ is a Schwartz function in $C^\infty([0,\infty))$ which does not belong to $V''$.
Under the identification $j$, the scaling action becomes the translation flow  on $C^\infty([0,\infty))$. So the operator $x \frac{d}{dx}$ is then exchanged for the usual differentiation operator $\frac{d}{dx}$.
You could try furthermore taking a Fourier transform at this point to get some subspace of the Hardy space of the line, and replace the operator with multiplication by $x$, but this seems a bit pointless because it is not very clear how to deal with the uniform norm under the Fourier transform.

Anyway, one upshot of these observations is that, if there does exist a subspace $W$ with the property you are after, then there will also exist an infinite-dimensional subspace $W'' \subset C^\infty([0,\infty))$ satisfying

$W''$ is invariant under translation (towards $0$).
The elements of $W''$ have Schwartz function behaviour at $\infty$.
$\frac{d}{dx}$ acts boundedly on $W''$ (with respect to the uniform norm).
Equivalently to (3), the translation semi-group is norm-continuous on $W''$. I guess that forces some kind of uniformly equicontinuity condition on $W''$, since the uniform norm is being used.

Let $W=<f_1,f_2,...,>$ then let $f=\sum_i c_i f_i$. $D(f) = \sum_i c_i D(f_i)$.
Take $f_i(x) = x^i$. This $W$ is flat and invariant under $D$ and infinite dimensional but if u wanted boundedness of $D$ then $deg(f_i)$ might cause issue but u might just be able to prove boundedness since $x^n < x^{n-1}$ for $0 \leq x < 1$ i.e., if u remove a small interval near $1$ from ur domain or norm.
On the other hand:
Let us apply $D$ several times. $D(D(f)) = D(xf^{(1)}(x)) = x^2 f^{(2)}(x)+xf^{(1)}(x)$. $D^3(f)=D(x^2 f^{(2)}(x)+xf^{(1)}(x)) = x^3 f^{(3)}(x)+3x^2f^{(2)}(x)+xf^{(1)}(x)$. You can see that the coefficient of $x^n f^{(n)}(x)$ keeps growing.
YOu can see that coefficient of $x^n f^{(n)}(x)$ is $O(n!)$ whereas we can only handle growth by $||D||^n$. This says how fast $f^{(n)}(x)$ must decay. Does this help ?


Is product of two closed sets closed? If $A$ is closed in $X$ and $B$ is closed in $Y$,

does it follow that $A \times B$ is closed in $X \times Y$ ?

$A$ is closed in $X$, so $A^c$ is open, likewise for $B$ in $Y$.

Moreover, $A^c \times Y$ and $X \times B^c$ are both open in $X \times Y$.

Thus $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c)  $$ is open. Hence $A \times B$ is closed.

Given $A$ and $B$ are closed sets, let $(x,y)$ be a limit point of $A \times B$. Then there exists a sequence $\{(x_n, y_n)\}$ in $A \times B$ such that $\{(x_n, y_n)\}$ converges to $(x, y)$.
Since $\{(x_n, y_n)\}$ is a sequence in $A \times B$, $\{x_n\}$ is a sequence in $A$ and $\{y_n\}$ is a sequence in $B$.
Also, since $\{(x_n, y_n)\}$ converges to $(x, y)$, $\{x_n\}$ converges to $x$ and $\{y_n\}$ converges to $y$.
We obtain a sequence $\{x_n\}$ in $A$ converging to $x$. This implies that $x$ is a limit point of $A$ and hence it is in the closure of $A$. Since $A$ is closed, $A$ equals it closure, i.e. $x$ is in $A$.
Similarly, we can prove $y$ is in $B$.
Therefore, $(x, y)$ is in $A \times B$, i.e. $A \times B$ contains all its limit points.
Hence $A \times B$ is closed.


Series expansion for $e^{- x^2}$ I was interested in one of the topics on this website, while I noticed a notation that I haven't been able to understand. This is the original question. One of the answers to the question states that
$$
e^{-x^2}=1-(1/2)x^2+(1/2)(1/4)x^4-(1/6)(1/8)x^6+(1/24)(1/16)x^8+\cdots
$$
 I don't see how my answer would be different from his. Mine involved simply substituting $-x^{2}$ instead of the $x$ in the regular $e^{x}$'s expansion, yielding :
$$
e^{-x^2}=1-x^2+(1/2)x^4-(1/6)x^6+(1/24)x^8+\cdots
$$
Thanks for taking the time to read it through.

PS : I would have liked simply commenting on the answer to ask the original writer, but I apparently need a certain amount of "reputation" to do that.

Plotting the three functions show that you are correct. The first one would be a series of $e^{-x^2/2}$

You can't put $-x^2$ at the place of x. Because the fact of the whole function will change because of the change of variable including the curvature.
You have to use the theorem of Maclaurin.
$f(x)= f(0) + xf'(0)/1! + x^2 f''(0)/2! + x^3 f'''(0)/3!+.............$


5.6 Axler's Linear Algebra Done Right Consider the following theorem:

  Theorem 5.6 Equivalent conditions to be an eigenvalue:
  Suppose V is finite-dimensional, $T \in \mathcal{L}(V),$ and $\lambda \in \mathbb{F}.$ Then the following are equivalent:

  (a) $\lambda$ is an eigenvalue of T ;

  (b) $T-\lambda I$ is not injective;

  (c) $T- \lambda I$ is not surjective;

  (d)$T- \lambda I$ is not invertible

Proof: Conditions (a) and (b) are equivalent because the equation $Tv = \lambda v$ is equivalent to the equation $(T - \lambda I)v = 0$.

From here I understand that $v \neq 0$, so that $(T - \lambda I)$ is a linear transformation that sends v to $Tv - \lambda v$. But from here I don't see why this implies that $(T - \lambda I)v = 0$ is not injective.

Can someone please explain? Thank you!

In order to tackle your question, we need the following theorems:

1- A linear map $A$ is injective, iff it has a trivial kernel. (That is, Ker($A$)={0})

2- Let $V,W$ be vector spaces, such that dim$V$ = dim$W$, $\varphi:V\rightarrow W$ a linear map. Then, the following statements are equivalent:

a. $\varphi$ is injective;

b.  $\varphi$ is surjective;

c.  $\varphi$ is bijective.

3- A map $f:X\rightarrow Y$ is bijective, iff $f$ is invertible.

Do you see how the rest of your theorem follows?
(Hint: $T-\lambda L$ is a linear transformation)

Not injective means that $\text{null}\ (T-\lambda I) \neq \{0\}$ (see 3.16). Thus $(T-\lambda I)$ is not injective if and only if there exists some $v \in \text{null}\ (T-\lambda I), v \neq 0$, so if and only if for some $v \neq 0$, $(T-\lambda I)v = 0$.


Exchanging limits and Riemann Integral Consider a function $f:\Theta \subseteq \mathbb{R}\rightarrow \mathbb{R}$ continuous at $\theta=\theta_0$ such that $f(\theta)\geq 0$ $\forall \theta \in \Theta$. Consider a sequence of real numbers $\{h_t\}_t$ such that $\lim_{t\rightarrow \infty}h_t=h$. Assume that $f(\theta)$ is Riemann integrable in a neighbourhood of $\theta_0$.

I want to show that $\lim_{t\rightarrow \infty} \int_{0}^1 h_t^2 f(\theta_0+u \frac{ h_t}{t})du=h^2 f(\theta_0)$.

  My attempt (incomplete):

  (i) if such a limit exists, then $\lim_{t\rightarrow \infty} \int_{0}^1
 h_t^2 f(\theta_0+u \frac{ h_t}{t})du=h^2 \lim_{t\rightarrow \infty} \int_{0}^1
 f(\theta_0+u \frac{ h_t}{t})du$

  (ii) By continuity of $f(\theta)$ at $\theta_0$,
   $\lim_{\theta\rightarrow \theta_0} f(\theta)=f(\theta_0)$; moreover,
   $\lim_{\theta\rightarrow \theta_0} f(\theta)=\lim_{t\rightarrow \infty}
 f(\theta_0+u \frac{ h_t}{t})$ since $\lim_{t\rightarrow \infty}\theta_0+u \frac{ h_t}{t}=\theta_0+u*0=\theta_0$; hence, $\lim_{t\rightarrow \infty}
 f(\theta_0+u \frac{ h_t}{t})=f(\theta_0)$ for any $u$

  (iii) Is there a way to exchange integrability and limit? The fact that $f(\theta)$ is Riemann integrable in a neighborhood of $\theta_0$ means that $f(\theta_0+u\frac{h_t}{t})$ is bounded when $u\in [0,1]$. Does this help?

Unless I mean missing something, what helps here is the uniform convergence of the sequence $\left(f\left(\theta_0+u  h_t/t\right)\right)_{t\geqslant 1}$ to the constant function equal to $f(\theta_0)$. This is due to the fact that $f$ is continuous at $\theta_0$: for a fixed positive $\varepsilon$, we can find a $\delta$ such that if $|x|\lt \delta$, then
$\left|f(\theta_0+x)-f(\theta_0)\right|\lt \delta$. For $t$ large enough and uniformly in $u\in [0,1]$, we have $|u  h_t/t|\leqslant |h_t|/t\lt \delta$.


Find the a < 0 for which the inequalities have at least one common solutions Find the $a < 0$ for which the inequalities $\hspace{0.5cm}2\sqrt{ax}<3a—x\hspace{0.5cm}$ and $\hspace{0.5cm}x-\sqrt{\frac{x}{a}}>\frac{6}{a}\hspace{0.5cm}$ have at least one solutions in common.

I don't know how to begin
Please give me a hint

For the first one to make sense, you need $x\lt 3a$. Then square both sides: $$4ax \lt 9a^2 +x^2 -6ax \\ x^2 -10ax +9a^2 \gt 0 \\ (x-9a)(x-a) \gt 0 $$  Since $x$ cannot be greater than $a$, it must be that case that $$x\lt 9a$$ The second one is $$\sqrt{ \frac xa} \lt x-\frac 6 a $$ For this to make sense, you need $x\gt \frac 6a $. Try to do a similar thing now as I did, keeping in mind the restriction. Once you get the two solution sets, you can see what $a$ must be for there to be non-zero intersection between them.

Notice that $x<0$ and remember that multiplying/dividing with negative number changes the inequality sign.
You can rewrite you sytem like this, multiply second by $a$ and you get $$xa-\sqrt{ax} -6>0$$
so $$(\sqrt{ax} -3)(\sqrt{ax} +2)<0 \implies \sqrt{ax} <3$$ First inequality is equivalent to $$( \sqrt{x} +3 \sqrt{a})( \sqrt{x} - \sqrt{a}) \implies  \sqrt{x} < \sqrt{a}$$
So $${9\over a}<x<a \implies a^2<9 \implies a\in (-3,0)$$


Proof: $A\subseteq (B\cup C)$ and $B\subseteq (A\cup C)$ then $(A - B) \subseteq C$ How do I prove this:

Let $A, B$ and $C$ be sets, $A \subseteq  (B \cup  C)$ and $B \subseteq  (A \cup  C)$ then $(A - B) \subseteq  C$

How about this:

Let $x \in A$ and $y \in  B.$

Since $A \subseteq  (B \cup  C)$ then $x \in  (B \cup  C).$

Since $B \subseteq  (A \cup  C)$ then $y \in  (A \cup  C)$

$x \in  B \cup  C$, so if $x \in  B$, then $x \notin  A - B.$
if $x \notin  B$, then $x \in  C.$

$y \in  A \cup  C$, so if $y \in  A$, then $y \notin  A - B.$
if $y \notin  B$, then $y \in  C.$

So in both cases, $A - B \subseteq  C$

Is it correct? if yes, the converse is false right?

The question only asks you to prove that $A-B$ is a subset of $C$. This means that we need to take $x\in A-B$ and show that $x\in C$.

Let $x$ be such element, then $x\in A$ and therefore $x\in B\cup C$. However, $x\notin B$ and therefore $x\in C$.

In the other direction, it is true that if $A-B\subseteq C$ then $A\subseteq B\cup C$. See if you can prove it.

(Hint: $x\in A$ then either $x\in B$ or $x\notin B$.)

Well... The converse will be this:
Let A,B and C be sets, (A−B)⊆C then A⊆(B∪C) and B⊆(A∪C)

I think this is false. Take the negation and proof:

Negation: Let A,B and C be sets, (A−B)⊆C then A Not⊆ (B∪C) or B Not⊆ (A∪C)

Here is a counter example:

A = {1, 2}
B = {1, 4}
C = {2, 5}

So A - B = {2} and is a subset of C.

But, A U C = {1, 2, 5} - but B = {1, 4}. So B is not a subset of AUC. Element 4 is missing in AUC.

Correct?


Closed subscheme of a projective scheme determined by homogeneous ideals So in Ravi Vakil's notes Ex 8.2C, I have to prove that if $\pi:X\hookrightarrow\text{Proj}\ S_{\cdot}$ is a closed subscheme (here $S_{\cdot}$ is a graded ring finitely generated by elements of degree 1), then there exists a homogeneous ideal $I$ such that $X=\text{Proj}\ S_{\cdot}/I$.

I have gone as far as proving the following:

Suppose $S_\cdot$ is generated by $x_i$'s for $i=0,...,n$. Then by definition of closed subscheme, on $\mathcal{O}_{\text{Proj}\ S_{\cdot}}(D(x_i ))$ we have an ideal $I_i$ of $((S_{\cdot})_{x_{i}})_{0}$. Then I let $J_i$ to be the homogeneous ideal in $S_{\cdot}$ such that

$$ J_{i}=\bigoplus_{d}\{a\in S_d: a/x_{i}^{d}\in I_{i}\}.$$

Then I let $I=\cap J_{i}$. The reason why I arrive at this is as follows and in fact I am not even sure it is correct: that each of the $I_i$ in $\mathcal{O}_{\text{Proj}\ S_{\cdot}}(D(x_i ))$ corresponds to a closed subscheme of $D(x_i )$. Then $X$ is supposed to be the union of these closed subschemes which translates to the intersection of ideals, and hence the result. Surprisingly, I got the correct homogeneous ideal by checking the sources, but I have a problem of proving that this corresponds to the desired $X$. I have searched several sources but it requires using coherent modules and some other techniques which are only proven in the later chapters. Is there a way to prove this using elementary way? Thanks!

I got the same $I$! Our definition is probably correct because that could be why Vakil says that generation in degree $1$ can be easier for some people, and we use this in our definition of $I$. I think now we are done because $\text{Proj}(S/I)$ and $X$ have the same data in terms of ideal sheaf. This is probably why several sources uses quasi-coherence. In fact, I believe Vakil is talking about this in 8.1.H. That is, affine-locally the two schemes $\text{Proj}(S/I)$ and $X$ have the same data, and their gluing are exactly the same. Please correct me if I am wrong!

Hints: 1) Closed immersions are projective. 2) The composition of projective morphisms is projective (Segre-embedding). 3) Surjectivity of a homomorphism of graded rings can be tested locally.


Is there a name for the "famous" inequality $1+x \leq e^x$? Is there a name for the "famous" inequality $1+x \leq e^x$? It has many variants depending on how you arrange the terms:

$$1 + x \leq e^x$$
$$e^{-x} -x - 1 \geq 0 $$
$$\ln(1+x) \leq x$$

Et cetera.

Perhaps the simplest mnemonic device is, "$e^x$ lies above its tangent line at the origin." This is at least a geometric instead of arbitrary algebraic expression of the fact.

It comes up in computer science and probability proofs quite frequently. In particular it is a lemma to Chernoff's bounds, and some results regarding the perceptron algorithm and Occam's razor in the PAC learning model. It is very easy to prove by drawing a graph or taking a derivative.

Does it have a name?

More generally I want to ask "why" it's so important, but this is an extremely soft question and I only hope to get used to it in time.

It is so important
because it provides
a nice lower bound
for $e^x$
for
small $x$.

The inequality
$\log(1+x) \ge 1+x$
for
$0 \le x$
follows easily from
$\log(x)
=\int_1^x \frac{dt}{t}
$.

It is a very special case
of the fact that
the tangent to a curve
at a point
is the best local linear
approximation
to the curve
at that point.

Analytically,
of course,
this is just
the first two terms in the
Taylor series:
$f(x+h)
\approx f(x)+hf'(x)+h^2f''(x)/2+...
$.
With the appropriate
remainder term,
this shows that,
if
$f''(x) \ge 0$,
then
$f(x+h)
\ge f(x)+hf'(x)
$.

The geometric intuition is certainly that $1+x$ is the tangent line to $e^x$ at $0$, so this amounts to the claim that "$e^x$ lies above its tangent line at the origin." The inequality is tight as $x \rightarrow 0$ in the sense that $e^x = 1 + x + o(x)$.

It's also the limit of the limit of Bernoulli $(1+x/n)^n \geq 1+x$, taking $n \rightarrow \infty$.

(Best answers were in question comments, so assembled here and self-accepted.)


Group under Additive modulo 6 Is the set $\{0,1,2,3,4,5,6\}$ a group under additive modulo $6$?

My Try:

The inverse of this group would be 0.
The Cayley-table entry for 6 would contain 0 at two locations
$6+_{6}0=0$ and $6+_{6}6=0$, but in a group the Cayley table entries are unique!!.
So this set is not a group.

Please let me know if I am correct?

When working in modulo $6$, notice that $0\equiv 6\bmod 6$; so actually your set in question is $\{0,1,2,3,4,5\}$.

Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions

The identity element of a group $G$, $e$ say, is an element such that $a\circ e=e\circ a=a$.
The inverse of an element $a$ in a group $G$ is an element $b$ such that $a\circ b=b\circ a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.

It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.


Finiteness of fixed points of a Lie group action Let $\psi: G\rightarrow \mathrm{Diff}(M)$ be a smooth non-trivial action of a compact connected Lie group $G$ on a compact connected smooth manifold $M$.

  Under which assumptions there will be a finite number of fixed points on $M$ by the action?

I don't have a complete answer, but I think the problem can be reduced to understanding Lefschetz fixed points of isometries. Here's why: The set of fixed points, $M^G$, is closed. A closed subset of a compact manifold is finite if and only if it's discrete, so we need to make $M^G$ discrete --- i.e. we want to make sure every $p \in M^G$ has a neighborhood disjoint from the rest of $M^G$. Fortunately, we can find a nice Riemannian metric on $M$ that is invariant under left multiplication $L_g: M \to M$, so each $L_g$ is an isometry. We can then use the Riemannian exponential map to find a neighborhood $U \subset M$ of fixed point $p \in M^G$ such that $q \in U$ lies in $M^G$ if and only if the velocity vector of the geodesic adjoining $p$ to $q$ is an eigenvector of the map $(dL_g)_p : T_p M \to T_p M$ with eigenvalue 1. In other words, we must make sure that if $p \in M^G$, then there exists $g\in G$ such that $p$ is a Lefschetz fixed point of the isometry $L_g : M \to M$. Some elaboration is below.

$\boldsymbol{M^G}$ is closed: Let $M^G:= \{ p \in M \mid L_g(p)=p \text{ for all } g \in G\}$, the set of points fixed by $G$. If we let $L_g : M \to M$ be $L_g(p)=g\cdot p$ and $\ell_g: M \to M \times M$ be $\ell_g(p)=(p,g\cdot p)$, we have  $$M^{\{g\}}:=\{p \in M \mid L_g(p)=p\}=\{ p \in M \mid \ell_g(p) \in \Delta(M)\}=\ell_g^{-1}(\Delta(M)),$$ where $\Delta(M)$ is the diagonal in $M \times M$. Since the diagonal is closed (since $M$ is Hausdorff) and $\ell_g$ is continuous, $M^{\{g\}}$ is closed. Therefore $M^G=\cap_{g \in G} M^{\{g\}}$ is closed.

How to get a nice neighborhood of a point in $\boldsymbol{M^G}$: We'll show that $M^G$ is a smooth submanifold. By compactness, $G$ possesses a left-invariant measure. Then we can take any Riemannian metric on $M$ and average it over $G$ to obtain a $G$-invariant Riemannian metric on $M$. Thus $L_g: M \to M$ is an isometry for all $g \in G$. Denote the corresponding exponential map on $M$ by $\exp_p: T_p M \to M$. Since $L_g$ is an isometry, it takes a geodesic at $p$ with initial velocity vector $v \in T_p M$ (with $v$ sufficiently small) to a geodesic at $L_g(p)$ with velocity $(dL_g)_p (v)$. This implies that $L_g \circ \exp_p(v)= \exp_{L_g(p)}((dL_g)_p v)$.
If $p \in M^G$, then it follows that $\exp_p (v)$ lies in $M^G$ if and only if $(dL_g)_p(v)=v$ for all $g \in G$. Thus the preimage $\exp_p^{-1}(M^G)$ is the intersection of a linear subspace with a neighborhood of $0 \in T_p M$, since $\exp_p (\alpha v +w ) \in M^G$ for any $\exp_p(v),\exp_p(w)\in M^G$ and $\alpha \in \mathbb{R}$. Since $\exp_p$ is a diffeomorphism near $0 \in T_p M$, it forms a local chart for $M^G$.

If G is discrete group. Because, this action is already proper (Lee Introduction to smooth manifolds, 2nd edition, Corollary 21.6). Then the stabilizer is a compact  subgroup of G (Lee again, Corollary 21.8). Hence G is finite.
I guess.


Radius of largest circle in an ellipse Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.

What I tried:

Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)

Then equation of ellipse is $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ where $a=5,b=4$

Then $\displaystyle b^2=a^2(1-e^2)$ . getting $\displaystyle e=3/5.$

coordinate of focus is $(\pm ae,0)=(\pm 3,0)$

Let equation of circle is $(x\pm 3)^2+y^2=r^2$

How do I solve it from here?

Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.


Exist vector field having only finitely many zeros, all lying in open set of compact connected manifold? Let $U$ be any open set on the compact connected manifold $X$. Does there exist a vector field having only finitely many zeros, all of which lie in $U$?

This is always possible. Start with a vector field on $X$ that has only finitely many zeros, and then apply a diffeomorphism of $X$ that takes all of the zeros into $U$. (For any one of the zeros, you can find a diffeomorphism of $X$ that takes it to a point in $U$; and if you've already put some of the zeros in $U$, you can arrange that the diffeomorphism doesn't move those points.)

This depends of the topology of $U$, if a vector field have finitely many zeros, their number are related to the Euler characteristic see

https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem

So you have to check that the Euler characteristic of the complement of $U$ (which is a manifold with boundary) is zero.


List all subsets of {a, b, c, d, e}, containing a but not containing b I wonder...can I solve this by just getting A = {a, c, d, e} minus those subsets which do not contain a?

So, let A be {a, c, d, e}. |A| is 2^4 = 16
And let B be {c, d, e}. |B| is 8
And let C be subset of subsets of {a, b, c, d, e}, containing a but not containing b

Can I conclude that that |C| will be 8, that is, |A| - |B|?

I could enumerate these:

{a}
{a, c}
{a, d}
{a, e}
{a, c, d}
{a,c, e}
{a, d, c}
{a, d, e}

Which are 8, not including the empty set so I guess there is some mistake with my reasoning. Could you people point that out for me?

P.S: this question is from the book Discrete Mathematics: Elementary and Beyond, by Lovász et al.

All you need to do is list all subsets of $\{c,d,e\}$, and then add $a$ to each of those.  There are eight of them.  (You've listed $\{a,c,d\}$ twice though, and you missed one of the eight.)

An important concept is something called a "power set". If SS is a set, then the power set of SS is the set of all subsets of SS, and may be written P(S)P(S), P(S)P(S), or P(S)P(S), depending on taste.

For example,
P(∅)={∅}
P(∅)={∅}
and
P({1,2})={∅,{1},{2},{1,2}}.
P({1,2})={∅,{1},{2},{1,2}}.
List out |P(S)||P(S)| for S={1,…,n}S={1,…,n} for a few small values of nn.

Now think about how you might represent the power set of SS as a finite sequence of 11s and 00s, that is, as a bit vector.

Can you see why the power set has the size it does for each nn?


Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$? Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?
If it exist, what is the function, and if it doesn't, what is the proof to that?
My intuition is that big O refers to the worst case, meaning worst x, so no constant should have a significant effect on the time complexity, but I'm not sure at all, and I'm struggling to write it formally.

As Christian Blatter commented, let $f(x)=e^{-x^2}$. Then
$$
\begin{align}
f(x-k)
&=e^{-(x-k)^2}\\
&=e^{2kx-k^2}e^{-x^2}\\
&=e^{2kx-k^2}f(x)
\end{align}
$$
That is, for $k\gt0$, $\frac{f(x-k)}{f(x)}$ grows without bound as $x\to\infty$.
This means that $f(x-k)\not\in O(f(x))$.

I got an idea, would love some verification:
I will prove that no such function exist and that for any function $f(x-k)=O(f(x))$.

To prove that, I will use proof by contradiction assuming that there is a function that for $f(x-k)\neq O(f(x))$ does hold.
There for $\lim \frac{f(x-k)}{f(x)}=\infty$
$\lim f(x)=\lim f(x-k)$
If so, then $\lim \frac{f(x-k)}{f(x)}=\frac{\lim f(x-k)}{\lim f(x)}=1$, in contradiction to our assumption that $f(x-k)\neq O(f(x))\Leftrightarrow \lim \frac{f(x-k)}{f(x)}=\infty$.
There for indeed $f(x-k)=O(f(x))\blacksquare$


Algebra: manipulation of expression How does this:

$$a\sqrt{2}=c$$

become this?:

$$\frac{a}{c}=\frac{a}{a\sqrt{2}}$$

If $c \neq 0$ you can divide both sides by $c$, you get:

$$\frac{a\sqrt{2}}{c}=\frac{c}{c}=1$$

Next divide both sides by $\sqrt{2}$

$$\frac{a}{c}=\frac{1}{\sqrt{2}}$$

You know that if $a \neq 0$ you have $\displaystyle 1=\frac{a}{a}$, so $\displaystyle \frac{1}{\sqrt{2}}=\frac{a}{a\sqrt{2}}$.

$$a\sqrt{2}=c$$

Divide both sides by $\sqrt{2}c$ ($c\ne0$) and clean up:

$$\frac{a\sqrt{2}}{\sqrt{2}c}=\frac{c}{\sqrt{2}c}$$

$$\frac{a}{c}=\frac{1}{\sqrt{2}}$$

By multiplying both numerator and denominator on the right hand side by $a$ ($a\ne0$) we get:

$$\frac{a}{c}=\frac{a}{a\sqrt{2}}$$


password lock problem There is a lock with the password between 000-999. But we could use only two digits to open the lock. For example, if the password is 123, the lock is open when you try 1*3, or *23, or 12*.
questions:
a) compose a strategy that the try number of opening the lock is the smallest one under the worst situation
b) Prove that your strategy is optimal

Assuming Gerry Myerson's interpretation of the question (see comments to OP), the answer is somewhere between 36 and 64. For 36, see Steven Stadnicki's comment to the OP. For 64:

000 (28) 111 (28) 222 (28) 333 (28) 444 (28)
555 (28) 666 (28) 777 (28) 888 (28) 999 (28)
012 (22) 021 (22) 102 (22) 120 (22) 201 (22)
210 (22) 345 (22) 354 (22) 435 (22) 453 (22)
534 (22) 543 (22) 678 (22) 687 (22) 768 (22)
786 (22) 867 (22) 876 (22) 039 (14) 093 (14)
309 (14) 390 (14) 903 (14) 930 (14) 149 (12)
194 (12) 419 (12) 491 (12) 914 (12) 941 (12)
256 (10) 265 (10) 526 (10) 562 (10) 625 (10)
652 (10) 004 (6)  040 (6)  113 (6)  131 (6)
227 (6)  228 (6)  272 (6)  282 (6)  311 (6)
400 (6)  557 (6)  558 (6)  575 (6)  585 (6)
722 (6)  755 (6)  822 (6)  855 (6)

Three-digit numbers are codes to try. Numbers in parentheses are the number of uncracked passwords that the three-digit code will crack. So, for instance, 491 (12) means that 491 will crack twelve passwords not cracked by earlier numbers in the list (read in the order 000, 111 etc.).

This table was generated by a simple greedy algorithm: at every step, try the code that will crack the greatest number of uncracked passwords. No claim for optimality is made.

Edited to add: I ran the same algorithm for four-digit numbers, and the number of codes required was 570.

Assuming:
(1) The lock requires a 3 digit key to open.

(2) Each of three positions can either be a digit (0-9) or an *

(3) The lock is cryptographically secure except for a crib for the 3 digit cipher in that one digit can be replaced by an "*".

(4) Each guess takes the same out effort/time (whatever we're optimizing...)

For 3 digits in the password, then 10^3 guesses would be required.

However the crib using the * reduces the key complexity by one digit. 2 digits is 100 permutations which generates sequence from 00 to 99.

But each sequence has to be tested three times. So 12 gets tested as 12*, 1*2, and *12.

99 digit sequences and 3 different positions for the * in each of the 100 digit permutation makes 300 permutations total.

Therefore 300 is the minimum number of permutations to test to insure opening the lock. In any individual case the expectation would be to try half.

edited wording to remove reference to NP-complete.

edited

Ok, I got so wrapped up in discussion about NP completeness that I blew by  TonyK's remark that 100 tries is enough. Obviously that is right. Since the * works in any position, just leave it in the third digit and do permutations of first two. You get the solution AD provided earlier.


Projective resolution of tensor product Let $M,N$ are $R$ modules and $P^\bullet, Q^\bullet$ are their projective resolutions. Can we obtain projective resolution $M\otimes N$ using $P^\bullet, Q^\bullet$. If i understand correctly homology groups of $P^\bullet\otimes Q^\bullet$ are $Tor(M,N)$ so its not a projective resolution.

It is not a projective resolution, see Mac Lane, Homology, Theorem 9.3 page 164.

There is a lot going on in your question, so I'll try and address it piece by piece. Lets work over a commutative ring so I dont need to be careful.

$Tor(M,N)$ is a bifunctor, and can be calculated as you probably know, by projective resolutions of $M$ or $N$, tensoring appropriately, and then taking the homology. However what you have described (taking a tensor product of complexes) is the proof of such a statement.

The tensor compled $P^*\otimes Q^*$ is a massive double complex. Depending on your familiarity with spectral sequences, it takes some or a lot of work, to show that since the rows of this double complex are exact, taking homology yields precisely $Tor(M,N)$. However, notice that this does not say that the tensor product complex is not projective or acyclic. Just as when you take a projective resolution, tensor and take homology, "philosophically" you are always left with at least one nonzero homology group (only the higher ones vanish in some cases), namely $Tor_0$ of what you want to calculate, in the case of a double complex, the $P^i$ and $Q^j$ dont contribute their own homology, so you dont need to worry that you're calculating stuff you dont want to.

Now to your actual question: Tensor product of flat objects remain flat. The tensor product of free things remains free (work this out!). The tensor product of projective objects, hence also remains projective!

This all bodes well for us. So now, we have a monstrous double complex, and we want to resolve the object $M\otimes N$. It doesnt quite make sense to say that "the double complex resolves $M\otimes N$", but its not far from the truth: take the total complex of this double complex. That is, direct sum up the objects of a given total degree and form a complex. To make it into a complex you have to be a bit careful: use the sign trick to correct the differentials (see here for instance) and get a complex. This complex is exact (work this out by hand, or use Kunneth). And each of the objects are projective.


Is $R=\left \{ (a,a),(a,b),(b,a),(b,b),(c,c),(c,d),(d,c),(d,d) \right \}$ an equivalence relation on $X$? Let $X= \left \{ a,b,c,d \right \}$ and $R=\left \{ (a,a),(a,b),(b,a),((b,b),(c,c),(c,d),(d,c),(d,d) \right \}$. I want to show that $R$ is an equivalence relation on $X$.

My work:

$R$ is reflexive: If $a\in X$, $aRa$ since $(a,a) \in R$.

$R$ is symmetric: If $a,b \in X$, then $(a,b) \in R \Leftrightarrow (b,a) \in R$.

$R$ is not transitive: If $a,b,c \in X$, then $aRb$ and $bRc$ implies $aRc$. But $(b,c)$ is not an element of $R$ and $(a,c)$ is not an element of $R$.

So $R$ is not an equivalence relation. But I am supposed to show that $R$ is an equivalence relation on $X$. For transitivity should $a\neq b\neq c$?

Please note that $R$ is transitive if $(x,y) \in R$ and $(y,z) \in R \Rightarrow (x,z) \in R$

Now as $(a,b) \in R$ and $(b,c) \notin R$ you need not check for $(a,c)$ for this pair.

So whenever you find any pair of elements $(x,y)$ and $(y,z)$ both belonging to $R$ then only you check whether $(x,z)$ is in $R$ or not.

for a relation to be an equivalence relation it must be reflexive, symmetric and transitive.

$R$ is reflexive if $ \forall x (x,x) \in R $ and $(a,a),(b,b),(c,c),(d,d) \in R$ so it is refexive

$R$ is symmetric if $ \forall x y \left( (x,y) \in R \to(y,x) \in R \right) $test this yourself.

$R$ is transitive if $ \forall x yz \left( \left((x,y) \in R \land (y,z) \in R \to (x,z) \in R \right) \right) $ test this also yourself

If your set pases all these tests it is an equivalence relation

Good luck


$ \lim_{x\to 0} \left(\frac{5}{2+\sqrt{9+x}}\right)^{\operatorname{cosec} x}$ I solved this question in two different ways and got two different answers to it.
Method #1:
The question is of the form $1^\infty$.
So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.

Now, according to the Maclauren series, $$\lim_{x\to 0} \left(\frac{\ln (x+1)}{x}\right) = 1$$
Using this result, then rationalising the power and then using the result $$\lim_{x\to 0} \left(\frac{\sin x}{x}\right) = 1$$
we get the answer as $e^{-1/30}$

Method #2:

In this method, I decided to go by the basics instead.
I took out the left-hand limit and the right-hand limit separately.

If we want the right-hand limit, then we will have to put $x = 0^+$.
Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1.
Now $cosec 0^+$ tends to  $\infty$
Also, a number less than 1 raised to the power of infinity tends to 0.

$\therefore$ The right-hand limit is zero.

In a similar manner when we take out the left-hand limit, that also comes out to be zero.

Hence, by this method, the answer should be 0.

Please help.

You need to work out what happens to
\begin{align}
\ln(2+\sqrt{9+x})
&=\ln(2+3\sqrt{1+(x/9)}\,)\\
&=\ln(2+3(1+x/18+o(x)))\\
&=\ln(5+x/6+o(x))\\
&=\ln 5+\ln(1+x/30+o(x))\\
&=\ln 5+x/30+o(x)
\end{align}
Thus the logarithm of your function is
$$
\frac{\ln 5-\ln(2+\sqrt{9+x})}{\sin x}=
\frac{\ln 5-\ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.

Your method #2 is flawed: for instance
$$
\lim_{x\to0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.

The limit is similar to

$$\lim_{t\to0}(1+t)^{1/t}.$$

In the positives, you can rewrite

$$\lim_{u\to\infty}\left(\frac{u+1}u\right)^u=e.$$

In the negatives,

$$\lim_{u\to-\infty}\left(\frac{u+1}u\right)^u=\lim_{u\to\infty}\left(\frac{u-1}u\right)^{-u}=\lim_{u\to\infty}\left(\frac v{v+1}\right)^{-v-1}
\\=\lim_{v\to-\infty}\left(\frac{v+1}v\right)^v\lim_{v\to\infty}\left(\frac v{v+1}\right)^{-1}=e.$$

So your second reasoning doesn't hold.


Prove that an element $(a,b,c)$ belongs to this set such that it satisfies a certain condition This particular question was asked in a masters entrance for which I am preparing.

Let $E =\{(x,y,z)\in \mathbb{R}^{3}\ |\ x,y,z >0 , xy+yz+zx=1 \}$. Prove that there exists $(a,b,c)\in E$ such that $\space abc\geq xyz \space$ for all $ \space(x,y,z)\in E$.

I have no idea what branch of mathematics should be used to solve it and so I can't provide anything in attempt.
Kindly guide me.

We just use AM-GM inequality.
$$(xyz)^{\frac{2}{3}}\leq\frac{xy+yz+zx}{3}=\frac{1}{3}\implies xyz\leq \frac{1}{3\sqrt{3}}$$ with equality if and only if $xy=yz=zx=\frac{1}{3}\implies x=y=z=\frac{1}{\sqrt{3}}$. So we can take $(a,b,c)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Done!

There may be an easier way to prove this, but it looks like a constrained optimization problem to me. You could reframe the problem in the following way:
Prove that $f(x,y,z) = xyz $ has a maximum when $(x,y,z)$ are subject to the constraint $h(x,y,z) = 0$, where $h(x,y,z) = xy + yz + zx - 1$.
Then, to prove it you could appeal to the KKT theorem (see https://en.wikipedia.org/wiki/Karush–Kuhn–Tucker_conditions), which in this case is just the idea of optimization using Lagrange multipliers.
First you might use the first-order necessary condition from KKT to find $(a,b,c)$:
$$ Df(a,b,c) - \lambda^* Dh(a,b,c) = 0, ~\text{subject to}~ h(a,b,c) = 0 .$$
Then, you can complete the proof by showing that the second-order sufficient condition is satisfied by $(a,b,c)$. Letting $$\mathscr{L}(x,y,z,\lambda) := f(x,y,z) - \lambda h(x,y,z), $$ you need to show $$ v^T D^2_\mathbf{x} \mathscr{L} (a,b,c,\lambda^*) v > 0 $$
for all $ v $ in the null space of $Dh(a,b,c)$.
Note: in my confusing notation for $D^2_\mathbf{x}  \mathscr{L}$, I mean that we only take the total derivative with respect to the arguments $(x,y,z)$, not with respect to $\lambda$.


What is the fastest way to factor polynomials algebraically? What is the fastest way that you can do by hand? For example: (just a simple example, the answer should be general of course)

$x^4 -4x^3 +8x$

or

$2x^2+3x+1$

Read up on Horner's Method. Ignore this part of the sentence as I just need to fill up the characters.


Homotopy lifting property: giving a counter example for $a \simeq b \Rightarrow \tilde{a} \simeq \tilde{b}$ The homotopy lifting property says that in case $a:I\rightarrow Y$ is homotopic to $b:I\rightarrow Y$ (denoting the homotopy in $F:I\times[0,1]\rightarrow Y )$ and $p:X\rightarrow Y$ is a covering map, for $\tilde{a} : I\rightarrow X$ a lifting of $a$ ($(p\circ\tilde{a}) (s)= a(s)$) then there exists a unique homotopy $G:I\times[0,1]\rightarrow X$ such that $p \circ G(s,t) = F(s,t)$. So as I understand it, it means that for $\tilde{a}$ there exists a unique lifting $\tilde{b}$ of $b$ such that $\tilde{a} \simeq \tilde{b}$.

I try to find an example of $a\simeq b$ such that $\tilde{a}$ is not homotopic to $\tilde{b}$, My intuition was creating a covering map: $p:X\rightarrow Y$ such that two homotopic loops $a,b$ in $(I,\partial I)\rightarrow (Y,y)$ where $a$ is lifted to a loop $\tilde a :(I,\partial I)\rightarrow (X,x\in P^{-1}(y))$ and $\tilde b :I\rightarrow X$ is not a loop in $X$ ($\tilde{b(0)},\tilde{b(1)}\in P^{-1}(y)$). Yet I faild to construct such spaces and covering map.

Any idea would be appreciated.

A covering projection $p : X \to Y$ is a fibration which means that it has the homotopy lifting property. This does neither mean that any map $f : Z \to Y$ has a lift $\tilde{f} : Z \to X$ nor that lifts are unique. We can only say that if $Y$ is connected and two lifts agree for some point of $Z$, then they are identical.

In particalar you cannot say that the homotopy $F$ has a unique lift $G$, but only that there exists a unique lift $G$ such that $G_0 = \tilde{a}$. In other words, you get a lift $\tilde{b}$ of $b$ such that $\tilde{b} \simeq \tilde{a}$, but in general there are other lifts which are not homotopic to $\tilde{a}$. See Tyrone's comment.

By the way, you have to make precise what you mean by a homotopy of paths. The usual approach is to consider only paths $u, v : I \to Y$ such that $u(0) = v(0) = y_0$ and $u(1) = u(1) = y_1$ and homotopies keeping $y_0, y_1$ fixed. The set of such homotopy classes is called the fundamental grupoid of $Y$.


How do I find a slice of the area of a semi-circle? A semi-circle around the origin, with radius $r$, is given by $$f(x) = \sqrt{r^2 - x^2}$$ The area of this semi-circle can be written as $$\int_{-r}^r \sqrt{r^2 - x^2} dx = \frac{\pi r^2}{2}$$
How do I find only a slice of this area, i.e. $$\int_a^b \sqrt{r^2 - x^2} dx$$ for $-r \le a < b \le r$?

There's probably a geometric approach to this, but I'm kinda stuck. If it's easier to calculate a slice of the area of a full circle that's also fine.

Yes, this can be done in a geometric way. Assume that $a\leqslant 0\leqslant b$ and consider this picture:

The area that you are interested in is the area of two triangles plus the area of a circular sector. The areas of those triangles are$$\frac12(-a)\sqrt{r^2-a^2}\text{ and }\frac12b\sqrt{r^2-b^2}.$$and the area of the circular sector is$$\frac{r^2}2\left(\arccos\left(\frac ar\right)-\arccos\left(\frac br\right)\right).$$So, the area that you are interested in is the sum of these three numbers.

You will have to change a bit this argument in order to deal with the other two cases ($0\leqslant a\leqslant b\leqslant r$ and $-r\leqslant a\leqslant b\leqslant 0$).

Take $x = r\sin\theta$ so that $r^{2}-x^{2}$ becomes $r^{2}(1-\sin^{2}\theta) = r^{2}\cos^{2}\theta$ and $\sqrt{r^{2}-x^{2}}$ becomes $r\sin\theta$. Now, with this change of variable, one have $dx \to r\cos\theta d\theta$ and, thus:
$$\int_{a}^{b}\sqrt{r^{2}-x^{2}}dx = \int_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}r^{2}\cos^{2}\theta d\theta  = r^{2}\bigg{(}\frac{1}{2}\theta+\frac{1}{4}\sin2\theta\bigg{)}\bigg{|}_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}$$


probability $2/4$ vs $3/6$ Recently I was asked the following in an interview:

  If you are a pretty good basketball player, and were betting on whether you could make $2$ out of $4$ or $3$ out of $6$ baskets, which would you take?

I said anyone since ratio is same. Any insights?

Depends on how good you are

The explanation is intuitive:

If you are not very good (probability that you make a single shot - p < 0.6), then your overall probability is not very high, but it is better to bet that you'll make 2 out of 4, because you may do it just by chance and your clumsiness has less chance to prove in 4 than in 6 attempts.
If you are really good (p > 0.6), then it is better to bet on 3 out of 6, because if you miss just by chance, you have better chance to correct yourself in 6 attempts.

The curves meet exactly at p = 0.6.

In general, the more attempts, the more of real skill reveals

This is best illustrated on the extreme case:

With more attempts, it is almost binary case - you either succeed or not, based on your skill. With high N, the result will be close to your original expectation.

Note that with high N and p = 0.5, the binomial distribution gets narrower and converges to normal distribution.

Everything here just revolves around binomial distribution,

which tells you that the probability that you will score exactly k shots out of n is

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$

The probability that you will score at least k = n/2 shots (and win the bet) is then

$$P(X \ge k) = \sum^{n}_{i=k} \binom{n}{i} p^i (1-p)^{n-i}$$

Why the curves don't meet at p = 0.5?

Look at the following plots:

These plots are for p = 0.5. The binomial distribution is symmetric for this value. Intuitivelly, you expect 2 of 4 or 3 of 6 to take half of the distribution. But if you look especially at the left plot, it is clear that the middle column (2 successful shots) goes far beyond the half of the distribution (dashed line), which is denoted by the red arrow. In the right plot (3/6), this proportion is much smaller.

If you sum the gold bars, you will get:

P(make at least   2 out of    4) = 0.6875
P(make at least   3 out of    6) = 0.65625
P(make at least 500 out of 1000) = 0.5126125

From these figures, as well as from the plots, is apparent that with high N, the proportion of the distribution "beyond the half" converges to zero, and the total probability converges to 0.5.

So, for the curves to meet for low Ns, p must be higher to compensate for this:

P(make at least   2 out of    4) = 0.8208
P(make at least   3 out of    6) = 0.8208

Full code in R:

f6 <- function(p) {
    dbinom(3, 6, p) +
    dbinom(4, 6, p) +
    dbinom(5, 6, p) +
    dbinom(6, 6, p)
}

f4 <- function(p) {
    dbinom(2, 4, p) +
    dbinom(3, 4, p) +
    dbinom(4, 4, p)
}

fN <- function(p, from, max) {
    #sum(sapply(from:max, function (x) dbinom(x, max, p)))
    s <- 0
    for (i in from:max) {
        s <- s + dbinom(i, max, p)
    }
    s
}
f1000 <- function (p) fN(p, 500, 1000)

plot(f6, xlim = c(0,1), col = "red", lwd = 2, ylab = "", main = "Probability that you will make ...", xlab = "p (probability you make a single shot)")
curve(f4, col = "green", add = TRUE, lwd = 2)
curve(f1000, add = TRUE, lwd = 2, col = "blue")
legend("topleft", c("2 out of 4", "3 out of 6", "500 out of 1000"), lwd = 2, col = c("green", "red", "blue"), bty = "n")

plotHist <- function (n, p) {
    plot(x=c(-0.5,n+0.5),y=c(0,0.41),type="n", xaxt="n", xlab = "successful shots", ylab = "probability",
        main = paste0(n/2, "/", n, ", p = ", p))
    axis(1, at=0:n, labels=0:n)
    x <- 0:n
    y <- dbinom(0:n, n, p)
    w <- 0.9
    #lines(0:4, dbinom(0:4, 4, 0.5), lwd = 50, type = "h", lend = "butt")
    rect(x-0.5*w, 0, x+0.5*w, y, col = "lightgrey")
    uind <- (n/2+1):(n+1)
    rect(x[uind]-0.5*w, 0, x[uind]+0.5*w, y[uind], col = "gold")
}

par(mfrow = c(1, 2))
plotHist(4, 0.5)
abline(v = 2, lty = 2)
arrows(2-0.5*0.9, 0.17, 2, 0.17, col = "red", code = 3, length = 0.1, lwd = 2)
plotHist(6, 0.5)

f4(0.5)
f6(0.5)
f1000(0.5)

par(mfrow = c(1, 2))
plotHist(4, 0.6)
plotHist(6, 0.6)

f4(0.6)
f6(0.6)

Assume you make your shots with probability $p$ independent of each other. Then, the probability of making $k$ shots out of $n$ is $\binom{n}{k} p^k (1-p)^{n-k}$ (That is, the number of shots you make out of $n$ baskets follows a Binomial(n,p) distribution).

The best free throw shooter in the NBA currently is Steve Nash, with $p=0.9041$. Calculate the probabilities for $n=4, k=2$ and $n=6,k=3$ and compare. You want the choice which gives you the higher probability.

octave:85> bincoeff(4,2)p^2(1-p)^2

ans =  0.045105

octave:86> bincoeff(6,3)p^3(1-p)^3

ans =  0.013036

So, you're better off taking the 2 out of the 4 shots in this model, if you're Steven Nash.

The problem is slightly different when you need to make at least 2 out of 4 shots versus at least 3 out of 6 shots. In this case, you need to sum the tail probabilities and compare them: The probability of making at least $k$ shots out of $n$ under our model is $\sum_{i=k}^n \binom{n}{i}p^i (1-p)^{n-i}$. I will leave you to calculate which one you're better off with in this case ($n=4,k=2$ and $n=6,k=3$ once again) [ The wording of the problem made it sound like you needed the exactly 2 out of 4 shots or exactly 3 out of 6 shots when I first read it ].


What is the area of the part of the surface $z=yx$ bounded by $x^2+y^2=1$? A parametrization of the part of the surface $z=yx$ bounded by $x^2+y^2=1$ is
\begin{align}
x &= u \cos v \\
y &= u \sin v \\
z &= \frac12 u^2 \sin 2v,
\end{align}
or
$$r(u,v)=u \cos v \, {\bf i} + u \sin v \, {\bf j} + \frac12 u^2 \sin 2v \, {\bf k}, \quad 0<v<2\pi, 0<u<1.$$

The norm of the cross product of $r_u$ and $r_v$ is $$\sqrt {u^4+u^2} = u\sqrt {u^2+1},$$
so
$$\int^{2\pi}_0\int_0^1u\sqrt {u^2+1}dudv=\int^{2\pi}u\sqrt {u^2+1}=\frac13((4\pi^2+1)^\frac23-1)$$

However, the answer showed in my text book is $\frac{2\pi}{3}(2\sqrt 2-1)$, so I don't know whether I did it wrong or there is a mistake in my textbook.

The first integral equality is not correct---you seem to be integrating with respect to $dv$ but over the interval for $u$.

Since the integrand does not depend on $v$, we can write
$$\int_0^{2 \pi} \int_0^1 u \sqrt{u^2 + 1} du\,dv = \int_0^{2 \pi} dv \int_0^1 u \sqrt{u^2 + 1} du = 2 \pi \int_0^1 u \sqrt{u^2 + 1} du$$

$I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$

Change of coordinates -

$x=r\cos \theta$

$y=r\sin \theta$

Jacobian -

$I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$

$\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&\sin\theta\\ -r\sin \theta&r\cos \theta\end{array}\right|=r$

$I=\int_0^1 \int_0^{2\pi} r^3\cos\theta \sin\theta\ d\theta\ dr=\int r^3\cdot0\ dr=0$

In retrospect, it is easy to see this is zero geometrically, since $xy+(-x)y=0$.